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latsm1x3m4CNbZCrVO1kmknb7se | maths | limits-continuity-and-differentiability | limits-of-algebric-function | The value of $$\mathop {\lim }\limits_{n \to \infty } {{[r] + [2r] + ... + [nr]} \over {{n^2}}}$$, where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to : | [{"identifier": "A", "content": "r"}, {"identifier": "B", "content": "$${r \\over 2}$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2r"}] | ["B"] | null | We know,<br><br>(x $$-$$ 1) $$ \le $$ [x] < x<br><br>$$ \therefore $$ (r $$-$$ 1) $$ \le $$ [r] < r <br><br>(2r $$-$$ 1) $$ \le $$ [2r] < 2r <br><br>.<br><br>.<br><br>.<br><br>(nr $$-$$ 1) $$ \le $$ [nr] < nr<br><br>Adding<br><br>$${{n(n + 1)} \over 2}r - n \le [r] + [2r] + .......[nr] < {{n(n + 1)} \over 2}r$$
<br><br>$${{{{n\left( {n + 1} \right)} \over 2}r - n} \over {{n^2}}} \le {{\left[ r \right] + \left[ {2r} \right] + .... + \left[ {nr} \right]} \over {{n^2}}} \le {{{{n\left( {n + 1} \right)} \over 2}r} \over {{n^2}}}$$
<br><br>$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{{n(n + 1)} \over 2}r - n} \over {{n^2}}}} \right) \le L < \mathop {\lim }\limits_{n \to \infty } {{n(n + 1)} \over 2}r$$<br><br>$$ \Rightarrow {r \over 2} \le L < {r \over 2}$$<br><br>$$ \Rightarrow L = {r \over 2}$$ | mcq | jee-main-2021-online-17th-march-evening-shift |
1krrqx6yp | maths | limits-continuity-and-differentiability | limits-of-algebric-function | If $$f:R \to R$$ is given by $$f(x) = x + 1$$, then the value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right]$$ is : | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${7 \\over 2}$$"}] | ["D"] | null | $$f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)$$<br><br>$$ \Rightarrow 1 + 1 + {5 \over n} + 1 + {{10} \over n} + .... + 1 + {{5(n - 1)} \over n}$$<br><br>$$ \Rightarrow n + {5 \over n}{{(n - 1)n} \over 2} = {{2n + 5n - 5} \over 2} = {{7n - 5} \over 2}$$<br><br>$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left( {{{7n - 5} \over 2}} \right) = {7 \over 2}$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1ks0994hx | maths | limits-continuity-and-differentiability | limits-of-algebric-function | Let f : R $$\to$$ R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of $$\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$$ is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "12"}] | ["D"] | null | This limit can be solved using L'Hopital's Rule, which states that for the limit of the form 0/0 or ±∞/±∞, the limit can be found by taking the derivative of the numerator and the derivative of the denominator separately.
<br/><br/>$$\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$$ is in the indeterminate form, and we are given that f(2) = 4 and f'(2) = 1, so we can apply L'Hopital's rule.
<br/><br/>Taking the derivative of the numerator and the denominator, we get :
<br/><br/>Numerator: derivative of $x^2 f(2) - 4f(x)$ is $2x f(2) - 4f'(x)$.
<br/><br/>Denominator: derivative of $x - 2$ is $1$.
<br/><br/>So, the limit becomes :
<br/><br/>$$\mathop {\lim }\limits_{x \to 2} {{{2xf(2) - 4f'(x)}} \over 1} = 2 \times 2 \times f(2) - 4 \times f'(2) = 16 - 4 = 12.$$
<br/><br/>Therefore, Option D, 12, is the correct answer. | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktd2oyng | maths | limits-continuity-and-differentiability | limits-of-algebric-function | $$\mathop {\lim }\limits_{x \to 2} \left( {\sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} } \right)$$ is equal to : | [{"identifier": "A", "content": "$${9 \\over {44}}$$"}, {"identifier": "B", "content": "$${5 \\over {24}}$$"}, {"identifier": "C", "content": "$${1 \\over 5}$$"}, {"identifier": "D", "content": "$${7 \\over {36}}$$"}] | ["A"] | null | $$S = \mathop {\lim }\limits_{x \to 2} \sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} $$<br><br>$$S = \sum\limits_{n = 1}^9 {{2 \over {4({n^2} + 3n + 2)}}} = {1 \over 2}\sum\limits_{n = 1}^9 {\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} $$<br><br>$$S = {1 \over 2}\left( {{1 \over 2} - {1 \over {11}}} \right) = {9 \over {44}}$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktg3wzp1 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | If $$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$$, then the ordered pair (a, b) is : | [{"identifier": "A", "content": "$$\\left( {1,{1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {1, - {1 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - 1,{1 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - 1, - {1 \\over 2}} \\right)$$"}] | ["B"] | null | $$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} } \right) - ax = b$$ ($$\infty$$ $$-$$ $$\infty$$)<br><br>Now, $$\mathop {\lim }\limits_{x \to \infty } {{({x^2} - x + 1 - {a^2}{x^2}}) \over {\sqrt {{x^2} - x + 1} + ax}} = b$$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {\sqrt {{x^2} - x + 1} + ax}} = b$$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$<br><br>$$ \Rightarrow 1 - {a^2} = 0 \Rightarrow a = 1$$<br><br>Now, $$\mathop {\lim }\limits_{x \to \infty } {{ - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$<br><br>$$ \Rightarrow {{ - 1} \over {1 + a}} = b \Rightarrow b = - {1 \over 2}$$<br><br>$$(a,b) = \left( {1, - {1 \over 2}} \right)$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktoau6kf | maths | limits-continuity-and-differentiability | limits-of-algebric-function | Let $$f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3$$, x $$\in$$ R. Then the natural number n for which $$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$$ is __________. | [] | null | 7 | $$f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3$$<br><br>$$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$$<br><br>$$\mathop {\lim }\limits_{x \to 1} {{9{x^n} - ({x^6} + 2{x^4} + {x^3} + 2x + 3)} \over {x - 1}} = 44$$<br><br>$$\mathop {\lim }\limits_{x \to 1} {{9n{x^{n - 1}} - (6{x^5} + 8{x^3} + 3{x^2} + 2)} \over 1} = 44$$<br><br>$$\Rightarrow$$ 9n $$-$$ (19) = 44<br><br>$$\Rightarrow$$ 9n = 63<br><br>$$\Rightarrow$$ n = 7 | integer | jee-main-2021-online-1st-september-evening-shift |
1l56u4jaa | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>Let [t] denote the greatest integer $$\le$$ t and {t} denote the fractional part of t. The integral value of $$\alpha$$ for which the left hand limit of the function</p>
<p>$$f(x) = [1 + x] + {{{\alpha ^{2[x] + {\{x\}}}} + [x] - 1} \over {2[x] + \{ x\} }}$$ at x = 0 is equal to $$\alpha - {4 \over 3}$$, is _____________.</p> | [] | null | 3 | <p>$$f(x) = [1 + x] + {{{a^{2[x] + \{ x\} }} + [x] - 1} \over {2[x] + \{ x\} }}$$</p>
<p>$$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \alpha - {4 \over 3}$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} 1 + [x] + {{{\alpha ^{x + [x]}} + [x] - 1} \over {x + [x]}} = \alpha - {4 \over 3}$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} 1 - 1 + {{{\alpha ^{ - h - 1}} - 1 - 1} \over { - h - 1}} = \alpha - {4 \over 3}$$</p>
<p>$$\therefore$$ $${{{\alpha ^{ - 1}} - 2} \over { - 1}} = \alpha - {4 \over 3}$$</p>
<p>$$ \Rightarrow 3{\alpha ^2} - 10\alpha + 3 = 0$$</p>
<p>$$\therefore$$ $$\alpha = 3$$ or $${1 \over 3}$$</p>
<p>$$\because$$ $$\alpha$$ in integer, hence $$\alpha$$ = 3</p> | integer | jee-main-2022-online-27th-june-evening-shift |
1l57o3waq | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>Let a be an integer such that $$\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$$ exists, where [t] is greatest integer $$\le$$ t. Then a is equal to :</p> | [{"identifier": "A", "content": "$$-$$6"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "6"}] | ["A"] | null | <p>$$\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$$ exist & $$a \in I$$.</p>
<p>$$ = \mathop {\lim }\limits_{x \to 7} {{17 - [ - x]} \over {[x] - 3a}}$$ exist</p>
<p>$$RHL = \mathop {\lim }\limits_{x \to {7^ + }} {{17 - [ - x]} \over {[x] - 3a}} = {{25} \over {7 - 3a}}$$ $$\left[ {a \ne {7 \over 3}} \right]$$</p>
<p>$$LHL = \mathop {\lim }\limits_{x \to {7^ - }} {{17 - [ - x]} \over {[x] - 3a}} = {{24} \over {6 - 3a}}$$ $$\left[ {a \ne 2} \right]$$</p>
<p>For limit to exist</p>
<p>$$LHL = RHL$$</p>
<p>$${{25} \over {7 - 3a}} = {{24} \over {6 - 3a}}$$</p>
<p>$$ \Rightarrow {{25} \over {7 - 3a}} = {8 \over {2 - a}}$$</p>
<p>$$\therefore$$ $$a = - 6$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
1l5ai4x49 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>Let f(x) be a polynomial function such that $$f(x) + f'(x) + f''(x) = {x^5} + 64$$. Then, the value of $$\mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x - 1}}$$ is equal to:</p> | [{"identifier": "A", "content": "$$-$$15"}, {"identifier": "B", "content": "$$-$$60"}, {"identifier": "C", "content": "60"}, {"identifier": "D", "content": "15"}] | ["A"] | null | <p>Given, $$f(x) + f'(x) + f''(x) = {x^5} + 64$$ .........(i)</p>
<p>$\Rightarrow f(x)$ is a polynomial in $x$ whose degree is 5.</p>
<p>Let $$f(x) = {x^5} + a{x^4} + b{x^3} + c{x^2} + dx + e$$</p>
<p>$$f'(x) = 5{x^4} + 4a{x^3} + 3b{x^2} + 2cx + d$$</p>
<p>$$f''(x) = 20{x^3} + 12a{x^2} + 6bx + 2c$$</p>
<p>On substituting the value of $f(x), f^{\prime}(x)$ and $f^{\prime \prime}(x)$ in Eq. (i), we get</p>
<p>$${x^5}+(a + 5){x^4} + (b + 4a + 20){x^3} + (c + 3b + 12a){x^2} + (d + 2c + 6b)x + e + d + 2c = {x^5} + 64$$</p>
<p>Now, equating the coefficient, we get</p>
<p>$$ \Rightarrow a + 5 = 0$$</p>
<p>$$b + 4a + 20 = 0$$</p>
<p>$$c + 3b + 12a = 0$$</p>
<p>$$d + 2c + 6b = 0$$</p>
<p>$$e + d + 2c = 64$$</p>
<p>$$\therefore$$ $$a = - 5,\,b = 0,\,c = 60,\,d = - 120,\,e = 64$$</p>
<p>$$\therefore$$ $$f(x) = {x^5} - 5{x^4} + 60{x^2} - 120x + 64$$</p>
<p>Now, $$\mathop {\lim }\limits_{x \to 1} {{{x^5} - 5{x^4} + 60{x^2} - 120x + 64} \over {x - 1}}$$ is ($${0 \over 0}$$ form)</p>
<p>By L' Hospital rule</p>
<p>$$\mathop {\lim }\limits_{x \to 1} {{5{x^4} - 20{x^3} + 120x - 120} \over 1}$$</p>
<p>$$ = - 15$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l6dv4a68 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>If $$\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$$, then $$8(\alpha+\beta)$$ is equal to :
</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$-$$8"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "8"}] | ["C"] | null | <p>$$\mathop {\lim }\limits_{n \to \alpha } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$$</p>
<p>[ This limit will be zero when $$\alpha$$ < 0 as when $$\alpha$$ > 0 then overall limit will be $$\infty $$. ]</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right)\left( {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {{n^2} - n - 1} \right) - {{\left( {n\alpha + \beta } \right)}^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2} - n - 1 - {n^2}{\alpha ^2} - 2n\alpha\beta - {\beta ^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2}\left( {1 - {\alpha ^2}} \right) - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + {\beta ^2}} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}}$$</p>
<p>Here power of "n" in the numerator is 2 and power of "n" in the denominator is 1.</p>
<p>To get the value of limit equal to zero power of "n" should be equal in both numerator and denominator, otherwise value of limit will be infinite ($$\infty $$).</p>
<p>$$\therefore$$ Coefficient of n<sup>2</sup> should be 0 in this case.</p>
<p>$$\therefore$$ $$1 - {\alpha ^2} = 0$$</p>
<p>$$ \Rightarrow \alpha = \, \pm \,1$$</p>
<p>But $$\alpha$$ should be < 0</p>
<p>$$\therefore$$ $$\alpha = \, + \,1$$ not possible</p>
<p>$$\therefore$$ $$\alpha = - 1$$.</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{0 - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + \beta } \right)} \over {n\left[ {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}} \right]}} = 0$$</p>
<p>Divide numerator and denominator by n then we get,</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{ - \left( {1 + 2\alpha \beta } \right) - {{(1 + \beta )} \over n}} \over {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}}} = 0$$</p>
<p>$$ \Rightarrow {{ - \left( {1 + 2\alpha \beta } \right) - 0} \over {\sqrt {1 - 0 - 0} - \alpha - 0}} = 0$$</p>
<p>$$ \Rightarrow {{ - \left( {1 + 2\alpha \beta } \right)} \over {1 - \alpha }} = 0$$</p>
<p>$$ \Rightarrow - \left( {1 + 2\alpha \beta } \right) = 0$$</p>
<p>$$ \Rightarrow 1 + 2\alpha \beta = 0$$</p>
<p>$$ \Rightarrow 2\alpha \beta = - 1$$</p>
<p>$$ \Rightarrow \beta = - {1 \over {2\alpha }} = - {1 \over {2( - 1)}} = {1 \over 2}$$</p>
<p>$$\therefore$$ $$8\left( {\alpha + \beta } \right)$$</p>
<p>$$ = 8\left( { - 1 + {1 \over 2}} \right)$$</p>
<p>$$ = 8 \times - {1 \over 2}$$</p>
<p>$$ = - 4$$</p> | mcq | jee-main-2022-online-25th-july-morning-shift |
ldoa27ax | maths | limits-continuity-and-differentiability | limits-of-algebric-function | $$
\lim\limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3
$$ | [{"identifier": "A", "content": "is equal to 9"}, {"identifier": "B", "content": "is equal to $\\frac{27}{2}$"}, {"identifier": "C", "content": "does not exist"}, {"identifier": "D", "content": "is equal to 27"}] | ["D"] | null | $\lim \limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}} x^{3}$
<br/><br/>$$
\begin{aligned}
& = \lim \limits_{x \rightarrow \infty} x^{3} \times\left\{\frac{x^{3}\left\{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^{6}+\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^{6}\right\}}{x^{6}\left\{\left(1+\sqrt{1-\frac{1}{x^{2}}}\right)^{6}+\left(1-\sqrt{1-\frac{1}{x^{2}}}\right)^{6}\right\}}\right\} \\\\
& =\frac{(2 \sqrt{3})^{6}+0}{2^{6}+0}=3^{3}=27
\end{aligned}
$$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldv1f3qc | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>The value of $$\mathop {\lim }\limits_{n \to \infty } {{1 + 2 - 3 + 4 + 5 - 6\, + \,.....\, + \,(3n - 2) + (3n - 1) - 3n} \over {\sqrt {2{n^4} + 4n + 3} - \sqrt {{n^4} + 5n + 4} }}$$ is :</p> | [{"identifier": "A", "content": "$${3 \\over {2\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${3 \\over 2}(\\sqrt 2 + 1)$$"}, {"identifier": "C", "content": "$$3(\\sqrt 2 + 1)$$"}, {"identifier": "D", "content": "$${{\\sqrt 2 + 1} \\over 2}$$"}] | ["B"] | null | $$
\begin{aligned}
& I=\lim _{n \rightarrow \infty} \frac{(1+2+3+\ldots+3 n)-2(3+6+9+. .+3 n)}{\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}} \\\\
& =\lim _{n \rightarrow \infty} \frac{\frac{3 n(3 n+1)}{2}-6 \frac{n(n+1)}{2}}{\left(\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}\right)} \\\\
& =\lim _{n \rightarrow \infty} \frac{3 n(n-1)\left[\sqrt{2 n^4+4 n+3}+\sqrt{n^4+5 n+4}\right]}{2 \cdot\left[\left(2 n^4+4 n-3\right)-\left(n^4+5 n+4\right)\right]} \\\\
& =\lim _{n \rightarrow \infty} \frac{3 \cdot 1 \cdot\left(1-\frac{1}{n}\right)\left[\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}+\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right]}{2\left[1-\frac{1}{n^3}-\frac{7}{n^4}\right]} \\\\
& =\frac{3(\sqrt{2}+1)}{2} \\
&
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldwwk58p | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>The set of all values of $$a$$ for which $$\mathop {\lim }\limits_{x \to a} ([x - 5] - [2x + 2]) = 0$$, where [$$\alpha$$] denotes the greatest integer less than or equal to $$\alpha$$ is equal to</p> | [{"identifier": "A", "content": "$$[-7.5,-6.5]$$"}, {"identifier": "B", "content": "$$(-7.5,-6.5]$$"}, {"identifier": "C", "content": "$$[-7.5,-6.5)$$"}, {"identifier": "D", "content": "$$(-7.5,-6.5)$$"}] | ["D"] | null | <p>$$\mathop {\lim }\limits_{x \to a} \left( {[x - 5] - [2x + 2]} \right) = 0$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {[x] - 5 - [2x] - 2} \right) = 0$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{x \to a} [x] - [2x] - 7 = 0$$</p>
<p>Case 1 :</p>
<p>If $$a \in \left[ {n,n + {1 \over 2}} \right)$$</p>
<p>then $$\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0$$</p>
<p>$$ \Rightarrow n - 2n - 7 = 0$$</p>
<p>$$ \Rightarrow n = - 7$$</p>
<p>$$\therefore$$ $$a \in \left[ { - 7, - 6.5} \right)$$</p>
<p>Case 2 :</p>
<p>If $$a \in \left[ {n + {1 \over 2},n + 1} \right)$$</p>
<p>then $$\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0$$</p>
<p>$$ \Rightarrow n - (2n + 1) - 7 = 0$$</p>
<p>$$ \Rightarrow - n - 8 = 0$$</p>
<p>$$ \Rightarrow n = - 8$$</p>
<p>$$\therefore$$ $$a \in \left[ { - 7.5, - 7} \right)$$</p>
<p>$$R.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ + }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right)$$</p>
<p>$$ = - 8 - \left( { - 15} \right) - 7$$</p>
<p>$$ = - 8 + 15 - 7 = 0$$</p>
<p>$$\eqalign{
& L.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ - }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right) \cr \\
& = - 8 - \left( { - 16} \right) - 7 \cr \\
& = - 8 + 16 - 7 = 1 \cr} $$</p>
<p>$$ \therefore $$ R.H.L $$ \ne $$ L.H.L</p>
<p>$$ \therefore $$ At x = -7.5, limit does not exists.</p>
<p>$$\therefore$$ $$a \in \left( { - 7.5, - 6.5} \right)$$</p> | mcq | jee-main-2023-online-24th-january-evening-shift |
1lh22m2xj | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>Let $$a_{1}, a_{2}, a_{3}, \ldots, a_{\mathrm{n}}$$ be $$\mathrm{n}$$ positive consecutive terms of an arithmetic progression. If $$\mathrm{d} > 0$$ is its common difference, then</p>
<p>$$\lim_\limits{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)$$ is</p> | [{"identifier": "A", "content": "$$\\frac{1}{\\sqrt{d}}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$\\sqrt{d}$$"}] | ["B"] | null | $$\lim_\limits{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)$$
<br/><br/>Now,
<br/><br/>$\begin{aligned} & \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}} \\\\ = & \frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\frac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+\ldots+\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{a_n-a_{n-1}} \\\\ = & \frac{\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+. .+\sqrt{a_n}-\sqrt{a_{n-1}}}{d}\end{aligned}$
<br/><br/>$\left(\because a_2-a_1=a_3-a_2=\ldots a_n-a_{n-1}=d\right)$
<br/><br/>$\begin{aligned} & =\frac{\sqrt{a_n}-\sqrt{a_1}}{d} \\\\ & =\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{d}\end{aligned}$
<br/><br/>$\therefore \lim\limits_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{d}\right)$
<br/><br/>$\begin{aligned} & =\lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{d}}\left(\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{\sqrt{n}}\right)\right] \\\\ & =\lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{d}}\left(\sqrt{\frac{a_1}{n}+\left(d-\frac{d}{n}\right)}-\sqrt{\frac{a_1}{n}}\right)\right] \\\\ & =\frac{1}{\sqrt{d}}(\sqrt{0+d-0}-\sqrt{0})=\frac{\sqrt{d}}{\sqrt{d}}=1\end{aligned}$ | mcq | jee-main-2023-online-6th-april-morning-shift |
lsamze84 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | Let $f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in \mathbf{N}\right.$.
<br/><br/> If for some $\mathrm{a} \in \mathbf{N}, f(f(f(\mathrm{a})))=21$, then $\lim\limits_{x \rightarrow \mathrm{a}^{-}}\left\{\frac{|x|^3}{\mathrm{a}}-\left[\frac{x}{\mathrm{a}}\right]\right\}$, where $[t]$ denotes the greatest integer less than or equal to $t$, is equal to : | [{"identifier": "A", "content": "169"}, {"identifier": "B", "content": "121"}, {"identifier": "C", "content": "225"}, {"identifier": "D", "content": "144"}] | ["D"] | null | $f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in \mathbf{N}\right.$
<br/><br/>Let $a$ is odd
<br/><br/>$$
\begin{aligned}
& \Rightarrow f(a)=2 a \\\\
& \Rightarrow f(f(a))=2 a-1 \\\\
& \Rightarrow f(f(f(a)))=2(2 a-1)
\end{aligned}
$$
<br/><br/>$2(2 a-1)=21$ Not possible for any $a \in N$
<br/><br/>Let $a$ is even
<br/><br/>$$
\begin{aligned}
& \Rightarrow f(a)=a-1 \\\\
& \Rightarrow f(f(a))=2(a-1) \\\\
& \Rightarrow f(f(f(a)))=2(a-1)-1=2 a-3 \\\\
& 2 a-3=21 \quad \Rightarrow a=12
\end{aligned}
$$
<br/><br/>Now
<br/><br/>$$
\begin{aligned}
& \lim _{x \rightarrow 12^{-}}\left(\frac{|x|^3}{2}-\left[\frac{x}{12}\right]\right) \\\\
& =\lim _{x \rightarrow 12^{-}} \frac{|x|^3}{12}-\lim _{x \rightarrow 12^{-}}\left[\frac{x}{12}\right] \\\\
& =144-0=144 .
\end{aligned}
$$ | mcq | jee-main-2024-online-1st-february-evening-shift |
lv0vxdko | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>If $$\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(x+5)^{1 / 3}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}=\frac{\mathrm{m} \sqrt{5}}{\mathrm{n}(2 \mathrm{n})^{2 / 3}}$$, where $$\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$$, then $$8 \mathrm{~m}+12 \mathrm{n}$$ is equal to _______.</p> | [] | null | 100 | <p>$$I=\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(+5)^{1 /}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}$$</p>
<p>From: $$\frac{0}{0}$$, using $$\mathrm{L}-\mathrm{H}$$ rule</p>
<p>$$\begin{aligned}
& I=\lim _{x \rightarrow 1} \frac{\frac{1}{3} \times 5(5 x+1)^{-2 / 3}--(+5)}{\frac{1}{2} \times 2(2 x+3)^{-1 / 2}-\frac{1}{2}(x+4)^{-1 / 2}} \\
& =\frac{\left(\left.\frac{5}{3}-\frac{1}{3} \right\rvert\, 6^{-2 / 3}\right.}{\frac{1}{2} 5^{-1 / 2}}=\frac{8}{3} \times \frac{5^{1 / 2}}{6^{2 / 3}}=\frac{m \sqrt{5}}{n(2 n)^{2 / 3}} \\
& \Rightarrow m=8, n=3 \\
& \Rightarrow 8 m+12 n=100
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-morning-shift |
lvb294dv | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>$$\lim _\limits{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\cdots+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\cdots \cdots+n^3\right)-\left(1^2+2^2+\cdots \cdots+n^2\right)}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{2}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{2}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{4}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{3}$$"}] | ["D"] | null | <p>$$\mathop {\lim }\limits_{n \to \infty } {{({1^2} - 1)(n - 1) + ({2^2} - 2)(n - 2) + ... + \left( {{{(n - 1)}^2} - (n - 1)} \right) \times 1} \over {({1^3} + {2^3} + ... + {n^3}) - ({1^2} + {2^2} + ... + {n^2})}}$$</p>
<p>$$\begin{aligned}
& \text { Numerator }=\sum_{r=1}^{n-1}\left((r-1)^2-(r-1)\right)(n-r) \\
& =\sum_{r=1}^{n-1}(r-1)-(r-2)(n-r) \\
& =\sum_{r=1}^{n-1}-r^3+(n+3) r^2-(2+3 n) r+2 n
\end{aligned}$$</p>
<p>We will take term with the greatest power of $$n$$</p>
<p>$$=\frac{-1}{4} n^4+\frac{1}{3} n^4=\frac{1}{12} n^4$$</p>
<p>$$\begin{aligned}
& \text { Denominator }=\sum_{r=1}^n r^3-\sum_{r=1}^n r^2 \\
& =\left(\frac{n(n+1)}{2}\right)^2-\left(\frac{n(n+1)(2 n+1)}{6}\right)
\end{aligned}$$</p>
<p>Greatest power of $$n$$ is $$\frac{n^4}{4}$$</p>
<p>$$\lim _\limits{n \rightarrow \infty} \frac{\frac{1}{12} n^4}{\frac{n^4}{4}}=\frac{1}{3}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
IKRUfBV0w2RAAt7D | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}$$ | [{"identifier": "A", "content": "$${e^4}$$"}, {"identifier": "B", "content": "$${e^2}$$"}, {"identifier": "C", "content": "$${e^3}$$"}, {"identifier": "D", "content": "$$1$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + {{4x + 1} \over {{x^2} + x + 2 }}} \right)^x}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + {{4x + 1} \over {{x^2} + x + 2}}} \right)}^{{{{x^2} + x + 2} \over {4x + 1}}}}} \right]^{{{\left( {4x + 1} \right)x} \over {{x^2} + x + 2}}}}$$
<br><br>=$$\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + {{4x + 1} \over {{x^2} + x + 2}}} \right)}^{{1 \over {{{4x + 1} \over {{x^2} + x + 2}}}}}}} \right]^{{{\left( {4x + 1} \right)x} \over {{x^2} + x + 2}}}}$$
<br><br>$$ = {e^{\mathop {\lim }\limits_{x \to \infty } {{4{x^2} + x} \over {{x^2} + x + 2}}}}$$
<br><br>[ As $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \lambda x} \right)^{{1 \over x}}} = {e^\lambda }$$ ]
<br><br>$$ = {e^{\mathop {\lim }\limits_{x \to \infty } {{4 + {1 \over x}} \over {1 + {1 \over x} + {2 \over {{x^2}}}}}}} = {e^4}$$ | mcq | aieee-2002 |
MCKjvmuPcTeZAzZD | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}$$, then the value of $$a$$ and $$b$$, are | [{"identifier": "A", "content": "$$a$$ = 1 and $$b$$ = 2"}, {"identifier": "B", "content": "$$a$$ = 1 and $$b$$ $$ \\in R$$"}, {"identifier": "C", "content": "$$a$$ $$ \\in R$$ and $$b$$ = 2"}, {"identifier": "D", "content": "$$a$$ $$ \\in R$$ and $$b$$ $$ \\in R$$"}] | ["B"] | null | We know that $$\mathop {\lim }\limits_{x \to \infty } \left( {1 + x{1 \over x}} \right) = e$$
<br><br>$$\therefore$$ $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}$$
<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left[ {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)\left( {{1 \over {{a \over x} + {b \over {{x^2}}}}}} \right)} \right]^{2x\left( {{a \over x} + {b \over {{x^2}}}} \right)}} = {e^2}$$
<br><br>$$ \Rightarrow {e^{\mathop {\lim }\limits_{x \to \infty } 2\left[ {a + {b \over x}} \right]}} = {e^2} \Rightarrow {e^{2a}} = e{}^2 \Rightarrow a = 1$$
<br><br>and $$b \in R$$ | mcq | aieee-2004 |
qkqEIPW5oUqkmilbNPzuL | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3},$$ then 'a' is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$${3 \\over 2}$$ "}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["B"] | null | Given,
<br><br>$$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3}$$
<br><br>So, $$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}}\,\left[ {{1^ \propto }\,\,\,form} \right]$$
<br><br>$$ = {e^{\mathop {\lim }\limits_{x \to \propto } \left[ {\left( {1 + {a \over x} - {4 \over {{x^2}}} - 1} \right)2x} \right]}}$$
<br><br>$$ = {e^{\mathop {\lim }\limits_{x \to \propto } \left( {2a - {8 \over x}} \right)}}$$
<br><br>$$ = {e^{2a}}$$
<br><br>$$ \therefore $$ e<sup>2a</sup> = e<sup>3</sup>
<br><br>$$ \therefore $$ 2a = 3
<br><br>$$ \Rightarrow $$ a $$=$$ $${3 \over 2}$$ | mcq | jee-main-2016-online-9th-april-morning-slot |
6GVqfQiZB8zwXJZLZS7k9k2k5e4r5pg | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}}$$ is equal to_______. | [] | null | 36 | $$\mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}}$$
<br><br>let 3<sup>x/2</sup> = t
<br><br>= $$\mathop {\lim }\limits_{t \to 3} {{{t^2} + {{27} \over {{t^2}}} - 12} \over {{1 \over t} - {3 \over {{t^2}}}}}$$
<br><br>= $$\mathop {\lim }\limits_{t \to 3} {{\left( {{t^2} - 9} \right)\left( {{t^2} - 3} \right)} \over {t - 3}}$$
<br><br>= $$\mathop {\lim }\limits_{t \to 3} \left( {t + 3} \right)\left( {{t^2} - 3} \right)$$
<br><br>= 6 $$ \times $$ 6
<br><br>= 36 | integer | jee-main-2020-online-7th-january-morning-slot |
6XlUY8FhC12Q5CmfPJ7k9k2k5gqyxk1 | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$ is equal to | [{"identifier": "A", "content": "e"}, {"identifier": "B", "content": "e<sup>2</sup>"}, {"identifier": "C", "content": "$${1 \\over {{e^2}}}$$"}, {"identifier": "D", "content": "$${1 \\over e}$$"}] | ["C"] | null | Given $$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$
<br><br>Putting x = 0 we get 1<sup>$$\infty $$</sup> form.
<br><br>$$ \therefore $$ $${e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{3{x^2} + 2} \over {7{x^2} + 2}} - 1} \right]}}$$
<br><br>= $${e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{ - 4{x^2}} \over {7{x^2} + 2}}} \right]}}$$
<br><br>= e<sup>-4/2</sup>
<br><br>= e<sup>-2</sup> = $${1 \over {{e^2}}}$$ | mcq | jee-main-2020-online-8th-january-morning-slot |
FBNwrp6wuHYjAyCO6Pjgy2xukezaisy8 | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$$ is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$e$$"}, {"identifier": "D", "content": "$$e$$<sup>2</sup>"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$$
<br><br>This is 1<sup>$$\infty $$</sup> form.
<br><br>= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {\tan \left( {{\pi \over 4} + x} \right) - 1} \right] \times {1 \over x}}}$$
<br><br>= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{1 + \tan x} \over {1 - \tan x}} - 1} \right] \times {1 \over x}}}$$
<br><br>= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{2\tan x} \over {x\left( {1 - \tan x} \right)}}} \right]}}$$
<br><br>= $${e^{2\mathop {\lim }\limits_{x \to 0} \left[ {{{\tan x} \over x} \times {1 \over {\left( {1 - \tan x} \right)}}} \right]}}$$
<br><br>= $${e^{2\mathop {\lim }\limits_{x \to 0} \left[ {1 \times {1 \over {\left( {1 - 0} \right)}}} \right]}}$$
<br><br>= e<sup>2</sup> | mcq | jee-main-2020-online-2nd-september-evening-slot |
t00z1tGGAJXa8Rxka5jgy2xukfqbtvgj | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$ | [{"identifier": "A", "content": "is equal to 0."}, {"identifier": "B", "content": "is equal to $$\\sqrt e $$."}, {"identifier": "C", "content": "is equal to 1."}, {"identifier": "D", "content": "does not exist."}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {1 + {x^2} + {x^4}} \right) - 1} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {1 + {x^2} + {x^4} - 1} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {x\left( {\sqrt {1 + 0 + 0} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + 0 + 0} + 1} \right)} \over {x + {x^3}}}$$
<br><br>= $$2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {2x}}}} - 1} \right]} \over {x + {x^3}}}$$
<br><br>= $$2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2} \times 2}}$$
<br><br>= $$2 \times {1 \over 2} \times 1$$
<br><br>= 1
<br><br><b>Note :</b> As from formula, $$\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2}}}$$ = 1 | mcq | jee-main-2020-online-5th-september-evening-slot |
yESaXThRJN9Sg4L4Rs1kls4zfqx | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{1 + {1 \over 2} + ........ + {1 \over n}} \over {{n^2}}}} \right)^n}$$ is equal to : | [{"identifier": "A", "content": "$${{1 \\over 2}}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$${{1 \\over e}}$$"}] | ["B"] | null | It is $${1^\infty }$$ form<br><br>$$L = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {{{1 + {1 \over 2} + {1 \over 3} + ...{1 \over n}} \over n}} \right)}}$$<br><br>$$S = 1 + \left( {{1 \over 2} + {1 \over 3}} \right) + \left( {{1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7}} \right) + \left( {{1 \over 8} + ......... + {1 \over {15}}} \right)$$<br><br>$$S < 1 + \left( {{1 \over 2} + {1 \over 2}} \right) + \left( {{1 \over 4} + {1 \over 4} + {1 \over 4} + {1 \over 4}} \right).......... + \underbrace {\left( {{1 \over {{2^P}}} + ............ + {1 \over {{2^P}}}} \right)}_{{2^P}times}$$<br><br>$$S < 1 + 1 + 1 + 1 + ....... + 1$$<br><br>$$S < P + 1$$<br><br>$$ \therefore $$ $$L = {e^{\mathop {\lim }\limits_{P \to \infty } {{(P + 1)} \over {{2^P}}}}}$$<br><br>$$ \Rightarrow L = {e^o} = 1$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
BJ4mdrETo7ewozsDTW1klta44r9 | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If $$\mathop {\lim }\limits_{x \to 0} {{ax - ({e^{4x}} - 1)} \over {ax({e^{4x}} - 1)}}$$ exists and is equal to b, then the value of a $$-$$ 2b is __________. | [] | null | 5 | $$\mathop {\lim }\limits_{x \to 0} {{ax - \left( {{e^{4x}} - 1} \right)} \over {ax\left( {{e^{4x}} - 1} \right)}}$$<br><br>Applying L' Hospital Rule<br><br>$$\mathop {\lim }\limits_{x \to 0} {{a - 4{e^{4x}}} \over {a\left( {{e^{4x}} - 1} \right) + ax\left( {4{e^{4x}}} \right)}}$$
<br><br>This is $${{a - 4} \over 0}$$.
<br><br> limit exist only when $$a - 4 = 0$$ $$ \Rightarrow $$ a = 4<br><br>Applying L' Hospital Rule<br><br>$$\mathop {\lim }\limits_{x \to 0} {{ - 16{e^{4x}}} \over {a\left( {4{e^{4x}}} \right) + a\left( {4{e^{4x}}} \right) + ax\left( {16{e^{4x}}} \right)}}$$<br><br>= $${{ - 16} \over {4a + 4a}} = {{ - 16} \over {32}} = - {1 \over 2} = b$$<br><br>$$a - 2b = 4 - 2\left( {{{ - 1} \over 2}} \right) = 4 + 1 = 5$$ | integer | jee-main-2021-online-25th-february-evening-slot |
1krq1tmci | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If the value of $$\mathop {\lim }\limits_{x \to 0} {(2 - \cos x\sqrt {\cos 2x} )^{\left( {{{x + 2} \over {{x^2}}}} \right)}}$$ is equal to e<sup>a</sup>, then a is equal to __________. | [] | null | 3 | $$\mathop {\lim }\limits_{x \to 0} {(2 - \cos x\sqrt {\cos 2x} )^{{{x + 2} \over {{x^2}}}}}$$<br><br>form : 1<sup>$$\infty$$</sup> $$ = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right) \times (x + 2)}}$$<br><br>Now, $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}} = \mathop {\lim }\limits_{x \to 0} {{\sin x\sqrt {\cos 2x} - \cos x \times {1 \over {2\sqrt {\cos 2x} }} \times ( - 2sin2x)} \over {2x}}$$ (by L' Hospital Rule)<br><br>$$\mathop {\lim }\limits_{x \to 0} {{\sin x\cos 2x + \sin 2x.\cos x} \over {2x}} = {1 \over 2} + 1 = {3 \over 2}$$<br><br>So, $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right)(x + 2)}}$$<br><br>$$ = {e^{{3 \over 2} \times 2}} = {e^3}$$<br><br>$$\Rightarrow$$ a = 3 | integer | jee-main-2021-online-20th-july-morning-shift |
1kten7puo | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If $$\alpha$$, $$\beta$$ are the distinct roots of x<sup>2</sup> + bx + c = 0, then <br/><br/>$$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$ is equal to : | [{"identifier": "A", "content": "b<sup>2</sup> + 4c"}, {"identifier": "B", "content": "2(b<sup>2</sup> + 4c)"}, {"identifier": "C", "content": "2(b<sup>2</sup> $$-$$ 4c)"}, {"identifier": "D", "content": "b<sup>2</sup> $$-$$ 4c"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to \beta } {{1\left( {1 + {{2({x^2} + bx + c)} \over {1!}} + {{{2^2}{{({x^2} + bx + c)}^2}} \over {2!}} + ...} \right) - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to \beta } {{2{{({x^2} + bx + 1)}^2}} \over {{{(x - \beta )}^2}}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to \beta } {{2{{(x - \alpha )}^2}{{(x - \beta )}^2}} \over {{{(x - \beta )}^2}}}$$<br><br>$$ = 2{(\beta - \alpha )^2} = 2({b^2} - 4c)$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1l6hy7n8m | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>Let $$\beta=\mathop {\lim }\limits_{x \to 0} \frac{\alpha x-\left(e^{3 x}-1\right)}{\alpha x\left(e^{3 x}-1\right)}$$ for some $$\alpha \in \mathbb{R}$$. Then the value of $$\alpha+\beta$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{14}{5}$$"}, {"identifier": "B", "content": "$$\\frac{3}{2}$$"}, {"identifier": "C", "content": "$$\\frac{5}{2}$$"}, {"identifier": "D", "content": "$$\\frac{7}{2}$$"}] | ["C"] | null | <p>$$\beta = \mathop {\lim }\limits_{x \to 0} {{\alpha x - ({e^{3x}} - 1)} \over {\alpha x({e^{3x}} - 1)}},\,\alpha \in R$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{{\alpha \over 3} - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {\alpha x\left( {{{{e^{3x - 1}}} \over {3x}}} \right)}}$$</p>
<p>So, $$\alpha = 3$$ (to make independent form)</p>
<p>$$\beta = \mathop {\lim }\limits_{x \to 0} {{1 - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {3x}} = {{1 - {{\left( {3x + {{9{x^2}} \over 2}\, + \,......} \right)} \over {3x}}} \over {3x}}$$</p>
<p>$$ = {{ - \left( {{9 \over 2}{x^2} + {{{{(3x)}^3}} \over {31}}\, + \,....} \right)} \over {9{x^2}}} = {{ - 1} \over 2}$$</p>
<p>$$\therefore$$ $$\alpha + \beta = 3 - {1 \over 2} = {5 \over 2}$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6m6yoet | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>$$\lim\limits_{x \rightarrow 0}\left(\frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}\right)^{\frac{100}{x}}$$ is equal to ___________.</p> | [] | null | 1 | <p>Let $$x + 2\cos x = a$$</p>
<p>$$x + 2 = b$$</p>
<p>as $$x \to 0$$, $$a \to 2$$ and $$b \to 2$$</p>
<p>$$\mathop {\lim }\limits_{x \to 0} {\left( {{{{a^3} + 2{a^2} + 3\sin a} \over {{b^3} + 2{b^2} + 3\sin b}}} \right)^{{{100} \over x}}}$$</p>
<p>$$ = {e^{\mathop {\lim }\limits_{x \to 0} \,.\,{{100} \over x}\,.\,{{({a^3} - {b^3}) + 2({a^2} - {b^2}) + 3(\sin a - \sin b)} \over {{b^3} + 2{b^2} + 3\sin b}}}}$$</p>
<p>$$\because$$ $$\mathop {\lim }\limits_{x \to 0} {{a - b} \over x} = \mathop {\lim }\limits_{x \to 0} {{2(\cos x - 1)} \over x} = 0$$</p>
<p>$$ = {e^0}$$</p>
<p>$$ = 1$$</p> | integer | jee-main-2022-online-28th-july-morning-shift |
1ldyaxaot | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>$$\mathop {\lim }\limits_{t \to 0} {\left( {{1^{{1 \over {{{\sin }^2}t}}}} + {2^{{1 \over {{{\sin }^2}t}}}}\, + \,...\, + \,{n^{{1 \over {{{\sin }^2}t}}}}} \right)^{{{\sin }^2}t}}$$ is equal to</p> | [{"identifier": "A", "content": "$${{n(n + 1)} \\over 2}$$"}, {"identifier": "B", "content": "n"}, {"identifier": "C", "content": "n$$^2$$ + n"}, {"identifier": "D", "content": "n$$^2$$"}] | ["B"] | null | $$
\begin{aligned}
& \lim _{t \rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots \ldots . .+n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t} \\\\
& =\lim _{t \rightarrow 0} n\left(\left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t}+\left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t}+\ldots \ldots . .+1\right)^{\sin ^2 t} \\\\
& =\mathrm{n}
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-morning-shift |
jaoe38c1lsd53rxf | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>If $$\lim _\limits{x \rightarrow 0} \frac{a x^2 e^x-b \log _e(1+x)+c x e^{-x}}{x^2 \sin x}=1$$, then $$16\left(a^2+b^2+c^2\right)$$ is equal to ________.</p> | [] | null | 81 | <p>$$\mathop {\lim }\limits_{x \to 0} {{a{x^2}\left( {1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ....} \right) - b\left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - .....} \right) + cx\left( {1 - x + {{{x^2}} \over {x!}} - {{{x^3}} \over {3!}} + .....} \right)} \over {{x^3}\,.\,{{\sin x} \over x}}}$$</p>
<p>$$=\lim _\limits{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots}{x^3}=1$$</p>
<p>$$\begin{aligned}
& \mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0 \\
& \mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2} \\
& \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} \\
& 16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse4rgly | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>$$\lim _\limits{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$$</p> | [{"identifier": "A", "content": "is equal to 1\n"}, {"identifier": "B", "content": "does not exist\n"}, {"identifier": "C", "content": "is equal to $$-1$$\n"}, {"identifier": "D", "content": "is equal to 2"}] | ["D"] | null | <p>$$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2} \\
& \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{|\sin x|^2} \times \frac{\sin ^2 x}{x^2}
\end{aligned}$$</p>
<p>Let $$|\sin \mathrm{x}|=\mathrm{t}$$</p>
<p>$$\begin{aligned}
& \lim _{t \rightarrow 0} \frac{e^{2 t}-2 t-1}{t^2} \times \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \\
& =\lim _{t \rightarrow 0} \frac{2 e^{2 t}-2}{2 t} \times 1=2 \times 1=2
\end{aligned}
$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
luxwdeld | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>$$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{-2}{e}$$\n"}, {"identifier": "B", "content": "$$e-e^2$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$e$$"}] | ["D"] | null | <p>$$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$$</p>
<p>Using expansion</p>
<p>$$\begin{aligned}
& =\lim _\limits{x \rightarrow 0} \frac{e-e\left[1-\frac{2 x}{2}+\frac{11 \times 4 x^2}{24}+\ldots\right]}{x} \\
& =\lim _\limits{x \rightarrow 0}\left(e-\frac{11 x}{6} e+\ldots\right)=e
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
lv3vec6m | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>If $$\alpha=\lim _\limits{x \rightarrow 0^{+}}\left(\frac{\mathrm{e}^{\sqrt{\tan x}}-\mathrm{e}^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)$$ and $$\beta=\lim _\limits{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}$$ are the roots of the quadratic equation $$\mathrm{a} x^2+\mathrm{b} x-\sqrt{\mathrm{e}}=0$$, then $$12 \log _{\mathrm{e}}(\mathrm{a}+\mathrm{b})$$ is equal to _________.</p> | [] | null | 6 | <p>$$\begin{aligned}
\alpha & =\lim _{x \rightarrow 0^{+}} \frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{(\sqrt{\tan x}-\sqrt{x})} \\
& =\lim _{x \rightarrow 0} \frac{e^{\sqrt{x}}\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{(\sqrt{\tan x}-\sqrt{x})}=1 \\
\beta & =\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}=\lim _{x \rightarrow 0} e^{(\sin x)\left(\frac{1}{2} \cot x\right)} \\
& =\lim _{x \rightarrow 0} e^{\frac{1}{2} \cos x}=e^{1 / 2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Product of roots }=\sqrt{e}=\frac{-\sqrt{e}}{a} \Rightarrow a=-1 \\
& \text { Sum of roots }=\frac{-b}{a}=1+\sqrt{e} \\
& \qquad=b=\sqrt{e}+1 \\
& \Rightarrow 12 \ln (a+b)=12 \ln (\sqrt{e}+1-1)=12 \ln \left(e^{1 / 2}\right)=6
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-evening-shift |
DZUSj4MdIWEgezAl | maths | limits-continuity-and-differentiability | limits-of-logarithmic-functions | $$\mathop {\lim }\limits_{x \to 0} {{\log {x^n} - \left[ x \right]} \over {\left[ x \right]}}$$, $$n \in N$$, ( [x] denotes the greatest integer less than or equal to x ) | [{"identifier": "A", "content": "has value $$ -1$$"}, {"identifier": "B", "content": "has value $$0$$"}, {"identifier": "C", "content": "has value $$1$$"}, {"identifier": "D", "content": "does not exist"}] | ["D"] | null | Since $$\mathop {\lim }\limits_{x \to 0} \left[ x \right]$$ does not exist, hence the required limit does not exist. | mcq | aieee-2002 |
gJ8W4ffbWbmqeEai | maths | limits-continuity-and-differentiability | limits-of-logarithmic-functions | If $$\mathop {\lim }\limits_{x \to 0} {{\log \left( {3 + x} \right) - \log \left( {3 - x} \right)} \over x}$$ = k, the value of k is | [{"identifier": "A", "content": "$$ - {2 \\over 3}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$ - {1 \\over 3}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 0} {{\log \left( {3 + x} \right) - \log \left( {3 - x} \right)} \over x} = K$$
<br><br>(by $$L'$$ Hospital rule)
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{{1 \over {3 + x}} - {{ - 1} \over {3 - x}}} \over 1} = K$$
<br><br>$$\therefore$$ $${2 \over 3} = K$$ | mcq | aieee-2003 |
1l6ggon0k | maths | limits-continuity-and-differentiability | limits-of-logarithmic-functions | <p>If the function $$f(x) = \left\{ {\matrix{
{{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr
k & , & {x = 0} \cr
} } \right.$$ is continuous at x = 0, then k is equal to:</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "e"}, {"identifier": "D", "content": "0"}] | ["A"] | null | <p>$$f(x) = \left\{ {\matrix{
{{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr
k & , & {x = 0} \cr
} } \right.$$</p>
<p>for continuity at $$x = 0$$</p>
<p>$$\mathop {\lim }\limits_{x \to 0} f(x) = k$$</p>
<p>$$\therefore$$ $$k = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {\sec x - \cos x}}\left( {{0 \over 0}\,\mathrm{form}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{\cos x{{\log }_e}({x^4} + {x^2} + 1)} \over {{{\sin }^2}x}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {{x^2}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{\ln (1 + {x^2} + {x^4})} \over {{x^2} + {x^4}}}\,.\,{{{x^2} + {x^4}} \over {{x^2}}}$$</p>
<p>$$ = 1$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
C3FEi2ihx23NKCcI | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{\sqrt {1 - \cos 2x} } \over {\sqrt 2 x}}$$ is | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$-1$$"}, {"identifier": "C", "content": "zero"}, {"identifier": "D", "content": "does not exist"}] | ["D"] | null | $$\lim {{\sqrt {1 - \cos \,2x} } \over {\sqrt 2 x}} \Rightarrow \lim {{\sqrt {1 - \left( {1 - 2\,{{\sin }^2}\,x} \right)} } \over {\sqrt 2 x}}$$
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{\sqrt {2\,{{\sin }^2}\,x} } \over {\sqrt {2x} }} \Rightarrow \mathop {\lim }\limits_{x \to 0} {{\left| {\sin x} \right|} \over x}$$
<br><br>The limit of above does not exist as
<br><br>$$LHS = - 1 \ne RHL = 1$$ | mcq | aieee-2002 |
jQEkUesKCe4vxz8X | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\left[ {1 - \tan \left( {{x \over 2}} \right)} \right]\left[ {1 - \sin x} \right]} \over {\left[ {1 + \tan \left( {{x \over 2}} \right)} \right]{{\left[ {\pi - 2x} \right]}^3}}}$$ is | [{"identifier": "A", "content": "$$\\infty $$"}, {"identifier": "B", "content": "$${1 \\over 8}$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$${1 \\over 32}$$"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\tan \left( {{\pi \over 4} - {x \over 2}} \right).\left( {1 - \sin x} \right)} \over {{{\left( {\pi - 2x} \right)}^3}}}$$
<br><br>Let $$x = {\pi \over 2} + x;\,\,y \to 0$$
<br><br>$$=$$ $$\mathop {\lim }\limits_{y \to 0} {{\tan \left( { - {y \over 2}} \right).\left( {1 - \cos \,y} \right)} \over {{{\left( { - 2y} \right)}^3}}}$$
<br><br>$$=$$ $$\mathop {\lim }\limits_{y \to 0} {{ - \tan {y \over 2}2{{\sin }^2}{y \over 2}} \over {\left( { - 8} \right).{{{y^3}} \over 8}.8}}$$
<br><br>$$ = \mathop {\lim }\limits_{y \to 0} {1 \over {32}}{{\tan {y \over 2}} \over {\left( {{y \over 2}} \right)}}.{\left[ {{{\sin \,y/2} \over {y/2}}} \right]^2} = {1 \over {32}}$$ | mcq | aieee-2003 |
NJEFjGuNFNugq1vY | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $$\alpha$$ and $$\beta$$ be the distinct roots of $$a{x^2} + bx + c = 0$$, then
<br/><br/>$$\mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a{x^2} + bx + c} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$ is equal to | [{"identifier": "A", "content": "$${{{a^2}{{\\left( {\\alpha - \\beta } \\right)}^2}} \\over 2}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$ - {{{a^2}{{\\left( {\\alpha - \\beta } \\right)}^2}} \\over 2}$$"}, {"identifier": "D", "content": "$${{{{\\left( {\\alpha - \\beta } \\right)}^2}} \\over 2}$$"}] | ["A"] | null | Given limit $$ = \mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \,a\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to \alpha } {{2{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to \alpha } {2 \over {{{\left( {x - \alpha } \right)}^2}}} \times {{{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}}} \times {{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}$$
<br><br>$$ = {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}.$$ | mcq | aieee-2005 |
PuCPbTskQL6G76Mb | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 2} \left( {{{\sqrt {1 - \cos \left\{ {2(x - 2)} \right\}} } \over {x - 2}}} \right)$$ | [{"identifier": "A", "content": "Equals $$\\sqrt 2 $$"}, {"identifier": "B", "content": "Equals $$-\\sqrt 2 $$"}, {"identifier": "C", "content": "Equals $${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "does not exist"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 2} {{\sqrt 2 \left| {\sin \left( {x - 2} \right)} \right|} \over {x - 2}}$$
<br><br>$$L.H.L. = \mathop {\lim }\limits_{x \to {2^ - }} {{\sqrt 2 \sin \left( {x - 2} \right)} \over {\left( {x - 2} \right)}} = - \sqrt 2 $$
<br><br>$$R.H.L. = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {x - 2} \right)} \over {\left( {x - 2} \right)}} = \sqrt 2 $$
<br><br>Thus $$L.H.L. \ne R.H.L.$$
<br><br>Hence, $$\mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}\,\,$$ does not exist. | mcq | aieee-2011 |
K2B3XCHyhhQxaDGa | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}$$ is equal to | [{"identifier": "A", "content": "$$ - {1 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["D"] | null | Multiply and divide by $$x$$ in the given expression, we get
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x_2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan 4x}}$$
<br><br>$$ = 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}$$
<br><br>$$ = 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2$$ | mcq | jee-main-2013-offline |
wmvCA4BOiiEDsFrU | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}$$ is equal to : | [{"identifier": "A", "content": "$$ - \\pi $$"}, {"identifier": "B", "content": "$$ \\pi $$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$"}, {"identifier": "D", "content": "1"}] | ["B"] | null | Consider $$\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {\pi \left( {1 - {{\sin }^2}x} \right)} \right]} \over {{x^2}}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} \sin {{\left( {\pi - \pi {{\sin }^2}x} \right)} \over {{x^2}}}$$
<br><br>$$\left[ \, \right.$$ As $$\sin \left( {\pi - \theta } \right) = \sin \theta $$ $$\left. \, \right]$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} \sin {{\left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}} \times {{\pi {{\sin }^2}x} \over {{x^2}}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} 1 \times \pi {\left( {{{\sin x} \over x}} \right)^2} = \pi $$ | mcq | jee-main-2014-offline |
R1Gk6O2WAxXViPUh | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}$$ is equal to | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["A"] | null | Multiply and divide by $$x$$ in the given expression, we get
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x^2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan \,4x}}$$
<br><br>$$ = 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}$$
<br><br>$$ = 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2$$ | mcq | jee-main-2015-offline |
nFuvyJlrfDgLfn04 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $$p = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}}$$ then $$log$$ $$p$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$${1 \\over 4}$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$1$$ "}] | ["A"] | null | $$\ln \,P = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {2x}}\ln \left( {1 + {{\tan }^2}\sqrt x } \right)$$
<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} {1 \over x}\ln \left( {\sec \sqrt x \,\,\,\,\,\,} \right)$$
<br><br>Applying $$L$$ Hospital's rule :
<br><br>$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\sec \sqrt x \tan \sqrt x } \over {\sec \sqrt x .2\sqrt x }}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\tan \sqrt x } \over {2\sqrt x }}$$
<br><br>$$ = {1 \over 2}$$ | mcq | jee-main-2016-offline |
59WFqj1zr1H1BHbGnna5X | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}}$$ is : | [{"identifier": "A", "content": "$$-$$ 2"}, {"identifier": "B", "content": "$$-$$ $${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "2"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to 0} {{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\tan x - x\tan 2x}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{{{\left( {2{{\sin }^2}x} \right)}^2}} \over {2x\left( {x + {{{x^3}} \over 3} + {{2{x^5}} \over {15}} + ....} \right) - x\left( {2x + {{{2^3}{x^3}} \over 3} + 2.{{{2^5}{x^5}} \over {15}}+...} \right)}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{4{{\left( {x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ....} \right)}^4}} \over {{x^4}\left( {{2 \over 3} - {8 \over 3}} \right) + {x^6}\left( {{4 \over {15}} - {{64} \over {15}}} \right) + ....}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{4{{\left( {1 - {{{x^2}} \over {3!}} + {{{3^4}} \over {5!}} - ....} \right)}^4}} \over { - 2 + {x^2}\left( { - {{60} \over {15}}} \right) + ....}}$$
<br><br>(By dividing numerator and denominator by x<sup>4</sup>)
<br><br>$$ = {{4{{\left( {1 - 0} \right)}^4}} \over { - 2 + 0}}$$
<br><br>$$ = {4 \over { - 2}}$$
<br><br>$$=$$ $$-$$ 2 | mcq | jee-main-2016-online-10th-april-morning-slot |
YSGWJySyjCgT1nDb | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$ equals | [{"identifier": "A", "content": "$${1 \\over {16}}$$"}, {"identifier": "B", "content": "$${1 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over {4}}$$"}, {"identifier": "D", "content": "$${1 \\over {24}}$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {1 \over 8}.{{\cos x\left( {1 - \sin x} \right)} \over {\sin x{{\left( {{\pi \over 2} - x} \right)}^3}}}$$
<br><br>Put $${{\pi \over 2} - x}$$ = t
<br><br>$$ \Rightarrow $$ x $$ \to $$ $${{\pi \over 2}}$$
<br><br>t $$ \to $$ 0
<br><br>= $$\mathop {\lim }\limits_{t \to 0} {1 \over 8}.{{\cos \left( {{\pi \over 2} - t} \right)\left( {1 - \sin \left( {{\pi \over 2} - t} \right)} \right)} \over {\sin \left( {{\pi \over 2} - t} \right){{\left( {{\pi \over 2} - {\pi \over 2} + t} \right)}^3}}}$$
<br><br>= $$\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {1 - \cos t} \right)} \over {\cos t{{\left( t \right)}^3}}}$$
<br><br>= $$\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {2{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$$
<br><br>= $$\mathop {{1 \over 4}\lim }\limits_{t \to 0} {{\sin t\left( {{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$$
<br><br>= $$\mathop {\lim }\limits_{t \to 0} \left( {{{\sin t} \over t}} \right){\left( {{{\sin {t \over 2}} \over {{t \over 2}}}} \right)^2}{1 \over {\cos t}}{1 \over 4}$$
<br><br>= $${1 \over 4} \times {1 \over 4}$$
<br><br>= $${1 \over {16}}$$ | mcq | jee-main-2017-offline |
JcX96L117UGDKvIsATAfR | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{x\tan 2x - 2x\tan x} \over {{{\left( {1 - \cos 2x} \right)}^2}}}$$ equals : | [{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$$-$$ $${1 \\over 2}$$"}] | ["C"] | null | Let, L = $$\mathop {\lim }\limits_{x \to 0} {{\left( {x\tan 2x - 2x\tan x} \right)} \over {{{\left( {1 - \cos 2x} \right)}^2}}}$$ = $$\mathop {\lim }\limits_{x \to 0} K$$ (say)
<br><br>$$ \Rightarrow $$ K = $${{x\left[ {{{2\tan x} \over {1 - {{\left( {\tan x} \right)}^2}}}} \right] - 2x\tan x} \over {{{\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right)}^2}}}$$
<br><br>= $${{2x\tan x - \left[ {2x\tan x - 2x{{\tan }^3}x} \right]} \over {4{{\sin }^4}x \times (1 - {{\tan }^2}x)}}$$
<br><br>= $${{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {1 - {{\tan }^2}x} \right)}}$$
<br><br>= $${{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}$$
<br><br>= $${{2x{{{{\sin }^3}x} \over {{{\cos }^3}x}}} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}$$
<br><br>$$ \Rightarrow $$K = $${x \over {2\sin x \times ({{\cos }^2}x - {{\sin }^2}x)\cos x}}$$
<br><br>$$\therefore\,\,\,$$ L = $$\mathop {\lim }\limits_{x \to 0} $$ $${x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos x({{\cos }^2}x - {{\sin }^2}x)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} $$ $${x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos 0\left( {{{\cos }^2}0 - {{\sin }^2}0} \right)}}$$
<br><br>= $${1 \over 2}$$ | mcq | jee-main-2018-online-15th-april-evening-slot |
M9bWcFHcnYf7eCFffvkjG | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}$$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} {{x{{\tan }^2}2x} \over {\tan 4x{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{x\left( {{{{{\tan }^2}2x} \over {4{x^2}}}} \right)4{x^2}} \over {\left( {{{\tan 4x} \over {4x}}} \right)4x\left( {{{{{\sin }^2}x} \over {{x^2}}}} \right){x^2}}} = 1$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
VrGyCUrha6FdMeKKOG3rsa0w2w9jxb3fiw6 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{x + 2\sin x} \over {\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}$$ is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267071/exam_images/hbbibexjhoou7gayvkwp.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Limits, Continuity and Differentiability Question 157 English Explanation"> | mcq | jee-main-2019-online-12th-april-evening-slot |
B0OUXwsIJuKbg7GgFHrzw | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$$ equals: | [{"identifier": "A", "content": "$$ \\sqrt 2$$"}, {"identifier": "B", "content": "$$2 \\sqrt 2$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$4 \\sqrt 2$$"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}} \right)\left( {{{\sqrt 2 + \sqrt {1 + \cos x} } \over {\sqrt 2 + \sqrt {1 + \cos x} }}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {1 - \cos x}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {2{{\sin }^2}\left( {{x \over 2}} \right)}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {1 \over 2}{\left( {{{\sin x} \over x}} \right)^2}{x^2}{\left( {{{{x \over 2}} \over {\sin {x \over 2}}}} \right)^2}{1 \over {{{{x^2}} \over 4}}}\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$$
<br><br>= $${1 \over 2} \times 4 \times 2\sqrt 2 $$
<br><br>= $$4\sqrt 2 $$ | mcq | jee-main-2019-online-8th-april-morning-slot |
HYXw2jZ1xzLywGWiYZrVQ | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }}$$ is equal to : | [{"identifier": "A", "content": "$$\\sqrt {{2 \\over \\pi }} $$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt {2\\pi } }}$$"}, {"identifier": "C", "content": "$$\\sqrt {{\\pi \\over 2}} $$"}, {"identifier": "D", "content": "$$\\sqrt \\pi $$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }} \times {{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}} } \over {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}$$
<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} {{2\left( {{\pi \over 2} - {{\sin }^{ - 1}}x} \right)} \over {\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$$
<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} }}.{1 \over {2\sqrt \pi }}$$
<br><br>Put $$x = \cos \theta $$
<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} {{2\theta } \over {\sqrt 2 \sin \left( {{\theta \over 2}} \right)}}.{1 \over {2\sqrt \pi }}$$
<br><br>$$ = \sqrt {{2 \over \pi }} $$ | mcq | jee-main-2019-online-12th-january-evening-slot |
ZhCzqfek1CZLC206tdsJx | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$ is : | [{"identifier": "A", "content": "$$8\\sqrt 2 $$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$4\\sqrt 2 $$"}, {"identifier": "D", "content": "8"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{1 \over {{{\tan }^3}x}} - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{1 - {{\tan }^4}x} \over {\left( {{{\tan }^3}x} \right)\cos \left( {x + {\pi \over 4}} \right)}}$$
<br><br>=$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}x} \right)} \over {\left( {{{\tan }^3}x} \right)\cos \left( {x + {\pi \over 4}} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}{\pi \over 4}} \right)} \over {\left( {{{\tan }^3}{\pi \over 4}} \right)\cos \left( {x + {\pi \over 4}} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + 1} \right)} \over {1.\cos \left( {x + {\pi \over 4}} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\left( {1 - {{\tan }^2}x} \right)} \over {\left( {{{\cos x - \sin x} \over {\sqrt 2 }}} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {1 - {{{{\sin }^2}x} \over {{{\cos }^2}x}}} \right)} \over {\left( {\cos x - \sin x} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {{{\cos }^2}x - {{\sin }^2}x} \right)} \over {{{\cos }^2}x\left( {\cos x - \sin x} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)} \over {{{\cos }^2}x\left( {\cos x - \sin x} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)} \over {{{\cos }^2}x}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos {\pi \over 4} + \sin {\pi \over 4}} \right)} \over {{{\cos }^2}{\pi \over 4}}}$$
<br><br>= $${{2\sqrt 2 \left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}$$
<br><br>= $${{2\sqrt 2 \left( {{2 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}$$
<br><br>= 8
<br><br><b>Other Method :</b>
<br><br>$$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$
<br><br>This is in $${0 \over 0}$$ form so L' Hospital rule is applicable.
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\cot }^3}x\left( { - \cos e{c^2}x} \right) - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}$$
<br><br>= $${{3 \times 1\left( { - {{\left( {\sqrt 2 } \right)}^2}} \right) - {{\left( {\sqrt 2 } \right)}^2}} \over { - 1}}$$
<br><br>= $${{ - 6 - 2} \over { - 1}}$$ = 8 | mcq | jee-main-2019-online-12th-january-morning-slot |
MnT35fZqunH4z67MqKLbB | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let [x] denote the greatest integer less than or equal to x. Then $$\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$ | [{"identifier": "A", "content": "equals $$\\pi $$ + 1"}, {"identifier": "B", "content": "equals 0"}, {"identifier": "C", "content": "does not exist "}, {"identifier": "D", "content": "equals $$\\pi $$"}] | ["C"] | null | R.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$
<br><br>(as x $$ \to $$ 0<sup>+</sup> $$ \Rightarrow $$ [x] $$=$$ 0)
<br><br>$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}$$
<br><br>$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$$
<br><br>L.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}$$
<br><br>(as x $$ \to $$ 0<sup>$$-$$</sup> $$ \Rightarrow $$ [x] $$=$$ $$-$$1)
<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi $$
<br><br>R.H.L. $$ \ne $$ L.H.L. | mcq | jee-main-2019-online-11th-january-morning-slot |
EvdzNUS9OvijXzGchQf93 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | For each t $$ \in $$ R , let [t] be the greatest integer less than or equal to t
<br/><br/>Then $$\mathop {\lim }\limits_{x \to 1^ + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}$$ | [{"identifier": "A", "content": "equals $$-$$ 1 "}, {"identifier": "B", "content": "equals 1"}, {"identifier": "C", "content": "equals 0"}, {"identifier": "D", "content": "does not exist"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|\left[ {1 - x} \right]}}$$
<br><br>$$=$$ $$\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - x} \right) + \sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)\left( { - 1} \right)}}$$ $$\sin \left( {{\pi \over 2}\left( { - 1} \right)} \right)$$
<br><br>$$=$$ $$\mathop {\lim }\limits_{x \to {1^ + }} \left( {1 - {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}} \right)\left( { - 1} \right) = \left( {1 - 1} \right)\left( { - 1} \right) = 0$$ | mcq | jee-main-2019-online-10th-january-morning-slot |
Mjg8JDXWkk9XV53FaN3r3 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | For each x$$ \in $$<b>R</b>, let [x] be the greatest integer less than or equal to x.
<br/><br/>Then $$\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$ is equal to : | [{"identifier": "A", "content": "$$-$$ sin 1"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "sin 1"}, {"identifier": "D", "content": "0"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to {0^ - }} {{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)} \over h}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} \left( { - 1 + h} \right)\sin \left( 1 \right) = - \sin 1$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
2x9c5HPz6lbLnD5rcAjgy2xukf0z9feu | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\}$$ = 2<sup>-k</sup>
<br/><br/>then the value of k is _______ . | [] | null | 8 | $$\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\} = {2^{ - k}}$$
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos {{{x^2}} \over 2}} \right)} \over {4{{\left( {{{{x^2}} \over 2}} \right)}^2}}}{{\left( {1 - \cos {{{x^2}} \over 4}} \right)} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} $$ = $${2^{ - k}}$$
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}{{{x^2}} \over 4}} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} \times {{2{{\sin }^2}{{{x^2}} \over 8}} \over {64{{\left( {{{{x^2}} \over 8}} \right)}^2}}}$$ = 2<sup>-k</sup>
<br><br>$$ \Rightarrow $$ $$ {1 \over 8} \times {1 \over {32}} = {2^{ - k}}$$
<br><br>$$ \Rightarrow $$ 2<sup>-8</sup> = 2<sup>-k</sup>
<br><br>$$ \Rightarrow $$ k = 8 | integer | jee-main-2020-online-3rd-september-morning-slot |
CbTrUE9Q2rPhXV8fxzjgy2xukfjjo939 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\alpha $$ is positive root of the equation, p(x) = x<sup>2</sup> - x - 2 = 0, then<br/><br/>
$$\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over \\sqrt2}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over \\sqrt2}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["C"] | null | $${x^2} - x - 2 = 0$$<br><br>roots are 2 & $$-$$1 $$ \Rightarrow $$ $$\alpha $$ = 2 (given $$\alpha$$ is positive)<br><br>Now $$ \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {1 - \cos ({x^2} - x - 2)} } \over {(x - 2)}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {2{{\sin }^2}{{({x^2} - x - 2)} \over 2}} } \over {(x - 2)}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {{{(x - 2)(x + 1)} \over 2}} \right)} \over {(x - 2)}}$$<br><br>$$ = {3 \over {\sqrt 2 }}$$ | mcq | jee-main-2020-online-5th-september-morning-slot |
K9MKlxYQfb5W0pceJy1klrisk9y | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{n \to \infty } \tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right\}$$ is equal to ______. | [] | null | 1 | $${\tan ^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)$$<br><br>$$ = {\tan ^{ - 1}}\left( {{{r + 1 - r} \over {1 + r(r + 1)}}} \right)$$<br><br>$$ = {\tan ^{ - 1}}(r + 1) - {\tan ^{ - 1}}r$$<br><br>$$ \therefore $$ $$\sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}(r + 1) - {{\tan }^{ - 1}}(r)} \right)} $$<br><br>$$ = {\tan ^{ - 1}}(2) - {\tan ^{ - 1}}(1) + ta{n^{ - 1}}(3) - {\tan ^1}(2) + ta{n^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n)$$<br><br>$$ = {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(1)$$<br><br>$$ = {\tan ^{ - 1}}\left( {{{n + 1 - 1} \over {1 + (n + 1)1}}} \right)$$<br><br>$$ = {\tan ^{ - 1}}\left( {{n \over {n + 2}}} \right)$$<br><br>$$\mathop {\lim }\limits_{n \to \infty } \tan \left( {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right)$$<br><br>$$ = \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}\left( {{n \over {n + 2}}} \right)} \right)$$<br><br>$$ = \mathop {\lim }\limits_{x \to \infty } {n \over {n + 2}}$$<br><br>$$ = 1$$ | integer | jee-main-2021-online-24th-february-morning-slot |
raNEv8qHb2QGtIhf9v1klug6ks1 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | The value of $$\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}$$ is : | [{"identifier": "A", "content": "$${4 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$"}] | ["A"] | null | Let L = $$\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}$$
<br><br>$$ \Rightarrow $$ L = $$\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{{{\sqrt 3 } \over 2}\sin \left( {{\pi \over 6} + h} \right) - {1 \over 2}\cos \left( {{\pi \over 6} + h} \right)} \over {2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } \right)}}} \right\}$$<br><br>$$ \Rightarrow $$ L = $$\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\cos {\pi \over 6}\sin \left( {{\pi \over 6} + h} \right) - \sin {\pi \over 6}\cos \left( {{\pi \over 6} + h} \right)} \over {2\sqrt 3 h\left( {\cos {\pi \over 6}\cosh - \sin {\pi \over 6}\sinh } \right)}}} \right\}$$<br><br>$$ \Rightarrow $$ L = $$\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\sin \left( {{\pi \over 6} + h - {\pi \over 6}} \right)} \over {2\sqrt 3 h\cos \left( {h + {\pi \over 6}} \right)}}} \right\}$$
<br><br>$$ \Rightarrow $$ L = $${4 \over {2\sqrt 3 }}\mathop {\lim }\limits_{h \to 0} \left\{ {{{\sin \left( h \right)} \over {h\cos \left( {h + {\pi \over 6}} \right)}}} \right\}$$
<br><br>$$ \Rightarrow $$ L = $${4 \over {2\sqrt 3 }} \times {2 \over {\sqrt 3 }}$$ = $${4 \over 3}$$ | mcq | jee-main-2021-online-26th-february-morning-slot |
2ZjqG1divjp9n1jn1B1kmhvs8fg | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $${S_k} = \sum\limits_{r = 1}^k {{{\tan }^{ - 1}}\left( {{{{6^r}} \over {{2^{2r + 1}} + {3^{2r + 1}}}}} \right)} $$. Then $$\mathop {\lim }\limits_{k \to \infty } {S_k}$$ is equal to : | [{"identifier": "A", "content": "$${\\cot ^{ - 1}}\\left( {{3 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "tan<sup>$$-$$1</sup> (3)"}, {"identifier": "D", "content": "$${\\tan ^{ - 1}}\\left( {{3 \\over 2}} \\right)$$"}] | ["A"] | null | S<sub>k</sub> = $$\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{6^r}(3 - 2)} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)$$<br><br>= $$\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{2^r}\,.\,{3^{r + 1}} - {3^r}{2^{r + 1}}} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)$$
<br><br>= $$\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{{\left( {{3 \over 2}} \right)}^{r + 1}} - {{\left( {{3 \over 2}} \right)}^r}} \over {1 + {{\left( {{3 \over 2}} \right)}^{r + 1}}{{\left( {{3 \over 2}} \right)}^r}}}} \right) $$
<br><br>= $$\sum\limits_{r = 1}^k {\left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{r + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^r}} \right]} $$
<br><br>= $${\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^2} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^1}$$
<br><br>+ $${\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^2}$$
<br><br>+ $${\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^4} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3}$$
<br>.
<br>.
<br>.
<br>+ $${{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^k}}$$
<br><br>= $${{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^1}}$$
<br><br>$$ \therefore $$ $$\mathop {\lim }\limits_{k \to \infty } {S_k}$$
<br><br>= $$\mathop {\lim }\limits_{k \to \infty } \left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^1}} \right]$$
<br><br>= $${{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}$$
<br><br>= $${{\pi \over 2} - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}$$
<br><br>= $${\cot ^{ - 1}}\left( {{3 \over 2}} \right)$$ | mcq | jee-main-2021-online-16th-march-morning-shift |
gr60nFENdvz6ErSCTo1kmhzpkqe | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\mathop {\lim }\limits_{x \to 0} {{a{e^x} - b\cos x + c{e^{ - x}}} \over {x\sin x}} = 2$$, then a + b + c is equal to ____________. | [] | null | 4 | $$\mathop {\lim }\limits_{x \to 0} {{\left\{ {a\left( {1 + x + {{{x^2}} \over {2!}} + .....} \right) - b\left( {1 - {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}}......} \right) + c\left( {1 - x + {{{x^2}} \over {2!}}......} \right)} \right\}} \over {x\left( {x - {{{x^3}} \over {3!}} + .....} \right)}} = 2$$<br><br>$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{(a - b + c) + x(a - c) + {x^2}\left( {{a \over 2} + {b \over 2} + {c \over 2}} \right) + ....} \over {{x^2}\left( {1 - {{{x^2}} \over 6}....} \right)}} = 2$$<br><br>For this limit to exist
<br><br>
a $$-$$ b + c = 0 & a $$-$$ c = 0<br><br>& $${a \over 2} + {b \over 2} + {c \over 2} = 2$$<br><br>$$ \Rightarrow $$ a + b + c = 4 | integer | jee-main-2021-online-16th-march-morning-shift |
owmdXqRf8g9VE2IoDB1kmja76au | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | The value of <br/>$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$$, where [ x ] denotes the greatest integer $$ \le $$ x is : | [{"identifier": "A", "content": "$$\\pi$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$"}, {"identifier": "D", "content": "0"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}\left( {x - {{[x]}^2}} \right).{{\sin }^{ - 1}}\left( {x - {{[x]}^2}} \right)} \over {x - {x^3}}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}x} \over {1 - {x^2}}}.{{{{\sin }^{ - 1}}x} \over x}$$<br><br>$$ = {\cos ^{ - 1}}0 = {\pi \over 2}$$ | mcq | jee-main-2021-online-17th-march-morning-shift |
7EgukTgHs5JcAZ2Ac41kmkn3tcl | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | The value of the limit <br/><br/>$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$-$$$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$$-$$$${1 \\over 4}$$"}] | ["B"] | null | Given,<br><br>$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$<br><br>$$ = \mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi - \pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ $$ \therefore $$ $$\left( {{{\cos }^2}\theta = 1 - {{\sin }^2}\theta } \right)$$<br><br>$$ = \mathop {\lim }\limits_{\theta \to 0} {{ - \tan (\pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ $$ \therefore $$ $$(\tan (\pi - \theta ) = - \tan \theta )$$<br><br>$$ = \mathop {\lim }\limits_{\theta \to 0} {{{{ - \tan (\pi {{\sin }^2}\theta )} \over {\pi {{\sin }^2}\theta }}} \over {{{\sin (2\pi {{\sin }^2}\theta )} \over {2\pi {{\sin }^2}\theta }} \times 2}}\left( \matrix{
As\,\theta \to 0 \hfill \cr
then \,{\sin ^2}\theta \to 0 \hfill \cr} \right)$$<br><br>$$ = -{1 \over 2}.$$ $$ \because $$ $$\left( \matrix{
\mathop {\lim }\limits_{\theta \to 0} {{\tan \theta } \over \theta } \to 1 \hfill \cr
\& \,\mathop {\lim }\limits_{\theta \to 0} {{\sin \theta } \over \theta } = 1 \hfill \cr} \right)$$ | mcq | jee-main-2021-online-17th-march-evening-shift |
CFfZTluMCLEQbMjgF01kmlihjxq | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x} \over {3{x^3}}}$$ is equal to L, then the value of (6L + 1) is | [{"identifier": "A", "content": "$${1 \\over 6}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["D"] | null | $$L = \mathop {\lim }\limits_{x \to 0} {{\left( {x + {{{x^3}} \over 6} + .....} \right) - \left( {x - {{{x^3}} \over 3}.....} \right)} \over {3{x^3}}}$$<br><br>$$L = {1 \over 3}\left( {{1 \over 6} + {1 \over 3}} \right) = {1 \over 6}$$<br><br>$$ \therefore $$ $$ 6L + 1 = 6.{1 \over 6} + 1 = 2$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
1krxkdox3 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | The value of <br/><br/>$$\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\root 8 \of {1 - \sin x} - \root 8 \of {1 + \sin x} }}} \right)$$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "$$-$$1"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\root 8 \of {1 - \sin x} - \root 8 \of {1 + \sin x} }}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {x \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}}}} \right]$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \right]} \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}}}} \right]$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ 2 \right]\left[ 2 \right]} \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}}}} \right]$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ 2 \right]\left[ 2 \right]\left[ 2 \right]} \over {\left[ {\left( {1 - \sin x} \right) - \left( {1 + \sin x} \right)} \right]}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} \left( { - {1 \over 2}} \right)(2)(2)(2)$$
<br><br> = -4
<br><br>$$\because$$ $$\left\{ {\mathop {\lim }\limits_{x \to 0} {{\sin x} \over x} = 1} \right\}$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1ktir1jy2 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$$ is equal to : | [{"identifier": "A", "content": "$${\\pi ^2}$$"}, {"identifier": "B", "content": "$$2{\\pi ^2}$$"}, {"identifier": "C", "content": "$$4{\\pi ^2}$$"}, {"identifier": "D", "content": "$$4\\pi $$"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$$<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi {{\cos }^4}x} \right)} \over {2{x^4}}}$$<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi - 2\pi {{\cos }^4}x} \right)} \over {{{\left[ {2\pi (1 - {{\cos }^4}x)} \right]}^2}}}4{\pi ^2}.{{{{\sin }^4}x} \over {2{x^4}}}{\left( {1 + {{\cos }^2}x} \right)^2}$$<br><br>$$ = {1 \over 2}.4{\pi ^2}.{1 \over 2}{(2)^2} = 4{\pi ^2}$$ | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktk6iv5f | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$ and $$\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}$$ are the roots of the equation, ax<sup>2</sup> + bx $$-$$ 4 = 0, then the ordered pair (a, b) is : | [{"identifier": "A", "content": "(1, $$-$$3)"}, {"identifier": "B", "content": "($$-$$1, 3)"}, {"identifier": "C", "content": "($$-$$1, $$-$$3)"}, {"identifier": "D", "content": "(1, 3)"}] | ["D"] | null | $$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}};{0 \over 0}$$ form<br><br>Using L Hospital rule<br><br>$$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\tan }^2}x{{\sec }^2}x - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}$$<br><br>$$\alpha$$ = $$-$$4<br><br>$$\beta = \mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}} = {e^{\mathop {\lim }\limits_{x \to 0} {{(\cos x - 1)} \over {\tan x}}}}$$<br><br>$$\beta = {e^{\mathop {\lim }\limits_{x \to 0} {{ - (1 - \cos x)} \over {{x^2}}}.{{{x^2}} \over {{{\left( {{{\tan x} \over x}} \right)}^x}}}}}$$<br><br>$$\beta = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{ - 1} \over 2}} \right).{x \over 1}}} = {e^0} \Rightarrow \beta = 1$$<br><br>$$\alpha$$ = $$-$$4; $$\beta$$ = 1<br><br>If ax<sup>2</sup> + bx $$-$$ 4 = 0 are the roots then<br><br>16a $$-$$ 4b $$-$$ 4 = 0 & a + b $$-$$ 4 = 0<br><br>$$\Rightarrow$$ a = 1 & b = 3 | mcq | jee-main-2021-online-31st-august-evening-shift |
1l54ar2pq | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>The value of $$\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^3} + 2x - 1}}$$ is equal to:</p> | [{"identifier": "A", "content": "$${{{\\pi ^2}} \\over 6}$$"}, {"identifier": "B", "content": "$${{{\\pi ^2}} \\over 3}$$"}, {"identifier": "C", "content": "$${{{\\pi ^2}} \\over 2}$$"}, {"identifier": "D", "content": "$$\\pi$$<sup>2</sup>"}] | ["D"] | null | <p>$$\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^2} + 2x - 1}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 1} {{({x^2} - 1)si{n^2}(\pi x)} \over {({x^2} - 1){{(x - 1)}^2}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 1} {{{{\sin }^2}(\pi x)} \over {{{(x - 1)}^2}}}$$</p>
<p>Let $$x = 1 + h$$</p>
<p>$$\therefore$$ when x $$\to$$ 1 then h $$\to$$ 0</p>
<p>$$ = \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi (1 + h))} \over {{{(1 + h - 1)}^2}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi h)} \over {{h^2}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{h \to 0} {\pi ^2} \times {{{{\sin }^2}(\pi h)} \over {{{(\pi h)}^2}}}$$</p>
<p>$$ = {\pi ^2} \times 1$$</p>
<p>$$ = {\pi ^2}$$</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l55iq9zy | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>The value of <br/><br/>$$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$$ is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <p>$$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{{(r + 2) - (r + 1)} \over {1 + (r + 2)(r + 1)}}} \right)} } \right\}$$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {({{\tan }^{ - 1}}(r + 2) - {{\tan }^{ - 1}}(r + 1))} } \right\}$$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {{{\tan }^{ - 1}}(n + 2) - {{\tan }^{ - 1}}2} \right\}$$</p>
<p>$$ = 6\tan \left\{ {{\pi \over 2} - {{\cot }^{ - 1}}\left( {{1 \over 2}} \right)} \right\}$$</p>
<p>$$ = 6\tan \left( {{{\tan }^{ - 1}}\left( {{1 \over 2}} \right)} \right)$$</p>
<p>$$ = 3$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l55j1jkd | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>If $$\mathop {\lim }\limits_{x \to 1} {{\sin (3{x^2} - 4x + 1) - {x^2} + 1} \over {2{x^3} - 7{x^2} + ax + b}} = - 2$$, then the value of (a $$-$$ b) is equal to ___________.</p> | [] | null | 11 | $$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\left(\frac{\sin \left(3 x^{2}-4 x+1\right)}{3 x^{2}-4 x+1}\right)\left(3 x^{2}-4 x+1\right)-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\
\Rightarrow & \lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+1-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\
\Rightarrow & \lim _{x \rightarrow 1} \frac{2(x-1)^{2}}{2 x^{3}-7 x^{2}+a x+b}=-2
\end{aligned}
$$
<br/><br/>
So $f(x)=2 x^{3}-7 x^{2}+a x+b=0$ has $x=1$ as repeated root, therefore $f(1)=0$ and $f^{\prime}(1)=0$ gives
<br/><br/>
$$
a+b+5 \text { and } a=8
$$
<br/><br/>
So, $a-b=11$ | integer | jee-main-2022-online-28th-june-evening-shift |
1l589fknh | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "$$ - \\sqrt 2 $$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$$ - {1 \\over {\\sqrt 2 }}$$"}] | ["D"] | null | <p>$$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$$</p>
<p>Let $${\cos ^{ - 1}}x = t$$</p>
<p>$$ \Rightarrow x = \cos t$$</p>
<p>When $$x \to {1 \over {\sqrt 2 }}$$, then $$t \to {\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right) \to {\pi \over 4}$$</p>
<p>$$\therefore$$ $$\mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - \tan (t)}}$$</p>
<p>$$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - {{\sin t} \over {\cos t}}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{(\sin t - \cos t)(\cos t)} \over {(\cos t - \sin t)}}$$</p>
<p>$$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} - \cos t$$</p>
<p>$$ = - \mathop {\lim }\limits_{t \to {\pi \over 4}} \cos t$$</p>
<p>$$ = - {1 \over {\sqrt 2 }}$$</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58f4816 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}}$$ is equal to :</p> | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 6}$$"}, {"identifier": "D", "content": "$${1 \\over 12}$$"}] | ["C"] | null | <p>$$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}} = \mathop {\lim }\limits_{x \to 0} {{2\sin (x + \sin x)\,.\,\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} 2\,.\,\left( {{{\left( {{{x + \sin x} \over 2}} \right)\left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {{{\left( {x + x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}}...} \right)\left( {x - x + {{{x^3}} \over {3!}}...} \right)} \over {{x^4}}}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {2 - {{{x^2}} \over {3!}} + {{{x^4}} \over {5!}}...} \right)\left( {{1 \over {3!}} - {{{x^2}} \over {5!}} - 1} \right)$$</p>
<p>$$ = {1 \over 6}$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59jt81c | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)$$ is equal to</p> | [{"identifier": "A", "content": "$${1 \\over {12}}$$"}, {"identifier": "B", "content": "$$-$$$${1 \\over {18}}$$"}, {"identifier": "C", "content": "$$-$$$${1 \\over {12}}$$"}, {"identifier": "D", "content": "$${1 \\over {6}}$$"}] | ["A"] | null | <p>$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {\tan ^2}x\left\{ {\sqrt {2{{\sin }^2}x + 3\sin x + 4} - \sqrt {{{\sin }^2}x + 6\sin x + 2} } \right\}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{{{\tan }^2}x({{\sin }^2}x - 3\sin x + 2)} \over {\sqrt {2{{\sin }^2}x + 3\sin x + 4} + \sqrt {{{\sin }^2}x + 6\sin x + 2} }}$$</p>
<p>$$ = {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(1 - \sin x)(2 - \sin x)} \over {{{\cos }^2}x}}\,.\,{\sin ^2}x$$</p>
<p>$$ = {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(2 - \sin x){{\sin }^2}x} \over {1 + \sin x}}$$</p>
<p>$$ = {1 \over {12}}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5w0zq08 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>Suppose $$\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$$ exists and is equal to L, where</p>
<p>$$F(x) = \left| {\matrix{
{a + \sin {x \over 2}} & { - b\cos x} & 0 \cr
{ - b\cos x} & 0 & {a + \sin {x \over 2}} \cr
0 & {a + \sin {x \over 2}} & { - b\cos x} \cr
} } \right|$$.</p>
<p>Then, $$-$$112 L is equal to ___________.</p> | [] | null | 14 | <p>Given,</p>
<p>$$F(x) = \left| {\matrix{
{a + \sin {x \over 2}} & { - b\cos x} & 0 \cr
{ - b\cos x} & 0 & {a + \sin {x \over 2}} \cr
0 & {a + \sin {x \over 2}} & { - b\cos x} \cr
} } \right|$$</p>
<p>$$ = \left( {a + \sin {x \over 2}} \right)\left( { - {{\left( {a + \sin {x \over 2}} \right)}^2}} \right) + b\cos x \times {b^2}{\cos ^2}x$$</p>
<p>$$ = - {\left( {a + \sin {x \over 2}} \right)^3} - {b^3}{\cos ^3}x$$</p>
<p>Now,</p>
<p>$$\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {a + \sin {x \over 2}} \right)}^3} - {b^3}{{\cos }^3}x} \over {{x^3}}}$$</p>
<p>Given limit exists, it only possible when a = 0 and b = 0.</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {\sin {x \over 2}} \right)}^3}} \over {{x^3}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2} \times \left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)} \right)^3}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2}} \right)^3} \times {\left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)^3}$$</p>
<p>$$ = - {1 \over 8} \times 1 = L$$</p>
<p>$$\therefore$$ $$ - 112L = - 112 \times - {1 \over 8} = 14$$</p> | integer | jee-main-2022-online-30th-june-morning-shift |
1l6f0xruh | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\lim\limits_{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$$ is equal to </p> | [{"identifier": "A", "content": "14"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "14$$\\sqrt2$$"}, {"identifier": "D", "content": "7$$\\sqrt2$$"}] | ["A"] | null | <p>$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{8\sqrt 2 - {{(\cos x + \sin x)}^7}} \over {\sqrt 2 - \sqrt 2 \sin 2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{0 \over 0}\,\mathrm{form}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 7{{(\cos x + \sin x)}^6}( - \sin x + \cos x)} \over { - 2\sqrt 2 \cos 2x}}$$ using $$\mathrm{L-H}$$ Rule</p>
<p>$$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{56(\cos x - \sin x)} \over {2\sqrt 2 \cos 2x}}\,\,\left( {{0 \over 0}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 56(\sin x + \cos x)} \over { - 4\sqrt 2 \sin 2x}}$$ using $$\mathrm{L-H}$$ Rule</p>
<p>$$ = 7\sqrt 2 \,.\,\sqrt 2 = 14$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6p17ava | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>If $$\lim\limits_{x \rightarrow 0} \frac{\alpha \mathrm{e}^{x}+\beta \mathrm{e}^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}$$, where $$\alpha, \beta, \gamma \in \mathbf{R}$$, then which of the following is NOT correct?</p> | [{"identifier": "A", "content": "$$\\alpha^{2}+\\beta^{2}+\\gamma^{2}=6$$"}, {"identifier": "B", "content": "$$\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha+1=0$$"}, {"identifier": "C", "content": "$$\\alpha\\beta^{2}+\\beta \\gamma^{2}+\\gamma \\alpha^{2}+3=0$$"}, {"identifier": "D", "content": "$$\\alpha^{2}-\\beta^{2}+\\gamma^{2}=4$$"}] | ["C"] | null | <p>$$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} + \gamma \sin x} \over {x{{\sin }^2}x}} = {2 \over 3}$$</p>
<p>$$ \Rightarrow \alpha + \beta = 0$$ (to make indeterminant form) ...... (i)</p>
<p>Now,</p>
<p>$$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} - \beta {e^{ - x}} + \gamma \cos x} \over {3{x^2}}} = {2 \over 3}$$ (Using L-H Rule)</p>
<p>$$ \Rightarrow \alpha - \beta + \gamma = 0$$ (to make indeterminant form) ...... (ii)</p>
<p>Now,</p>
<p>$$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} - \gamma \sin x} \over {6x}} = {2 \over 3}$$ (Using L-H Rule)</p>
<p>$$ \Rightarrow {{\alpha - \beta + \gamma } \over 6} = {2 \over 3}$$</p>
<p>$$ \Rightarrow \alpha - \beta + \gamma = 4$$ ...... (iii)</p>
<p>$$ \Rightarrow \gamma = - 2$$</p>
<p>and (i) + (ii)</p>
<p>$$2\alpha = - \gamma $$</p>
<p>$$ \Rightarrow \alpha = 1$$ and $$\beta = - 1$$</p>
<p>and $$\alpha {\beta ^2} + \beta {\gamma ^2} + \gamma {\alpha ^2} + 3 = 1 - 4 - 2 + 3 = - 2$$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
ldqw6aac | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $f, g$ and $h$ be the real valued functions defined on $\mathbb{R}$ as
<br/><br/>$f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}\right.$
<br/><br/>$g(x)=\left\{\begin{array}{cc}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{array}\right.$
<br/><br/>and $h(x)=2[x]-f(x)$, where $[x]$ is the greatest integer $\leq x$.
Then the <br/><br/>value of $\lim\limits_{x \rightarrow 1} g(h(x-1))$ is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$-1$"}, {"identifier": "C", "content": "$\\sin (1)$"}, {"identifier": "D", "content": "0"}] | ["A"] | null | <p>$$f(x) = {\mathop{\rm sgn}} (x)$$</p>
<p>$$h(x) = 2[x] - {\mathop{\rm sgn}} (x)$$</p>
<p>If $$x \to {1^ + }$$ then $$h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$$</p>
<p>$$ = 0 - 1 = - 1$$</p>
<p>& if $$x \to {1^ - }$$ then $$h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$$</p>
<p>$$ = - 2 + 1 = - 1$$</p>
<p>$$\therefore$$ $$\mathop {\lim }\limits_{x \to {1^ + }} g(h(x - 1)) = \mathop {\lim }\limits_{x \to {1^ + }} {{\sin (h(x + 1) + 1)} \over {h(x - 1) + 1}} = 1$$</p>
<p>$$\mathop {\lim }\limits_{x \to {1^ - }} g(h(x - 1)) = \mathop {\lim }\limits_{x \to {1^ - }} {{\sin (h(x - 1) + 1)} \over {h(x - 1) + 1}} = 1$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldswip6v | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>Let $$x=2$$ be a root of the equation $$x^2+px+q=0$$ and $$f(x) = \left\{ {\matrix{
{{{1 - \cos ({x^2} - 4px + {q^2} + 8q + 16)} \over {{{(x - 2p)}^4}}},} & {x \ne 2p} \cr
{0,} & {x = 2p} \cr
} } \right.$$</p>
<p>Then $$\mathop {\lim }\limits_{x \to 2{p^ + }} [f(x)]$$, where $$\left[ . \right]$$ denotes greatest integer function, is</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$-1$$"}] | ["C"] | null | $\lim \limits_{x \rightarrow 2^{+}}\left(\frac{1-\cos \left(x^{2}-4 p x+q^{2}+8 q+16\right)}{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}\right)\left(\frac{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}{(x-2 p)^{2}}\right)$
<br/><br/>
$\lim \limits_{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^{2}-4 p(2 p+h)+q^{2}+82+16}{h^{2}}\right)^{2}=\frac{1}{2}$
<br/><br/>
Using L'Hospital's
<br/><br/>
$\lim _{x \rightarrow 2 p^{+}}[f(x)]=0$ | mcq | jee-main-2023-online-29th-january-morning-shift |
1lgow7xsy | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>If $$\lim_\limits{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{cx e^{-c x}}{2}}{1-\cos (2 x)}=17$$, then $$5 a^{2}+b^{2}$$ is equal to</p> | [{"identifier": "A", "content": "64"}, {"identifier": "B", "content": "68"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "76"}] | ["B"] | null | <p>The Taylor series for $e^x$, $\cos x$, and $e^{-x}$ around $x=0$ are:</p>
<p>$$e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \ldots,$$
<br/><br/>$$\cos(bx) = 1 - \frac{(bx)^2}{2!} + \ldots,$$
<br/><br/>$$e^{-cx} = 1 - cx + \frac{(cx)^2}{2!} - \ldots.$$</p>
<p>Substituting these into the limit and simplifying:</p>
<p>$$\lim\limits_{x \rightarrow 0} \frac{(1+ax+\frac{(ax)^2}{2!})-(1-\frac{(bx)^2}{2!})-\frac{cx}{2}(1-(cx)+\frac{(cx)^2}{2!})}{1-\cos 2x}.$$</p>
<p>This simplifies to:</p>
<p>$$\lim\limits _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{\left(\frac{1-\cos 2 x}{(2 x)^2}\right) \times 4 x^2}.$$</p>
<p>Let's evaluate this expression:</p>
<p>$$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2.$$</p>
<p>We know that $1 - \cos 2x$ can be rewritten using the double-angle formula for cosine, which is $\cos 2x = 1 - 2\sin^2x$. So, $1 - \cos 2x = 2\sin^2x$.</p>
<p>Substituting this into our expression we get:</p>
<p>$$(\frac{2\sin^2x}{(2x)^2}) \times 4x^2.$$</p>
<p>This simplifies to:</p>
<p>$${1 \over 2}$$$$(\frac{\sin x}{x})^2 \times 4x^2 = {1 \over 2} \times 4x^2 = 2x^2.$$</p>
<p>So, </p>
<p>$$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2 = 2x^2.$$ </p>
<p>$$ \therefore $$ $$\lim _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{2x^2} = 17.$$</p>
<p>Now, for the limit to exist, the coefficient of $x$ in the numerator must be zero, as the limit would be undefined otherwise. This gives us the equation:</p>
<p>$$a-\frac{c}{2} = 0 \Rightarrow c=2a.$$</p>
<p>Then, for the limit to equal $17$, the coefficient of $x^2$ in the numerator must equal $17 \times 2 = 34$. This gives us the equation:</p>
<p>$$\frac{a^2+b^2+c^2}{2} = 34 \Rightarrow a^2+b^2+c^2 = 68.$$</p>
<p>Since $c=2a$, we can substitute $c$ in the equation to get:</p>
<p>$$a^2 + b^2 + (2a)^2 = 68 \Rightarrow 5a^2 + b^2 = 68.$$</p> | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgylxsd8 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>If $$\alpha > \beta > 0$$ are the roots of the equation $$a x^{2}+b x+1=0$$, and $$\lim_\limits{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right), \text { then } \mathrm{k} \text { is equal to }$$ :</p> | [{"identifier": "A", "content": "$$2 \\beta$$"}, {"identifier": "B", "content": "$$\\beta$$"}, {"identifier": "C", "content": "$$\\alpha$$"}, {"identifier": "D", "content": "$$2 \\alpha$$"}] | ["D"] | null | Since, $\alpha, \beta$ are roots of $a x^2+b x+1=0$
<br/><br/>Replace $x \rightarrow \frac{1}{x}$
<br/><br/>$$
\frac{a}{x^2}+\frac{b}{x}+1=0 \Rightarrow x^2+b x+a=0
$$
<br/><br/>So, $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots.
<br/><br/>Now, $\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}}$
<br/><br/>$$
=\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2\left(\frac{x^2+b x+a}{2}\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}}
$$
<br/><br/>$$
\begin{aligned}
& =\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2 \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{4 \times 2 \alpha^2 \frac{\left(x-\frac{1}{\alpha}\right)^2\left(x-\frac{1}{\beta}\right)^2}{4}\left(x-\frac{1}{\beta}\right)^2}\right]^{\frac{1}{2}} \\\\
& =\lim _{x \rightarrow \frac{1}{\alpha}}\left[ \pm \frac{1}{2} \frac{\sin \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{\alpha \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}\left(x-\frac{1}{\beta}\right)\right] \\\\
& =\frac{1}{2 \alpha}\left[\frac{-1}{\alpha}+\frac{1}{\beta}\right) \\\\
& \Rightarrow \frac{1}{k}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right]=\frac{1}{2 \alpha}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right] \\\\
& \Rightarrow k=2 \alpha
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lgzy95yj | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\lim_\limits{x \rightarrow 0}\left(\left(\frac{\left(1-\cos ^{2}(3 x)\right.}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)$$ is equal to _____________.</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "24"}] | ["B"] | null | $$
\begin{aligned}
& \lim _{x \rightarrow 0}\left[\left(\frac{1-\cos ^2(3 x)}{\cos ^3(4 x)}\right)\left(\frac{\sin ^3(4 x)}{\left(\log _e(2 x+1)\right)^5}\right)\right] \\\\
& =\lim _{x \rightarrow 0}\left[\frac{1-\cos ^2(3 x)}{9 x^2} \times \frac{9 x^2}{\cos ^3(4 x)}\right] \times \frac{\frac{\sin ^3 4 x}{(4 x)^3} \times 64 x^3}{\left[\frac{\log _e(2 x+1)}{2 x}\right]^5 \times(2 x)^5} \\\\
& =\left[\frac{1 \times 9 \times 1}{(1)}\right] \times\left[\frac{1 \times 64}{1 \times 32}\right] \\\\
& =9 \times 2=18
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
lsaq12m0 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $\{x\}$ denote the fractional part of $x$ and $f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0$. If $\mathrm{L}$ and $\mathrm{R}$ respectively denotes the left hand limit and the right hand limit of $f(x)$ at $x=0$, then $\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)$ is equal to ___________. | [] | null | 18 | Finding right hand limit
<br/><br/>R = $$
\begin{gathered}
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)
=\lim _{h \rightarrow 0} f(h)
\end{gathered}
$$
<br/><br/>$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)} \\\\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{\cos ^{-1}\left(1-\mathrm{h}^2\right)}{\mathrm{h}}\left(\frac{\sin ^{-1} 1}{1}\right)\end{aligned}$
<br/><br/>Let $\cos ^{-1}\left(1-h^2\right)=\theta \Rightarrow \cos \theta=1-h^2$
<br/><br/>$$
\begin{aligned}
& =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}} \\\\
& =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1-\cos \theta}{\theta^2}}} \\\\
& =\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}} \\\\
& \therefore \mathrm{R}=\frac{\pi}{\sqrt{2}}
\end{aligned}
$$
<br/><br/>Now finding left hand limit
<br/><br/>$$
\begin{aligned}
& L=\lim _{x \rightarrow 0^{-}} f(x) =\lim _{h \rightarrow 0} f(-h)
\end{aligned}
$$
<br/><br/>$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-\{-h\}^2\right) \sin ^{-1}(1-\{-h\})}{\{-h\}-\{-h\}^3} \\\\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-(-h+1)^2\right) \sin ^{-1}(1-(-h+1))}{(-h+1)-(-h+1)^3} \\\\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(-h^2+2 h\right) \sin ^{-1} h}{(1-h)\left(1-(1-h)^2\right)}\end{aligned}$
<br/><br/>$\begin{aligned} & =\lim _{h \rightarrow 0}\left(\frac{\pi}{2}\right) \frac{\sin ^{-1} h}{\left(1-(1-h)^2\right)} \\\\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{-h^2+2 h}\right) \\\\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{h}\right)\left(\frac{1}{-h+2}\right) \\\\ & \quad L=\frac{\pi}{4}\end{aligned}$
<br/><br/>$\begin{aligned} & \frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)=\frac{32}{\pi^2}\left(\frac{\pi^2}{2}+\frac{\pi^2}{16}\right) \\\\ & =16+2 \\\\ & =18\end{aligned}$ | integer | jee-main-2024-online-1st-february-morning-shift |
lsbkqsz3 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $\mathrm{a}=\lim\limits_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $\mathrm{b}=\lim\limits _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $a b^3$ is : | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "30"}] | ["C"] | null | <p>$$\begin{aligned}
a= & \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} \\
& =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)} \\
& =\lim _{x \rightarrow 0} \frac{x^4}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)\left(\sqrt{1+x^4}+1\right)}
\end{aligned}$$</p>
<p>Applying limit $$\mathrm{a}=\frac{1}{4 \sqrt{2}}$$</p>
<p>$$\begin{aligned}
& b=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}} \\
& =\lim _{x \rightarrow 0} \frac{\left(1-\cos ^2 x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} \\
& b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})
\end{aligned}$$</p>
<p>Applying limits $$b=2(\sqrt{2}+\sqrt{2})=4 \sqrt{2}$$</p>
<p>Now, $$a b^3=\frac{1}{4 \sqrt{2}} \times(4 \sqrt{2})^3=32$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscnb2gp | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\text { If } \lim _\limits{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3} \text {, then } 2 \alpha-\beta \text { is equal to : }$$</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "7"}] | ["C"] | null | <p>$$\lim _\limits{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3}$$</p>
<p>$$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^3}{3 !}+\ldots .\right]+\beta\left[1-\frac{x^2}{2 !}+\frac{x^4}{4 !} \ldots .\right]+\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)}{3 \tan ^2 x}=\frac{1}{3}$$</p>
<p>$$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^2+\ldots .}{3 x^2} \times \frac{x^2}{\tan ^2 x}=\frac{1}{3}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \beta+3=0, \alpha-1=0 \text { and } \frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3} \\
& \Rightarrow \beta=-3, \alpha=1 \\
& \Rightarrow 2 \alpha-\beta=2+3=5
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
lv5grwgp | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>The value of $$\lim _\limits{x \rightarrow 0} 2\left(\frac{1-\cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \ldots \ldots . \sqrt[10]{\cos 10 x}}{x^2}\right)$$ is __________.</p> | [] | null | 55 | <p>$$\mathop {\lim }\limits_{x \to 0} 2\left( {{{1 - \cos x{{(\cos 2x)}^{{1 \over 2}}}{{(\cos 3x)}^{{1 \over 3}}}\,...\,{{(\cos 10x)}^{{1 \over {10}}}}} \over {{x^2}}}} \right)$$ $$\left(\frac{0}{0} \text { form }\right)$$</p>
<p>Using L' hospital</p>
<p>$$2 \lim _\limits{x \rightarrow 0} \frac{\sin x(\cos 2 x)^{\frac{1}{2}} \ldots(\cos 10 x)^{\frac{1}{10}} \ldots(\sin 2 x)(\cos x)(\cos 3 x)^{\frac{1}{3}}+\ldots}{2 x}$$</p>
<p>$$\begin{aligned}
\Rightarrow & \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}+\frac{\sin 2 x}{x}+\ldots+\frac{\sin 10 x}{x}\right) \\
\quad & =1+2+\ldots+10=55
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-morning-shift |
lv9s20md | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>Let $$\mathrm{a}>0$$ be a root of the equation $$2 x^2+x-2=0$$. If $$\lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2}=\alpha+\beta \sqrt{17}$$, where $$\alpha, \beta \in Z$$, then $$\alpha+\beta$$ is equal to _________.</p> | [] | null | 170 | <p>$$\because 2 x^2+x-2=0$$ has two roots where
$$a=\frac{\sqrt{17}-1}{4}$$ and another root is $$\frac{-\sqrt{17}-1}{4}$$</p>
<p>And $$2+x-2 x^2=-2\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$$</p>
<p>Now $$\lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2}$$</p>
<p>$$\begin{aligned}
& =\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin ^2\left(\frac{2+x-2 x^2}{2}\right)}{a^2\left(\frac{1}{a}-x\right)^2} \\
& =\lim _{x \rightarrow \frac{1}{a}} \frac{\left(x+\frac{4}{\sqrt{17}+1}\right)^2 32 \cdot\left(\sin \left(\frac{1}{2} \cdot(-2)\right)\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2}{a^2\left(\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2} \\
& =2 \cdot\left(\frac{1}{a}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\
& =2\left(\frac{4}{\sqrt{17}-1}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\
& =\frac{17 \times 4}{18-2 \sqrt{17}}=\frac{68}{9-\sqrt{17}} \\
& =17(9+\sqrt{17}) \\
& \alpha+\beta=170
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-evening-shift |
1lguw56u8 | maths | logarithm | logarithm-inequalities | <p>The number of integral solutions $$x$$ of $$\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0$$ is :</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["D"] | null | $$\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0$$
<br/><br/><b>Domain :</b>
<br/><br/>$$
\begin{aligned}
& x+\frac{7}{2}>0 \\\\
& x>\frac{-7}{2} \\\\
& x+\frac{7}{2} \neq 1 \\\\
& x \neq \frac{-5}{2} \\\\
& \frac{x-7}{2 x-3} \neq 0 \\\\
& x \neq 7 \\\\
& x \neq \frac{3}{2}
\end{aligned}
$$
<br/><br/>$$
\text { Taking intersection : } x \in\left(\frac{-7}{2}, \infty\right)-\left\{-\frac{5}{2}, \frac{3}{2}, 7\right\}
$$
<br/><br/>Now $\log _{\mathrm{a}} \mathrm{b} \geq 0$ if $\mathrm{a}>1$ and $\mathrm{b} \geq 1$
<br/><br/>$$
\begin{gathered}
\text { Or } \\\\
a \in(0,1) \text { and } b \in(0,1)
\end{gathered}
$$
<br/><br/><b>Case I :</b> $$
x+\frac{7}{2}>1 \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \geq 1
$$
<br/><br/>$$ \therefore $$ $$
x > -\frac{5}{2}
$$
<br/><br/>and
<br/><br/>$$
\begin{aligned}
& (2 x-3)^2-(x-7)^2 \leq 0 \\\\
& (2 x-3+x-7)(2 x-3-x+7) \leq 0 \\\\
& (3 x-10)(x+4) \leq 0
\end{aligned}
$$
<br/><br/>$$
x \in\left[-4, \frac{10}{3}\right]
$$
<br/><br/>$$
\text { Intersection : } \mathrm{x} \in\left(\frac{-5}{2}, \frac{10}{3}\right]
$$
<br/><br/><b>Case II :</b> $$
x+\frac{7}{2} \in(0,1) \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \in(0,1)
$$
<br/><br/>$$
\begin{aligned}
& \therefore 0 < x+\frac{7}{2}<1 \\\\
& -\frac{7}{2} < x < \frac{-5}{2}
\end{aligned}
$$
<br/><br/>and
<br/><br/>$$
\begin{aligned}
& \left(\frac{x-7}{2 x-3}\right)^2<1 \\\\
& \Rightarrow (x-7)^2 < (2 x-3)^2
\end{aligned}
$$
<br/><br/>$$ \therefore $$ $$
x \in(-\infty,-4) \cup\left(\frac{10}{3}, \infty\right)
$$
<br/><br/>No common values of $\mathrm{x}$.
<br/><br/>Hence intersection with feasible region.
<br/><br/>We get $x \in\left(\frac{-5}{2}, \frac{10}{3}\right]-\left\{\frac{3}{2}\right\}$
<br/><br/>Integral value of $x$ are $\{-2,-1,0,1,2,3\}$
<br/><br/>No. of integral values $=6$ | mcq | jee-main-2023-online-11th-april-morning-shift |
L7moS1Zt7WzMgKYdTL7k9k2k5itzmge | maths | logarithm | logarithmic-equations | The number of distinct solutions of the equation<br/>
$${\log _{{1 \over 2}}}\left| {\sin x} \right| = 2 - {\log _{{1 \over 2}}}\left| {\cos x} \right|$$ in the interval
[0, 2$$\pi $$], is ____. | [] | null | 8 | $${\log _{{1 \over 2}}}\left| {\sin x} \right| = 2 - {\log _{{1 \over 2}}}\left| {\cos x} \right|$$
<br><br>$$ \Rightarrow $$ $${\log _{{1 \over 2}}}\left| {\sin x} \right|$$ + $${\log _{{1 \over 2}}}\left| {\cos x} \right|$$ = 2
<br><br>$$ \Rightarrow $$ $${\log _{{1 \over 2}}}\left( {\left| {\sin x\cos x} \right|} \right)$$ = 2
<br><br>$$ \Rightarrow $$ $${\left| {\sin x\cos x} \right| = {1 \over 4}}$$
<br><br>$$ \Rightarrow $$ sin 2x = $$ \pm $$ $${1 \over 2}$$
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267636/exam_images/ndn0paa8angdbmk4hb5t.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th January Morning Slot Mathematics - Logarithm Question 7 English Explanation">
<br><br>$$ \therefore $$ Number of distinct solution = 8 | integer | jee-main-2020-online-9th-january-morning-slot |
Eqw2IGWkAs3wRXidEM1kluhm49z | maths | logarithm | logarithmic-equations | The number of solutions of the equation log<sub>4</sub>(x $$-$$ 1) = log<sub>2</sub>(x $$-$$ 3) is _________. | [] | null | 1 | $${\log _4}(x - 1) = {\log _2}(x - 3)$$<br><br>$$ \Rightarrow {1 \over 2}{\log _2}(x - 1) = {\log _2}(x - 3)$$<br><br>$$ \Rightarrow {\log _2}{(x - 1)^{1/2}} = {\log _2}(x - 3)$$<br><br>$$ \Rightarrow {(x - 1)^{1/2}} = {\log _2}(x - 3)$$<br><br>$$ \Rightarrow {(x - 1)^{1/2}} = x - 3$$<br><br>$$ \Rightarrow x - 1 = {x^2} + 9 - 6x$$<br><br>$$ \Rightarrow {x^2} - 7x + 10 = 0$$<br><br>$$ \Rightarrow (x - 2)(x - 5) = 0$$<br><br>$$ \Rightarrow x = 2,5$$<br><br>But x $$ \ne $$ 2 because it is not satisfying the domain of given equation i.e. log<sub>2</sub>(x $$-$$ 3) $$ \to $$ its domain x > 3<br><br>finally x is 5<br><br>$$ \therefore $$ No. of solutions = 1. | integer | jee-main-2021-online-26th-february-morning-slot |
1krrvdewm | maths | logarithm | logarithmic-equations | The number of solutions of the equation <br/><br/>$${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$$, x > 0, is : | [] | null | 1 | $${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$$<br><br>$${\log _{(x + 1)}}(2x + 5)(x + 1) + 2{\log _{(2x + 5)}}(x + 1) = 4$$<br><br>$${\log _{(x + 1)}}(2x + 5) + 1 + 2{\log _{(2x + 5)}}(x + 1) = 4$$<br><br>Put $${\log _{(x + 1)}}(2x + 5) = t$$<br><br>$$t + {2 \over t} = 3 \Rightarrow {t^2} - 3t + 2 = 0$$<br><br>t = 1, 2<br><br>$${\log _{(x + 1)}}(2x + 5) = 1$$ & $${\log _{(x + 1)}}(2x + 5) = 2$$<br><br>$$x + 1 = 2x + 3$$ & $$2x + 5 = {(x + 1)^2}$$<br><br>$$x = - 4$$ (rejected)<br><br>$${x^2} = 4 \Rightarrow x = 2, - 2$$ (rejected)<br><br>So, x = 2<br><br>No. of solution = 1 | integer | jee-main-2021-online-20th-july-evening-shift |
1ldr78hpn | maths | logarithm | logarithmic-equations | <p>If the solution of the equation $$\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1, x \in\left(0, \frac{\pi}{2}\right)$$, is $$\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right)$$, where $$\alpha$$, $$\beta$$ are integers, then $$\alpha+\beta$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "5"}] | ["C"] | null | <p>$${\log _{\cos x}}\cot x + 4{\log _{\sin x}}\tan x = 1$$</p>
<p>$$ \Rightarrow {\log _{\cos x}}\cot x - 4{\log _{\sin x}}\cot x = 1$$</p>
<p>$$ \Rightarrow 1 - {\log _{\cos x}}\sin x - 4 - 4{\log _{\sin x}}\cos x = 1$$</p>
<p>Let $${\log _{\cos x}}\sin x = t$$</p>
<p>$$t + {4 \over t} = 4$$</p>
<p>$$ \Rightarrow t = 2$$</p>
<p>$$\sin x = {\cos ^2}x$$</p>
<p>$$ \Rightarrow \sin x = 1 - {\sin ^2}x$$</p>
<p>$$ \Rightarrow {\sin ^2}x + \sin {x^{ - 1}} = 0$$</p>
<p>$$ \Rightarrow \sin x = {{ - 1\, \pm \,\sqrt 5 } \over 2}$$</p>
<p>as $$x \in \left( {0,{\pi \over 2}} \right)$$</p>
<p>$$\sin x = {{\sqrt 5 - 1} \over 2}$$</p>
<p>$$x = {\sin ^{ - 1}}\left( {{{ - 1 + \sqrt 5 } \over 2}} \right)$$</p>
<p>$$ \Rightarrow \alpha = - 1,\beta = 5$$</p>
<p>$$\alpha + \beta = 4$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldv30uxy | maths | logarithm | logarithmic-equations | <p>Let $$S = \left\{ {\alpha :{{\log }_2}({9^{2\alpha - 4}} + 13) - {{\log }_2}\left( {{5 \over 2}.\,{3^{2\alpha - 4}} + 1} \right) = 2} \right\}$$. Then the maximum value of $$\beta$$ for which the equation $${x^2} - 2{\left( {\sum\limits_{\alpha \in s} \alpha } \right)^2}x + \sum\limits_{\alpha \in s} {{{(\alpha + 1)}^2}\beta = 0} $$ has real roots, is ____________.</p> | [] | null | 25 | $$
\begin{aligned}
& \log _2\left(9^{2 \alpha-4}+13\right)-\log _2\left(\frac{5}{2} \cdot 3^{2 \alpha-4}+1\right)=2 \\\\
& \Rightarrow \frac{9^{2 \alpha-4}+13}{\frac{5}{2} 3^{2 \alpha-4}+1}=4 \\\\
& \Rightarrow \alpha=2 \quad \text { or } \quad 3 \\\\
& \sum_{\alpha \in \mathrm{S}} \alpha=5 \text { and } \sum_{\alpha \in \mathrm{S}}(\alpha+1)^2=25 \\\\
& \Rightarrow x^2-50 x+25 \beta=0 \text { has real roots } \\\\
& \Rightarrow \beta \leq 25 \\\\
& \Rightarrow \beta_{\max }=25
\end{aligned}
$$ | integer | jee-main-2023-online-25th-january-morning-shift |
1lgxwj4h5 | maths | logarithm | logarithmic-equations | <p>Let a, b, c be three distinct positive real numbers such that $${(2a)^{{{\log }_e}a}} = {(bc)^{{{\log }_e}b}}$$ and $${b^{{{\log }_e}2}} = {a^{{{\log }_e}c}}$$.<br/><br/>Then, 6a + 5bc is equal to ___________.</p> | [] | null | 8 | Given, $(2 a)^{\ln a}=(b c)^{\ln b}$, where $2 a>0, b c>0$
<br/><br/>$$
\Rightarrow \ln a(\ln 2+\ln a)=\ln b(\ln b+\ln c)
$$ ..........(i)
<br/><br/>and
$(b)^{\ln 2}=(a)^{\ln c}$
<br/><br/>$$
\Rightarrow \ln 2 \cdot \ln b=\ln c \cdot \ln a
$$ ..........(ii)
<br/><br/>Now, let $\ln a=x, \ln b=y$
<br/><br/>$$
\ln 2=p, \ln c=z
$$
<br/><br/>Now, from Eqs. (i) and (ii), we get
<br/><br/>$$
p \cdot y=x z \text { and } x(p+x)=y(y+z)
$$
<br/><br/>$$
\begin{aligned}
& \therefore p=\frac{x z}{y} \\\\
& \therefore x\left(\frac{x z}{y}+x\right)=y(y+z) \\\\
& \Rightarrow x^2 z+x^2 y=y^2(y+z) \\\\
& \Rightarrow \left(x^2-y^2\right)(y+z)=0
\end{aligned}
$$
<p>$$ \therefore $$ $ x^2 = y^2 $
<br/><br/>$ x = \pm y $</p>
<p>So, from the equations, there are two cases :</p>
<p>Case 1 :
<br/><br/>$ x = y $
<br/><br/>In this case, since x and y are natural logarithms of positive numbers a and b respectively, this implies that a = b. However, this cannot be true as a, b, and c are given to be distinct positive real numbers.</p>
<p>Case 2 :
<br/><br/>x = -y
<br/><br/>In this case, $ \ln a = -\ln b $
<br/><br/>$ \Rightarrow a \times b = 1$
<br/><br/>$ \Rightarrow b = \frac{1}{a} $</p>
<p>Also, y + z = 0 (from the equations above)
<br/><br/>$ \Rightarrow \ln b + \ln c = 0 $
<br/><br/>$ \Rightarrow \ln (b \times c) = 0 $
<br/><br/>$ \Rightarrow b \times c = 1 $</p>
<p>Given $ b = \frac{1}{a} $ and $ b \times c = 1 $
$ \Rightarrow c = a $</p>
<p>Thus, in the case where x = -y, the possible values are :
<br/><br/>$b = \frac{1}{a} $
<br/><br/>$c = a$</p>
<br/>$$
\begin{aligned}
& \text { If } b c=1 \Rightarrow(2 a)^{\ln a}=1 \\\\
& \Rightarrow a=1 / 2 \\\\
& \text { So, } 6 a+5 b c=6\left(\frac{1}{2}\right)+5=3+5=8
\end{aligned}
$$ | integer | jee-main-2023-online-10th-april-morning-shift |
3YLjdLeFsjhnI2wc | maths | mathematical-induction | mathematical-induction | If $${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } } $$ having $$n$$ radical signs then by methods of mathematical induction which is true | [{"identifier": "A", "content": "$${a_n} > 7\\,\\,\\forall \\,\\,n \\ge 1$$ "}, {"identifier": "B", "content": "$${a_n} < 7\\,\\,\\forall \\,\\,n \\ge 1$$ "}, {"identifier": "C", "content": "$${a_n} < 4\\,\\,\\forall \\,\\,n \\ge 1$$ "}, {"identifier": "D", "content": "$${a_n} > 3\\,\\,\\forall \\,\\,n \\ge 1$$ "}] | ["D"] | null | Given $${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } } $$
<br><br>$$\therefore$$ $${a_n} = \sqrt {7 + {a_n}} $$
<br><br>$$ \Rightarrow $$ $$a_n^2 = 7 + {a_n}$$
<br><br>$$ \Rightarrow $$ $$a_n^2 - {a_n} - 7 = 0$$
<br><br>$$ \Rightarrow {a_n} = {{1 \pm \sqrt {1 - 4 \times 1 \times - 7} } \over 2}$$
<br><br>$$ \Rightarrow {a_n} = {{1 \pm \sqrt {29} } \over 2}$$
<br><br>As $${a_n}$$ > 0,
<br><br>$$\therefore$$ $${a_n} = {{1 + \sqrt {29} } \over 2}$$ = 3.19
<br><br>So $${a_n} > 3\,\,\forall \,\,n \ge 1$$ | mcq | aieee-2002 |
hj54pq327GNyXr3n | maths | mathematical-induction | mathematical-induction | Let $$S(K)$$ $$ = 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}.$$ Then which of the following is true | [{"identifier": "A", "content": "Principle of mathematical induction can be used to prove the formula"}, {"identifier": "B", "content": "$$S\\left( K \\right) \\Rightarrow S\\left( {K + 1} \\right)$$ "}, {"identifier": "C", "content": "$$S\\left( K \\right) \\ne S\\left( {K + 1} \\right)$$ "}, {"identifier": "D", "content": "$$S\\left( 1 \\right)$$ is correct "}] | ["B"] | null | Given $$S(K)$$ $$ = 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}$$
<br><br>When k = 1, S(1): 1 = 3 + 1,
<br><br>L.H.S of S(k) $$ \ne $$ R.H.S of S(k)
<br><br>So S(1) is not true.
<br><br>As S(1) is not true so principle of mathematical induction can not be used.
<br><br>S(K+1) = 1 + 3 + 5... + (2K - 1) + (2K + 1) = 3 + (k + 1)<sup>2</sup>
<br><br>Now let S(k) is true
<br><br>$$\therefore$$ 1 + 3 + 5 +........(2k - 1) = 3 + k<sup>2</sup>
<br><br>$$ \Rightarrow $$ 1 + 3 + 5 +........(2k - 1) + (2k + 1) = 3 + k<sup>2</sup> + 2k +1
<br><br>= 3 + (k + 1)<sup>2</sup>
<br><br>$$ \Rightarrow $$ S(k + 1) is true.
<br><br>$$\therefore$$ S(k) $$ \Rightarrow $$ S(k + 1) | mcq | aieee-2004 |
9WbUqXiIlcooNReg | maths | mathematical-induction | mathematical-induction | If $$A = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]$$ and $$I = \left[ {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right],$$ then which one of the following holds for all $$n \ge 1,$$ by the principle of mathematical induction? | [{"identifier": "A", "content": "$${A^n} = nA - \\left( {n - 1} \\right){\\rm I}$$ "}, {"identifier": "B", "content": "$${A^n} = {2^{n - 1}}A - \\left( {n - 1} \\right){\\rm I}$$ "}, {"identifier": "C", "content": "$${A^n} = nA + \\left( {n - 1} \\right){\\rm I}$$"}, {"identifier": "D", "content": "$${A^n} = {2^{n - 1}}A + \\left( {n - 1} \\right){\\rm I}$$"}] | ["A"] | null | Given $$A = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]$$
<br><br>$$\therefore$$ $$A \times A$$ = $${A^2}$$ = $$\left[ {\matrix{
1 & 0 \cr
2 & 1 \cr
} } \right]$$
<br><br>and $${A^3}$$ = $${A^2} \times A$$ = $$\left[ {\matrix{
1 & 0 \cr
3 & 1 \cr
} } \right]$$
<br><br>So we can say $${A^n}$$ = $$\left[ {\matrix{
1 & 0 \cr
n & 1 \cr
} } \right]$$
<br><br>Now $$nA - \left( {n - 1} \right){\rm I}$$
<br><br>= $$\left[ {\matrix{
n & 0 \cr
n & n \cr
} } \right]$$ - $$\left[ {\matrix{
{n - 1} & 0 \cr
0 & {n - 1} \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
1 & 0 \cr
n & 1 \cr
} } \right]$$ = $${A^n}$$
<br><br>$$\therefore$$ $${A^n} = nA - \left( {n - 1} \right){\rm I}$$ | mcq | aieee-2005 |
x3tbE32XVPE9RRjV | maths | mathematical-reasoning | logical-connectives | The statement $$p \to \left( {q \to p} \right)$$ is equivalent to | [{"identifier": "A", "content": "$$p \\to \\left( {p \\leftrightarrow q} \\right)$$"}, {"identifier": "B", "content": "$$p \\to \\left( {p \\to q} \\right)$$"}, {"identifier": "C", "content": "$$p \\to \\left( {p \\vee q} \\right)$$"}, {"identifier": "D", "content": "$$p \\to \\left( {p \\wedge q} \\right)$$"}] | ["C"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 551px">
<colgroup>
<col style="width: 52px">
<col style="width: 63px">
<col style="width: 81px">
<col style="width: 122px">
<col style="width: 101px">
<col style="width: 132px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">q</th>
<th class="tg-baqh">p</th>
<th class="tg-baqh">$$q \to p$$</th>
<th class="tg-baqh">$$p \to (q \to p)$$</th>
<th class="tg-baqh">$$(p \vee q)$$</th>
<th class="tg-baqh">$$p \to (p \vee q)$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
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<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table></p>
<p>$$\therefore$$ Statement p $$\to$$ (q $$\to$$ p) is equivalent to p $$\to$$ (p $$\vee$$ q).</p> | mcq | aieee-2008 |
U4V9b6lrngYsbSHC | maths | mathematical-reasoning | logical-connectives | <b>Statement-1 :</b> $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ is equivalent to $${p \leftrightarrow q}$$.
<br/><b>Statement-2 :</b> $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ is a tautology. | [{"identifier": "A", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1"}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1"}, {"identifier": "C", "content": "Statement-1 is true, Statement-2 is false"}, {"identifier": "D", "content": "Statement-1 is false, Statement-2 is true"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3fdb05p/c9d4c926-25a7-47f2-b2d6-27e5c7d3dd72/0e89a3d0-d8be-11ec-bed7-8f98fee06619/file-1l3fdb05q.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3fdb05p/c9d4c926-25a7-47f2-b2d6-27e5c7d3dd72/0e89a3d0-d8be-11ec-bed7-8f98fee06619/file-1l3fdb05q.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="AIEEE 2009 Mathematics - Mathematical Reasoning Question 112 English Explanation">
<p>Therefore, S1 is true and S2 is false.</p> | mcq | aieee-2009 |
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