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1l5w0n374 | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Let $$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$ and $$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$$ where the inverse trigonometric functions take principal values. Then, the equation whose roots are $$\alpha$$ and $$\beta$$ is :</p> | [{"identifier": "A", "content": "$$15{x^2} - 8x - 7 = 0$$"}, {"identifier": "B", "content": "$$5{x^2} - 12x + 7 = 0$$"}, {"identifier": "C", "content": "$$25{x^2} - 18x - 7 = 0$$"}, {"identifier": "D", "content": "$$25{x^2} - 32x + 7 = 0$$"}] | ["C"] | null | <p>Given,</p>
<p>$$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$</p>
<p>We know, $$2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1)$$</p>
<p>$$\therefore$$ $$2{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) = {\cos ^1}\left( {2 \times {1 \over 5} - 1} \right) = {\cos ^{ - 1}}\left( { - {3 \over 5}} \right)$$</p>
<p>$$\therefore$$ $$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( { - {3 \over 5}} \right)} \right.} \right)$$</p>
<p>$$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {\pi - {{\cos }^{ - 1}}\left( { {3 \over 5}} \right)} \right.} \right)$$</p>
<p>$$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right.} \right)$$</p>
<p>$$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right)} \right.} \right)$$</p>
<p>$$ = \tan \left( {{{5\pi } \over {16}} \times {4 \over 5}} \right)$$</p>
<p>$$ = \tan \left( {{\pi \over 4}} \right)$$</p>
<p>$$ = 1$$</p>
<p>$$\therefore$$ $$\alpha = 1$$</p>
<p>Also given,</p>
<p>$$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$$</p>
<p>$$ = \cos \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$$</p>
<p>$$ = \cos \left( {2{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$$</p>
<p>$$ = \cos \left( {{{\cos }^{ - 1}}\left( {2\left. {{{\left( {{3 \over 5}} \right)}^2} - 1} \right)} \right.} \right)$$</p>
<p>$$ = \cos \left( {{{\cos }^{-1}}\left( {{{18} \over {25}} - 1} \right.} \right)$$</p>
<p>$$ = {{18} \over {25}} - 1$$</p>
<p>$$ = {{18 - 25} \over {25}}$$</p>
<p>$$ = - {7 \over {25}}$$</p>
<p>$$\therefore$$ $$\beta = - {7 \over {25}}$$</p>
<p>$$\therefore$$ The quadratic equation with roots $$\alpha$$ and $$\beta$$ is</p>
<p>$${x^2} - (\alpha + \beta )x + \alpha \beta = 0$$</p>
<p>$$ \Rightarrow {x^2} - \left( {1 - {7 \over {25}}} \right)x + 1 \times \left( { - {7 \over {25}}} \right) = 0$$</p>
<p>$$ \Rightarrow 25{x^2} - 18x - 7 = 0$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6jduff0 | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>For $$k \in \mathbb{R}$$, let the solutions of the equation $$\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k, 0<|x|<\frac{1}{\sqrt{2}}$$ be $$\alpha$$ and $$\beta$$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $$x^{2}-b x-5=0$$ are $$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$$ and $$\frac{\alpha}{\beta}$$, then $$\frac{b}{k^{2}}$$ is equal to ____________.</p> | [] | null | 12 | <p>$$\cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}} \right)} \right)} \right)} \right)} \right) = k$$</p>
<p>$$ \Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\sqrt {1 - {x^2}} } \right)} \right)} \right) = k$$</p>
<p>$$ \Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {{x \over {\sqrt {1 - {x^2}} }}} \right)} \right) = k$$</p>
<p>$$ \Rightarrow {{\sqrt {1 - 2{x^2}} } \over {\sqrt {1 - {x^2}} }} = k$$</p>
<p>$$ \Rightarrow {{1 - 2{x^2}} \over {1 - {x^2}}} = {k^2}$$</p>
<p>$$ \Rightarrow 1 - 2{x^2} = {k^2} - {k^2}{x^2}$$</p>
<p>$$\therefore$$<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7ps10fx/454d4b24-e609-4a62-af81-65229ca491c4/559cd2d0-2da8-11ed-8542-f96181a425b5/file-1l7ps10fy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7ps10fx/454d4b24-e609-4a62-af81-65229ca491c4/559cd2d0-2da8-11ed-8542-f96181a425b5/file-1l7ps10fy.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Morning Shift Mathematics - Inverse Trigonometric Functions Question 25 English Explanation"></p>
<p>$${1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} = 2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right)$$ ...... (1)</p>
<p>and $${\alpha \over \beta } = - 1$$ ...... (2)</p>
<p>$$\therefore$$ $$2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right)( - 1) = - 5$$</p>
<p>$$ \Rightarrow {k^2} = {1 \over 3}$$</p>
<p>and $$b = S.R = 2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right) - 1 = 4$$</p>
<p>$$\therefore$$ $${b \over {{k^2}}} = {4 \over {{1 \over 3}}} = 12$$</p> | integer | jee-main-2022-online-27th-july-morning-shift |
1l6m56q48 | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Considering only the principal values of the inverse trigonometric functions, the domain of the function $$f(x)=\cos ^{-1}\left(\frac{x^{2}-4 x+2}{x^{2}+3}\right)$$ is :</p> | [{"identifier": "A", "content": "$$\\left(-\\infty, \\frac{1}{4}\\right]$$"}, {"identifier": "B", "content": "$$\\left[-\\frac{1}{4}, \\infty\\right)$$"}, {"identifier": "C", "content": "$$(-1 / 3, \\infty)$$"}, {"identifier": "D", "content": "$$\\left(-\\infty, \\frac{1}{3}\\right]$$"}] | ["B"] | null | <p>$$ - 1 \le {{{x^2} - 4x + 2} \over {{x^2} + 3}} \le 1$$</p>
<p>$$ \Rightarrow - {x^2} - 3 \le {x^2} - 4x + 2 \le {x^2} + 3$$</p>
<p>$$ \Rightarrow 2{x^2} - 4x + 5 \ge 0$$ & $$ - 4x \le 1$$</p>
<p>$$x \in R$$ & $$x \ge - {1 \over 4}$$</p>
<p>So domain is $$\left[ { - {1 \over 4},\infty } \right)$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6m5b57u | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $$\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x)$$ is equal to :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "$$-\\frac{1}{2}$$"}] | ["A"] | null | <p>$${\cos ^{ - 1}}x - 2{\sin ^{ - 1}}x = {\cos ^{ - 1}}2x$$</p>
<p>For Domain : $$x \in \left[ {{{ - 1} \over 2},{1 \over 2}} \right]$$</p>
<p>$${\cos ^{ - 1}}x - 2\left( {{\pi \over 2} - {{\cos }^{ - 1}}x} \right) = {\cos ^{ - 1}}(2x)$$</p>
<p>$$ \Rightarrow {\cos ^{ - 1}}x + 2{\cos ^{ - 1}}x = \pi + {\cos ^{ - 1}}2x$$</p>
<p>$$ \Rightarrow \cos (3{\cos ^{ - 1}}x) = - \cos ({\cos ^{ - 1}}2x)$$</p>
<p>$$ \Rightarrow 4{x^3} = x$$</p>
<p>$$ \Rightarrow x = 3,\, \pm \,{1 \over 2}$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1ldyay4yl | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$${\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["B"] | null | <p>$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$</p>
<p>$$= {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{16 + 8\sqrt 3 } \over {12 + 6\sqrt 3 }}} \right)^{{1 \over 2}}}$$</p>
<p>$$ = {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {\sqrt 3 (\sqrt 3 + 1)}}} \right) + {\sec ^{ - 1}}{\left( {{{4({1^2} + {{(\sqrt 3 )}^2} + 2\,.\,1\,.\,\sqrt 3 } \over {{3^2}{{(\sqrt 3 )}^2} + 2\,.\,3\,.\,\sqrt 3 }}} \right)^{{1 \over 2}}}$$</p>
<p>$$ = {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{4{{(\sqrt 3 + 1)}^2}} \over {{{(3 + \sqrt 3 )}^2}}}} \right)^{{1 \over 2}}}$$</p>
<p>$$ = {\pi \over 6} + {\sec ^{ - 1}}\left( {{{2(\sqrt 3 + 1)} \over {\sqrt 3 (\sqrt 3 + 1)}}} \right)$$</p>
<p>$$ = {\pi \over 6} + {\sec ^{ - 1}}\left( {{2 \over {\sqrt 3 }}} \right)$$</p>
<p>$$ = {\pi \over 6} + {\cos ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right)$$</p>
<p>$$ = {\pi \over 6} + {\pi \over 6}$$</p>
<p>$$ = {\pi \over 3}$$</p> | mcq | jee-main-2023-online-24th-january-morning-shift |
jaoe38c1lscmv27h | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Considering only the principal values of inverse trigonometric functions, the number of positive real values of $$x$$ satisfying $$\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}$$ is :</p> | [{"identifier": "A", "content": "more than 2"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <p>$$\begin{aligned}
& \tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4} ; x>0 \\
& \Rightarrow \tan ^{-1} 2 x=\frac{\pi}{4}-\tan ^{-1} x
\end{aligned}$$</p>
<p>Taking tan both sides</p>
<p>$$\begin{aligned}
& \Rightarrow 2 \mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}} \\
& \Rightarrow 2 \mathrm{x}^2+3 \mathrm{x}-1=0 \\
& \mathrm{x}=\frac{-3 \pm \sqrt{9+8}}{8}=\frac{-3 \pm \sqrt{17}}{8}
\end{aligned}$$</p>
<p>Only possible $$x=\frac{-3+\sqrt{17}}{8}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd4basi | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>If $$a=\sin ^{-1}(\sin (5))$$ and $$b=\cos ^{-1}(\cos (5))$$, then $$a^2+b^2$$ is equal to</p> | [{"identifier": "A", "content": "25"}, {"identifier": "B", "content": "$$4 \\pi^2+25$$\n"}, {"identifier": "C", "content": "$$8 \\pi^2-40 \\pi+50$$\n"}, {"identifier": "D", "content": "$$4 \\pi^2-20 \\pi+50$$"}] | ["C"] | null | <p>$$\begin{aligned}
& a=\sin ^{-1}(\sin 5)=5-2 \pi \\
& \text { and } b=\cos ^{-1}(\cos 5)=2 \pi-5 \\
& \therefore a^2+b^2=(5-2 \pi)^2+(2 \pi-5)^2 \\
& =8 \pi^2-40 \pi+50
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
luxwcnzi | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $$2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5}$$, is __________.</p> | [] | null | 0 | <p>$$\begin{aligned}
& 2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5} \\
& \frac{\pi}{2}+\cos ^{-1} x=\frac{2 \pi}{5} \\
& \cos ^{-1} x=\frac{2 \pi}{5}-\frac{\pi}{2} \\
& \cos ^{-1} x=\frac{-\pi}{10}
\end{aligned}$$</p>
<p>Which is not possible as $$\cos ^{-1} x \in[0, \pi]$$</p>
<p>$$\therefore \quad$$ No solution</p> | integer | jee-main-2024-online-9th-april-evening-shift |
lv2erz4o | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Given that the inverse trigonometric function assumes principal values only. Let $$x, y$$ be any two real numbers in $$[-1,1]$$ such that $$\cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi$$.
Then, the minimum value of $$x^2+y^2+2 x y \sin \alpha$$ is</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "$$\\frac{-1}{2}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \cos ^{-1} x-\frac{\pi}{2}+\cos ^{-1} y=\alpha \\
& \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\
& \because \quad \alpha \in\left[\frac{-\pi}{2}, \pi\right] \\
& \text { then } \frac{\pi}{2}+\alpha \in\left(0, \frac{3 \pi}{2}\right) \\
& \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\frac{\pi}{2}+\alpha \\
& x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha \\
& x y+\sin \alpha=\sqrt{1-x^2} \sqrt{1-y^2} \\
& x^2 y^2+\sin ^2 \alpha+2 x y \sin \alpha=1-x^2-y^2+x^2 y^2 \\
& \underbrace{x^2+y^2+2 x y \sin \alpha}_E=\cos ^2 \alpha
\end{aligned}$$</p>
<p>Now, minimum value of $$E$$ is 0.</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
KkxLSqZpDRwCLbbj | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | $${\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x,$$ then sin x is equal to : | [{"identifier": "A", "content": "$${\\tan ^2}\\left( {{\\alpha \\over 2}} \\right)$$ "}, {"identifier": "B", "content": "$${\\cot ^2}\\left( {{\\alpha \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\tan \\alpha $$ "}, {"identifier": "D", "content": "$$cot\\left( {{\\alpha \\over 2}} \\right)$$ "}] | ["A"] | null | Given that,
<br><br>$${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( \sqrt{\cos x} \right) = x\,\,\,\,...\left( 1 \right)$$
<br><br>We know,
<br><br>$${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = {\pi \over 2}$$
<br><br>$$\therefore$$ $$\,\,\,$$ $${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) + {\tan ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {\pi \over 2}\,\,\,...\left( 2 \right)$$
<br><br>Adding $$(1)$$ and $$(2),$$ we get,
<br><br>$$2{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = x + {\pi \over 2}$$
<br><br>$$ \Rightarrow \,\,{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {x \over 2} + {\pi \over 4}$$
<br><br>$$ \Rightarrow \,\,\sqrt {\cos x} = \cot \left( {{x \over 2} + {\pi \over 4}} \right)$$
<br><br>$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cot {x \over 2} - 1} \over {1 + \cot {x \over 2}}}$$
<br><br>$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}}$$
<br><br>Squaring both sides we get,
<br><br>$$ \Rightarrow \,\,\cos x = {{1 - 2\sin {x \over 2}\cos {x \over 2}} \over {1 + 2\sin {x \over 2}\cos {x \over 2}}}$$
<br><br>$$ \Rightarrow \,\,\cos x = {{1 - \sin x} \over {1 + \sin x}}$$
<br><br>$$ \Rightarrow \,\,{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}} = {{1 - \sin x} \over {1 + \sin x}}$$
<br><br>Applying compounds and dividendo rule,
<br><br>$$ \Rightarrow \,\,{{2\sin x} \over 2} = {{2{{\tan }^2}{x \over 2}} \over 2}$$
<br><br>$$ \Rightarrow \,\,\sin x = {\tan ^2}{x \over 2}$$
<br/><br/><b>Other Method :</b>
<br/><br/>$$
\begin{aligned}
& \text { Since, } \cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\
& \Rightarrow \tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\
& \Rightarrow \tan ^{-1}\left(\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}\right)=x \\\\
& \Rightarrow \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=\tan x \\\\
& \Rightarrow \cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha} \\\\
& \because \operatorname{cosec} x=\sqrt{1+\cot { }^2 x} \\\\
& \therefore \operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha} \\\\
& \Rightarrow \sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\tan ^2 \frac{\alpha}{2}
\end{aligned}
$$ | mcq | aieee-2002 |
VgRPBRM1vhbktxrx | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha ,$$ then $$4{x^2} - 4xy\cos \alpha + {y^2}$$ is equal to : | [{"identifier": "A", "content": "$$2\\sin 2\\alpha $$ "}, {"identifier": "B", "content": "$$4$$ "}, {"identifier": "C", "content": "$$4{\\sin ^2}\\alpha $$ "}, {"identifier": "D", "content": "$$-4{\\sin ^2}\\alpha $$"}] | ["C"] | null | As we know,
<br><br>$${\cos ^{ - 1}}A - {\cos ^{ - 1}}B$$
<br><br>$$ = {\cos ^{ - 1}}\left( {AB + \sqrt {1 - {A^2}} .\sqrt {1 - {B^2}} } \right)$$
<br><br>Given, $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$
<br><br>$$ \Rightarrow {\cos ^{ - 1}}\left( {x.{y \over 2} + \sqrt {1 - {x^2}} .\sqrt {1 - {{{y^2}} \over 4}} } \right) = \alpha $$
<br><br>$$ \Rightarrow {{xy} \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{y{}^2} \over 4}} = \cos \,x$$
<br><br>$$ \Rightarrow {\left( {\cos x - {{xy} \over 2}} \right)^2} = \left( {1 - {x^2}} \right)\left( {1 - {{{y^2}} \over 4}} \right)$$
<br><br>$$ \Rightarrow {\cos ^2} + {{{x^2}{y^2}} \over 4} - 2.\cos x.{{xy} \over 2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - {x^2} - {{{y^2}} \over 4} + {{{x^2}{y^2}} \over 4}$$
<br><br>$$ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\,\cos x = 1 - {\cos ^2}x$$
<br><br>$$ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\cos x = {\sin ^2}x$$
<br><br>$$ \Rightarrow 4{x^2} + y{}^2 - 4xy\cos x = 4{\sin ^2}x$$ | mcq | aieee-2005 |
1TDQHlhEobjkoJLP | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ + cosec<sup>-1</sup>$$\left( {{5 \over 4}} \right)$$ = $${\pi \over 2}$$, then the value of x is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Given sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ + cosec<sup>-1</sup>$$\left( {{5 \over 4}} \right)$$ = $${\pi \over 2}$$
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ + sin<sup>-1</sup>$$\left( {{4 \over 5}} \right)$$ = $${\pi \over 2}$$
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ = $${\pi \over 2}$$ - sin<sup>-1</sup>$$\left( {{4 \over 5}} \right)$$
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ = cos<sup>-1</sup>$$\left( {{4 \over 5}} \right)$$
<br><br>$$ \Rightarrow $$ $${x \over 5}$$ = sin(cos<sup>-1</sup>$$ {{4 \over 5}}$$)
<br><br>$$ \Rightarrow $$ $${x \over 5}$$ = sin(sin<sup>-1</sup>$$ {{3 \over 5}}$$)
<br><br>$$ \Rightarrow $$ $${x \over 5}$$ = $${3 \over 5}$$
<br><br>$$ \Rightarrow $$ x = 3 | mcq | aieee-2007 |
IgvCQgf8GDVXARHa | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of $$cot\left( {\cos e{c^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$ is : | [{"identifier": "A", "content": "$${{6 \\over 17}}$$"}, {"identifier": "B", "content": "$${{3 \\over 17}}$$"}, {"identifier": "C", "content": "$${{4 \\over 17}}$$"}, {"identifier": "D", "content": "$${{5 \\over 17}}$$"}] | ["A"] | null | Given,
<br><br>$$Cot\left( {so{{\sec }^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$
<br><br>$$ = \cot \left( {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$
<br><br>$$ = cot\left( {{{\tan }^{ - 1}}{{{3 \over 4} + {2 \over 3}} \over {1 - {3 \over 4} - {2 \over 3}}}} \right)$$
<br><br>$$ = \cot \left( {{{\tan }^{ - 1}}\left( {{{17} \over 6}} \right)} \right)$$
<br><br>$$ = \cot \left( {{{\cot }^{ - 1}}{6 \over {17}}} \right)$$
<br><br>$$ = {6 \over {17}}$$ | mcq | aieee-2008 |
uhYKiIhxgTCzlkLr | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $$x, y, z$$ are in A.P. and $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$$ and $${\tan ^{ - 1}}z$$ are also in A.P., then : | [{"identifier": "A", "content": "$$x=y=z$$ "}, {"identifier": "B", "content": "$$2x=3y=6z$$ "}, {"identifier": "C", "content": "$$6x=3y=2z$$ "}, {"identifier": "D", "content": "$$6x=4y=3z$$"}] | ["A"] | null | Given that, $$x,y,z\,\,$$ are in $$AP$$
<br><br>So, $$\,\,\,$$ $$2y = x + y$$
<br><br>Also given that,
<br><br> $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y\,\,$$ and $$\,\,\,{\tan ^{ - 1}}z\,\,$$ are in $$AP$$
<br><br>So, $$2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z$$
<br><br>$$ \Rightarrow {\tan ^{ - 1}}\left( {{{2y} \over {1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {{{x + 7} \over {1 - xz}}} \right)$$
<br><br>$$ \Rightarrow {{2y} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$$
<br><br>$$ \Rightarrow {{x + z} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$$
<br><br>[ as $$\,\,\,\,$$ $$2y = x + z$$]
<br><br>$$ \Rightarrow 1 - {y^2} = 1 - xz$$
<br><br>$$ \Rightarrow $$ $${y^2} = xz$$
<br><br>As we get $${y^2} = xz,$$ so, $$x,y,z$$ are in $$GP.$$
<br><br>According to the question $$x,y,z$$ are $$AP.$$
<br><br>$$x,y,z$$ both can be $$AP$$ as well as $$GP$$
<br><br>when $$x=y=z.$$ | mcq | jee-main-2013-offline |
tb7Srq7xtkaCBPWe | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let $${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right),$$
<br/> where $$\left| x \right| < {1 \over {\sqrt 3 }}.$$ Then a value of $$y$$ is : | [{"identifier": "A", "content": "$${{3x - {x^3}} \\over {1 + 3{x^2}}}$$ "}, {"identifier": "B", "content": "$${{3x + {x^3}} \\over {1 + 3{x^2}}}$$"}, {"identifier": "C", "content": "$${{3x - {x^3}} \\over {1 - 3{x^2}}}$$ "}, {"identifier": "D", "content": "$${{3x + {x^3}} \\over {1 - 3{x^2}}}$$ "}] | ["C"] | null | Given,
<br><br>$${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)$$
<br><br>$$ \Rightarrow {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} \right)}}} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\tan ^{ - 1}}\left( {{{x - {x^3} + 2x} \over {1 - {x^2} - 2{x^2}}}} \right)$$
<br><br>$$\therefore$$ $$\,\,\,$$ $${\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{3x - {x^2}} \over {1 - 3{x^2}}}} \right)$$
<br><br>$$ \Rightarrow y = {{3x - {x^3}} \over {1 - 3{x^2}}}$$ | mcq | jee-main-2015-offline |
wFBhtX1szbvhwRLii253K | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of tan<sup>-1</sup> $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right],$$ $$\left| x \right| < {1 \over 2},x \ne 0,$$ is equal to : | [{"identifier": "A", "content": "$${\\pi \\over 4} + {1 \\over 2}{\\cos ^{ - 1}}\\,{x^2}$$"}, {"identifier": "B", "content": "$${\\pi \\over 4} + {\\cos ^{ - 1}}\\,{x^2}$$ "}, {"identifier": "C", "content": "$${\\pi \\over 4} - {1 \\over 2}{\\cos ^{ - 1}}\\,{x^2}$$"}, {"identifier": "D", "content": "$${\\pi \\over 4} - {\\cos ^{ - 1}}\\,{x^2}$$"}] | ["A"] | null | Given,
<br><br>tan<sup>-1</sup> $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$$
<br><br>Let x<sup>2</sup> = cos $$\theta $$
<br><br>= tan<sup>-1</sup> $$\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \cos \theta } - \sqrt {1 - \cos \theta } }}} \right]$$
<br><br>= tan<sup>-1</sup> $$\left[ {{{\sqrt {2{{\cos }^2}{\theta \over 2}} + \sqrt {2{{\sin }^2}{\theta \over 2}} } \over {\sqrt {2{{\cos }^2}{\theta \over 2}} - \sqrt {2{{\sin }^2}{\theta \over 2}} }}} \right]$$
<br><br>= tan<sup>-1</sup> $$\left[ {{{\sqrt 2 \cos {\theta \over 2} + \sqrt 2 \sin {\theta \over 2}} \over {\sqrt 2 \cos {\theta \over 2} - \sqrt 2 \sin {\theta \over 2}}}} \right]$$
<br><br>= tan<sup>-1</sup> $$\left[ {{{\cos {\theta \over 2} + \sin {\theta \over 2}} \over {\cos {\theta \over 2} - \sin {\theta \over 2}}}} \right]$$
<br><br>= tan<sup>-1</sup> $$\left[ {{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}}} \right]$$
<br><br>= tan<sup>-1</sup> $$\left[ {{{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4} + \tan {\theta \over 2}}}} \right]$$
<br><br>= tan<sup>-1</sup> $$\left( {\tan \left( {{\pi \over 4} + {\theta \over 2}} \right)} \right)$$
<br><br>= $${\pi \over 4} + {\theta \over 2}$$
<br><br>= $${\pi \over 4} + {1 \over 2}$$ cos<sup>-1</sup> x<sup>2</sup> | mcq | jee-main-2017-online-8th-april-morning-slot |
1AVHKdvqPiPIR7eBWvXeh | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | A value of x satisfying the equation sin[cot<sup>−1</sup> (1+ x)] = cos [tan<sup>−1 </sup>x], is : | [{"identifier": "A", "content": "$$ - {1 \\over 2}$$ "}, {"identifier": "B", "content": "$$-$$ 1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$ {1 \\over 2}$$"}] | ["A"] | null | <p>Let, $${\cot ^{ - 1}}(1 + x) = \alpha $$</p>
<p>$$ \Rightarrow 1 + x = \cot \alpha $$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0v2ql/1fa8d111-320f-4de3-8dab-6d23cbe7b0e2/436fe4d0-d65a-11ec-9a06-bd4ec5b93eb4/file-1l3b0v2qm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3b0v2ql/1fa8d111-320f-4de3-8dab-6d23cbe7b0e2/436fe4d0-d65a-11ec-9a06-bd4ec5b93eb4/file-1l3b0v2qm.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2017 (Online) 9th April Morning Slot Mathematics - Inverse Trigonometric Functions Question 68 English Explanation 1"></p>
<p>Now, let $${\tan ^{ - 1}}(x) = \beta $$</p>
<p>$$ \Rightarrow x = \tan \beta $$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0vpg8/606245c5-bf1b-4633-a0a9-2c4dec6f8121/54fb5180-d65a-11ec-9a06-bd4ec5b93eb4/file-1l3b0vpg9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3b0vpg8/606245c5-bf1b-4633-a0a9-2c4dec6f8121/54fb5180-d65a-11ec-9a06-bd4ec5b93eb4/file-1l3b0vpg9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2017 (Online) 9th April Morning Slot Mathematics - Inverse Trigonometric Functions Question 68 English Explanation 2"></p>
<p>Given, $$\sin ({\cot ^{ - 1}}(x)) = \cos ({\tan ^{ - 1}}(x))$$</p>
<p>$$ \Rightarrow \sin \alpha = \cos \beta $$</p>
<p>From $$\Delta$$ABC, $$\sin \alpha = {1 \over {\sqrt {1 + {x^2} + 2x + 1} }}$$</p>
<p>$$ = {1 \over {\sqrt {{x^2} + 2x + 2} }}$$</p>
<p>From $$\Delta$$MNO, $$\cos \beta = {1 \over {\sqrt {{x^2} + 1} }}$$</p>
<p>$$\therefore$$ $${1 \over {\sqrt {{x^2} + 2x + 2} }} = {1 \over {\sqrt {{x^2} + 1} }}$$</p>
<p>$$ \Rightarrow {x^2} + 2x + 2 = {x^2} + 1$$</p>
<p>$$ \Rightarrow 2x + 2 = 1$$</p>
<p>$$ \Rightarrow 2x = - 1$$</p>
<p>$$ \Rightarrow x = -{1 \over 2}$$</p> | mcq | jee-main-2017-online-9th-april-morning-slot |
XPZSMXZi1ZlGHmdnVX3rsa0w2w9jx2gxojq | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$,where –1 $$ \le $$ x $$ \le $$ 1, – 2 $$ \le $$ y $$ \le $$ 2, x $$ \le $$ $${y \over 2}$$
, then for all x, y, 4x<sup>2</sup>
– 4xy cos $$\alpha $$ + y<sup>2</sup>
is equal
to : | [{"identifier": "A", "content": "4 sin<sup>2</sup> $$\\alpha $$"}, {"identifier": "B", "content": "2 sin<sup>2</sup> $$\\alpha $$"}, {"identifier": "C", "content": "4 sin<sup>2</sup> $$\\alpha $$ - 2x<sup>2</sup>y<sup>2</sup>"}, {"identifier": "D", "content": "4 cos<sup>2</sup> $$\\alpha $$ + 2x<sup>2</sup>y<sup>2</sup>"}] | ["A"] | null | $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$<br><br>
$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha $$<br><br>
$$ \Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha $$<br><br>
$$\left( {\cos \alpha - {{xy} \over 2}} \right) = \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} $$<br><br>
squaring both sides<br><br>
$${x^2} + {{{y^2}} \over 4} - xy\cos \alpha = 1 - {\cos ^2}\alpha = {\sin ^2}\alpha $$<br><br>
$$ \therefore $$ 4x<sup>2</sup>
– 4xy cos $$\alpha $$ + y<sup>2</sup> = 4 $${\sin ^2}\alpha $$ | mcq | jee-main-2019-online-10th-april-evening-slot |
8ggGtIclV7LzC4NVgY3rsa0w2w9jx6ge0vw | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of $${\sin ^{ - 1}}\left( {{{12} \over {13}}} \right) - {\sin ^{ - 1}}\left( {{3 \over 5}} \right)$$ is equal to : | [{"identifier": "A", "content": "$$\\pi - {\\sin ^{ - 1}}\\left( {{{63} \\over {65}}} \\right)$$"}, {"identifier": "B", "content": "$${\\pi \\over 2} - {\\sin ^{ - 1}}\\left( {{{56} \\over {65}}} \\right)$$"}, {"identifier": "C", "content": "$${\\pi \\over 2} - {\\cos ^{ - 1}}\\left( {{9 \\over {65}}} \\right)$$"}, {"identifier": "D", "content": "$$\\pi - {\\cos ^{ - 1}}\\left( {{{33} \\over {65}}} \\right)$$"}] | ["B"] | null | $${\sin ^{ - 1}}{{12} \over {13}} - {\sin ^{ - 1}}{3 \over 5} = {\sin ^{ - 1}}\left( {{{12} \over {13}}.{4 \over 5}.{3 \over 5}.{5 \over {13}}} \right)$$<br><br>
$$ \Rightarrow {\sin ^{ - 1}}{{33} \over {65}} = {\pi \over 2} - {\cos ^{ - 1}}{{33} \over {65}}$$<br><br>
$$ \Rightarrow {\pi \over 2} - {\sin ^{ - 1}}{{56} \over {65}}$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
LzROsT3Flxd6M6C9U01L3 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $$\alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$, $$\beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right)$$ where $$0 < \alpha ,\beta < {\pi \over 2}$$ , then $$\alpha $$ - $$\beta $$ is equal to : | [{"identifier": "A", "content": "$${\\tan ^{ - 1}}\\left( {{9 \\over {14 }}} \\right)$$"}, {"identifier": "B", "content": "$${\\sin ^{ - 1}}\\left( {{9 \\over {5\\sqrt {10} }}} \\right)$$"}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{9 \\over {5\\sqrt {10} }}} \\right)$$"}, {"identifier": "D", "content": "$${\\tan ^{ - 1}}\\left( {{9 \\over {5\\sqrt {10} }}} \\right)$$"}] | ["B"] | null | Here $$\cos \alpha = {3 \over 5}$$
<br><br>$$ \therefore $$ $$\tan \alpha = {4 \over 3}$$
<br><br>and $$\tan \beta = {1 \over 3}$$
<br><br>We know,
<br><br>$$\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}$$
<br><br>= $${{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}}$$ = $${9 \over {13}}$$
<br><br>$$ \therefore $$ $$\left( {\alpha - \beta } \right)$$ = $${\tan ^{ - 1}}\left( {{9 \over {13}}} \right)$$ = $${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$$
| mcq | jee-main-2019-online-8th-april-morning-slot |
0fOD1sLduw0qCWTfaKgYw | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | All x satisfying the inequality (cot<sup>–1</sup>
x)<sup>2</sup>– 7(cot<sup>–1</sup> x) + 10 > 0, lie in the interval : | [{"identifier": "A", "content": "(cot 2, $$\\infty $$)"}, {"identifier": "B", "content": "(\u2013$$\\infty $$, cot 5) $$ \\cup $$ (cot 2, $$\\infty $$)"}, {"identifier": "C", "content": "(cot 5, cot 4)"}, {"identifier": "D", "content": "(\u2013 $$\\infty $$, cot 5) $$ \\cup $$ (cot 4, cot 2)"}] | ["A"] | null | cot<sup>$$-$$1</sup> x > 5, cot<sup>$$-$$1</sup> x < 2
<br><br>$$ \Rightarrow $$ x < cot5, x > cot2 | mcq | jee-main-2019-online-11th-january-evening-slot |
6LQjR16XBRAFABZMYrmun | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If x = sin<sup>$$-$$1</sup>(sin10) and y = cos<sup>$$-$$1</sup>(cos10), then y $$-$$ x is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "7$$\\pi $$"}, {"identifier": "D", "content": "$$\\pi $$"}] | ["D"] | null | x = sin<sup>$$-$$1</sup> sin 10 = 3$$\pi $$ $$-$$ 10
<br><br>y = cos<sup>$$-$$1</sup>cos 10 = 4$$\pi $$ $$-$$ 10
<br><br>y $$-$$ x = (4$$\pi $$ $$-$$ 10) $$-$$ (3$$\pi $$ $$-$$ 10) = $$\pi $$ | mcq | jee-main-2019-online-9th-january-evening-slot |
zmr9GaOF2EgS90jIGdF4s | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$ (x > $$3 \over 4$$), then x is equal to : | [{"identifier": "A", "content": "$${{\\sqrt {145} } \\over {10}}$$"}, {"identifier": "B", "content": "$${{\\sqrt {145} } \\over {11}}$$"}, {"identifier": "C", "content": "$${{\\sqrt {145} } \\over {12}}$$"}, {"identifier": "D", "content": "$${{\\sqrt {146} } \\over {12}}$$"}] | ["C"] | null | Given,
<br><br>$${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$
<br><br>$$ \Rightarrow {\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right)$$
<br><br>$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {{2 \over {3x}}} \right)} \right) = \cos \left[ {{\pi \over 2} - {{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right]$$
<br><br>$$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right\}$$
<br><br>$$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\sin }^{ - 1}}{{\sqrt {16{x^2} - 9} } \over {4x}}} \right\}$$
<br><br>$$ \Rightarrow {2 \over {3x}} = {{16{x^2} - 9} \over {4x}}$$
<br><br>$$ \Rightarrow 64 = 9\left( {16{x^2} - 9} \right)$$
<br><br>$$ \Rightarrow 16{x^2} - 9 = {{64} \over 9}$$
<br><br>$$ \Rightarrow 16{x^2} = {{64} \over 9} + 9$$
<br><br>$$ \Rightarrow 16{x^2} = - {{145} \over 9}$$
<br><br>$$ \Rightarrow x = \pm {{\sqrt {145} } \over {4 \times 3}}$$
<br><br>$$ \Rightarrow x = \pm {{\sqrt {145} } \over {12}}$$
<br><br>as given that $$x > {3 \over 4}$$
<br><br>$$ \therefore $$ x $$ \ne $$ $$ - {{\sqrt {145} } \over {12}}$$
<br><br>$$ \therefore $$ x $$ = {{\sqrt {145} } \over {12}}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
UKJTY6DgHPShPaIPIOjgy2xukf0y124i | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | 2$$\pi $$ - $$\left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${{7\\pi } \\over 4}$$"}, {"identifier": "B", "content": "$${{5\\pi } \\over 4}$$"}, {"identifier": "C", "content": "$${{3\\pi } \\over 2}$$"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$"}] | ["C"] | null | $$2\pi - \left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$$<br><br>$$ = 2\pi - \left( {{{\tan }^{ - 1}}{4 \over 3} + {{\tan }^{ - 1}}{5 \over {12}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$$<br><br>$$ = 2\pi - \left\{ {{{\tan }^{ - 1}}\left( {{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3}.{5 \over {12}}}}} \right) + {{\tan }^{ - 1}}{{16} \over {63}}} \right\}$$<br><br>$$ = 2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$$<br><br>= $$2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\cot }^{ - 1}}{{63} \over {16}}} \right)$$<br><br>$$ = 2\pi - {\pi \over 2}$$<br><br>$$ = {{3\pi } \over 2}$$ | mcq | jee-main-2020-online-3rd-september-morning-slot |
KKoj2g0io0AWLE2BYi1klrm0vh5 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | A possible value of $$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$$ is : | [{"identifier": "A", "content": "$$\\sqrt 7 - 1$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 7 }}$$"}, {"identifier": "C", "content": "$$2\\sqrt 2 - 1$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt 2 }}$$"}] | ["B"] | null | $$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$$<br><br>$${\sin ^{ - 1}}\left( {{{\sqrt {63} } \over 8}} \right) = \theta $$ $$\sin \theta = {{\sqrt {63} } \over 8}$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265142/exam_images/vmbo5oe8fbvyyyiih8ml.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Evening Shift Mathematics - Inverse Trigonometric Functions Question 56 English Explanation"><br><br>$$\cos \theta = {1 \over 8}$$<br><br>$$2{\cos ^2}{\theta \over 2} - 1 = {1 \over 8}$$<br><br>$${\cos ^2}{\theta \over 2} = {9 \over {16}}$$<br><br>$$\cos {\theta \over 2} = {3 \over 4}$$<br><br>$${{1 - {{\tan }^2}{\theta \over 4}} \over {1 + {{\tan }^2}{\theta \over 4}}} = {3 \over 4}$$<br><br>$$\tan {\theta \over 4} = {1 \over {\sqrt 7 }}$$ | mcq | jee-main-2021-online-24th-february-evening-slot |
cBLZQ3EuvzBZDVfyOI1klt7uc51 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | cosec$$\left[ {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right]$$ is equal to : | [{"identifier": "A", "content": "$${{75} \\over {56}}$$"}, {"identifier": "B", "content": "$${{65} \\over {56}}$$"}, {"identifier": "C", "content": "$${{56} \\over {33}}$$"}, {"identifier": "D", "content": "$${{65} \\over {33}}$$"}] | ["B"] | null | $$\cos ec\left( {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$$<br><br>$$\cos ec\left( {2{{\tan }^{ - 1}}\left( {{1 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$$<br><br>$$ = \cos ec\left( {{{\tan }^{ - 1}}\left( {{{2\left( {{1 \over 5}} \right)} \over {1 - {{\left( {{1 \over 5}} \right)}^2}}}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$$<br><br>$$ = \cos ec\left( {{{\tan }^{ - 1}}\left( {{5 \over {12}}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$$<br><br>Let $${\tan ^{ - 1}}(5/12) = \theta \Rightarrow \sin \theta = {5 \over {13}},\cos \theta = {{12} \over {13}}$$<br><br>and $${\cos ^{ - 1}}\left( {{4 \over 5}} \right) = \phi \Rightarrow \cos \phi = {4 \over 5}$$ and $$\sin \phi = {3 \over 5}$$<br><br>$$ = \cos ec(\theta + \phi )$$<br><br>$$ = {1 \over {\sin \theta \cos \phi + \cos \theta \sin \phi }}$$<br><br>$$ = {1 \over {{5 \over {13}}.{4 \over 5} + {{12} \over {13}}.{3 \over 5}}} = {{65} \over {56}}$$ | mcq | jee-main-2021-online-25th-february-evening-slot |
pVJ6lPiMpUPawu7uY81klugzvt7 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${{{{\sin }^1}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$; $$0 < x < 1$$, <br/>then the value of $$\cos \left( {{{\pi c} \over {a + b}}} \right)$$ is : | [{"identifier": "A", "content": "$${{1 - {y^2}} \\over {2y}}$$"}, {"identifier": "B", "content": "$${{1 - {y^2}} \\over {y\\sqrt y }}$$"}, {"identifier": "C", "content": "$$1 - {y^2}$$"}, {"identifier": "D", "content": "$${{1 - {y^2}} \\over {1 + {y^2}}}$$"}] | ["D"] | null | $${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$<br><br>$${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}$$<br><br>Now, $${{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + b)}}$$<br><br>$$2{\tan ^{ - 1}}y = {{\pi c} \over {a + b}}$$<br><br>$$ \Rightarrow \cos \left( {{{\pi c} \over {a + b}}} \right) = \cos (2{\tan ^{ - 1}}y) = {{1 - {y^2}} \over {1 + {y^2}}}$$ | mcq | jee-main-2021-online-26th-february-morning-slot |
mbltkG0zPddqjhSIul1kluw1xwf | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If 0 < a, b < 1, and tan<sup>$$-$$1</sup>a + tan<sup>$$-$$1</sup>b = $${\pi \over 4}$$, then the value of <br/><br/>$$(a + b) - \left( {{{{a^2} + {b^2}} \over 2}} \right) + \left( {{{{a^3} + {b^3}} \over 3}} \right) - \left( {{{{a^4} + {b^4}} \over 4}} \right) + .....$$ is : | [{"identifier": "A", "content": "$${\\log _e}$$2"}, {"identifier": "B", "content": "e"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{e \\over 2}} \\right)$$"}, {"identifier": "D", "content": "e<sup>2</sup> = 1"}] | ["A"] | null | tan<sup>$$-$$1</sup>a + tan<sup>$$-$$1</sup>b = $${\pi \over 4}$$ 0 < a, b < 1<br><br>$$ \Rightarrow {{a + b} \over {1 - ab}} = 1$$<br><br>a + b = 1 $$-$$ ab<br><br>(a + 1)(b + 1) = 2<br><br>Now $$\left[ {a - {{{a^2}} \over 2} + {{{a^3}} \over 3} + ....} \right] + \left[ {b - {{{b^2}} \over 2} + {{{b^3}} \over 3} + ....} \right]$$<br><br>$$ = {\log _e}(1 + a) + {\log _e}(1 + b)$$<br><br>($$ \because $$ expansion of log<sub>e</sub>(1 + x))<br><br>$$ = {\log _e}[(1 + a)(1 + b)]$$<br><br>$$ = {\log _e}2$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
1nIy9Pldg3q5vC5Kur1kmjakoz4 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The sum of possible values of x for <br/><br/>tan<sup>$$-$$1</sup>(x + 1) + cot<sup>$$-$$1</sup>$$\left( {{1 \over {x - 1}}} \right)$$ = tan<sup>$$-$$1</sup>$$\left( {{8 \over {31}}} \right)$$ is : | [{"identifier": "A", "content": "$$-$$$${{{32} \\over 4}}$$"}, {"identifier": "B", "content": "$$-$$$${{{33} \\over 4}}$$"}, {"identifier": "C", "content": "$$-$$$${{{31} \\over 4}}$$"}, {"identifier": "D", "content": "$$-$$$${{{30} \\over 4}}$$"}] | ["A"] | null | tan<sup>$$-$$1</sup>(x + 1) + cot<sup>$$-$$1</sup>$$\left( {{1 \over {x - 1}}} \right)$$ = tan<sup>$$-$$1</sup>$$\left( {{8 \over {31}}} \right)$$
<br><br>$$ \Rightarrow $$ tan<sup>$$-$$1</sup>(x + 1) + tan<sup>$$-$$1</sup>(x - 1) = tan<sup>$$-$$1</sup>$$\left( {{8 \over {31}}} \right)$$
<br><br>$$ \Rightarrow $$ $${\tan ^{ - 1}}\left( {{{\left( {x + 1} \right) + \left( {x - 1} \right)} \over {1 - \left( {x + 1} \right)\left( {x - 1} \right)}}} \right)$$ = tan<sup>$$-$$1</sup>$$\left( {{8 \over {31}}} \right)$$
<br><br>$$ \Rightarrow $$ $${{(1 + x) + (x - 1)} \over {1 - (1 + x)(x - 1)}} = {8 \over {31}}$$<br><br>$$ \Rightarrow {{2x} \over {2 - {x^2}}} = {8 \over {31}}$$<br><br>$$ \Rightarrow 4{x^2} + 31x - 8 = 0$$<br><br>$$ \Rightarrow x = - 8,{1 \over 4}$$<br><br>but at $$x = {1 \over 4}$$<br><br>$$LHS > {\pi \over 2}$$ and $$RHS < {\pi \over 2}$$<br><br>So, only solution is x = $$-$$ 8 = $$-$$$${{{32} \over 4}}$$ | mcq | jee-main-2021-online-17th-march-morning-shift |
1krrolzup | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of $$\tan \left( {2{{\tan }^{ - 1}}\left( {{3 \over 5}} \right) + {{\sin }^{ - 1}}\left( {{5 \over {13}}} \right)} \right)$$ is equal to : | [{"identifier": "A", "content": "$${{ - 181} \\over {69}}$$"}, {"identifier": "B", "content": "$${{220} \\over {21}}$$"}, {"identifier": "C", "content": "$${{ - 291} \\over {76}}$$"}, {"identifier": "D", "content": "$${{151} \\over {63}}$$"}] | ["B"] | null | $$2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) = {\tan ^{ - 1}}\left( {{{6/5} \over {1 - {9 \over {{2^5}}}}}} \right) = {\tan ^{ - 1}}\left( {{{{6 \over 5}} \over {{{16} \over {25}}}}} \right) = {\tan ^{ - 1}}{{15} \over 8}$$<br><br>$$\therefore$$ $$2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) + {\sin ^{ - 1}}\left( {{5 \over {13}}} \right) = {\tan ^{ - 1}}\left( {{{15} \over 8}} \right) + {\tan ^{ - 1}}\left( {{5 \over {12}}} \right)$$<br><br>$$ = {\tan ^{ - 1}}\left( {{{{{15} \over 8} + {5 \over {12}}} \over {1 - {{15} \over 8},{5 \over {12}}}}} \right)$$<br><br>$$ = {\tan ^{ - 1}}\left( {{{180 + 40} \over {21}}} \right) = {\tan ^{ - 1}}\left( {{{220} \over {21}}} \right)$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1ktei3u10 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$$; 0 < x < 1, a $$\ne$$ 0, then the value of 2x<sup>2</sup> $$-$$ 1 is : | [{"identifier": "A", "content": "$$\\cos \\left( {{{4a} \\over \\pi }} \\right)$$"}, {"identifier": "B", "content": "$$\\sin \\left( {{{2a} \\over \\pi }} \\right)$$"}, {"identifier": "C", "content": "$$\\cos \\left( {{{2a} \\over \\pi }} \\right)$$"}, {"identifier": "D", "content": "$$\\sin \\left( {{{4a} \\over \\pi }} \\right)$$"}] | ["B"] | null | Given $$a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}$$<br><br>$$ = ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$$<br><br>$$ = {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)$$<br><br>$$ \Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }$$<br><br>$$ \Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi \over 2} - {{2a} \over \pi }$$<br><br>$$ \Rightarrow 2{x^2} - 1 = \cos \left( {{\pi \over 2} - {{2a} \over \pi }} \right)$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktfwajo8 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let M and m respectively be the maximum and minimum values of the function <br/>f(x) = tan<sup>$$-$$1</sup> (sin x + cos x) in $$\left[ {0,{\pi \over 2}} \right]$$, then the value of tan(M $$-$$ m) is equal to : | [{"identifier": "A", "content": "$$2 + \\sqrt 3 $$"}, {"identifier": "B", "content": "$$2 - \\sqrt 3 $$"}, {"identifier": "C", "content": "$$3 + 2\\sqrt 2 $$"}, {"identifier": "D", "content": "$$3 - 2\\sqrt 2 $$"}] | ["D"] | null | Let g(x) = sin x + cos x = $$\sqrt 2 $$ sin$$\left( {x + {\pi \over 4}} \right)$$<br><br>g(x)$$\in$$ $$\left[ {1,\sqrt 2 } \right]$$ for x$$\in$$ [0, $$\pi$$/2]<br><br>f(x) = tan<sup>$$-$$1</sup> (sin x + cos x) $$\in$$ $$\left[ {{\pi \over 4},{{\tan }^{ - 1}}\sqrt 2 } \right]$$<br><br>tan$$({\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}) = {{\sqrt 2 - 1} \over {1 + \sqrt 2 }} \times {{\sqrt 2 - 1} \over {\sqrt 2 - 1}} = 3 - 2\sqrt 2 $$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1l5464te0 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>$$50\tan \left( {3{{\tan }^{ - 1}}\left( {{1 \over 2}} \right) + 2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right) + 4\sqrt 2 \tan \left( {{1 \over 2}{{\tan }^{ - 1}}(2\sqrt 2 )} \right)$$ is equal to ____________.</p> | [] | null | 29 | $50 \tan \left(\tan ^{-1} \frac{1}{2}+2 \tan ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(2)\right)$
$$
+4 \sqrt{2} \tan \left(\frac{\tan ^{-1}}{2}(2 \sqrt{2})\right)
$$
<br/><br/>
$\Rightarrow 50 \tan \left(\pi+\tan ^{-1}\left(\frac{1}{2}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$
<br/><br/>
$\Rightarrow \quad 50\left(\frac{1}{2}\right)+4 \sqrt{2} \tan \alpha$
<br/><br/>
Where $2 \alpha=\tan ^{-1} 2 \sqrt{2}$
<br/><br/>
$\Rightarrow \frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=2 \sqrt{2} \quad$.. (i)
<br/><br/>
$\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+2 \tan \alpha-2 \sqrt{2}=0$
<br/><br/>
$\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+4 \tan \alpha-2 \tan \alpha-2 \sqrt{2}=0$
<br/><br/>
$\Rightarrow(2 \sqrt{2} \tan \alpha-2)(\tan \alpha-\sqrt{2})=0$
<br/><br/>
$\Rightarrow \tan \alpha=\sqrt{2}$ or $\frac{1}{\sqrt{2}}$
<br/><br/>
$\Rightarrow \tan \alpha=\frac{1}{\sqrt{2}}$
<br/><br/>
$(\tan \alpha=\sqrt{2}$ doesn't satisfy (i))
<br/><br/>
$\Rightarrow \quad 25+4 \sqrt{2} \frac{1}{\sqrt{2}}=29$ | integer | jee-main-2022-online-29th-june-morning-shift |
1l56rlph0 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The value of $$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$$ is :</p> | [{"identifier": "A", "content": "$${{26} \\over {25}}$$"}, {"identifier": "B", "content": "$${{25} \\over {26}}$$"}, {"identifier": "C", "content": "$${{50} \\over {51}}$$"}, {"identifier": "D", "content": "$${{52} \\over {51}}$$"}] | ["A"] | null | <p>$$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$$</p>
<p>$$ = \cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(n + 1) - n} \over {1 + (n + 1)n}}} \right)} } \right)$$</p>
<p>$$ = \cot \left( {\sum\limits_{n = 1}^{50} {({{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}n} } \right)$$</p>
<p>$$ = \cot ({\tan ^{ - 1}}51 - {\tan ^{ - 1}}1)$$</p>
<p>$$ = \cot \left( {{{\tan }^{ - 1}}\left( {{{51 - 1} \over {1 + 51}}} \right)} \right)$$</p>
<p>$$ = \cot \left( {{{\cot }^{ - 1}}\left( {{{52} \over {50}}} \right)} \right)$$</p>
<p>$$ = {{26} \over {25}}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l59kzd00 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The value of $${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$ - {\\pi \\over 4}$$"}, {"identifier": "B", "content": "$$ - {\\pi \\over 8}$$"}, {"identifier": "C", "content": "$$ - {{5\\pi } \\over {12}}$$"}, {"identifier": "D", "content": "$$ - {{4\\pi } \\over 9}$$"}] | ["B"] | null | <p>$${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin {\pi \over 4}}}} \right)$$</p>
<p>$$ = {\tan ^{ - 1}}\left( {{{{1 \over {\sqrt 2 }} - 1} \over {{1 \over {\sqrt 2 }}}}} \right)$$</p>
<p>$$ = {\tan ^{ - 1}}(1 - \sqrt 2 ) = - {\tan ^{ - 1}}(\sqrt 2 - 1)$$</p>
<p>$$ = - {\pi \over 8}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5b7rjam | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$x * y = {x^2} + {y^3}$$ and $$(x * 1) * 1 = x * (1 * 1)$$.</p>
<p>Then a value of $$2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$$ is :</p> | [{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$"}, {"identifier": "D", "content": "$${\\pi \\over 6}$$"}] | ["B"] | null | The star "*" in this context represents a binary operation, similar to addition (+), subtraction (-), multiplication (×), and division (÷). It is a custom operation defined by the problem statement, and the specific rules of the operation are provided in the problem.
<br/><br/>
In this case, the operation "*" is defined by the equation $x * y = x^2 + y^3$, which means if you have two numbers $x$ and $y$, then the result of applying the "*" operation to them is $x^2 + y^3$.
<br/><br/>
The problem also specifies an additional rule for this operation: $(x * 1) * 1 = x * (1 * 1)$, which needs to be taken into account when solving the problem. This is a type of "associativity" condition.
<p>Given,</p>
<p>$$x\, * \,y = {x^2} + {y^3}$$</p>
<p>$$\therefore$$ $$x\, * \,1 = {x^2} + {1^3} = {x^2} + 1$$</p>
<p>Now, $$(x\, * \,1)\, * \,1 = ({x^2} + 1)\, * \,1$$</p>
<p>$$ \Rightarrow (x\, * \,1)\, * \,1 = {({x^2} + 1)^2} + {1^3}$$</p>
<p>$$ \Rightarrow (x\, * \,1)\, * \,1 = {x^4} + 1 + 2{x^2} + 1$$</p>
<p>Also, $$x\, * \,(1\, * \,1)$$</p>
<p>$$ = x\, * \,({1^2} + {1^3})$$</p>
<p>$$ = x\, * \,2$$</p>
<p>$$ = {x^2} + {2^3}$$</p>
<p>$$ = {x^2} + 8$$</p>
<p>Given that,</p>
<p>$$(x\, * \,1)\, * \,1 = x\, * \,(1\, * \,1)$$</p>
<p>$$\therefore$$ $${x^4} + 1 + 2{x^2} + 1 = {x^2} + 8$$</p>
<p>$$ \Rightarrow {x^4} + {x^2} - 6 = 0$$</p>
<p>$$ \Rightarrow {x^4} + 3{x^2} - 2{x^2} - 6 = 0$$</p>
<p>$$ \Rightarrow {x^2}({x^2} + 3) - 2({x^3} + 3) = 0$$</p>
<p>$$ \Rightarrow ({x^2} + 3)({x^2} - 2) = 0$$</p>
<p>$$ \Rightarrow {x^2} = 2,\, - 3$$</p>
<p>[$${x^2} = -3$$ not possible as square of anything should be always positive]
<p>$$\therefore$$ $${x^2} = 2$$</p>
<p>$$\therefore$$ Now,</p>
<p>$$2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$$</p>
<p>$$ = 2{\sin ^{ - 1}}\left( {{{{2^2} + 2 - 2} \over {{2^2} + 2 + 2}}} \right)$$</p>
<p>$$ = 2{\sin ^{ - 1}}\left( {{4 \over 8}} \right)$$</p>
<p>$$ = 2{\sin ^{ - 1}}\left( {{1 \over 2}} \right)$$</p>
<p>$$ = 2 \times {\pi \over 6}$$</p>
<p>$$ = {\pi \over 3}$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5c13084 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The set of all values of k for which <br/><br/>$${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3},\,x \in R$$, is the interval :</p> | [{"identifier": "A", "content": "$$\\left[ {{1 \\over {32}},{7 \\over 8}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{1 \\over {24}},{{13} \\over {16}}} \\right)$$"}, {"identifier": "C", "content": "$$\\left[ {{1 \\over {48}},{{13} \\over {16}}} \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {{1 \\over {32}},{9 \\over 8}} \\right)$$"}] | ["A"] | null | <p>$${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3}$$</p>
<p>Let $$f(t) = {t^3} + {\left( {{\pi \over 2} - t} \right)^3}$$</p>
<p>Where $$t = {\tan ^{ - 1}}x$$ ; $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$</p>
<p>$$ = {t^3} + {\left( {{\pi \over 2}} \right)^3} - {{3{\pi ^2}t} \over 4} + {{3\pi } \over 2}{t^2} - {t^3}$$</p>
<p>$$f(t) = {{3\pi } \over 2}{t^2} - {{3{\pi ^2}} \over 4}\,.\,t + {{{\pi ^3}} \over 8}$$</p>
<p>This is a quadratic equation of t.</p>
<p>Here, coefficient of t<sup>2</sup> term is $${{3\pi } \over 2}$$ which is > 0.</p>
<p>$$\therefore$$ It is a upward parabola.</p>
<p>Now, $$f'(t) = 3\pi t - {{3{\pi ^2}} \over 4}$$</p>
<p>$$f''(t) = 3\pi > 0$$</p>
<p>$$\therefore$$ $$3\pi t - {{3{\pi ^2}} \over 4} = 0$$</p>
<p>$$ \Rightarrow t = {\pi \over 4}$$ (minima)</p>
<p>$$\therefore$$ vertex of graph at $${\pi \over 4}$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5d349j2/93a986e3-81fb-491a-9045-ba91b8ee8ccd/15480400-ff15-11ec-8a3d-a39bde0e21e9/file-1l5d349j5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5d349j2/93a986e3-81fb-491a-9045-ba91b8ee8ccd/15480400-ff15-11ec-8a3d-a39bde0e21e9/file-1l5d349j5.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Morning Shift Mathematics - Inverse Trigonometric Functions Question 32 English Explanation"> </p>
<p>$$\therefore$$ Minimum value at $${\pi \over 4}$$ and maximum value at $$-$$$${\pi \over 2}$$.</p>
<p>$$\therefore$$ $$f\left( {{\pi \over 4}} \right) = {{{\pi ^3}} \over {64}} + {\left( {{\pi \over 2} - {\pi \over 4}} \right)^3} = {{{\pi ^3}} \over {32}}$$</p>
<p>$$f\left( { - {\pi \over 2}} \right) = - {{{\pi ^3}} \over 8} + {\pi ^3}$$</p>
<p>$$ = {{7{\pi ^3}} \over 8}$$</p>
<p>$$\therefore$$ $$k{\pi ^3} \in \left[ {{{{\pi ^3}} \over {32}},\,{{7{\pi ^3}} \over 8}} \right)$$</p>
<p>$$ \Rightarrow k \in \left[ {{1 \over {32}},\,{7 \over 8}} \right)$$</p> | mcq | jee-main-2022-online-24th-june-morning-shift |
1l5vzrho2 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let m and M respectively be the minimum and the maximum values of $$f(x) = {\sin ^{ - 1}}2x + \sin 2x + {\cos ^{ - 1}}2x + \cos 2x,\,x \in \left[ {0,{\pi \over 8}} \right]$$. Then m + M is equal to :</p> | [{"identifier": "A", "content": "$$1 + \\sqrt 2 + \\pi $$"}, {"identifier": "B", "content": "$$\\left( {1 + \\sqrt 2 } \\right)\\pi $$"}, {"identifier": "C", "content": "$$\\pi + \\sqrt 2 $$"}, {"identifier": "D", "content": "$$1 + \\pi $$"}] | ["A"] | null | <p>$$f(x) = {\sin ^{ - 1}}(2x) + \sin 2x + {\cos ^{ - 1}}(2x) + \cos 2x$$</p>
<p>$$ = {\sin ^{ - 1}}(2x) + {\cos ^{ - 1}}(2x) + \sin 2x + \cos 2x$$</p>
<p>$$ = {\pi \over 2} + \sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2x + {1 \over {\sqrt 2 }}\cos 2x} \right)$$</p>
<p>$$ = {\pi \over 2} + \sqrt 2 \left( {\cos {\pi \over 4}\sin 2x + \sin {\pi \over 4}\cos 2x} \right)$$</p>
<p>$$ = {\pi \over 2} + \sqrt 2 \,.\,\sin \left( {2x + {\pi \over 4}} \right)$$</p>
<p>f(x) is maximum when $$\sin \left( {2x + {\pi \over 4}} \right)$$ is maximum means $$x = {\pi \over 8}$$ or $$\sin \left( {2 \times {\pi \over 8} + {\pi \over 4}} \right) = \sin {\pi \over 2} = 1$$</p>
<p>$$\therefore$$ $${\left[ {f(x)} \right]_{\max }} = {\pi \over 2} + \sqrt 2 \,.\,1 = {\pi \over 2} + \sqrt 2 = M$$</p>
<p>f(x) is minimum when $$\sin \left( {2x + {\pi \over 4}} \right)$$ is minimum means $$x = 0$$ or $$\sin \left( {2 \times 0 + {\pi \over 4}} \right) = {1 \over {\sqrt 2 }}$$</p>
<p>$$\therefore$$ $${\left[ {f(x)} \right]_{\min }} = {\pi \over 2} + \sqrt 2 \,.\,{1 \over {\sqrt 2 }} = {\pi \over 2} + 1 = m$$</p>
<p>$$\therefore$$ $$m + M = {\pi \over 2} + \sqrt 2 + {\pi \over 2} + 1 = \pi + \sqrt 2 + 1$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6f3v5u9 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$x = \sin (2{\tan ^{ - 1}}\alpha )$$ and $$y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right)$$. If $$S = \{ a \in R:{y^2} = 1 - x\} $$, then $$\sum\limits_{\alpha \in S}^{} {16{\alpha ^3}} $$ is equal to _______________.</p> | [] | null | 130 | <p>$$\because$$ $$x = \sin \left( {2{{\tan }^{ - 1}}\alpha } \right) = {{2\alpha } \over {1 + {\alpha ^2}}}$$ ...... (i)</p>
<p>and $$y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right) = \sin \left( {{{\sin }^{ - 1}}{1 \over {\sqrt 5 }}} \right) = {1 \over {\sqrt 5 }}$$</p>
<p>Now, $${y^2} = 1 - x$$</p>
<p>$${1 \over 5} = 1 - {{2\alpha } \over {1 + {\alpha ^2}}}$$</p>
<p>$$ \Rightarrow 1 + {\alpha ^2} = 5 + 5{\alpha ^2} - 10\alpha $$</p>
<p>$$ \Rightarrow 2{\alpha ^2} - 5\alpha + 2 = 0$$</p>
<p>$$\therefore$$ $$\alpha = 2,{1 \over 2}$$</p>
<p>$$\therefore$$ $$\sum\limits_{\alpha \in S} {16{\alpha ^3} = 16 \times {2^3} + 16 \times {1 \over {{2^3}}} = 130} $$</p> | integer | jee-main-2022-online-25th-july-evening-shift |
1l6gj38zp | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>$$\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$\\frac{1}{4}$$"}, {"identifier": "D", "content": "$$\\frac{5}{4}$$"}] | ["B"] | null | <p>$$\tan \left( {2{{\tan }^{ - 1}}{1 \over 5} + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2} + 2{{\tan }^{ - 1}}{1 \over 8}} \right)$$</p>
<p>$$ = \tan \left( {2{{\tan }^{ - 1}}\left( {{{{1 \over 5} + {1 \over 8}} \over {1 - {1 \over 5}\,.\,{1 \over 8}}}} \right) + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2}} \right)$$</p>
<p>$$ = \tan \left[ {2{{\tan }^{ - 1}}{1 \over 3} + {{\tan }^{ - 1}}{1 \over 2}} \right]$$</p>
<p>$$ = \tan \left[ {{{\tan }^{ - 1}}{{{2 \over 3}} \over {1 - {1 \over 9}}} + {{\tan }^{ - 1}}{1 \over 2}} \right]$$</p>
<p>$$ = \tan \left[ {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{1 \over 2}} \right]$$</p>
<p>$$ = \tan \left[ {{{\tan }^{ - 1}}{{{3 \over 4} + {1 \over 2}} \over {1 - {3 \over 8}}}} \right] = \tan \left[ {{{\tan }^{ - 1}}{{{5 \over 4}} \over {{5 \over 8}}}} \right]$$</p>
<p>$$ = \tan [{\tan ^{ - 1}}2] = 2$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6hz53dp | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If $$0 < x < {1 \over {\sqrt 2 }}$$ and $${{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta }$$, then the value of $$\sin \left( {{{2\pi \alpha } \over {\alpha + \beta }}} \right)$$ is :</p> | [{"identifier": "A", "content": "$$4 \\sqrt{\\left(1-x^{2}\\right)}\\left(1-2 x^{2}\\right)$$"}, {"identifier": "B", "content": "$$4 x \\sqrt{\\left(1-x^{2}\\right)}\\left(1-2 x^{2}\\right)$$"}, {"identifier": "C", "content": "$$2 x \\sqrt{\\left(1-x^{2}\\right)}\\left(1-4 x^{2}\\right)$$"}, {"identifier": "D", "content": "$$4 \\sqrt{\\left(1-x^{2}\\right)}\\left(1-4 x^{2}\\right)$$"}] | ["B"] | null | <p>Let $${{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta } = k \Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = k(\alpha + \beta )$$</p>
<p>$$ \Rightarrow \alpha + \beta = {\pi \over {2k}}$$</p>
<p>Now, $${{2\pi \,\alpha } \over {\alpha + \beta }} = {{2\pi \,\alpha } \over {{\pi \over {2k}}}} = 4k\alpha = 4{\sin ^{ - 1}}x$$</p>
<p>Here $$\sin \left( {{{2\pi \,\alpha } \over {\alpha + \beta }}} \right) = \sin (4{\sin ^{ - 1}}x)$$</p>
<p>Let $${\sin ^{ - 1}}x = \theta $$</p>
<p>$$\because$$ $$x \in \left( {0,{1 \over {\sqrt 2 }}} \right) \Rightarrow \theta \in \left( {0,{\pi \over 4}} \right)$$</p>
<p>$$ \Rightarrow x = \sin \theta $$</p>
<p>$$ \Rightarrow \cos \theta = \sqrt {1 - {x^2}} $$</p>
<p>$$ \Rightarrow \sin 2\theta = 2x\,.\,\sqrt {1 - {x^2}} $$</p>
<p>$$ \Rightarrow \cos 2\theta = \sqrt {1 - 4{x^2}(1 - {x^2})} = \sqrt {{{(2{x^2} - 1)}^2}} = 1 - 2{x^2}$$</p>
<p>$$\because$$ $$\left( {\cos 2\theta > 0\,\mathrm{as}\,2\theta \in \left( {0,{\pi \over 2}} \right)} \right)$$</p>
<p>$$ \Rightarrow \sin 4\theta = 2\,.\,2x\sqrt {1 - {x^2}} (1 - 2{x^2})$$</p>
<p>$$ = 4x\sqrt {1 - {x^2}} (1 - 2{x^2})$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6nm2vtd | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The sum of the absolute maximum and absolute minimum values of the function $$f(x)=\tan ^{-1}(\sin x-\cos x)$$ in the interval $$[0, \pi]$$ is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$\\tan ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)-\\frac{\\pi}{4}$$"}, {"identifier": "C", "content": "$$\\cos ^{-1}\\left(\\frac{1}{\\sqrt{3}}\\right)-\\frac{\\pi}{4}$$"}, {"identifier": "D", "content": "$$\\frac{-\\pi}{12}$$"}] | ["C"] | null | <p>$$f(x) = {\tan ^{ - 1}}(\sin x - \cos x),\,\,\,\,\,[0,\pi ]$$</p>
<p>Let $$g(x) = \sin x - \cos x$$</p>
<p>$$ = \sqrt 2 \sin \left( {x - {\pi \over 4}} \right)$$ and $$x - {\pi \over 4} \in \left[ {{{ - \pi } \over 4},\,{{3\pi } \over 4}} \right]$$</p>
<p>$$\therefore$$ $$g(x) \in \left[ { - 1,\,\sqrt 2 } \right]$$</p>
<p>and $${\tan ^{ - 1}}x$$ is an increasing function</p>
<p>$$\therefore$$ $$f(x) \in \left[ {{{\tan }^{ - 1}}( - 1),\,{{\tan }^{ - 1}}\sqrt 2 } \right]$$</p>
<p>$$ \in \left[ { - {\pi \over 4},\,{{\tan }^{ - 1}}\sqrt 2 } \right]$$</p>
<p>$$\therefore$$ Sum of $${f_{\max }}$$ and $${f_{\min }} = {\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}$$</p>
<p>$$ = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) - {\pi \over 4}$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldo6tynu | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$$.</p>
<p>If $$\mathrm{n(S)}$$ denotes the number of elements in $$\mathrm{S}$$ then :</p> | [{"identifier": "A", "content": "$$\\mathrm{n}(\\mathrm{S})=0$$"}, {"identifier": "B", "content": "$$\\mathrm{n}(\\mathrm{S})=1$$ and only one element in $$\\mathrm{S}$$ is less than $$\\frac{1}{2}$$."}, {"identifier": "C", "content": "$$\\mathrm{n}(\\mathrm{S})=1$$ and the elements in $$\\mathrm{S}$$ is more than $$\\frac{1}{2}$$."}, {"identifier": "D", "content": "$$\\mathrm{n}(\\mathrm{S})=1$$ and the element in $$\\mathrm{S}$$ is less than $$\\frac{1}{2}$$."}] | ["D"] | null | $$ {\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)}$$
<br/><br/>$\begin{aligned} & \text { Put } x=\tan \theta \quad \theta \in\left(0, \frac{\pi}{4}\right) \\\\ & 2 \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\\\ & 2 \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\cos ^{-1}[\cos (2 \theta)] \\\\ & \Rightarrow 2\left(\frac{\pi}{4}-\theta\right)=2 \theta \Rightarrow \theta=\frac{\pi}{8} \\\\ & \Rightarrow x=\tan \frac{\pi}{8}=\sqrt{2}-1 \simeq 0.414\end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift |
ldo7xp9f | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let (a, b) $\subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)$,
holds. <br/><br/>If $\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to : | [{"identifier": "A", "content": "$\\frac{\\pi}{16}$\n"}, {"identifier": "B", "content": "$\\frac{\\pi}{48}$\n"}, {"identifier": "C", "content": "$\\frac{\\pi}{8}$\n"}, {"identifier": "D", "content": "$\\frac{\\pi}{12}$"}] | ["D"] | null | $\sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0$
<br/><br/>$\Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4}$
<br/><br/>$\Rightarrow \sin \theta>\frac{1}{\sqrt{2}}$
<br/><br/>So, $\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$
<br/><br/>$\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=(\mathrm{a}, \mathrm{b})$
<br/><br/>$b-a=\frac{\pi}{2}=\alpha-\beta$
<br/><br/>$\Rightarrow \beta=\alpha-\frac{\pi}{2}$
<br/><br/>$\Rightarrow \alpha x^{2}+\beta \mathrm{x}+\sin ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]+\cos ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]=0$
<br/><br/>$x=3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0 \\\\
& \Rightarrow 12 \alpha-\pi=0 \\\\
& \alpha=\frac{\pi}{12}
\end{aligned}
$$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldom652p | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$S$$ be the set of all solutions of the equation $$\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$$. Then $$\sum_\limits{x \in S} 2 \sin ^{-1}\left(x^{2}-1\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\pi-2 \\sin ^{-1}\\left(\\frac{\\sqrt{3}}{4}\\right)$$"}, {"identifier": "B", "content": "$$\\pi-\\sin ^{-1}\\left(\\frac{\\sqrt{3}}{4}\\right)$$"}, {"identifier": "C", "content": "$$\\frac{-2 \\pi}{3}$$"}, {"identifier": "D", "content": "None"}] | ["D"] | null | $$
\begin{aligned}
& \cos ^{-1}(2 \mathrm{x})=\pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2} \\\\
& \text { Since } \cos ^{-1}(2 \mathrm{x}) \in[0, \pi] \\\\
& \text { R.H.S. } \geq \pi \\\\
& \pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2}=\pi \\\\
& \Rightarrow \cos ^{-1} \sqrt{1-\mathrm{x}^2}=0 \\\\
& \Rightarrow \sqrt{1-\mathrm{x}^2}=1 \\\\
& \Rightarrow \mathrm{x}=0 \\\\
& \text { but at } \mathrm{x}=0 \\\\
& \cos ^{-1}(2 \mathrm{x})=\cos ^{-1}(0)=\frac{\pi}{2}
\end{aligned}
$$
<br/><br/>$$ \therefore $$ No solution possible for given equation.
<br/><br/>$$
\mathrm{x} \in \phi
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldprk1i1 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If $${\sin ^{ - 1}}{\alpha \over {17}} + {\cos ^{ - 1}}{4 \over 5} - {\tan ^{ - 1}}{{77} \over {36}} = 0,0 < \alpha < 13$$, then $${\sin ^{ - 1}}(\sin \alpha ) + {\cos ^{ - 1}}(\cos \alpha )$$ is equal to :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "$$\\pi$$"}, {"identifier": "C", "content": "16 $$-$$ 5$$\\pi$$"}, {"identifier": "D", "content": "0"}] | ["B"] | null | $\sin ^{-1}\left(\frac{\alpha}{17}\right)=-\cos ^{4}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{77}{36}\right)$
<br/><br/>Let $\cos ^{-1}\left(\frac{4}{5}\right)=p$ and $\tan ^{-1}\left(\frac{77}{36}\right)=q$
<br/><br/>$\Rightarrow \sin \left(\sin ^{-1} \frac{\alpha}{17}\right)=\sin (q-p)$
<br/><br/>$=\sin q \cdot \cos p-\cos q \cdot \sin p$
<br/><br/>$\Rightarrow \frac{\alpha}{17}=\frac{77}{85} \cdot \frac{4}{5}-\frac{36}{85} \cdot \frac{3}{5}$
<br/><br/>$\Rightarrow \alpha=\frac{200}{25}=8$
<br/><br/>$\sin ^{-1} \sin 8+\cos ^{-1} \cos 8$
<br/><br/>$= -8+3 \pi+8-2 \pi$
<br/><br/>$=\pi$ | mcq | jee-main-2023-online-31st-january-morning-shift |
ldqv4ewv | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let $a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .$. be consecutive natural numbers.
<br/><br/>Then $\tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots . .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$ is equal to : | [{"identifier": "A", "content": "$\\frac{\\pi}{4}-\\cot ^{-1}(2022)$"}, {"identifier": "B", "content": "$\\frac{\\pi}{4}-\\tan ^{-1}(2022)$"}, {"identifier": "C", "content": "$\\cot ^{-1}(2022)-\\frac{\\pi}{4}$"}, {"identifier": "D", "content": "$\\tan ^{-1}(2022)-\\frac{\\pi}{4}$"}] | null | null | $a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .$. be consecutive natural numbers.
<br/><br/>$$
\begin{aligned}
& \therefore \quad a_2=2, a_3=3, \ldots ., a_{2021}=2021, a_{2022}=2022 \\\\
& \tan ^{-1}\left[\frac{1}{1+a_1 a_2}\right]=\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2}\right) \\\\
& \tan ^{-1}\left[\frac{1}{1+a_2 a_3}\right]=\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]=\tan ^{-1}\left(\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{3}\right)
\end{aligned}
$$
<br/><br/>.
<br/>.
<br/>.
<br/>.
<br/><br/>$$
\begin{aligned}
\tan ^{-1}\left[\frac{1}{1+a_{2021} a_{2022}}\right] & =\tan ^{-1}\left[\frac{1}{1+2021 \cdot 2022}\right] \\\\
& =\tan ^{-1}\left(\frac{1}{2021}\right)-\tan ^{-1}\left(\frac{1}{2022}\right)
\end{aligned}
$$
<br/><br/>$$ \therefore $$ $\tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots . .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$
<br/><br/>$$
=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2022}\right)=\frac{\pi}{4}-\cot ^{-1}(2022)
$$
<br/><br/>$$
=\frac{\pi}{4}-\left(\frac{\pi}{2}-\tan ^{-1}(2022)\right)=\tan ^{-1}(2022)-\frac{\pi}{4}
$$ | mcqm | jee-main-2023-online-30th-january-evening-shift |
1ldv37zch | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If the sum of all the solutions of $${\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) + {\cot ^{ - 1}}\left( {{{1 - {x^2}} \over {2x}}} \right) = {\pi \over 3}, - 1 < x < 1,x \ne 0$$, is $$\alpha - {4 \over {\sqrt 3 }}$$, then $$\alpha$$ is equal to _____________.</p> | [] | null | 2 | <b>Case-I</b>
<br/><br/>
$-1 < x < 0$
<br/><br/>
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\pi+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$
<br/><br/>
$\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{-\pi}{3}$
<br/><br/>
$2 \tan ^{-1} x=\frac{-\pi}{3}$
<br/><br/>
$\tan ^{-1} x=\frac{-\pi}{6}$
<br/><br/>
$x=\frac{-1}{\sqrt{3}}$
<br/><br/>
<b>Case-II</b><br/><br/>
$0 < x < 1$
<br/><br/>
$\tan ^{-1} \frac{2 x}{1-x^{2}}+\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3}$
<br/><br/>
$$
\begin{aligned}
& \tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{6} \\\\
& 2 \tan ^{-1} x=\frac{\pi}{6} \\\\
& \tan ^{-1} x=\frac{\pi}{12} \\\\
& x=2-\sqrt{3} \\\\
& \text { Sum }=\frac{-1}{\sqrt{3}}+2-\sqrt{3}=2-\frac{4}{\sqrt{3}} \\\\
& \Rightarrow \alpha=2
\end{aligned}
$$ | integer | jee-main-2023-online-25th-january-morning-shift |
1lgoy4b0g | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>For $$x \in(-1,1]$$, the number of solutions of the equation $$\sin ^{-1} x=2 \tan ^{-1} x$$ is equal to __________.</p> | [] | null | 2 | <p>We're given the equation $\sin^{-1}x = 2\tan^{-1}x$ for $x$ in the interval $(-1, 1]$. We want to find the number of solutions.</p>
<p>Step 1: Apply the sine and tangent functions to both sides :</p>
<p>We can rewrite the equation by applying the sine function to both sides :</p>
<p>$$\sin(\sin^{-1}x) = \sin(2\tan^{-1}x).$$</p>
<p>This simplifies to:</p>
<p>$$x = \sin(2\tan^{-1}x).$$</p>
<p>Step 2: Use the double-angle identity for sine :</p>
<p>Recall that $\sin(2y) = 2\sin(y)\cos(y)$. Applying this identity to the right-hand side gives :</p>
<p>$$x = 2\sin(\tan^{-1}x)\cos(\tan^{-1}x).$$</p>
<p>Step 3: Use the identities for sine and cosine of an inverse tangent :</p>
<p>Recall that $\sin(\tan^{-1}x) = \frac{x}{\sqrt{1 + x^2}}$ and $\cos(\tan^{-1}x) = \frac{1}{\sqrt{1 + x^2}}$. Substituting these into the equation gives :</p>
<p>$$x = 2 \cdot \frac{x}{\sqrt{1 + x^2}} \cdot \frac{1}{\sqrt{1 + x^2}}.$$</p>
<p>This simplifies to :</p>
<p>$$x = \frac{2x}{1 + x^2}.$$</p>
<p>Step 4: Solve for $x$ :</p>
<p>We have :</p>
<p>$$x = \frac{2x}{1 + x^2}.$$</p>
<p>Cross-multiplying gives :</p>
<p>$$x(1 + x^2) = 2x.$$</p>
<p>This simplifies to :</p>
<p>$$x^3 + x - 2x = 0.$$</p>
<p>Rearranging terms gives :</p>
<p>$$x^3 - x = 0.$$</p>
<p>This factors to:</p>
<p>$$x(x^2 - 1) = 0.$$</p>
<p>Setting each factor equal to zero gives the solutions $x = 0$, $x = -1$, and $x = 1$.</p>
<p>However, we are given that $x \in (-1, 1]$. Therefore, the only solutions in this interval are $x = 0$ and $x = 1$.</p>
<p>So there are 2 solutions to the equation $\sin ^{-1} x=2 \tan ^{-1} x$ in the interval $x \in(-1,1]$.</p>
| integer | jee-main-2023-online-13th-april-evening-shift |
1lgq0x4gc | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If $$S=\left\{x \in \mathbb{R}: \sin ^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}\right\}$$, then $$\sum_\limits{x \in s}\left(\sin \left(\left(x^{2}+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^{2}+x+5\right) \pi\right)\right)$$ is equal to ____________.</p> | [] | null | 4 | Given equation is
<br/><br/>$$
\sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}.
$$
<br/><br/>Let's denote:
<br/><br/>$$A = \sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right),$$
<br/><br/>$$B = \sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right).$$
<br/><br/>So, we have the equation $A - B = \frac{\pi}{4}$.
<br/><br/>We can also write this as $A = B + \frac{\pi}{4}$.
<br/><br/>This gives us
<br/><br/>$$\sin(A) = \sin\left(B + \frac{\pi}{4}\right).$$
<br/><br/>We can use the identity $\sin(a + b) = \sin a \cos b + \cos a \sin b$ and rewrite this equation as:
<br/><br/>$$\frac{x+1}{\sqrt{(x+1)^2+1}} = \frac{x}{\sqrt{x^2+1}} \cos\left(\frac{\pi}{4}\right) + \sqrt{1-\left(\frac{x}{\sqrt{x^2+1}}\right)^2} \sin\left(\frac{\pi}{4}\right).$$
<br/><br/>After simplifying, we get:
<br/><br/>$$\frac{x+1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{2}}\left(\frac{x}{\sqrt{x^2 + 1}} + \sqrt{1 - \frac{x^2}{x^2 + 1}}\right).$$
<br/><br/>Let's square both sides to remove the square roots:
<br/><br/>On the left side, squaring gives:
<br/><br/>$$\left(\frac{x+1}{\sqrt{x^2 + 2x + 2}}\right)^2 = \frac{(x+1)^2}{x^2 + 2x + 2}.$$
<br/><br/>On the right side, squaring gives:
<br/><br/>$$\left(\frac{1}{\sqrt{2}}\left(\frac{x}{\sqrt{x^2 + 1}} + \sqrt{1 - \frac{x^2}{x^2 + 1}}\right)\right)^2 = \frac{1}{2}\left(\frac{x^2}{x^2+1} + 2\frac{x}{\sqrt{x^2+1}}\sqrt{1-\frac{x^2}{x^2+1}} + 1 - \frac{x^2}{x^2+1}\right).$$
<br/><br/>$$ \therefore $$ $$\frac{(x+1)^2}{x^2 + 2x + 2}$$ = $$\frac{1}{2}\left(\frac{x^2}{x^2+1} + 2\frac{x}{\sqrt{x^2+1}}\sqrt{1-\frac{x^2}{x^2+1}} + 1 - \frac{x^2}{x^2+1}\right)$$
<br/><br/>$$ \Rightarrow $$ $$\frac{(x+1)^2}{x^2 + 2x + 2}$$ = $${1 \over 2}\left( {2 \times {x \over {\sqrt {{x^2} + 1} }}\sqrt {{{{x^2} + 1 - {x^2}} \over {{x^2} + 1}}} + 1} \right)$$
<br/><br/>$$ \Rightarrow $$ $$\frac{(x+1)^2}{x^2 + 2x + 2}$$ = $${1 \over 2}\left( {2 \times {x \over {\sqrt {{x^2} + 1} }}{1 \over {\sqrt {{x^2} + 1} }} + 1} \right)$$
<br/><br/>$$ \Rightarrow $$ $$\frac{(x+1)^2}{x^2 + 2x + 2}$$ = $${1 \over 2}\left( {2 \times {x \over {{x^2} + 1}} + 1} \right)$$
<br/><br/>$$ \Rightarrow $$ $$\frac{(x+1)^2}{x^2 + 2x + 2}$$ = $${1 \over 2}\left( {{{2x + {x^2} + 1} \over {{x^2} + 1}}} \right)$$
<br/><br/>$$ \Rightarrow $$ $$\frac{(x+1)^2}{x^2 + 2x + 2}$$ = $${1 \over 2}\left( {{{{{\left( {x + 1} \right)}^2}} \over {{x^2} + 1}}} \right)$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{x+1}{\sqrt{x^2+2x+2}}=\frac{x+1}{\sqrt{2} \sqrt{x^2+1}} \\\\
& \Rightarrow x=-1 \text { OR } \sqrt{x^2+2x+2}=\sqrt{2} \cdot \sqrt{x^2+1} \\\\
& \Rightarrow x=0, x=2 \text { (Rejected) } \\\\
& S=\{0,-1\}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \sum_{x \in R}\left(\sin \left(\left(x^2+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^2+x+5\right) \pi\right)\right) \\\\
& =\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right]+\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right] \\\\
& = (1 -(-1)) + (1 -(-1))\\\\
& = 2 + 2 \\\\
& = 4
\end{aligned}
$$ | integer | jee-main-2023-online-13th-april-morning-shift |
jaoe38c1lse4tf9q | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>For $$\alpha, \beta, \gamma \neq 0$$, if $$\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi$$ and $$(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta$$, then $$\gamma$$ equals</p> | [{"identifier": "A", "content": "$$\\sqrt{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\sqrt{3}}{2}$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{2}}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{3}-1}{2 \\sqrt{2}}$$"}] | ["B"] | null | <p>Let $$\sin ^{-1} \alpha=A, \sin ^{-1} \beta=B, \sin ^{-1} \gamma=C$$</p>
<p>$$\begin{aligned}
& \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\
& (\alpha+\beta)^2-\gamma^2=3 \alpha \beta \\
& \alpha^2+\beta^2-\gamma^2=\alpha \beta \\
& \frac{\alpha^2+\beta^2-\gamma^2}{2 \alpha \beta}=\frac{1}{2} \\
& \Rightarrow \cos \mathrm{C}=\frac{1}{2} \\
& \sin \mathrm{C}=\gamma \\
& \cos \mathrm{C}=\sqrt{1-\gamma^2}=\frac{1}{2} \\
& \gamma=\frac{\sqrt{3}}{2}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsfkxule | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$x=\frac{m}{n}$$ ($$m, n$$ are co-prime natural numbers) be a solution of the equation $$\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$$ and let $$\alpha, \beta(\alpha >\beta)$$ be the roots of the equation $$m x^2-n x-m+ n=0$$. Then the point $$(\alpha, \beta)$$ lies on the line</p> | [{"identifier": "A", "content": "$$3 x-2 y=-2$$\n"}, {"identifier": "B", "content": "$$3 x+2 y=2$$\n"}, {"identifier": "C", "content": "$$5 x+8 y=9$$\n"}, {"identifier": "D", "content": "$$5 x-8 y=-9$$"}] | ["C"] | null | <p>Assume $$\sin ^{-1} x=\theta$$</p>
<p>$$\begin{aligned}
& \cos (2 \theta)=\frac{1}{9} \\
& \sin \theta= \pm \frac{2}{3}
\end{aligned}$$</p>
<p>as $$\mathrm{m}$$ and $$\mathrm{n}$$ are co-prime natural numbers,</p>
<p>$$\mathrm{x}=\frac{2}{3}$$</p>
<p>i.e. $$m=2, n=3$$</p>
<p>So, the quadratic equation becomes $$2 x^2-3 x+1=0$$ whose roots are $$\alpha=1, \beta=\frac{1}{2}$$</p>
<p>$$\left(1, \frac{1}{2}\right)$$ lies on $$5 x+8 y=9$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
lvc57nxl | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>For $$n \in \mathrm{N}$$, if $$\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$$, then $$n$$ is equal to ________.</p> | [] | null | 47 | <p>For $ n \in \mathbb{N} $, if $ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4} $, then $ n $ is equal to <strong></strong><strong></strong>.</p>
<p>Given the equation:</p>
<p>$ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1}(n) = \frac{\pi}{4} $</p>
<p>we can use the identity for the sum of inverse cotangents. Starting with the first two terms:</p>
<p>$ \cot^{-1} 3 + \cot^{-1} 4 = \cot^{-1}\left(\frac{3 \times 4 - 1}{3 + 4}\right) = \cot^{-1}\left(\frac{11}{7}\right) $</p>
<p>Now, adding the third term:</p>
<p>$ \cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}(5) $</p>
<p>we apply the identity again:</p>
<p>$ \cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}\left(\frac{n \times 5 - 1}{5 + n}\right) = \frac{\pi}{4} $</p>
<p>Rewriting this to isolate the sum of the terms, we proceed as follows:</p>
<p>$ \cot^{-1}\left( \frac{\left(\frac{11}{7} \times \frac{5n-1}{5+n} - 1\right)}{\left(\frac{11}{7} + \frac{5n-1}{5+n} \right)} \right) = \frac{\pi}{4} $</p>
<p>This simplifies to:</p>
<p>$ \frac{11}{7} \left(\frac{5n-1}{5+n}\right) - 1 = \frac{11}{7} + \frac{5n-1}{5+n} $</p>
<p>Solving the equation:</p>
<p>$ \frac{55n - 11}{5 + n} - 1 = \frac{11}{7} + \frac{5n - 1}{5 + n} $</p>
<p>Further simplification yields:</p>
<p>$ 55n - 11 - 35 - 7n = 55 + 11n + 35n - 7 $</p>
<p>Bringing the terms together, we get:</p>
<p>$ 48n - 46 = 48 $</p>
<p>Therefore:</p>
<p>$ 2n = 94 $</p>
<p>So finally:</p>
<p>$ n = 47 $</p> | integer | jee-main-2024-online-6th-april-morning-shift |
wQwDqvp4mARIPRuzLBD6U | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | The value of $$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \right)$$ is : | [{"identifier": "A", "content": "$${{22} \\over {23}}$$"}, {"identifier": "B", "content": "$${{23} \\over {22}}$$"}, {"identifier": "C", "content": "$${{21} \\over {19}}$$"}, {"identifier": "D", "content": "$${{19} \\over {21}}$$"}] | ["C"] | null | $$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right.} } \right)$$
<br><br>$$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} } \right)$$
<br><br>$$\sum\limits_{n = 1}^{19} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)} $$
<br><br>$$\cot \left( {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right) = {{\cot A\cot \beta + 1} \over {\cot \beta - \cot A}}$$
<br><br>(Where tanA $$=$$ 20, tanB $$=$$ 1) $${{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
P82eExl9uoiR298CyCjgy2xukfg71qct | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | If S is the sum of the first 10 terms of the series <br/><br/>
$${\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....$$<br/><br/>
then tan(S) is equal to : | [{"identifier": "A", "content": "$${10 \\over {11}}$$"}, {"identifier": "B", "content": "$${5 \\over {11}}$$"}, {"identifier": "C", "content": "-$${6 \\over {5}}$$"}, {"identifier": "D", "content": "$${5 \\over {6}}$$"}] | ["D"] | null | S = $${\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....$$
<br><br>= $${\tan ^{ - 1}}\left( {{1 \over {1 + 1 \times 2}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {1 + 2 \times 3}}} \right) + ...$$
<br><br>$$ \therefore $$ T<sub>r</sub> = $${\tan ^{ - 1}}\left( {{1 \over {1 + r \times \left( {r + 1} \right)}}} \right)$$
<br><br>= tan<sup>–1</sup>(r + 1) – tan<sup>–1</sup>r
<br><br>$$ \therefore $$ T<sub>1</sub>
= tan<sup>–1</sup>2 – tan<sup>–1</sup>1
<br><br>T<sub>2</sub>
= tan<sup>–1</sup>3 – tan<sup>–1</sup>2
<br><br>T<sub>3</sub>
= tan<sup>–1</sup>4 – tan<sup>–1</sup>3
<br>.
<br>.
<br>.
<br><br>T<sub>10</sub> = tan<sup>-1</sup>11 – tan<sup>–1</sup>10
<br><br>$$ \therefore $$ S = tan<sup>–1</sup>11 – tan<sup>–1</sup>1 = $${\tan ^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)$$
<br><br>$$ \therefore $$ tan(S) = $$\tan \left( {{{\tan }^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)} \right)$$
<br><br>= $${{{11 - 1} \over {1 + 11}}}$$ = $${{10} \over {12}} = {5 \over 6}$$ | mcq | jee-main-2020-online-5th-september-morning-slot |
j9QMKExBDBcaXztP8N1kmjbau49 | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | If cot<sup>$$-$$1</sup>($$\alpha$$) = cot<sup>$$-$$1</sup> 2 + cot<sup>$$-$$1</sup> 8 + cot<sup>$$-$$1</sup> 18 + cot<sup>$$-$$1</sup> 32 + ...... upto 100 terms, then $$\alpha$$ is : | [{"identifier": "A", "content": "1.02"}, {"identifier": "B", "content": "1.03"}, {"identifier": "C", "content": "1.01"}, {"identifier": "D", "content": "1.00"}] | ["C"] | null | $${\cot ^{ - 1}}(\alpha ) = co{t^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....100$$ terms<br><br>$$ = {\tan ^{ - 1}}{1 \over 2} + {\tan ^{ - 1}}{1 \over 8} + {\tan ^{ - 1}}{1 \over {18}} + {\tan ^{ - 1}}{1 \over {32}} + ....100$$ term<br><br>$$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{1 \over {2{k^2}}}} $$<br><br>$$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{2 \over {4{k^2}}} = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}{{(2k + 1) - (2k - 1)} \over {1 + (2k - 1)(2k + 1)}}} } $$<br><br>$$ = \sum\limits_{k = 1}^{100} {\left( {{{\tan }^{ - 1}}(2k + 1) - {{\tan }^{ - 1}}(2k - 1)} \right)} $$<br><br>$$ = {\tan ^{ - 1}}201 - {\tan ^{ - 1}}1$$<br><br>$$ = {\tan ^{ - 1}}{{200} \over {202}}$$<br><br>$$ = {\cot ^{ - 1}}(1.01)$$<br><br>Hence $$\alpha = 1.01$$ | mcq | jee-main-2021-online-17th-march-morning-shift |
1ktd06pxr | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | If $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}{1 \over {2{r^2}}} = p} $$, then the value of tan p is : | [{"identifier": "A", "content": "$${{101} \\over {102}}$$"}, {"identifier": "B", "content": "$${{50} \\over {51}}$$"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "$${{51} \\over {50}}$$"}] | ["B"] | null | $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{2 \over {4{r^2}}}} \right) = \sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(2r + 1) - (2r - 1)} \over {1 + (2r + 1)(2r - 1)}}} \right)} } $$<br><br>= $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}(2r + 1) - {{\tan }^{ - 1}}(2r - 1)} $$<br><br>= $${\tan ^{ - 1}}(101) - {\tan ^{ - 1}}1 = {\tan ^{ - 1}}{{50} \over {51}}$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
qaoJWPlc6NQngtMw | maths | limits-continuity-and-differentiability | continuity | $$f$$ is defined in $$\left[ { - 5,5} \right]$$ as
<br/><br/>$$f\left( x \right) = x$$ if $$x$$ is rational
<br/><br/>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = - x$$ if $$x$$ is irrational. Then | [{"identifier": "A", "content": "$$f(x)$$ is continuous at every x, except $$x = 0$$"}, {"identifier": "B", "content": "$$f(x)$$ is discontinuous at every $$x,$$ except $$x = 0$$"}, {"identifier": "C", "content": "$$f(x)$$ is continuous everywhere"}, {"identifier": "D", "content": "$$f(x)$$ is discontinuous everywhere"}] | ["B"] | null | <b>Let a is a rational number</b> other than $$0,$$ in
<br><br>$$\left[ { - 5,5} \right],$$ then
<br><br>$$f\left( a \right) = a$$ and $$\mathop {\lim }\limits_{x \to a} \,\,f\left( x \right) = - a$$
<br><br>[ As in the immediate neighbourhood of a rational -
<br><br>number, we find irrational numbers ]
<br><br>$$\therefore$$ $$f(x)$$ is not continuous at any rational number
<br><br><b>If a is irrational number,</b> then
<br><br>$$\,f\left( a \right) = - a$$ and $$\mathop {\lim }\limits_{x \to a} \,f\left( x \right) = a$$
<br><br>$$\therefore$$ $$f(x)$$ is not continuous at any irrational number clearly
<br><br>$$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) = 0$$
<br><br>$$\therefore$$ $$f(x)$$ is continuous at $$x=0$$
| mcq | aieee-2002 |
0qO7zPDbm3sJgFcz | maths | limits-continuity-and-differentiability | continuity | Let $$f(x) = {{1 - \tan x} \over {4x - \pi }}$$, $$x \ne {\pi \over 4}$$, $$x \in \left[ {0,{\pi \over 2}} \right]$$.
<br/><br/>If $$f(x)$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$, then $$f\left( {{\pi \over 4}} \right)$$ is | [{"identifier": "A", "content": "$$-1$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$-{1 \\over 2}$$"}, {"identifier": "D", "content": "$$1$$"}] | ["C"] | null | $$f\left( x \right) = {{1 - \tan x} \over {4x - \pi }}$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$
<br><br>$$\therefore$$ $$f\left( {{\pi \over 4}} \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} f\left( x \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} \, + f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {{\pi \over 4} + h} \right)$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{1 - \tan \left( {{\pi \over 4} + h} \right)} \over {4\left( {{\pi \over 4} + h} \right) - \pi }},h > 0 = \mathop {\lim }\limits_{h \to 0} {{1 - {{1 + \tan \,h} \over {1 - \tan \,h}}} \over {4h}}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} \,{{ - 2} \over {1 - \tan \,h}}.{{\tan \,h} \over {4h}} = {{ - 2} \over 4} = - {1 \over 2}$$ | mcq | aieee-2004 |
jUvwcp4tbFocqlD1 | maths | limits-continuity-and-differentiability | continuity | The function $$f:R/\left\{ 0 \right\} \to R$$ given by
<br/><br/>$$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
<br/><br/>can be made continuous at $$x$$ = 0 by defining $$f$$(0) as | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$-1$$"}] | ["B"] | null | Given, $$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
<br><br>$$ \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{e^{2x}} - 1} \right) - 2x} \over {x\left( {{e^{2x}} - 1} \right)}}$$ $$\left[ \, \right.$$ $${0 \over 0}$$ form $$\left. \, \right]$$
<br><br>$$\therefore$$ using, $$L$$'Hospital rule
<br><br>$$f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {2\left( {x{e^{2x}}2 + {e^{2x}}.1} \right) + {e^{2x}}.2}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4x{e^{2x}} + 2{e^{2x}} + 2{e^{2x}}}}\,\,$$ $$\left[ {\,\,} \right.$$ $${0 \over 0}$$ form $$\left. {\,\,} \right]$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4\left( {x{e^{2x}} + {e^{2x}}} \right)}} = {{4.{e^0}} \over {4\left( {0 + {e^0}} \right)}} = 1$$ | mcq | aieee-2007 |
FvET6cmrFqjSbbCw | maths | limits-continuity-and-differentiability | continuity | The value of $$p$$ and $$q$$ for which the function
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr
q & {,x = 0} \cr
{{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{3/2}}}}} & {,x > 0} \cr
} } \right.$$
<br/><br/>is continuous for all $$x$$ in R, are | [{"identifier": "A", "content": "$$p =$$ $${5 \\over 2}$$, $$q = $$ $${1 \\over 2}$$"}, {"identifier": "B", "content": "$$p =$$ $$-{3 \\over 2}$$, $$q = $$ $${1 \\over 2}$$"}, {"identifier": "C", "content": "$$p =$$ $${1 \\over 2}$$, $$q = $$ $${3 \\over 2}$$"}, {"identifier": "D", "content": "$$p =$$ $${1 \\over 2}$$, $$q = $$ $$-{3 \\over 2}$$"}] | ["B"] | null | $$L.H.L. = \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sin \left\{ {\left( {p + 1} \right)\left( { - h} \right)} \right\} - \sin \left( { - h} \right)} \over { - h}}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{ - \sin \left( {p + 1} \right)h} \over { - h}} + {{\sin \left( { - h} \right)} \over { - h}} = p + 1 + 1 = p + 2$$
<br><br>$$R.H.L.$$ $$ = \mathop {\lim }\limits_{x \to {\sigma ^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {1 + h} - 1} \over h}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {1 \over {\left( {\sqrt {1 + h} + 1} \right)}} = {1 \over 2}$$
<br><br>and $$f\left( 0 \right) = q \Rightarrow p = - {3 \over 2},q = {1 \over 2}$$ | mcq | aieee-2011 |
kuyx43Dm2TedQW6y | maths | limits-continuity-and-differentiability | continuity | If $$f:R \to R$$ is a function defined by
<br/><br/>$$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$,
<br/><br/>where [x] denotes the greatest integer function, then $$f$$ is | [{"identifier": "A", "content": "continuous for every real $$x$$"}, {"identifier": "B", "content": "discontinuous only at $$x=0$$"}, {"identifier": "C", "content": "discontinuous only at non-zero integral values of $$x$$"}, {"identifier": "D", "content": "continuous only at $$x=0$$"}] | ["A"] | null | Let $$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$$
<br><br>Doubtful points are $$x = n,n \in I$$
<br><br>$$L.H.L$$ $$ = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$
<br><br>$$ = \left( {n - 1} \right)\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0$$
<br><br>( As $$\left[ x \right]$$ is the greatest integer function )
<br><br>$$R.H.L. = \mathop {\lim }\limits_{x \to {n^ + }} \,\left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi = n\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0$$
<br><br>Now, value of the function at $$x = n$$ is $$f(n)=0$$
<br><br>Since, $$L.H.L.$$ $$ = R.H.L. = f\left( n \right)$$
<br><br>$$\therefore$$ $$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$$ is continuous for every real $$x.$$ | mcq | aieee-2012 |
ktLYuFU6rv2nIELWJagcv | maths | limits-continuity-and-differentiability | continuity | Let a, b $$ \in $$ <b>R</b>, (a $$ \ne $$ 0). If the function <i>f</i> defined as
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \cr
{a\,\,\,,} & {1 \le x < \sqrt 2 } \cr
{{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \cr
} } \right.$$
<br/><br/>is continuous in the interval [0, $$\infty $$), then an ordered pair ( a, b) is : | [{"identifier": "A", "content": "$$\\left( {\\sqrt 2 ,1 - \\sqrt 3 } \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - \\sqrt 2 ,1 + \\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {\\sqrt 2 , - 1 + \\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - \\sqrt 2 ,1 - \\sqrt 3 } \\right)$$"}] | ["A"] | null | f(x) is continuous at x = 1
<br><br>$$ \therefore $$ $${{2{{\left( 1 \right)}^2}} \over a} = a$$
<br><br>$$ \Rightarrow $$ a<sup>2</sup> = 2
<br><br>$$ \Rightarrow $$ a = $$ \pm $$ $$\sqrt 2 $$
<br><br>Also f(x) is continuous at x = $$\sqrt 2 $$
<br><br>$$ \therefore $$ a = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$
<br><br>Now, when a = $$\sqrt 2 ,$$ then
<br><br>4 = 2b<sup>2</sup> $$-$$ 4b
<br><br>$$ \Rightarrow $$ b<sup>2</sup> $$-$$ 2b = 2
<br><br>$$ \Rightarrow $$ b<sup>2</sup> $$-$$ 2b $$-$$ 2 = 0
<br><br>$$ \Rightarrow $$ b = $${{2 \mp \sqrt {4 + 4.2} } \over 2}$$
<br><br>= 1 $$ \pm $$ $$\sqrt 3 $$
<br><br>$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$$
<br><br>When a = $$-$$ $$\sqrt 2 $$, then
<br><br>$$-$$ $$\sqrt 2 $$ = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$
<br><br>$$ \Rightarrow $$ $$-$$ 4 = 2b<sup>2</sup> $$-$$ 4b
<br><br>$$ \Rightarrow $$ b<sup>2</sup> $$-$$ 2b + 2 = 0
<br><br>$$ \therefore $$ b = $${{2 \pm \sqrt {4 - 8} } \over 2}$$ = 1 $$ \pm $$ i
<br><br>As b $$ \in $$ Real number so,
<br><br>b = 1 $$ \pm $$ i is not accepted.
<br><br>$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 + \sqrt 3 } \right)$$ or $$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$ | mcq | jee-main-2016-online-10th-april-morning-slot |
oKPTJbDBgTv4QOlEtDpJg | maths | limits-continuity-and-differentiability | continuity | The value of k for which the function
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr
{k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \cr
} } \right.$$
<br/><br/>is continuous at x = $${\pi \over 2},$$ is : | [{"identifier": "A", "content": "$${{17} \\over {20}}$$ "}, {"identifier": "B", "content": "$${{2} \\over {5}}$$"}, {"identifier": "C", "content": "$${{3} \\over {5}}$$"}, {"identifier": "D", "content": "$$-$$ $${{2} \\over {5}}$$"}] | ["C"] | null | $f(x)=\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}$<br/><br/>
$\lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}=k+\frac{2}{5}$<br/><br/>
$\Rightarrow \lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\frac{2}{5}$<br/><br/>
$\Rightarrow\left(\frac{4}{5}\right)^{\lim\limits_{x \rightarrow \frac{x}{2}}(\tan 4 x \cdot \cot (5 x))}=k+\frac{2}{5}$<br/><br/>
$\Rightarrow\left(\frac{4}{5}\right)^{0 \times \cos \left(2 x+\frac{x}{2}\right)}=k+\frac{2}{5}$<br/><br/>
$\Rightarrow\left(\frac{4}{5}\right)^0=k+\frac{2}{5}$<br/><br/>
$\Rightarrow k+\frac{2}{5}=1 \Rightarrow k=1-\frac{2}{5} \Rightarrow k=\frac{3}{5}$ | mcq | jee-main-2017-online-9th-april-morning-slot |
RWZ8zTUsm0YzRTvqcemcN | maths | limits-continuity-and-differentiability | continuity | Let f(x) = $$\left\{ {\matrix{
{{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr
{k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr
} } \right.$$
<br/><br/>Thevaue of k for which f s continuous at x = 2 is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "e"}, {"identifier": "C", "content": "e<sup>-1</sup>"}, {"identifier": "D", "content": "e<sup>-2</sup>"}] | ["C"] | null | Since f(x) is continuous at x = 2.
<br><br>$$\therefore\,\,\,$$$$\mathop {\lim }\limits_{x \to 2} $$ f(x) = f(2)
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} $$ $${\left( {x - 1} \right)^{{1 \over {2 - x}}}}$$ = k ($${{1^\infty }}$$ form)
<br><br>$$ \therefore $$ $${{e^l}}$$ = k
<br><br>Where $$l$$ = $$\mathop {\lim }\limits_{x \to 2} $$(x $$-$$ $$l$$ $$-$$ 1) $$ \times $$ $${1 \over {2 - x}}$$ = $$\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 2} \left( {{{x - 2} \over {x - 2}}} \right)$$
<br><br>$$ \Rightarrow $$ k = e<sup>$$-$$1</sup> | mcq | jee-main-2018-online-15th-april-evening-slot |
6hRA7o4FY7sWQFr59ThCp | maths | limits-continuity-and-differentiability | continuity | If the function f defined as
<br/><br/>$$f\left( x \right) = {1 \over x} - {{k - 1} \over {{e^{2x}} - 1}},x \ne 0,$$ is continuous at
<br/><br/> x = 0, then the ordered pair (k, f(0)) is equal to : | [{"identifier": "A", "content": "(3, 2)"}, {"identifier": "B", "content": "(3, 1)"}, {"identifier": "C", "content": "(2, 1)"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 3},\\,2} \\right)$$"}] | ["B"] | null | If the function is continuous at x = 0, then
<br>$$\mathop {\lim }\limits_{x \to 0} $$ f(x) will exist and f(0) = $$\mathop {\lim }\limits_{x \to 0} $$ f(x)
<br><br>Now, $$\mathop {\lim }\limits_{x \to 0} $$ f(x) = $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {{k - 1} \over {{e^{2x}} - 1}}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{e^{2x}} - 1 - kx + x} \over {\left( x \right)\left( {{e^{2x}} - 1} \right)}}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ....} \right) - 1 - kx + x} \over {\left( x \right)\left( {\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ...} \right) - 1} \right)}}} \right]$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {3 - k} \right)x + {{4{x^2}} \over {2!}} + {{8{x^3}} \over {3!}} + ...} \over {\left( {2{x^2} + {{4{x^3}} \over {2!}} + {{8{x^3}} \over {3!}} + ....} \right)}}} \right]$$
<br><br>For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to zero
<br><br>$$ \Rightarrow $$ 3 $$-$$ k = 0
<br><br>$$ \Rightarrow $$ k = 3
<br><br>So the limit reduces to
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{\left( {{x^2}} \right)\left( {{4 \over {2!}} + {{8x} \over {3!}} + ...} \right)} \over {\left( {{x^2}} \right)\left( {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{{4 \over {2!}} + {{8x} \over {3!}} + ...} \over {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...}}$$ = 1
<br><br>Hence, f(0) = 1 | mcq | jee-main-2018-online-16th-april-morning-slot |
m9GFj6XbKauU5vPk0j9WR | maths | limits-continuity-and-differentiability | continuity | Let f : R $$ \to $$ R be a function defined as
<br/>$$f(x) = \left\{ {\matrix{
5 & ; & {x \le 1} \cr
{a + bx} & ; & {1 < x < 3} \cr
{b + 5x} & ; & {3 \le x < 5} \cr
{30} & ; & {x \ge 5} \cr
} } \right.$$
<br/><br/>Then, f is | [{"identifier": "A", "content": "continuous if a = 0 and b = 5"}, {"identifier": "B", "content": "continuous if a = \u20135 and b = 10"}, {"identifier": "C", "content": "continuous if a = 5 and b = 5 "}, {"identifier": "D", "content": "not continuous for any values of a and b"}] | ["D"] | null | Checking
<br><br>if f(x) is continuous at x = 1 :
<br><br>f(1<sup>$$-$$</sup>) = 5
<br><br>f(1) = 5
<br><br>f(1<sup>+</sup>) = a + b
<br><br>if f(x) is continuous at x = 1,
<br><br>then
<br><br>f(1<sup>$$-$$</sup>) = f(1) = f(1<sup>+</sup>)
<br><br>$$ \Rightarrow $$ 5 = 5 = a + b
<br><br>$$ \therefore $$ a + b = 5 . . . . . . . . (1)
<br><br>checking if f(x) is continuous at x = 3 :
<br><br>f(3<sup>$$-$$</sup>) = a + 3b
<br><br>f(3) = b + 15
<br><br>f(3<sup>+</sup>) = b + 15
<br><br>if f(x) = is continuous at x = 3
<br><br>then,
<br><br>f(3<sup>$$-$$</sup>) = f(3) = f(3<sup>+</sup>)
<br><br>$$ \Rightarrow $$ a + 3b = b + 15 = b + 15
<br><br>$$ \Rightarrow $$ a + 2b = 15 . . . . . (2)
<br><br>checking if f(x) is continuous at x = 5 :
<br><br>f(5<sup>$$-$$</sup>) = b + 25
<br><br>f(5) = 30
<br><br>f(5<sup>+</sup>) = 30
<br><br>if f(x) is continuous at x = 5 then,
<br><br>f(5<sup>$$-$$</sup>) = f(5) = f(5<sup>+</sup>)
<br><br>$$ \Rightarrow $$ b + 25 = 30 = 30
<br><br>$$ \Rightarrow $$ b = 5
<br><br>By putting this value in equation (2), we get,
<br><br>a + 2(5) = 15
<br><br>$$ \Rightarrow $$ a = 5
<br><br>when a = 5 and b = 5 then equation (1)
<br><br>a + b = 5 does not satisfy.
<br><br>$$ \therefore $$ f is not continuous for any value of a and b. | mcq | jee-main-2019-online-9th-january-morning-slot |
yrQAhYb5CYLwaNHm7fLB7 | maths | limits-continuity-and-differentiability | continuity | Let ƒ : [–1,3] $$ \to $$ R be defined as<br/><br/>
$$f(x) = \left\{ {\matrix{
{\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr
{x + \left| x \right|} & , & {1 \le x < 2} \cr
{x + \left[ x \right]} & , & {2 \le x \le 3} \cr
} } \right.$$<br/><br/>
where [t] denotes the greatest integer less than
or equal to t. Then, ƒ is discontinuous at: | [{"identifier": "A", "content": "only three points"}, {"identifier": "B", "content": "four or more points"}, {"identifier": "C", "content": "only two points"}, {"identifier": "D", "content": "only one point"}] | ["A"] | null | $$f(x) = \left\{ {\matrix{
{\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr
{x + \left| x \right|} & , & {1 \le x < 2} \cr
{x + \left[ x \right]} & , & {2 \le x \le 3} \cr
} } \right.$$
<br><br>= $$ = \left\{ {\matrix{
{ - x - 1,} & { - 1 \le x < 0} \cr
{x,} & {0 \le x < 1} \cr
{2x,} & {1 \le x < 2} \cr
{x + 2,} & {2 \le x < 3} \cr
{6,} & {x = 3} \cr
} } \right.$$
<br><br>f(-1) = $$\mathop {\lim }\limits_{x \to - 1} \left( { - x - 1} \right)$$ = -( -1) - 1 = 0
<br><br>f(-1<sup>+</sup>) = $$\mathop {\lim }\limits_{x \to - {1^ + }} \left( { - x - 1} \right)$$ = 0
<br><br>$$ \therefore $$ f(x) is continuous at x = - 1
<br><br>f(0<sup>-</sup>) =$$\mathop {\lim }\limits_{x \to {0^ - }} \left( { - x - 1} \right)$$
<br><br> = -(0) - 1 = - 1
<br><br>f(0) = $$\mathop {\lim }\limits_{x \to 0} \left( x \right)$$ = 0
<br><br>f(0<sup>+</sup>) = $$\mathop {\lim }\limits_{x \to {0^ + }} \left( x \right)$$ = 0
<br><br>$$ \therefore $$ f(x) is discontinuous at x = 0
<br><br>f(1<sup>-</sup>) = $$\mathop {\lim }\limits_{x \to {1^ - }} \left( x \right)$$ = 1
<br><br>f(1) = $$\mathop {\lim }\limits_{x \to 1} \left( {2x} \right)$$ = 2
<br><br>f(1<sup>+</sup>) = $$\mathop {\lim }\limits_{x \to {1^ + }} \left( {2x} \right)$$ = 2
<br><br>$$ \therefore $$ f(x) is discontinuous at x = 1
<br><br>f(3<sup>-</sup>) = $$\mathop {\lim }\limits_{x \to {3^ - }} \left( {x + 2} \right)$$ = 3 + 2 = 5
<br><br>f(3) = 6
<br><br>$$ \therefore $$ f(x) is discontinuous at x = 3
<br><br>So, f(x) is discontinuous at x = {0, 1, 3}.
| mcq | jee-main-2019-online-8th-april-evening-slot |
8fazrTwsz2FrcJd12a18hoxe66ijvwpqk6h | maths | limits-continuity-and-differentiability | continuity | If the function ƒ defined on , $$\left( {{\pi \over 6},{\pi \over 3}} \right)$$ by
$$$f(x) = \left\{ {\matrix{
{{{\sqrt 2 {\mathop{\rm cosx}\nolimits} - 1} \over {\cot x - 1}},} & {x \ne {\pi \over 4}} \cr
{k,} & {x = {\pi \over 4}} \cr
} } \right.$$$
is continuous, then
k is equal to | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "1 / $$\\sqrt 2$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "2"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$$ = f($${{\pi \over 4}}$$) = k
<br><br>$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$$ ($${0 \over 0}$$ form) = k
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - \sqrt 2 \sin x} \over {-\cos e{c^2}x}}$$ (Using L Hospital Rule)
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {\pi \over 4}} \sqrt 2 {\sin ^3}x$$ = k
<br><br>$$ \Rightarrow $$ k = $$\sqrt 2 {\left( {{1 \over {\sqrt 2 }}} \right)^3}$$ = $${1 \over 2}$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
jhCiGFJptUGSuZbsqt18hoxe66ijvwvnf7c | maths | limits-continuity-and-differentiability | continuity | If $$f(x) = [x] - \left[ {{x \over 4}} \right]$$ ,x $$ \in $$
4
, where [x] denotes the
greatest integer function, then | [{"identifier": "A", "content": "Both $$\\mathop {\\lim }\\limits_{x \\to 4 - } f(x)$$ and $$\\mathop {\\lim }\\limits_{x \\to 4 + } f(x)$$ exist but are not\nequal"}, {"identifier": "B", "content": "f is continuous at x = 4"}, {"identifier": "C", "content": "$$\\mathop {\\lim }\\limits_{x \\to 4 + } f(x)$$ exists but $$\\mathop {\\lim }\\limits_{x \\to 4 - } f(x)$$ does not exist"}, {"identifier": "D", "content": "$$\\mathop {\\lim }\\limits_{x \\to 4 - } f(x)$$ exists but $$\\mathop {\\lim }\\limits_{x \\to 4 + } f(x)$$ does not exist"}] | ["B"] | null | $$f(x) = [x] - \left[ {{x \over 4}} \right]$$
<br><br>Here check continuty at x = 4
<br><br>LHL = $$\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to {4^ - }} \left[ x \right] - \left[ {{x \over 4}} \right]$$
<br><br>= $$\mathop {\lim }\limits_{h \to 0} \left[ {4 - h} \right] - \left[ {{{4 - h} \over 4}} \right]$$
<br><br>= 3 - 0 = 3
<br><br>h $$ \to $$ 0 means h > 0 (very small positive value)
<br><br>So 4 - h = 3.something (means less than 4)
<br><br>$$ \therefore $$ [4 - h] = [3.something] = 3
<br><br>and $$\left[ {{{4 - h} \over 4}} \right]$$ = [0.something] = 0
<br><br>RHL = $$\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to {4^ + }} \left[ x \right] - \left[ {{4 \over x}} \right]$$
<br><br>= $$\mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] - \left[ {{{4 + h} \over 4}} \right]$$
<br><br>= 4 - 1 = 3
<br><br>Here [4 + h] = [4.something] = 4
<br><br>and $$\left[ {{{4 + h} \over 4}} \right]$$ = [1.something] = 1
<br><br>And f(4) = $$\left[ 4 \right] - \left[ {{4 \over 4}} \right]$$ = 4 - 1 = 3
<br><br>As LHL = RHL = f(4)
<br><br>$$ \therefore $$ f is continuous at x = 4. | mcq | jee-main-2019-online-9th-april-evening-slot |
TCbtDz0RxmdD0SVTI218hoxe66ijvwvjoqb | maths | limits-continuity-and-differentiability | continuity | If the function $$f(x) = \left\{ {\matrix{
{a|\pi - x| + 1,x \le 5} \cr
{b|x - \pi | + 3,x > 5} \cr
} } \right.$$<br/>
is
continuous at x = 5, then the value of a – b is :- | [{"identifier": "A", "content": "$${2 \\over {\\pi - 5 }}$$"}, {"identifier": "B", "content": "$${2 \\over {5 - \\pi }}$$"}, {"identifier": "C", "content": "$${-2 \\over {\\pi + 5 }}$$"}, {"identifier": "D", "content": "$${2 \\over {\\pi + 5 }}$$"}] | ["B"] | null | As f(x) is continuous at x = 5 then
<br><br>$$\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = f\left( 5 \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)$$
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} f\left( {5 - h} \right) = f\left( 5 \right) = \mathop {\lim }\limits_{h \to 0} f\left( {5 + h} \right)$$
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} \left( {a\left| {\pi - \left( {5 - h} \right)} \right| + 1} \right) = a\left| {\pi - 5} \right| + 1$$
<br><br> = $$\mathop {\lim }\limits_{h \to 0} \left( {b\left| {5 + h - \pi } \right| + 3} \right)$$
<br><br>$$ \Rightarrow $$ $${a\left| {\pi - 5} \right| + 1}$$ = $$a\left| {\pi - 5} \right| + 1$$
<br><br>= $${b\left| {5 - \pi } \right| + 3}$$
<br><br><b>Note :</b> As $$\pi $$ = 3.14, then $$\pi $$ - 5 = 3.14 - 5 = -1.84 < 0
<br><br>$$ \therefore $$ $${\left| {\pi - 5} \right|}$$ = - ($$\pi $$ - 5) and $${\left| {5 - \pi } \right|}$$ = (5 - $$\pi $$)
<br><br>$$ \Rightarrow $$ $$ - a\left( {\pi - 5} \right) + 1$$ = $$ - a\left( {\pi - 5} \right) + 1$$ = $${b\left( {5 - \pi } \right) + 3}$$
<br><br>$$ \Rightarrow $$ $$ - a\left( {\pi - 5} \right) + 1$$ = $${b\left( {5 - \pi } \right) + 3}$$
<br><br>$$ \Rightarrow $$ $$\left( {a - b} \right)\left( {5 - \pi } \right) = 2$$
<br><br>$$ \Rightarrow $$ $$\left( {a - b} \right) = {2 \over {\left( {5 - \pi } \right)}}$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
b2zSpcAT0sllT3pXyX3rsa0w2w9jwxsw3fk | maths | limits-continuity-and-differentiability | continuity | If$$f(x) = \left\{ {\matrix{
{{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr
q & {,x = 0} \cr
{{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{{\raise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}}}} & {,x > 0} \cr
} } \right.$$
<br/>is continuous at x = 0, then the ordered pair (p, q) is equal to | [{"identifier": "A", "content": "$$\\left( { - {3 \\over 2}, - {1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over 2},{3 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {3 \\over 2}, {1 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { {5 \\over 2}, {1 \\over 2}} \\right)$$"}] | ["C"] | null | $$f(x) = \left\{ {\matrix{
{{{\sin (p + 1)x + \sin x} \over x}} & {x < 0} \cr
q & {x = 0} \cr
{{{\sqrt {{x^2} + x} - \sqrt x } \over {{x^{{3 \over 2}}}}}} & {x > 0} \cr
} } \right.$$<br><br>
is continuous at x = 0<br><br>
So f(0<sup>–</sup>) = f(0) = f (0<sup>+</sup>) ... (1)<br><br>
$$f({0^ - }) = \mathop {lt}\limits_{h \to 0} f(0 - h)$$<br><br>
$$ \Rightarrow \mathop {lt}\limits_{h \to 0} {{\sin (p + 1)( - h) + \sin ( - h)} \over { - h}}$$<br><br>
$$ \Rightarrow \mathop {lt}\limits_{h \to 0} \left[ {{{ - \sin (p + 1)h} \over { - h}} + {{\sinh } \over h}} \right]$$<br><br>
$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sin (p + 1)h} \over {h(p + 1)}} \times (p + 1) + \mathop {\lim }\limits_{h \to 0} {{\sinh } \over h}$$<br><br>
= (p + 1) + 1 = p + 2 ...... (2)<br><br>
Now $$f({0^ + }) = \mathop {\lim }\limits_{h \to 0} (0 + h) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {{h^2} + h} - \sqrt h } \over {{h^{3/2}}}}$$<br><br>
$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(h)}^{{1 \over 2}}}\left[ {\sqrt {h + 1} - 1} \right]} \over {h\left( {{h^{{1 \over 2}}}} \right)}}$$<br><br>
$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sqrt {h + 1} - 1} \over h} \times {{\sqrt {h + 1} + 1} \over {\sqrt {h + 1} + 1}}$$<br><br>
$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{h + 1 - 1} \over {h\left( {\sqrt {h + 1} + 1} \right)}}$$<br><br>
$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {1 \over {\sqrt {h + 1} + 1}} = {1 \over {1 + 1}} = {1 \over 2}$$ ..... (3)<br><br>
Now, from equation (1)<br>
f(0<sup>–</sup>) = f(0) = f(0<sup>+</sup>)<br><br>
p + 2 = q = 1/2<br><br>
So, $$q = {1 \over 2}$$ and $$p = {1 \over 2} - 2 = {{ - 3} \over 2}$$<br><br>
$$(p,q) \equiv \left( { - {3 \over 2},{1 \over 2}} \right)$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
LCqQPEx1rXZyyJNIHG7k9k2k5fp3lto | maths | limits-continuity-and-differentiability | continuity | If the function ƒ defined on $$\left( { - {1 \over 3},{1 \over 3}} \right)$$ by
<br/><br/>f(x) = $$\left\{ {\matrix{
{{1 \over x}{{\log }_e}\left( {{{1 + 3x} \over {1 - 2x}}} \right),} & {when\,x \ne 0} \cr
{k,} & {when\,x = 0} \cr
} } \right.$$
<br/><br/>is continuous, then
k is equal to_______. | [] | null | 5 | $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {{{\ln \left( {1 + 3x} \right)} \over x} - {{\ln \left( {1 - 2x} \right)} \over x}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {3{{\ln \left( {1 + 3x} \right)} \over {3x}} - \left( { - 2} \right){{\ln \left( {1 - 2x} \right)} \over { - 2x}}} \right)$$
<br><br>= 3 + 2 = 5
<br><br>f(x) is continuous
<br><br>$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$ = f(0)
<br><br>So f(0) = 5 = k | integer | jee-main-2020-online-7th-january-evening-slot |
9ZFCv3vcgoXxoGMIksjgy2xukfjjuex1 | maths | limits-continuity-and-differentiability | continuity | Let $$f(x) = x.\left[ {{x \over 2}} \right]$$, for -10< x < 10, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to _____. | [] | null | 8 | $$x \in ( - 10,10)$$<br><br>$$ \Rightarrow $$ $${x \over 2} \in ( - 5,5) \to 9$$ integers<br><br>check continuity at x = 0<br><br>$$\left. {\matrix{
f & {(0) = } & 0 \cr
f & {({0^ + }) = } & 0 \cr
f & {({0^ - }) = } & 0 \cr
} } \right\}continuous\,at\,x = 0$$<br><br>function will be discontinuous when<br><br>$${x \over 2} = \pm 4, \pm 3, \pm 2, \pm 1$$
<br><br>For example checking continuity at x = 4<br><br>$$\left. {\matrix{
f & {(4) = } & 4 \cr
f & {({4^ + }) = } & 4 \cr
f & {({4^ - }) = } & 3 \cr
} } \right\}discontinuous\,at\,x = 4$$
<br><br>$$ \therefore $$ 8 points of discontinuity. | integer | jee-main-2020-online-5th-september-morning-slot |
IDBHWW16iY3z0Pmutmjgy2xukewms5il | maths | limits-continuity-and-differentiability | continuity | If a function f(x) defined by
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr
{c{x^2},} & {1 \le x \le 3} \cr
{a{x^2} + 2cx,} & {3 < x \le 4} \cr
} } \right.$$
<br/><br/>be continuous for some $$a$$, b, c $$ \in $$ R and f'(0) + f'(2) = e, then the value of of $$a$$ is : | [{"identifier": "A", "content": "$${e \\over {{e^2} - 3e - 13}}$$"}, {"identifier": "B", "content": "$${1 \\over {{e^2} - 3e + 13}}$$"}, {"identifier": "C", "content": "$${e \\over {{e^2} - 3e + 13}}$$"}, {"identifier": "D", "content": "$${e \\over {{e^2} + 3e + 13}}$$"}] | ["C"] | null | Given function,
<br>$$f\left( x \right) = \left\{ {\matrix{
{a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr
{c{x^2},} & {1 \le x \le 3} \cr
{a{x^2} + 2cx,} & {3 < x \le 4} \cr
} } \right.$$
<br><br>For continuity at x = 1
<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$$
<br><br>$$ \Rightarrow $$ $$ae + b{e^{ - 1}} = c$$
<br><br>$$ \Rightarrow $$ b = ce - $$a$$e<sup>2</sup> .....(1)
<br><br>For continuity at x = 3
<br><br>$$\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)$$
<br><br>$$ \Rightarrow $$ 9c = 9a + 6c
<br><br>$$ \Rightarrow $$ c = 3a .......(2)
<br><br>Also given, f'(0) + f'(2) = e
<br><br>$$ \Rightarrow $$ (ae<sup>x</sup>
– be<sup>x</sup>
)<sub>x=0</sub> + (2c<sup>x</sup>)<sub>x=2</sub> = e
<br><br>$$ \Rightarrow $$ a – b + 4c = e ........(3)
<br><br>From (1), (2) & (3)
<br><br>a – 3ae + ae<sup>2</sup>
+ 12a = e
<br><br>$$ \Rightarrow $$ a(e<sup>2</sup>
+ 13 – 3e) = e
<br><br>$$ \Rightarrow $$ a = $${e \over {{e^2} - 3e + 13}}$$ | mcq | jee-main-2020-online-2nd-september-morning-slot |
605rQcjA9lhdt5uZLM7k9k2k5k6lz22 | maths | limits-continuity-and-differentiability | continuity | Let [t] denote the greatest integer $$ \le $$ t
and $$\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right] = A$$. <br/>Then the function,
f(x) = [x<sup>2</sup>]sin($$\pi $$x) is discontinuous, when x is
equal to : | [{"identifier": "A", "content": "$$\\sqrt {A + 1} $$"}, {"identifier": "B", "content": "$$\\sqrt {A + 5} $$"}, {"identifier": "C", "content": "$$\\sqrt {A + 21} $$"}, {"identifier": "D", "content": "$$\\sqrt {A} $$"}] | ["A"] | null | A = $$\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right]$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} x\left( {{4 \over x} - \left\{ {{4 \over x}} \right\}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {4 - \left\{ {{4 \over x}} \right\}} \right)$$
<br><br>= 4
<br><br>Now, when x = $$\sqrt {A + 1} $$ = $$\sqrt 5 $$, f(x) = [x<sup>2</sup>]sin($$\pi $$x) is discontinuous at this non integer point.
<br><br>But at x = 2, 3 and 5, f(x) is continuous.
| mcq | jee-main-2020-online-9th-january-evening-slot |
zx9Jvq6OgUw6bi7xYi7k9k2k5itocz7 | maths | limits-continuity-and-differentiability | continuity | If $$f(x) = \left\{ {\matrix{
{{{\sin (a + 2)x + \sin x} \over x};} & {x < 0} \cr
{b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;} & {x = 0} \cr
{{{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{ {1 \over 3}}}} \over {{x^{{4 \over 3}}}}};} & {x > 0} \cr
} } \right.$$<br/>
is continuous at x = 0, then a + 2b is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "-2"}, {"identifier": "D", "content": "1"}] | ["A"] | null | f(0<sup>-</sup>) = $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x + \sin x} \over x}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x} \over {\left( {a + 2} \right)x}} \times \left( {a + 2} \right)$$ + $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin x} \over x}$$
<br><br>= $$\left( {a + 2} \right)$$ + 1
<br><br>= $$\left( {a + 3} \right)$$
<br><br>f(0<sup>+</sup>) = $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{{1 \over 3}}}} \over {{x^{{4 \over 3}}}}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\left( {1 + 3x} \right)}^{{1 \over 3}}} - 1} \over {{x^{{1 \over 3}}}}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {0^ + }} {{1 + x - 1} \over x}$$
<br><br>= 1
<br><br>And f(0) = b
<br><br>As f(x) is continuous at x = 0, then
<br><br>f(0<sup>-</sup>) = f(0) = f(0<sup>+</sup>)
<br><br>$$ \Rightarrow $$ $$a + 3$$ = b = 1
<br><br>$$ \therefore $$ $$a$$ = -2 and b = 1
<br><br>$$ \therefore $$ $$a$$ + 2b = -2 + 2 = 0 | mcq | jee-main-2020-online-9th-january-morning-slot |
ha7ang0gyEpK5LIsYn1klrekjnm | maths | limits-continuity-and-differentiability | continuity | If f : R $$ \to $$ R is a function defined by f(x)= [x - 1] $$\cos \left( {{{2x - 1} \over 2}} \right)\pi $$, where [.] denotes the greatest
integer function, then f is : | [{"identifier": "A", "content": "continuous for every real x"}, {"identifier": "B", "content": "discontinuous at all integral values of x except at x = 1"}, {"identifier": "C", "content": "discontinuous only at x = 1"}, {"identifier": "D", "content": "continuous only at x = 1"}] | ["A"] | null | Given, $$f(x) = [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$ where [ . ] is greatest integer function and f : R $$\to$$ R<br/><br/>$$\because$$ It is a greatest integer function then we need to check its continuity at x $$\in$$ I except these it is continuous.<br/><br/>Let, x = n where n $$\in$$ I<br/><br/>Then <br/><br/>LHL = $$\mathop {\lim }\limits_{x \to {n^ - }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$<br/><br/>$$ = (n - 2)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0$$<br/><br/>RHL = $$\mathop {\lim }\limits_{x \to {n^ + }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$<br/><br/>$$ = (n - 1)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0$$<br/><br/>and f(n) = 0<br/><br/>Here, $$\mathop {\lim }\limits_{x \to {n^ - }} f(x) = \mathop {\lim }\limits_{x \to {n^ + }} f(x) = f(n)$$<br/><br/>$$\therefore$$ It is continuous at every integers.<br/><br/>Therefore, the given function is continuous for all real x. | mcq | jee-main-2021-online-24th-february-morning-slot |
0uHZpf1JbjDDpEOZKe1kluxgk8f | maths | limits-continuity-and-differentiability | continuity | Let f : R $$ \to $$ R be defined as <br/><br/>$$f(x) = \left\{ \matrix{
2\sin \left( { - {{\pi x} \over 2}} \right),if\,x < - 1 \hfill \cr
|a{x^2} + x + b|,\,if - 1 \le x \le 1 \hfill \cr
\sin (\pi x),\,if\,x > 1 \hfill \cr} \right.$$ If f(x) is continuous on R, then a + b equals : | [{"identifier": "A", "content": "$$-$$3"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $$f( - {1^ - }) = 2$$<br><br>$$f( - {1^ + }) = |a + b - 1|$$<br><br>$$|a + b - 1|\, = 2$$ ... (i)<br><br>$$f({1^ - }) = |a + b + 1|$$<br><br>$$f({1^ + }) = 0$$<br><br>$$|a + b + 1| = 0 \Rightarrow a + b + 1 = 0$$<br><br>$$ \Rightarrow a + b = - 1$$ .... (ii) | mcq | jee-main-2021-online-26th-february-evening-slot |
OthbyCL4SlClGBaQfD1kmix3tnd | maths | limits-continuity-and-differentiability | continuity | Let $$\alpha$$ $$\in$$ R be such that the function $$f(x) = \left\{ {\matrix{
{{{{{\cos }^{ - 1}}(1 - {{\{ x\} }^2}){{\sin }^{ - 1}}(1 - \{ x\} )} \over {\{ x\} - {{\{ x\} }^3}}},} & {x \ne 0} \cr
{\alpha ,} & {x = 0} \cr
} } \right.$$ is continuous at x = 0, where {x} = x $$-$$ [ x ] is the greatest integer less than or equal to x. Then : | [{"identifier": "A", "content": "no such $$\\alpha$$ exists"}, {"identifier": "B", "content": "$$\\alpha$$ = 0"}, {"identifier": "C", "content": "$$\\alpha$$ = $${\\pi \\over 4}$$"}, {"identifier": "D", "content": "$$\\alpha$$ = $${\\pi \\over {\\sqrt 2 }}$$"}] | ["A"] | null | $$RHL = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2}){{\sin }^{ - 1}}(1 - x)} \over {x(1 - {x^2})}} $$
<br><br>$$= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2})} \over x}$$<br><br>$$ = {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{ - 1} \over {\sqrt {1 - {{(1 - {x^2})}^2}} }}( - 2x)$$ (L' Hospital Rule)<br><br>$$ = \pi \mathop {\lim }\limits_{x \to {0^ + }} {x \over {\sqrt {2{x^2} - {x^4}} }} = \pi \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {\sqrt {2 - {x^2}} }} = {\pi \over {\sqrt 2 }}$$<br><br>$$LHL = \mathop {\lim }\limits_{x \to {0^ - }} {{{{\cos }^{ - 1}}(1 - {{(1 + x)}^2}){{\sin }^{ - 1}}( - x)} \over {(1 + x) - {{(1 + x)}^3}}} $$
<br><br>$$= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ - }} {{{{\sin }^{ - 1}}x} \over {(1 - x)\left[ {{{(1 + x)}^2} - 1} \right]}} = {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ - }} {{{{\sin }^{ - 1}}x} \over {{x^2} + 2x}}$$<br><br>$$ = {\pi \over 2}\left( {{1 \over 2}} \right) = {\pi \over 4}$$<br><br>As LHL $$ \ne $$ RHL so f(x) is not continuous at x = 0 | mcq | jee-main-2021-online-16th-march-evening-shift |
dTxhqnR9k40ZyMN5n31kmizqg2q | maths | limits-continuity-and-differentiability | continuity | Let f : R $$ \to $$ R and g : R $$ \to $$ R be defined as <br/><br/>$$f(x) = \left\{ {\matrix{
{x + a,} & {x < 0} \cr
{|x - 1|,} & {x \ge 0} \cr
} } \right.$$ and <br/><br/>$$g(x) = \left\{ {\matrix{
{x + 1,} & {x < 0} \cr
{{{(x - 1)}^2} + b,} & {x \ge 0} \cr
} } \right.$$,<br/><br/> where a, b are non-negative real numbers. If (gof) (x) is continuous for all x $$\in$$ R, then a + b is equal to ____________. | [] | null | 1 | $$g[f(x)] = \left[ {\matrix{
{f(x) + 1} & {f(x) < 0} \cr
{{{(f(x) - 1)}^2} + b} & {f(x) \ge 0} \cr
} } \right.$$<br><br>$$g[f(x)] = \left[ {\matrix{
{x + a + 1} & {x + a < 0\& x < 0} \cr
{|x - 1| + 1} & {|x - 1| < 0\& x \ge 0} \cr
{{{(x + a - 1)}^2} + b} & {x + a \ge 0\& x < 0} \cr
{{{(|x - 1| - 1)}^2} + b} & {|x - 1| \ge 0\& x \ge 0} \cr
} } \right.$$<br><br>$$g[f(x)] = \left[ {\matrix{
{x + a + 1} & {x \in ( - \infty , - a)\& x \in ( - \infty ,0)} \cr
{|x - 1| + 1} & {x \in \phi } \cr
{{{(x + a - 1)}^2} + b} & {x \in [ - a,\infty )\& x \in [0,\infty )} \cr
{{{(|x - 1| - 1)}^2} + b} & {x \in R\& x \in [0,\infty )} \cr
} } \right.$$<br><br>$$g[f(x)] = \left[ {\matrix{
{x + a + 1} & {x \in ( - \infty , - a)} \cr
{{{(x + a - 1)}^2} + b} & {x \in [ - a,0)} \cr
{{{(|x - 1| - 1)}^2} + b} & {x \in [0,\infty )} \cr
} } \right.$$<br><br>g(f(x)) is continuous.
<br><br>At x = $$-$$a
<br><br>-a + a + 1 = (-a + a - 1)<sup>2</sup> + b
<br><br>$$ \Rightarrow $$ 1 = b + 1
<br><br>$$ \Rightarrow $$ b = 0
<br><br>at x = 0
<br><br>(a $$-$$1)<sup>2</sup> + b = (|0 - 1| - 1)<sup>2</sup> + b
<br><br>$$ \Rightarrow $$ (a $$-$$1)<sup>2</sup> + b = b
<br><br>$$ \Rightarrow $$ a = 1<br><br>$$ \Rightarrow $$ a + b = 1 | integer | jee-main-2021-online-16th-march-evening-shift |
IiJVI2bm82owjrynl21kmjcls4g | maths | limits-continuity-and-differentiability | continuity | If the function $$f(x) = {{\cos (\sin x) - \cos x} \over {{x^4}}}$$ is continuous at each point in its domain and $$f(0) = {1 \over k}$$, then k is ____________. | [] | null | 6 | $$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {{\cos \left( {\sin x} \right) - \cos x} \over {{x^4}}}$$
<br><br>$$ \Rightarrow $$ $${1 \over k} = \mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{\sin x + x} \over 2}} \right)\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{x + \sin x} \over 2}} \right)} \over {\left( {{{x + \sin x} \over 2}} \right)}} \times {{\sin \left( {{{x - \sin x} \over 2}} \right)} \over {\left( {{{x - \sin x} \over 2}} \right)}} \times {{{x^2} - {{\sin }^2}x} \over {4{x^4}}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{x + \sin x} \over x}} \right)\left( {{{x - \sin x} \over {{x^3}}}} \right) \times {1 \over 4}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + \cos x} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 + \sin x} \over {6x}}} \right) \times {1 \over 4}$$
<br><br>= $$2 \times 2 \times {1 \over 6} \times {1 \over 4}$$ = $${1 \over 6}$$ | integer | jee-main-2021-online-17th-march-morning-shift |
h98WMUthi0WrOatQRp1kmm32325 | maths | limits-continuity-and-differentiability | continuity | Let f : R $$ \to $$ R be a function defined as<br/><br/>$$f(x) = \left\{ \matrix{
{{\sin (a + 1)x + \sin 2x} \over {2x}},if\,x < 0 \hfill \cr
b,\,if\,x\, = 0 \hfill \cr
{{\sqrt {x + b{x^3}} - \sqrt x } \over {b{x^{5/2}}}},\,if\,x > 0 \hfill \cr} \right.$$<br/><br/>If f is continuous at x = 0, then the value of a + b is equal to : | [{"identifier": "A", "content": "$$-$$3"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "$$ - {5 \\over 2}$$"}, {"identifier": "D", "content": "$$ - {3 \\over 2}$$<br/>"}] | ["D"] | null | Given, $f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+\sin 2 x}{2 x}, & x<0 \\ b, & x=0 \\ \frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}, & x>0\end{array}\right.$<br/><br/>
$$
\begin{array}{ll}
\because & f(x) \text { is continuous at } x=0 . \\\\
\therefore & \lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{+}} f(x)=f(0) \\\\
\because & f(0)=b
\end{array}
$$<br/><br/>
Now, $\lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x+\sin 2 x}{2 x}\right)$<br/><br/>
$$
\begin{aligned}
\Rightarrow \quad \lim _\limits{x \rightarrow 0^{-}} f(x) & =\lim _\limits{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x}{2 x}+\frac{\sin 2 x}{2 x}\right) \\\\
& =\lim _\limits{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x}{(a+1) x} \times\left(\frac{a+1}{2}\right)+\frac{\sin 2 x}{2 x}\right) \\\\
& =\frac{a+1}{2}+1
\end{aligned}
$$<br/><br/>
Again, $\lim _\limits{x \rightarrow 0^{+}} f(x)=\lim _\limits{x \rightarrow 0^{+}}\left(\frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}\right)$<br/><br/>
$$
=\lim _\limits{x \rightarrow 0^{+}} \frac{\left(\sqrt{x+b x^3}-\sqrt{x}\right)\left(\sqrt{x+b x^3}+\sqrt{x}\right)}{b x^{5 / 2}\left(\sqrt{x+b x^3}+\sqrt{x}\right)}
$$<br/><br/>
$$
\begin{aligned}
& =\lim _{x \rightarrow 0^{+}} \frac{\left(x+b x^3-x\right)}{b x^{5 / 2}\left(\sqrt{x+b x^3}+\sqrt{x}\right)} \\\\
& =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1+b x^2}+1\right)}
\end{aligned}
$$<br/><br/>
$$
\Rightarrow \quad \lim _\limits{x \rightarrow 0^{+}} f(x)=1 / 2
$$<br/><br/>
From Eq. (i), (ii), (iii) and (iv)<br/><br/>
$$
\begin{aligned}
&\frac{1}{2} =b=\frac{a+1}{2}+1 \Rightarrow b=\frac{1}{2}, a=-2 \\\\
&\therefore \quad a+b =\frac{-3}{2}
\end{aligned}
$$ | mcq | jee-main-2021-online-18th-march-evening-shift |
1krpzg126 | maths | limits-continuity-and-differentiability | continuity | Let a function f : R $$\to$$ R be defined as $$f(x) = \left\{ {\matrix{
{\sin x - {e^x}} & {if} & {x \le 0} \cr
{a + [ - x]} & {if} & {0 < x < 1} \cr
{2x - b} & {if} & {x \ge 1} \cr
} } \right.$$ <br/><br/>where [ x ] is the greatest integer less than or equal to x. If f is continuous on R, then (a + b) is equal to: | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "5"}] | ["B"] | null | Continuous x = 0<br><br>f(0<sup>+</sup>) = f(0<sup>$$-$$</sup>) $$\Rightarrow$$ a $$-$$ 1 = 0 $$-$$ e<sup>0</sup><br><br>$$\Rightarrow$$ a = 0<br><br>Continuous at x = 1<br><br>f(1<sup>+</sup>) = f(1<sup>$$-$$</sup>)<br><br>$$\Rightarrow$$ 2(1) $$-$$ b = a + ($$-$$1)<br><br>$$\Rightarrow$$ b = 2 $$-$$ a + 1 $$\Rightarrow$$ b = 3<br><br>$$\therefore$$ a + b = 3 | mcq | jee-main-2021-online-20th-july-morning-shift |
1kru9vuuq | maths | limits-continuity-and-differentiability | continuity | Let f : R $$\to$$ R be defined as $$f(x) = \left\{ {\matrix{
{{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr
{\alpha ,} & {x = 0} \cr
} } \right.$$<br/><br/>If f is continuous at x = 0, then $$\alpha$$ is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2"}] | ["A"] | null | For continuity <br><br>$$\mathop {\lim }\limits_{x \to 0} {{{x^3}} \over {4{{\sin }^4}x}}(\ln (1 + 2x{e^{ - 2x}}) - 2\ln (1 - x{e^{ - x}})) = \alpha $$<br><br>$$\mathop {\lim }\limits_{x \to 0} {1 \over {4x}}[2x{e^{ - 2x}} + 2x{e^{ - x}}] = \alpha $$<br><br>$$ = {1 \over 4}(4) = \alpha = 1$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krvv39l9 | maths | limits-continuity-and-differentiability | continuity | Let f : R $$\to$$ R be defined as<br/><br/>$$f(x) = \left\{ {\matrix{
{{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x < 2} \cr
{{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x > 2} \cr
{\mu ,} & {x = 2} \cr
} } \right.$$<br/><br/>where [x] is the greatest integer is than or equal to x. If f is continuous at x = 2, then $$\lambda$$ + $$\mu$$ is equal to : | [{"identifier": "A", "content": "e($$-$$e + 1)"}, {"identifier": "B", "content": "e(e $$-$$ 2)"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2e $$-$$ 1"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} {e^{{{\tan (x - 2)} \over {x - 2}}}} = {e^1}$$<br><br>$$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} {{ - \lambda (x - 2)(x - 3)} \over {\mu (x - 2)(x - 3)}} = - {\lambda \over \mu }$$<br><br>For continuity $$\mu = e = - {\lambda \over \mu } \Rightarrow \mu = e,\lambda = - {e^2}$$<br><br>$$\lambda + \mu = e( - e + 1)$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1krzqujtw | maths | limits-continuity-and-differentiability | continuity | Consider the function<br/><br/><br/>where P(x) is a polynomial such that P'' (x) is always a constant and P(3) = 9. If f(x) is continuous at x = 2, then P(5) is equal to _____________.<img src="data:image/png;base64,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"/> | [] | null | 39 | $$f(x) = \left\{ {\matrix{
{{{P(x)} \over {\sin (x - 2)}},} & {x \ne 2} \cr
{7,} & {x = 2} \cr
} } \right.$$<br><br>P''(x) = const. $$\Rightarrow$$ P(x) is a 2 degree polynomial<br><br>f(x) is cont. at x = 2<br><br>f(2<sup>+</sup>) = f(2<sup>$$-$$</sup>)<br><br>$$\mathop {\lim }\limits_{x \to {2^ + }} {{P(x)} \over {\sin (x - 2)}} = 7$$<br><br>$$\mathop {\lim }\limits_{x \to {2^ + }} {{(x - 2)(ax + b)} \over {\sin (x - 2)}} = 7 \Rightarrow 2a + b = 7$$<br><br>P(x) = (x $$-$$ 2)(ax + b)<br><br>P(3) = (3 $$-$$ 2)(3a + b) = 9 $$\Rightarrow$$ 3a + b = 9<br><br>a = 2, b = 3<br><br>P(5) = (5 $$-$$ 2)(2.5 + 3) = 3.13 = 39 | integer | jee-main-2021-online-25th-july-evening-shift |
1ks08g1ia | maths | limits-continuity-and-differentiability | continuity | Let $$f:\left( { - {\pi \over 4},{\pi \over 4}} \right) \to R$$ be defined as $$f(x) = \left\{ {\matrix{
{{{(1 + |\sin x|)}^{{{3a} \over {|\sin x|}}}}} & , & { - {\pi \over 4} < x < 0} \cr
b & , & {x = 0} \cr
{{e^{\cot 4x/\cot 2x}}} & , & {0 < x < {\pi \over 4}} \cr
} } \right.$$<br/><br/>If f is continuous at x = 0, then the value of 6a + b<sup>2</sup> is equal to : | [{"identifier": "A", "content": "1 $$-$$ e"}, {"identifier": "B", "content": "e $$-$$ 1"}, {"identifier": "C", "content": "1 + e"}, {"identifier": "D", "content": "e"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} f(x) = b$$<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} x{e^{{{\cot 4x} \over {\cot 2x}}}} = {e^{{1 \over 2}}} = b$$<br><br>$$\mathop {\lim }\limits_{x \to {0^ - }} {(1 + |\sin x|)^{{{3a} \over {|\sin x|}}}} = {e^{3a}} = {e^{{1 \over 2}}}$$<br><br>$$a = {1 \over 6} \Rightarrow 6a = 1$$<br><br>$$ \therefore $$ $$(6a + {b^2}) = (1 + e)$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktbj2omr | maths | limits-continuity-and-differentiability | continuity | Let a, b $$\in$$ R, b $$\in$$ 0, Define a function <br/><br/>$$f(x) = \left\{ {\matrix{
{a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr
{{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr
} } \right.$$. <br/><br/>If f is continuous at x = 0, then 10 $$-$$ ab is equal to ________________. | [] | null | 14 | $$f(x) = \left\{ {\matrix{
{a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr
{{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr
} } \right.$$<br><br>For continuity at '0'<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{\tan 2x - \sin 2x} \over {b{x^3}}} = - a$$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{{{8{x^3}} \over 3} + {{8{x^3}} \over {3!}}} \over {b{x^3}}} = - a$$<br><br>$$ \Rightarrow 8\left( {{1 \over 3} + {1 \over {3!}}} \right) = - ab$$<br><br>$$ \Rightarrow 4 = - ab$$<br><br>$$ \Rightarrow 10 - ab = 14$$ | integer | jee-main-2021-online-26th-august-morning-shift |
1ktiqpn6h | maths | limits-continuity-and-differentiability | continuity | If the function <br/>$$f(x) = \left\{ {\matrix{
{{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \cr
k & , & {x = 0} \cr
{{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \cr
} } \right.$$ is continuous <br/><br/>at x = 0, then $${1 \over a} + {1 \over b} + {4 \over k}$$ is equal to : | [{"identifier": "A", "content": "$$-$$5"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "4"}] | ["A"] | null | If f(x) is continuous at x = 0, RHL = LHL = f(0)<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}.{{\sqrt {{x^2} + 1} + 1} \over {\sqrt {{x^2} + 1} + 1}}$$ (Rationalisation)<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} - {{2{{\sin }^2}x} \over {{x^2}}}.\left( {\sqrt {{x^2} + 1} + 1} \right) = - 4$$<br><br>$$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} {1 \over x}\ln \left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)$$<br><br>$$\mathop {\lim }\limits_{x \to {0^ - }} {{\ln \left( {1{x \over a}} \right)} \over {\left( {{x \over a}} \right).\,a}} + {{\ln \left( {1 - {x \over b}} \right)} \over {\left( { - {x \over b}} \right)\,.\,b}}$$$$ = {1 \over a} + {1 \over b}$$<br><br>So, $${1 \over a} + {1 \over b} = - 4 = k$$<br><br>$$ \Rightarrow {1 \over a} + {1 \over b} + {4 \over k} = - 4 - 1 = - 5$$ | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktoc7wu0 | maths | limits-continuity-and-differentiability | continuity | Let [t] denote the greatest integer $$\le$$ t. The number of points where the function $$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1],x \in ( - 2,2)$$ is not continuous is _____________. | [] | null | 2 | $$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1]$$<br><br>$$f\left( x \right) = \left\{ {\matrix{
{ - 2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr
{ - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr
{\sin {\pi \over 3} + 1,} & {0 \le x < 1} \cr
{\left| {{x^2} - 1} \right| + {1 \over {\sqrt 2 }} - 2,} & {1 \le x < 2} \cr
} } \right.$$
<br><br>$$ \therefore $$ at x = -1, $$\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = 1$$ and $$\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = 1$$
<br><br>Hence continuous at x = –1
<br><br>Similarly check at x = 0,
<br><br>$$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$$ and $$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$$
<br><br>So, f(x) discontinuous
and at x = 0
<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$$ and $$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = {1 \over {\sqrt 2 }} - 2$$
<br><br>So, f(x) discontinuous
and at x = 1
<br><br>Hence 2 points of discontinuity. | integer | jee-main-2021-online-1st-september-evening-shift |
1l55hacch | maths | limits-continuity-and-differentiability | continuity | <p>Let f, g : R $$\to$$ R be functions defined by</p>
<p>$$f(x) = \left\{ {\matrix{
{[x]} & , & {x < 0} \cr
{|1 - x|} & , & {x \ge 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{{e^x} - x} & , & {x < 0} \cr
{{{(x - 1)}^2} - 1} & , & {x \ge 0} \cr
} } \right.$$ where [x] denote the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly :</p> | [{"identifier": "A", "content": "one point"}, {"identifier": "B", "content": "two points"}, {"identifier": "C", "content": "three points"}, {"identifier": "D", "content": "four points"}] | ["B"] | null | $f(x)=\left\{\begin{array}{ll}{[x],} & x<0 \\ |1-x|, & x \geq 0\end{array}\right.$ and $g(x)= \begin{cases}e^{x}-x, & x<0 \\ (x-1)^{2}-1, & x \geq 0\end{cases}$
<br><br>
$$
f \circ g(x)= \begin{cases}{[g(x)],} & g(x)<0 \\ |1-g(x)|, & g(x) \geq 0\end{cases}
$$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l99t3fun/ee0d4d39-3197-4ca1-984e-cfb744e32a27/66a131e0-4c78-11ed-b94d-45a8040c2a81/file-1l99t3fuo.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l99t3fun/ee0d4d39-3197-4ca1-984e-cfb744e32a27/66a131e0-4c78-11ed-b94d-45a8040c2a81/file-1l99t3fuo.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Evening Shift Mathematics - Limits, Continuity and Differentiability Question 78 English Explanation"><br>
$$
=\left\{\begin{array}{cc}
\left|1+x-e^{x}\right|, & x<0 \\
1, & x=0 \\
{\left[(x-1)^{2}-1\right],} & 0 < x < 2 \\
\left|2-(x-1)^{2}\right|, & x \geq 2
\end{array}\right.
$$
<br><br>
So, $x=0,2$ are the two points where fog is discontinuous. | mcq | jee-main-2022-online-28th-june-evening-shift |
1l566fwdl | maths | limits-continuity-and-differentiability | continuity | <p>Let f : R $$\to$$ R be defined as</p>
<p>$$f(x) = \left[ {\matrix{
{[{e^x}],} & {x < 0} \cr
{a{e^x} + [x - 1],} & {0 \le x < 1} \cr
{b + [\sin (\pi x)],} & {1 \le x < 2} \cr
{[{e^{ - x}}] - c,} & {x \ge 2} \cr
} } \right.$$</p>
<p>where a, b, c $$\in$$ R and [t] denotes greatest integer less than or equal to t. Then, which of the following statements is true?</p> | [{"identifier": "A", "content": "There exists a, b, c $$\\in$$ R such that f is continuous on R."}, {"identifier": "B", "content": "If f is discontinuous at exactly one point, then a + b + c = 1"}, {"identifier": "C", "content": "If f is discontinuous at exactly one point, then a + b + c $$\\ne$$ 1"}, {"identifier": "D", "content": "f is discontinuous at at least two points, for any values of a, b and c"}] | ["C"] | null | <p>$$f(x) = \left\{ {\matrix{
0 & {x < 0} \cr
{a{e^x} - 1} & {0 \le x < 1} \cr
b & {x = 1} \cr
{b - 1} & {1 < x < 2} \cr
{ - c} & {x \ge 2} \cr
} } \right.$$</p>
<p>To be continuous at x = 0</p>
<p>a $$-$$ 1 = 0</p>
<p>to be continuous at x = 1</p>
<p>ae $$-$$ 1 = b = b $$-$$ 1 $$\Rightarrow$$ not possible</p>
<p>to be continuous at x = 2</p>
<p>b $$-$$ 1 = $$-$$ c $$\Rightarrow$$ b + c = 1</p>
<p>If a = 1 and b + c = 1 then f(x) is discontinuous at exactly one point.</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l59l625z | maths | limits-continuity-and-differentiability | continuity | <p>Let $$f(x) = \left[ {2{x^2} + 1} \right]$$ and $$g(x) = \left\{ {\matrix{
{2x - 3,} & {x < 0} \cr
{2x + 3,} & {x \ge 0} \cr
} } \right.$$, where [t] is the greatest integer $$\le$$ t. Then, in the open interval ($$-$$1, 1), the number of points where fog is discontinuous is equal to ______________.</p> | [] | null | 62 | $$
\mathrm{f}(\mathrm{g}(\mathrm{x}))=\left[2 \mathrm{~g}^2(\mathrm{x})\right]+1
$$<br/><br/>
$$
=\left\{\begin{array}{l}
{\left[2(2 x-3)^2\right]+1 ; x<0} \\
{\left[2(2 x+3)^2\right]+1 ; x \geq 0}
\end{array}\right.
$$<br/><br/>
$\therefore$ fog is discontinuous whenever $2(2 x-3)^2$ or $2(2 x+3)^2$ belongs to integer except $x=0$<br/><br/>
$\therefore 62$ points of discontinuity. | integer | jee-main-2022-online-25th-june-evening-shift |
1l5c2bcte | maths | limits-continuity-and-differentiability | continuity | <p>The number of points where the function</p>
<p>$$f(x) = \left\{ {\matrix{
{|2{x^2} - 3x - 7|} & {if} & {x \le - 1} \cr
{[4{x^2} - 1]} & {if} & { - 1 < x < 1} \cr
{|x + 1| + |x - 2|} & {if} & {x \ge 1} \cr
} } \right.$$</p>
<p>[t] denotes the greatest integer $$\le$$ t, is discontinuous is _____________.</p> | [] | null | 7 | $\because f(-1)=2$ and $f(1)=3$
<br/><br/>
For $x \in(-1,1),\left(4 x^{2}-1\right) \in[-1,3)$
<br/><br/>
hence $f(x)$ will be discontinuous at $x=1$ and also
<br/><br/>
whenever $4 x^{2}-1=0,1$ or 2
<br/><br/>
$$
\Rightarrow x=\pm \frac{1}{2}, \pm \frac{1}{\sqrt{2}} \text { and } \pm \frac{\sqrt{3}}{2}
$$
<br/><br/>
So there are total 7 points of discontinuity. | integer | jee-main-2022-online-24th-june-morning-shift |
1l6ggcf4g | maths | limits-continuity-and-differentiability | continuity | <p>Let f : R $$\to$$ R be a continuous function such that $$f(3x) - f(x) = x$$. If $$f(8) = 7$$, then $$f(14)$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "16"}] | ["B"] | null | <p>$$f(3x) - f(x) = x$$ ...... (1)</p>
<p>$$x \to {x \over 3}$$</p>
<p>$$f(x) - f\left( {{x \over 3}} \right) = {x \over 3}$$ ....... (2)</p>
<p>Again $$x \to {x \over 3}$$</p>
<p>$$f\left( {{x \over 3}} \right) - f\left( {{x \over 9}} \right) = {x \over {{3^2}}}$$ ...... (3)</p>
<p>Similarly</p>
<p>$$f\left( {{x \over {{3^{n - 2}}}}} \right) - f\left( {{x \over {{3^{n - 1}}}}} \right) = {x \over {{3^{n - 1}}}}\,.....\,(n)$$</p->
<p>Adding all these and applying $$n \to \infty $$</p>
<p>$$\mathop {\lim }\limits_{n \to \infty } \left( {f(3x) - f\left( {{x \over {{3^{n - 1}}}}} \right)} \right) = x\left( {1 + {1 \over 3} + {1 \over {{3^2}}}\, + \,....} \right)$$</p>
<p>$$f(3x) - f(0) = {{3x} \over 2}$$</p>
<p>Putting $$x = {8 \over 3}$$</p>
<p>$$f(8) - f(0) = 4$$</p>
<p>$$ \Rightarrow f(0) = 3$$</p>
<p>Putting $$x = {{14} \over 3}$$</p>
<p>$$f(14) - 3 = 7 \Rightarrow f(14) = 10$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6ggrtzq | maths | limits-continuity-and-differentiability | continuity | <p>If $$f(x) = \left\{ {\matrix{
{x + a} & , & {x \le 0} \cr
{|x - 4|} & , & {x > 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{x + 1} & , & {x < 0} \cr
{{{(x - 4)}^2} + b} & , & {x \ge 0} \cr
} } \right.$$ are continuous on R, then $$(gof)(2) + (fog)( - 2)$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$10"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "$$-$$8"}] | ["D"] | null | <p>$$f(x) = \left\{ {\matrix{
{x + a} & , & {x \le 0} \cr
{|x - 4|} & , & {x > 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{x + 1} & , & {x < 0} \cr
{{{(x - 4)}^2} + b} & , & {x \ge 0} \cr
} } \right.$$</p>
<p>$$\because$$ $$f(x)$$ and $$g(x)$$ are continuous on R</p>
<p>$$\therefore$$ $$a = 4$$ and $$b = 1 - 16 = - 15$$</p>
<p>then $$(gof)(2) + (fog)( - 2)$$</p>
<p>$$ = g(2) + f( - 1)$$</p>
<p>$$ = - 11 + 3 = - 8$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6gh1jfg | maths | limits-continuity-and-differentiability | continuity | <p>Let $$f(x) = \left\{ {\matrix{
{{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr
{ - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr
} } \right.$$.</p>
<p>Then the set of all values of b, for which f(x) has maximum value at x = 1, is :</p> | [{"identifier": "A", "content": "($$-$$6, $$-$$2)"}, {"identifier": "B", "content": "(2, 6)"}, {"identifier": "C", "content": "$$[ - 6, - 2) \\cup (2,6]$$"}, {"identifier": "D", "content": "$$\\left[ {-\\sqrt 6 , - 2} \\right) \\cup \\left( {2,\\sqrt 6 } \\right]$$"}] | ["C"] | null | <p>$$f(x) = \left\{ {\matrix{
{{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr
{ - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr
} } \right.$$</p>
<p>If $$f(x)$$ has maximum value at $$x = 1$$ then $$f(1 + ) \le f(1)$$</p>
<p>$$ - 2 + {\log _2}({b^2} - 4) \le 1 - 1 + 10 - 7$$</p>
<p>$${\log _2}({b^2} - 4) \le 5$$</p>
<p>$$0 < {b^2} - 4 \le 32$$</p>
<p>(i) $${b^2} - 4 > 0 \Rightarrow b \in ( - \infty , - 2) \cup (2,\infty )$$</p>
<p>(ii) $${b^2} - 36 \le 0 \Rightarrow b \in [ - 6,6]$$</p>
<p>Intersection of above two sets</p>
<p>$$b \in [ - 6, - 2) \cup (2,6]$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6kimxb4 | maths | limits-continuity-and-differentiability | continuity | <p>If for $$\mathrm{p} \neq \mathrm{q} \neq 0$$, the function $$f(x)=\frac{\sqrt[7]{\mathrm{p}(729+x)}-3}{\sqrt[3]{729+\mathrm{q} x}-9}$$ is continuous at $$x=0$$, then :</p> | [{"identifier": "A", "content": "$$7 p q \\,f(0)-1=0$$"}, {"identifier": "B", "content": "$$63 q \\,f(0)-\\mathrm{p}^{2}=0$$"}, {"identifier": "C", "content": "$$21 q \\,f(0)-\\mathrm{p}^{2}=0$$"}, {"identifier": "D", "content": "$$7 p q \\,f(0)-9=0$$"}] | ["B"] | null | <p>$$f(x) = {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$$</p>
<p>for continuity at $$x = 0$$, $$\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$$</p>
<p>Now, $$\therefore$$ $$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$$</p>
<p>$$ \Rightarrow p = 3$$ (To make indeterminant form)</p>
<p>So, $$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{{{\left( {{3^7} + 3x} \right)}^{{1 \over 7}}} - 3} \over {{{\left( {729 + qx} \right)}^{{1 \over 3}}} - 9}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{3\left[ {{{\left( {1 + {x \over {{3^6}}}} \right)}^{{1 \over 7}}} - 1} \right]} \over {9\left[ {{{\left( {1 + {q \over {729}}x} \right)}^{{1 \over 3}}} - 1} \right]}} = {1 \over 3}\,.\,{{{1 \over 7}\,.\,{1 \over {{3^6}}}} \over {{1 \over 3}\,.\,{q \over {729}}}}$$</p>
<p>$$\therefore$$ $$f(0) = {1 \over {7q}}$$</p>
<p>$$\therefore$$ Option (B) is correct.</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6nlhfp1 | maths | limits-continuity-and-differentiability | continuity | <p>The function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ defined by <br/><br/>$$f(x)=\lim\limits_{n \rightarrow \infty} \frac{\cos (2 \pi x)-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}}$$ is continuous for all x in :</p> | [{"identifier": "A", "content": "$$R-\\{-1\\}$$"}, {"identifier": "B", "content": "$$ \\mathbb{R}-\\{-1,1\\}$$"}, {"identifier": "C", "content": "$$R-\\{1\\}$$"}, {"identifier": "D", "content": "$$R-\\{0\\}$$"}] | ["B"] | null | <p>$$f(x) = \mathop {\lim }\limits_{n \to \infty } {{\cos (2\pi x) - {x^{2n}}\sin (x - 1)} \over {1 + {x^{2n + 1}} - {x^{2n}}}}$$</p>
<p>For $$|x| < 1,\,f(x) = \cos 2\pi x$$, continuous function</p>
<p>$$|x| > 1,\,f(x) = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {{x^{2n}}}}\cos 2\pi x - \sin (x - 1)} \over {{1 \over {{x^{2n}}}} + x - 1}}$$</p>
<p>$$ = {{ - \sin (x - 1)} \over {x - 1}}$$, continuous</p>
<p>For $$|x| = 1,\,f(x) = \left\{ {\matrix{
1 & {\mathrm{if}} & {x = 1} \cr
{ - (1 + \sin 2)} & {\mathrm{if}} & {x = - 1} \cr
} } \right.$$</p>
<p>Now,</p>
<p>$$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = - 1,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$$, so discontinuous at $$x = 1$$</p>
<p>$$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1,\,\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = - {{\sin 2} \over 2}$$, so discontinuous at $$x = - 1$$</p>
<p>$$\therefore$$ $$f(x)$$ is continuous for all $$x \in R - \{ - 1,1\} $$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
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