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1l5aijk91 | maths | functions | composite-functions | <p>Let $$f:R \to R$$ and $$g:R \to R$$ be two functions defined by $$f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$$ and $$g(x) = {{1 - 2{e^{2x}}} \over {{e^x}}}$$. Then, for which of the following range of $$\alpha$$, the inequality $$f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha -{5 \over 3}} \right)} \right)$$ holds ?</p> | [{"identifier": "A", "content": "(2, 3)"}, {"identifier": "B", "content": "($$-$$2, $$-$$1)"}, {"identifier": "C", "content": "(1, 2)"}, {"identifier": "D", "content": "($$-$$1, 1)"}] | ["A"] | null | <p>$$f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$$</p>
<p>$$f'(x) = {{2x} \over {{x^2} + 1}} + {e^{ - x}}$$</p>
<p>$$ = {2 \over {x + {1 \over x}}} + {e^{ - x}} > 0\,\,\forall x \in R$$</p>
<p>$$g(x) = {e^{ - x}} - 2{e^x}$$</p>
<p>$$g'(x) - - {e^{ - x}} - 2{e^x} < 0\,\,\,\,\forall x \in R$$</p>
<p>$$\Rightarrow$$ f(x) is increasing and g(x) is decreasing function.</p>
<p>$$f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha - {5 \over 3}} \right)} \right)$$</p>
<p>$$ \Rightarrow {{{{(\alpha - 1)}^2}} \over 3} < \alpha - {5 \over 3}$$</p>
<p>$$ = {\alpha ^2} - 5\alpha + 6 < 0$$</p>
<p>$$ = (\alpha - 2)(\alpha - 3) < 0$$</p>
<p>$$ = \alpha \in (2,\,3)$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5ajukpk | maths | functions | composite-functions | <p>Let $$f:R \to R$$ be a function defined by <br/><br/>$$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)(2 + {x^{25}})} \right)^{{1 \over {50}}}}$$. If the function $$g(x) = f(f(f(x))) + f(f(x))$$, then the greatest integer less than or equal to g(1) is ____________.</p> | [] | null | 2 | <p>Given,</p>
<p>$$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)\left( {2 + {x^{25}}} \right)} \right)^{{1 \over {50}}}}$$</p>
<p>and $$g(x) = f\left( {f\left( {f\left( x \right)} \right)} \right) + f\left( {f\left( x \right)} \right)$$</p>
<p>$$\therefore$$ $$g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$$</p>
<p>Now, $$f(1) = {\left( {2\left( {1 - {{{1^{25}}} \over 2}} \right)\left( {2 + {1^{25}}} \right)} \right)^{{1 \over {50}}}}$$</p>
<p>$$ = {\left( {2\left( {1 - {1 \over 2}} \right)\left( {2 + 1} \right)} \right)^{{1 \over {50}}}}$$</p>
<p>$$ = {\left( 3 \right)^{{1 \over {50}}}}$$</p>
<p>$$\therefore$$ $$f\left( {f\left( 1 \right)} \right) = f\left( {{3^{{1 \over {50}}}}} \right)$$</p>
<p>$$ = {\left( {2\left( {1 - {{{{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \over 2}} \right)\left( {2 + {{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \right)} \right)^{{1 \over {50}}}}$$</p>
<p>$$ = {\left( {2\left( {1 - {{{3^{{1 \over 2}}}} \over 2}} \right)\left( {2 + {3^{{1 \over 2}}}} \right)} \right)^{{1 \over {50}}}}$$</p>
<p>$$ = {\left( {2 \times \left( {{{2 - \sqrt 3 } \over 2}} \right)\left( {2 + \sqrt 3 } \right)} \right)^{{1 \over {50}}}}$$</p>
<p>$$ = {\left[ {\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)} \right]^{{1 \over {50}}}}$$</p>
<p>$$ = {\left( {4 - 3} \right)^{{1 \over {50}}}}$$</p>
<p>$$ = {1^{{1 \over {50}}}} = 1$$</p>
<p>Now, $$f\left( {f\left( {f\left( 1 \right)} \right)} \right) = f(1) = {3^{{1 \over {50}}}}$$</p>
<p>$$\therefore$$ $$g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$$</p>
<p>$$ = {3^{{1 \over {50}}}} + 1$$</p>
<p>Now, greatest integer less than or equal to $$g(1)$$</p>
<p>$$ = \left[ {g(1)} \right]$$</p>
<p>$$ = \left[ {{3^{{1 \over {50}}}} + 1} \right]$$</p>
<p>$$ = \left[ {{3^{{1 \over {50}}}}} \right] + \left[ 1 \right]$$</p>
<p>$$ = [1.02] + 1$$</p>
<p>$$ = 1 + 1 = 2$$</p> | integer | jee-main-2022-online-25th-june-morning-shift |
1l6f37b3x | maths | functions | composite-functions | <p>Let $$f(x)$$ be a quadratic polynomial with leading coefficient 1 such that $$f(0)=p, p \neq 0$$, and $$f(1)=\frac{1}{3}$$. If the equations $$f(x)=0$$ and $$f \circ f \circ f \circ f(x)=0$$ have a common real root, then $$f(-3)$$ is equal to ________________.</p> | [] | null | 25 | <p>Let $$f(x) = (x - \alpha )(x - \beta )$$</p>
<p>It is given that $$f(0) = p \Rightarrow \alpha \beta = p$$</p>
<p>and $$f(1) = {1 \over 3} \Rightarrow (1 - \alpha )(1 - \beta ) = {1 \over 3}$$</p>
<p>Now, let us assume that, $$\alpha$$ is the common root of $$f(x) = 0$$ and $$fofofof(x) = 0$$</p>
<p>$$fofofof(x) = 0$$</p>
<p>$$ \Rightarrow fofof(0) = 0$$</p>
<p>$$ \Rightarrow fof(p) = 0$$</p>
<p>So, $$f(p)$$ is either $$\alpha$$ or $$\beta$$.</p>
<p>$$(p - \alpha )(p - \beta ) = \alpha $$</p>
<p>$$(\alpha \beta - \alpha )(\alpha \beta - \beta ) = \alpha \Rightarrow (\beta - 1)(\alpha - 1)\beta = 1$$ ($$\because$$ $$\alpha \ne 0$$)</p>
<p>So, $$\beta = 3$$</p>
<p>$$(1 - \alpha )(1 - 3) = {1 \over 3}$$</p>
<p>$$\alpha = {7 \over 6}$$</p>
<p>$$f(x) = \left( {x - {7 \over 6}} \right)(x - 3)$$</p>
<p>$$f( - 3) = \left( { - 3 - {7 \over 6}} \right)(3 - 3) = 25$$</p> | integer | jee-main-2022-online-25th-july-evening-shift |
1l6jb0fby | maths | functions | composite-functions | <p>Let $$f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}$$ be functions defined by $$f(a)=\alpha$$, where $$\alpha$$ is the maximum of the powers of those primes $$p$$ such that $$p^{\alpha}$$ divides $$a$$, and $$g(a)=a+1$$, for all $$a \in \mathbb{N}-\{1\}$$. Then, the function $$f+g$$ is</p> | [{"identifier": "A", "content": "one-one but not onto"}, {"identifier": "B", "content": "onto but not one-one"}, {"identifier": "C", "content": "both one-one and onto"}, {"identifier": "D", "content": "neither one-one nor onto"}] | ["D"] | null | <p>$$f,g:N - \{ 1\} \to N$$ defined as</p>
<p>$$f(a) = \alpha $$, where $$\alpha$$ is the maximum power of those primes p such that p<sup>$$\alpha$$</sup> divides a.</p>
<p>$$g(a) = a + 1$$,</p>
<p>Now,</p>
<p>$$\matrix{
{f(2) = 1,} & {g(2) = 3} & \Rightarrow & {(f + g)\,(2) = 4} \cr
{f(3) = 1,} & {g(3) = 4} & \Rightarrow & {(f + g)\,(3) = 5} \cr
{f(4) = 2,} & {g(4) = 5} & \Rightarrow & {(f + g)\,(4) = 7} \cr
{f(5) = 1,} & {g(5) = 6} & \Rightarrow & {(f + g)\,(5) = 7} \cr
} $$</p>
<p>$$\because$$ $$(f + g)\,(5) = (f + g)\,(4)$$</p>
<p>$$\therefore$$ $$f + g$$ is not one-one</p>
<p>Now, $$\because$$ $${f_{\min }} = 1,\,{g_{\min }} = 3$$</p>
<p>So, there does not exist any $$x \in N - \{ 1\} $$ such that $$(f + g)(x) = 1,2,3$$</p>
<p>$$\therefore$$ $$f + g$$ is not onto</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6m66fbv | maths | functions | composite-functions | <p>Let $$\alpha, \beta$$ and $$\gamma$$ be three positive real numbers. Let $$f(x)=\alpha x^{5}+\beta x^{3}+\gamma x, x \in \mathbf{R}$$ and $$g: \mathbf{R} \rightarrow \mathbf{R}$$ be such that $$g(f(x))=x$$ for all $$x \in \mathbf{R}$$. If $$\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}$$ be in arithmetic progression with mean zero, then the value of $$f\left(g\left(\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right)$$ is equal to :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "27"}] | ["A"] | null | <p>$$f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right)$$</p>
<p>$${{{a_1} + {a_2} + {a_3}\, + \,......\, + \,{a_n}} \over n} = 0$$</p>
<p>$$\therefore$$ First and last term, second and second last and so on are equal in magnitude but opposite in sign.</p>
<p>$$f(x) = \alpha {x^5} + \beta {x^3} + \gamma x$$</p>
<p>$$\sum\limits_{i = 1}^n {f({a_i}) = \alpha \left( {a_1^5 + a_2^5 + a_3^5\, + \,.....\, + \,a_n^5} \right) + \beta \left( {a_1^3 + a_2^3\, + \,....\, + \,a_n^3} \right) + \gamma \left( {{a_1} + {a_2}\, + \,....\, + \,{a_n}} \right)} $$</p>
<p>$$ = 0\alpha + 0\beta + 0\gamma $$</p>
<p>$$ = 0$$</p>
<p>$$\therefore$$ $$f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right) = {1 \over n}\sum\limits_{i = 1}^n {f({a_i}) = 0} $$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1ldv2y8dq | maths | functions | composite-functions | <p>For some a, b, c $$\in\mathbb{N}$$, let $$f(x) = ax - 3$$ and $$\mathrm{g(x)=x^b+c,x\in\mathbb{R}}$$. If $${(fog)^{ - 1}}(x) = {\left( {{{x - 7} \over 2}} \right)^{1/3}}$$, then $$(fog)(ac) + (gof)(b)$$ is equal to ____________.</p> | [] | null | 2039 | $f(x)=a x-3$
<br/><br/>
$g(x)=x^{b}+c$
<br/><br/>
$(fog)^{-1}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
<br/><br/>
$(fog)^{-1}(x)=\left(\frac{x+3-c a}{a}\right)^{\frac{1}{b}}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
<br/><br/>
$\Rightarrow a=2, b=3, c=5$
<br/><br/>
$fog(a c)+gof(b)$
<br/><br/>
$\because f(x)=2 x-3$
<br/><br/>
$g(x)=x^{3}+5$
<br/><br/>
$fog(10)+g o f(3)$
<br/><br/>
$=2007+32$
<br/><br/>
$=2039$ | integer | jee-main-2023-online-25th-january-morning-shift |
1lgpyd2rn | maths | functions | composite-functions | <p>For $$x \in \mathbb{R}$$, two real valued functions $$f(x)$$ and $$g(x)$$ are such that, $$g(x)=\sqrt{x}+1$$ and $$f \circ g(x)=x+3-\sqrt{x}$$. Then $$f(0)$$ is equal to</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$-$$3"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $$
\begin{aligned}
& g(x)=\sqrt{x}+1 \\\\
& \operatorname{fog}(x)=x+3-\sqrt{x} \\\\
& =(\sqrt{x}+1)^2-3(\sqrt{x}+1)+5 \\\\
& =g^2(x)-3 g(x)+5 \\\\
& \Rightarrow f(x)=x^2-3 x+5 \\\\
& \therefore f(0)=5
\end{aligned}
$$
<br/><br/>But, if we consider the domain of the composite function $f \circ g(x)$ then in that case $f(0)$ will be not defined as $\mathrm{g}(\mathrm{x})$ cannot be equal to zero. | mcq | jee-main-2023-online-13th-april-morning-shift |
lsaoplpj | maths | functions | composite-functions | Let $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be defined as
<br/><br/>$f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.$ and
<br/><br/>$g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.$. Then, gof : $\mathbf{R} \rightarrow \mathbf{R}$ is : | [{"identifier": "A", "content": "one-one but not onto"}, {"identifier": "B", "content": "neither one-one nor onto"}, {"identifier": "C", "content": "onto but not one-one"}, {"identifier": "D", "content": "both one-one and onto"}] | ["B"] | null | Given, $f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.$ and
<br><br>$g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.$
<br><br>then $g \circ f(x)$ $=g(f(x))$
<br><br>$\begin{aligned} & \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}f(x), f(x) \geq 0 \\ e^{f(x)}, f(x)<0\end{array}\right. \\\\ & \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}e^{-x},(-\infty, 0] \\ e^{\ln x},(0,1) \\ \ln x,[1, \infty)\end{array}\right.\end{aligned}$
<br><br>$g(f(x))=\left\{\begin{array}{c}e^{-x},(-\infty, 0] \\ x,(0,1) \\ \ln x,[1, \infty)\end{array}\right.$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ltao3w7v/14f352dd-3dd6-4237-8545-8edf0028c3df/01db21c0-d8e6-11ee-a868-79761dafd483/file-6y3zli1ltao3w7w.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ltao3w7v/14f352dd-3dd6-4237-8545-8edf0028c3df/01db21c0-d8e6-11ee-a868-79761dafd483/file-6y3zli1ltao3w7w.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Morning Shift Mathematics - Functions Question 21 English Explanation">
<br><br> From the graph of $g(f(x))$, we can say
<br><br>$\mathrm{g}(\mathrm{f}(\mathrm{x})) \Rightarrow$ Many one into | mcq | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lscna9if | maths | functions | composite-functions | <p>Let $$f: \mathbf{R}-\left\{\frac{-1}{2}\right\} \rightarrow \mathbf{R}$$ and $$g: \mathbf{R}-\left\{\frac{-5}{2}\right\} \rightarrow \mathbf{R}$$ be defined as $$f(x)=\frac{2 x+3}{2 x+1}$$ and $$g(x)=\frac{|x|+1}{2 x+5}$$. Then, the domain of the function fog is :</p> | [{"identifier": "A", "content": "$$\\mathbf{R}-\\left\\{-\\frac{7}{4}\\right\\}$$\n"}, {"identifier": "B", "content": "$$\\mathbf{R}$$\n"}, {"identifier": "C", "content": "$$\\mathbf{R}-\\left\\{-\\frac{5}{2},-\\frac{7}{4}\\right\\}$$\n"}, {"identifier": "D", "content": "$$\\mathbf{R}-\\left\\{-\\frac{5}{2}\\right\\}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& f(x)=\frac{2 x+3}{2 x+1} ; x \neq-\frac{1}{2} \\
& g(x)=\frac{|x|+1}{2 x+5}, x \neq-\frac{5}{2}
\end{aligned}$$</p>
<p>Domain of $$f(g(x))$$</p>
<p>$$f(g(x))=\frac{2 g(x)+3}{2 g(x)+1}$$</p>
<p>$$x \neq-\frac{5}{2}$$ and $$\frac{|x|+1}{2 x+5} \neq-\frac{1}{2}$$</p>
<p>$$x \in R-\left\{-\frac{5}{2}\right\}$$ and $$x \in R$$</p>
<p>$$\therefore$$ Domain will be $$\mathrm{R}-\left\{-\frac{5}{2}\right\}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lse52f1r | maths | functions | composite-functions | <p>If $$f(x)=\frac{4 x+3}{6 x-4}, x \neq \frac{2}{3}$$ and $$(f \circ f)(x)=g(x)$$, where $$g: \mathbb{R}-\left\{\frac{2}{3}\right\} \rightarrow \mathbb{R}-\left\{\frac{2}{3}\right\}$$, then $$(g ogog)(4)$$ is equal to</p> | [{"identifier": "A", "content": "$$-4$$"}, {"identifier": "B", "content": "$$\\frac{19}{20}$$"}, {"identifier": "C", "content": "$$-\\frac{19}{20}$$"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <p>To find $$(g \circ g \circ g)(4),$$ we first need to understand the composition of $$f$$ with itself, i.e., $$(f \circ f)(x) = f(f(x)) = g(x).$$ We can then repeatedly apply $$g$$ to get the given expression.</p>
<p>First, let's calculate $$(f \circ f)(x) = g(x):$$</p>
<p>$$g(x) = (f \circ f)(x) = f(f(x))$$</p>
<p>$$= f\left(\frac{4x+3}{6x-4}\right)$$</p>
<p>To evaluate this expression, we substitute $$\frac{4x+3}{6x-4}$$ for $$x$$ in the function $$f(x):$$</p>
<p>$$g(x) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}$$</p>
<p>Now, we simplify the expression:</p>
<p>$$g(x) = \frac{4(4x+3) + 3(6x-4)}{6(4x+3) - 4(6x-4)}$$</p>
<p>$$= \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16}$$</p>
<p>$$= \frac{34x}{34}$$</p>
<p>$$= x$$</p>
<p>So, $$g(x) = x$$ for all $$x$$ in the domain of $$g$$, which is $$\mathbb{R}-\left\{\frac{2}{3}\right\}$$. It's important to note that the domain restriction is preserved through the composition because $$f(x)$$ has a vertical asymptote at $$x = \frac{2}{3}$$ which doesn't intersect the graph.</p>
<p>So, $$g(x)$$ is the identity function on its domain, which means that applying $$g$$ any number of times will result in the same input for $$x$$ in the given domain. Hence, we have:</p>
<p>$$ (g \circ g \circ g \circ g)(4) = g(g(g(g(4)))) = g(g(g(4))) = g(g(4)) = g(4) = 4 $$</p>
<p>This corresponds to option D, which is $$4$$.</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lseypzri | maths | functions | composite-functions | <p>If $$f(x)=\left\{\begin{array}{cc}2+2 x, & -1 \leq x < 0 \\ 1-\frac{x}{3}, & 0 \leq x \leq 3\end{array} ; g(x)=\left\{\begin{array}{cc}-x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1\end{array}\right.\right.$$, then range of $$(f o g)(x)$$ is</p> | [{"identifier": "A", "content": "$$[0,1)$$\n"}, {"identifier": "B", "content": "$$[0,3)$$\n"}, {"identifier": "C", "content": "$$(0,1]$$\n"}, {"identifier": "D", "content": "$$[0,1]$$"}] | ["D"] | null | <p>$$f(g(x)) = \left\{ {\matrix{
{2 + 2g(x),} & { - 1 \le g(x) < 0} & {.....(1)} \cr
{1 - {{g(x)} \over 3},} & {0 \le g(x) \le 3} & {.....(2)} \cr
} } \right.$$</p>
<p>$$\text { By (1) } x \in \phi$$</p>
<p>And by (2) $$x \in[-3,0]$$ and $$x \in[0,1]$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt2y3n7n/57ad7e01-e1c9-4f79-b41c-30dff98ab61e/d0fe9930-d4a6-11ee-bdd1-01c80c3e2d9a/file-1lt2y3n7o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt2y3n7n/57ad7e01-e1c9-4f79-b41c-30dff98ab61e/d0fe9930-d4a6-11ee-bdd1-01c80c3e2d9a/file-1lt2y3n7o.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Morning Shift Mathematics - Functions Question 17 English Explanation"></p>
<p>Range of $$\mathrm{f(g(x))}$$ is $$[0,1]$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
lv2erz8l | maths | functions | composite-functions | <p>Consider the function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ defined by $$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$$. If the composition of $$f, \underbrace{(f \circ f \circ f \circ \cdots \circ f)}_{10 \text { times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}$$, then the value of $$\sqrt{3 \alpha+1}$$ is equal to _______.</p> | [] | null | 1024 | <p>$$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$$</p>
<p>$$(f \circ f)(x)=\frac{2 f(x)}{\sqrt{1+9(f(x))^2}}=\frac{\frac{4 x}{\sqrt{1+9 x^2}}}{\sqrt{1+9 \times \frac{4 x^2}{1+9 x^2}}}=\frac{4 x}{\sqrt{1+45 x^2}}$$</p>
<p>$$(f \circ f \circ f)(x)=\frac{4 \times \frac{2 x}{\sqrt{1+9 x^2}}}{\sqrt{1+45 \times \frac{4 x^2}{1+9 x^2}}}=\frac{8 x}{\sqrt{1+21 \times 9 x^2}}$$</p>
<p>$$(f \circ f \circ f \circ f)(x)=\frac{16 x}{\sqrt{1+85 \times 9 x^2}}$$</p>
<p>$$\Rightarrow \alpha$$ is $$10^{\text {th }}$$ term of $$1,5,21,85, \ldots \alpha$$ is $$10^{\text {th }}$$ term of</p>
<p>$$\begin{aligned}
& \frac{\left(2^1\right)^2-1}{3}, \frac{\left(2^2\right)^2-1}{3}, \frac{\left(2^3\right)^2-1}{3}, \frac{\left(2^4\right)^2-1}{3}, \ldots \\
\Rightarrow \quad & \alpha=\frac{\left(2^{10}\right)^2-1}{3} \\
\Rightarrow \quad & \sqrt{3 \alpha+1}=2^{10}=1024
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-evening-shift |
lv9s1zso | maths | functions | composite-functions | <p>Let $$f, g: \mathbf{R} \rightarrow \mathbf{R}$$ be defined as :</p>
<p>$$f(x)=|x-1| \text { and } g(x)= \begin{cases}\mathrm{e}^x, & x \geq 0 \\ x+1, & x \leq 0 .\end{cases}$$</p>
<p>Then the function $$f(g(x))$$ is</p> | [{"identifier": "A", "content": "neither one-one nor onto.\n"}, {"identifier": "B", "content": "one-one but not onto.\n"}, {"identifier": "C", "content": "both one-one and onto.\n"}, {"identifier": "D", "content": "onto but not one-one."}] | ["A"] | null | <p>$$\begin{aligned}
& f(x)= \begin{cases}x-1, & x \geq 1 \\
1-x & x<0\end{cases} \\
& g(x)=\left\{\begin{array}{cc}
e^x & ; \quad x \geq 0 \\
x+1 & ; \quad x \leq 0
\end{array}\right.
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwelk3uw/46e00973-b9cb-4a02-b3d4-b1b1a19ec9fa/d20cd080-1673-11ef-97a1-518a89cfb45d/file-1lwelk3ux.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwelk3uw/46e00973-b9cb-4a02-b3d4-b1b1a19ec9fa/d20cd080-1673-11ef-97a1-518a89cfb45d/file-1lwelk3ux.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Evening Shift Mathematics - Functions Question 4 English Explanation 1"></p>
<p>$$\begin{aligned}
f(g(x)) & = \begin{cases}g(x)-1, & g(x) \geq 1 \\
1-g(x) & g(x)<1\end{cases} \\
& =\left\{\begin{array}{cl}
e^x-1 & x \geq 0 \\
-x & x<0
\end{array}\right.
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwemmte7/0b8f7144-df01-42b2-8694-08cedef5dd1e/06860df0-1678-11ef-922a-09b27e16ac69/file-1lwemmte8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwemmte7/0b8f7144-df01-42b2-8694-08cedef5dd1e/06860df0-1678-11ef-922a-09b27e16ac69/file-1lwemmte8.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Evening Shift Mathematics - Functions Question 4 English Explanation 2"></p> | mcq | jee-main-2024-online-5th-april-evening-shift |
xkNnw5MPjtXPCYQ0 | maths | functions | domain | The domain of $${\sin ^{ - 1}}\left[ {{{\log }_3}\left( {{x \over 3}} \right)} \right]$$ is | [{"identifier": "A", "content": "[1, 9]"}, {"identifier": "B", "content": "[-1, 9]"}, {"identifier": "C", "content": "[9, 1]"}, {"identifier": "D", "content": "[-9, -1]"}] | ["A"] | null | $$f\left( x \right) = {\sin ^{ - 1}}\left( {{{\log }_3}\left( {{x \over 3}} \right)} \right)$$ exists
<br><br>if $$\,\,\,\, - 1 \le {\log _3}\left( {{x \over 3}} \right) \le 1$$
<br><br>$$ \Leftrightarrow {3^{ - 1}} \le {x \over 3} \le {3^1}$$
<br><br>$$ \Leftrightarrow 1 \le x \le 9$$
<br><br>or $$\,\,\,\,x \in \left[ {1,9} \right]$$ | mcq | aieee-2002 |
jKHkNWI2VphTKzW4 | maths | functions | domain | Domain of definition of the function f(x) = $${3 \over {4 - {x^2}}}$$ + $${\log _{10}}\left( {{x^3} - x} \right)$$, is | [{"identifier": "A", "content": "(-1, 0)$$ \\cup $$(1, 2)$$ \\cup $$(2, $$\\infty $$)"}, {"identifier": "B", "content": "(1, 2)"}, {"identifier": "C", "content": "(-1, 0) $$ \\cup $$ (1, 2)"}, {"identifier": "D", "content": "(1, 2)$$ \\cup $$(2, $$\\infty $$)"}] | ["A"] | null | $$f\left( x \right) = {3 \over {4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} \right)$$
<br><br>$$4 - {x^2} \ne 0;\,\,\,{x^3} - x > 0;$$
<br><br>$$x \ne \pm \sqrt 4 $$ and $$ - 1 < x < 0$$
<br><br>or $$\,\,\,1 < x < \infty $$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264992/exam_images/p877upzzzrjlz3axmszg.webp" loading="lazy" alt="AIEEE 2003 Mathematics - Functions Question 136 English Explanation">
<br><br>$$\therefore$$ $$D = \left( { - 1,0} \right) \cup \left( {1,\infty } \right) - \left\{ {\sqrt 4 } \right\}$$
<br><br>$$D = \left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right).$$ | mcq | aieee-2003 |
htONhxhNRHk21CjA | maths | functions | domain | The domain of the function
<br/>$$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$$ | [{"identifier": "A", "content": "[1, 2]"}, {"identifier": "B", "content": "[2, 3)"}, {"identifier": "C", "content": "[1, 2)"}, {"identifier": "D", "content": "[2, 3]"}] | ["B"] | null | $$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$$ is defined
<br><br>if $$(i)$$ $$\,\,\, - 1 \le x - 3 \le 1 \Rightarrow 2 \le x \le 4$$
<br><br>and $$(ii)$$ $$9 - {x^2} > 0 \Rightarrow - 3 < x < 3$$
<br><br>Taking common solution of $$\left( i \right)$$ and $$\left( {ii} \right),$$
<br><br>we get $$2 \le x < 3$$
<br><br>$$\therefore$$ Domain $$ = \left[ {2,\left. 3 \right)} \right.$$ | mcq | aieee-2004 |
d90DFXOHvtSgLUje | maths | functions | domain | The largest interval lying in $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$ for which the function
<br/><br/>$$f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right)$$$$ + \log \left( {\cos x} \right)$$,
<br/><br/>is defined, is | [{"identifier": "A", "content": "$$\\left[ { - {\\pi \\over 4},{\\pi \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left[ {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\left[ {0,\\pi } \\right]$$"}, {"identifier": "D", "content": "$$\\left( { - {\\pi \\over 2},{\\pi \\over 2}} \\right)$$"}] | ["B"] | null | $$f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right) + \log \left( {\cos \,x} \right)$$
<br><br>$$f\left( x \right)$$ is defined if $$ - 1 \le \left( {{x \over 2} - 1} \right) \le 1$$ and $$\cos \,x > 0$$
<br><br>or $$\,\,\,\,0 \le {x \over 2} \le 2\,\,$$ and $$\,\, - {\pi \over 2} < x < {\pi \over 2}$$
<br><br>or $$\,\,\,$$ $$0 \le x \le 4$$ $$\,\,$$ and $$\,\, - {\pi \over 2} < x < {\pi \over 2}$$
<br><br>$$\therefore$$ $$\,\,\,\,\,x\, \in \left[ {0,{\pi \over 2}} \right)$$ | mcq | aieee-2007 |
IuZ2hfWBOzl3se7H | maths | functions | domain | The domain of the function f(x) = $${1 \over {\sqrt {\left| x \right| - x} }}$$ is | [{"identifier": "A", "content": "$$\\left( {0,\\infty } \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - \\infty ,0} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - \\infty ,\\infty } \\right) - \\left\\{ 0 \\right\\}$$"}, {"identifier": "D", "content": "$$\\left( { - \\infty ,\\infty } \\right)$$"}] | ["B"] | null | $$f\left( x \right) = {1 \over {\sqrt {\left| x \right| - x} }},\,\,$$ define if $$\,\,\,\left| x \right| - x > 0$$
<br><br>$$ \Rightarrow \left| x \right| > x,\,\,\, \Rightarrow x < 0$$
<br><br>Hence domain of $$f\left( x \right)$$ is $$\left( { - \infty ,0} \right)$$ | mcq | aieee-2011 |
w8Zuk7uFMhknNRsWwF18hoxe66ijvww16jt | maths | functions | domain | The domain of the definition of the function
<br/><br/>$$f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$$ is | [{"identifier": "A", "content": "(-1, 0) $$ \\cup $$ (1, 2) $$ \\cup $$ (2, $$\\infty $$)"}, {"identifier": "B", "content": "(-2, -1) $$ \\cup $$ (-1,0) $$ \\cup $$ (2, $$\\infty $$)"}, {"identifier": "C", "content": "(1, 2) $$ \\cup $$ (2, $$\\infty $$)"}, {"identifier": "D", "content": "(-1, 0) $$ \\cup $$ (1,2) $$ \\cup $$ (3, $$\\infty $$)"}] | ["A"] | null | Given $$f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$$
<br><br>Let f<sub>1</sub>(x) = $${1 \over {4 - {x^2}}}$$ and f<sub>2</sub>(x) = $${\log _{10}}({x^3} - x)$$
<br><br>Here in f<sub>1</sub>(x) denominator $$ \ne $$ 0
<br><br>4 - x<sup>2</sup> $$ \ne $$ 0
<br><br>$$ \Rightarrow $$ x $$ \ne $$ $$ \pm $$ 2 ...........(1)
<br><br>and x<sup>3</sup> - x > 0
<br><br>$$ \Rightarrow $$ x(x<sup>2</sup> - 1) > 0
<br><br>$$ \Rightarrow $$ x(x + 1)(x - 1) > 0
<br><br><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264003/exam_images/hktvoypsx5mtyfkdrglr.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265916/exam_images/r7fx8y3xs6bnq5mkq0us.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266956/exam_images/cwlfbycr4bty1astkwr9.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Functions Question 105 English Explanation"></picture>
<br><br>x $$ \in $$ (-1, 0) $$ \cup $$ (1, $$\infty $$) ........(2)
<br><br>Hence domain is intersection of (1) and (2)
<br><br>x $$ \in $$ (-1, 0) $$ \cup $$ (1, 2) $$ \cup $$ (2, $$\infty $$) | mcq | jee-main-2019-online-9th-april-evening-slot |
gQxMfOaeoMVTTKD7A61kmlivagb | maths | functions | domain | The real valued function <br/>$$f(x) = {{\cos e{c^{ - 1}}x} \over {\sqrt {x - [x]} }}$$, where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to : | [{"identifier": "A", "content": "all real except integers"}, {"identifier": "B", "content": "all non-integers except the interval [ $$-$$1, 1 ]"}, {"identifier": "C", "content": "all integers except 0, $$-$$1, 1"}, {"identifier": "D", "content": "all real except the interval [ $$-$$1, 1 ]"}] | ["B"] | null | Domain of $$\cos e{c^{ - 1}}x$$ :<br><br>$$x \in ( - \infty , - 1] \cup [1,\infty )$$<br><br>and, $$x - [x] > 0$$<br><br>$$ \Rightarrow \{ x\} > 0$$<br><br>$$ \Rightarrow x \ne I$$<br><br>$$ \therefore $$ Required domain = $$( - \infty , - 1] \cup [1,\infty ) - I$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
1krpv18s8 | maths | functions | domain | Let [ x ] denote the greatest integer $$\le$$ x, where x $$\in$$ R. If the domain of the real valued function $$f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}} $$ is ($$-$$ $$\infty$$, a) $$]\cup$$ [b, c) $$\cup$$ [4, $$\infty$$), a < b < c, then the value of a + b + c is : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$2"}, {"identifier": "D", "content": "$$-$$3"}] | ["C"] | null | For domain,<br><br>$${{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$$ $$\ge$$ 0<br><br>Case I :<br><br>When $${\left| {[x]} \right| - 2}$$ $$\ge$$ 0<br><br>and $${\left| {[x]} \right| - 3}$$ > 0<br><br>$$\therefore$$ x $$\in$$ ($$-$$ $$\infty$$, $$-$$3) $$\cup$$ [4, $$\infty$$) ...... (1)<br><br>Case II :<br><br>When $${\left| {[x]} \right| - 2}$$ $$\le$$ 0<br><br>and $${\left| {[x]} \right| - 3}$$ < 0<br><br>$$\therefore$$ x $$\in$$ [$$-$$2, 3) ..... (2)<br><br>So, from (1) and (2) we get<br><br>Domain of function<br><br>= ($$-$$ $$\infty$$, $$-$$3) $$\cup$$ [$$-$$2, 3) $$\cup$$ [4, $$\infty$$)<br><br>$$\therefore$$ (a + b + c) = $$-$$3 + ($$-$$2) + 3 = $$-$$2 (a < b < c)<br><br>$$\Rightarrow$$ Option (3) is correct. | mcq | jee-main-2021-online-20th-july-morning-shift |
1ldsvexaq | maths | functions | domain | <p>The domain of $$f(x) = {{{{\log }_{(x + 1)}}(x - 2)} \over {{e^{2{{\log }_e}x}} - (2x + 3)}},x \in \mathbb{R}$$ is</p> | [{"identifier": "A", "content": "$$( - 1,\\infty ) - \\{ 3\\} $$"}, {"identifier": "B", "content": "$$\\mathbb{R} - \\{ - 1,3)$$"}, {"identifier": "C", "content": "$$(2,\\infty ) - \\{ 3\\} $$"}, {"identifier": "D", "content": "$$\\mathbb{R} - \\{ 3\\} $$"}] | ["C"] | null | $x-2>0 \Rightarrow x>2$
<br/><br/>
$\mathrm{x}+1>0 \Rightarrow \mathrm{x}>-1$
<br/><br/>
$x+1 \neq 1 \Rightarrow x \neq 0$ and $x>0$
<br/><br/>
Denominator
<br/><br/>
$\mathrm{x}^{2}-2 \mathrm{x}-3 \neq 0$
<br/><br/>
$(x-3)(x+1) \neq 0$
<br/><br/>
$\mathrm{x} \neq-1,3$
<br/><br/>
So Ans $(2, \infty)-\{3\}$ | mcq | jee-main-2023-online-29th-january-morning-shift |
1lgrgf67d | maths | functions | domain | <p>Let $$\mathrm{D}$$ be the domain of the function $$f(x)=\sin ^{-1}\left(\log _{3 x}\left(\frac{6+2 \log _{3} x}{-5 x}\right)\right)$$. If the range of the function $$\mathrm{g}: \mathrm{D} \rightarrow \mathbb{R}$$ defined by $$\mathrm{g}(x)=x-[x],([x]$$ is the greatest integer function), is $$(\alpha, \beta)$$, then $$\alpha^{2}+\frac{5}{\beta}$$ is equal to</p> | [{"identifier": "A", "content": "45"}, {"identifier": "B", "content": "136"}, {"identifier": "C", "content": "46"}, {"identifier": "D", "content": " nearly 135"}] | ["D"] | null | <p>First, the function $f(x) = \sin^{-1}(\log_{3x}(\frac{6 + 2 \log_3{x}}{-5x}))$ has several restrictions :</p>
<ol>
<li><p>Since the arcsine function $\sin^{-1}(x)$ is only defined for $-1\leq x\leq 1$, this means that $\log_{3x}(\frac{6 + 2 \log _3 x}{-5 x})$ must be between -1 and 1.</p>
</li>
<li><p>For the logarithm to be defined, $3x > 0 \Rightarrow x > 0$ .......(1)<br/><br/> because the base of a logarithm must be greater than 0 and not equal to 1. Also, $x \neq \frac{1}{3}$ .......(2)<br/><br/> as the base cannot be 1.</p>
</li>
<li><p>Moreover, the inner function of the logarithm $\frac{6 + 2 \log_3{x}}{-5x}$ must be greater than 0. <br/><br/>$$ \Rightarrow $$ $6+2 \log _3 x<0 \quad(\because x>0)$
<br/><br/>$$
\begin{aligned}
\Rightarrow & \log _3 x<-3 \\\\
\Rightarrow & x<3^{-3} \\\\
\Rightarrow & x<\frac{1}{27} .........(3)
\end{aligned}
$$
<br/><br/>$$
\Rightarrow \text { From (1), (2), (3), } 0 < x < \frac{1}{27}
$$</p>
</li>
</ol>
<p>The next step is to solve the inequality for $-1 \leq \log_{3x}(\frac{6+2 \log _3 x}{-5 x}) \leq 1$. To do this, we make the observation that for the logarithmic part to be within $[-1,1]$, it must be true that $3x \leq \frac{6+2 \log_3 x}{-5x} \leq \frac{1}{3x}$.</p>
<p>Solving the inequality $15x^2 + 6 + 2 \log_3 x \geq 0$, we find that $x \in(0, \frac{1}{27})$ ........(4)</p>
<p>Likewise, solving the inequality $6+2 \log_3 x + \frac{5}{3} \geq 0$, we find that $x \geq 3^{-\frac{23}{6}}$ ........(5)</p>
<p>Combining Equations (3), (4), and (5), the intersection of all these intervals is $x \in[3^{-\frac{23}{6}}, \frac{1}{27})$.</p>
<p>Now, consider the function $g(x) = x - [x]$, where $[x]$ is the greatest integer function. For this function, the range is the fractional part of $x$. In this case, the range $(\alpha, \beta)$ is given by the minimum and maximum possible values of $x$ in its domain. Hence, $\alpha = 3^{-\frac{23}{6}}$ and $\beta = \frac{1}{27}$.</p>
<p>Finally, substitute these values into the equation $\alpha^{2}+\frac{5}{\beta}$:</p>
<p>$\alpha^{2}+\frac{5}{\beta} = (3^{-\frac{23}{6}})^2 + \frac{5}{\frac{1}{27}} = 3^{-\frac{23}{3}} + 135$ = 135.0002198.</p>
<p>Since $3^{-\frac{23}{3}}$ = 0.0002198 is an extremely small number, it's approximately 0. So, $\alpha^{2}+\frac{5}{\beta} \approx 135$.</p>
| mcq | jee-main-2023-online-12th-april-morning-shift |
1lgsw05gr | maths | functions | domain | <p>The domain of the function $$f(x)=\frac{1}{\sqrt{[x]^{2}-3[x]-10}}$$ is : ( where $$[\mathrm{x}]$$ denotes the greatest integer less than or equal to $$x$$ )</p> | [{"identifier": "A", "content": "$$(-\\infty,-2) \\cup[6, \\infty)$$"}, {"identifier": "B", "content": "$$(-\\infty,-3] \\cup[6, \\infty)$$"}, {"identifier": "C", "content": "$$(-\\infty,-2) \\cup(5, \\infty)$$"}, {"identifier": "D", "content": "$$(-\\infty,-3] \\cup(5, \\infty)$$"}] | ["A"] | null | $$
f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}
$$
<br/><br/>For Domain $[x]^2-3[x]-10>0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow ([x]-5)([x]+2)>0 \\\\
& \Rightarrow [x] \in(-\infty,-2) \cup(5, \infty) \\\\
& \therefore x \in(-\infty,-2) \cup[6, \infty)
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-evening-shift |
1lgyq24t0 | maths | functions | domain | <p>If domain of the function $$\log _{e}\left(\frac{6 x^{2}+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^{2}-3 x+4}{3 x-5}\right)$$ is $$(\alpha, \beta) \cup(\gamma, \delta]$$, then $$18\left(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\right)$$ is equal to ______________.</p> | [] | null | 20 | Domain of $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)$
<br/><br/>So, $\frac{6 x^2+5 x+1}{2 x-1}>0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{(3 x+1)(2 x+1)}{2 x-1}>0 \\\\
& \Rightarrow x \in\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left(\frac{1}{2}, \infty\right)
\end{aligned}
$$
<br/><br/>Domain of
$$
\cos ^{-1} x \rightarrow[-1,1]
$$
<br/><br/>$$
\text { For domain of } \cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right)
$$
<br/><br/>$$
\begin{aligned}
& -1 \leq \frac{2 x^2-3 x+4}{3 x-5} \leq 1 \\\\
& \frac{2 x^2-1}{3 x-5} \geq 0 \text { and } \frac{2 x^2-6 x+9}{3 x-5} \leq 0
\end{aligned}
$$
<br/><br/>$$
\Rightarrow x \in\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \cup\left(\frac{5}{3}, \infty\right)
$$
<br/><br/>So, common domain is $\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left[\frac{1}{2}, \frac{1}{\sqrt{2}}\right]$
<br/><br/>$$
\begin{aligned}
& \therefore 18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)=18\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{4}+\frac{1}{2}\right) \\\\
& =18\left(\frac{9+4+9+18}{36}\right)=\frac{1}{2}(40)=20
\end{aligned}
$$ | integer | jee-main-2023-online-8th-april-evening-shift |
lsam05oo | maths | functions | domain | If the domain of the function
<br/><br/>$f(x)=\frac{\sqrt{x^2-25}}{\left(4-x^2\right)}+\log _{10}\left(x^2+2 x-15\right)$ is $(-\infty, \alpha) \cup[\beta, \infty)$, then $\alpha^2+\beta^3$ is equal to : | [{"identifier": "A", "content": "140"}, {"identifier": "B", "content": "175"}, {"identifier": "C", "content": "125"}, {"identifier": "D", "content": "150"}] | ["D"] | null | <p>To find the domain of the function
$$f(x) = \frac{\sqrt{x^2-25}}{(4-x^2)}+\log_{10}(x^2+2x-15),$$
we need to consider the domain conditions for both the square root function and the logarithmic function.</p>
<p>The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, so
$$x^2 - 25 \geq 0.$$
This inequality is satisfied when
$$x \leq -5 \quad \text{or} \quad x \geq 5.$$</p>
<p>The denominator of the rational part of $f(x)$, $(4-x^2)$, cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have
$$4 - x^2 \neq 0.$$
This inequality is violated when
$$x = \pm2.$$</p>
<p>Combining these conditions gives us the domain for the rational part of the function:
$$x \in (-\infty, -5] \cup (5, \infty) \quad \text{and} \quad x \neq 2,-2.$$</p>
<p>Moving on to the logarithmic function, $\log_{10}(x^2+2x-15)$, the argument must be positive:
$$x^2 + 2x - 15 > 0.$$
This is a quadratic inequality, which we can factor to find the solution:
$$(x+5)(x-3) > 0.$$
From this, we see that the inequality is satisfied for
$$x < -5 \quad \text{or} \quad x > 3.$$</p>
<p>The overall domain of $f(x)$ is the intersection of the domains for each piece. Taking the intersection of the two sets gives us:
$$x \in (-\infty, -5) \cup (5, \infty),$$</p>
<p>Since the question states that the domain is of the form $(-\infty, \alpha) \cup [\beta, \infty)$, we can infer that
$$\alpha = -5 \quad \text{and} \quad \beta = 5.$$</p>
<p>We calculate $\alpha^2 + \beta^3$ as follows:
$$\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150.$$</p>
<p>So the correct answer, representing the sum of $\alpha^2$ and $\beta^3$, is:
Option D $150$.</p>
| mcq | jee-main-2024-online-1st-february-evening-shift |
1lsg4mhuk | maths | functions | domain | <p>If the domain of the function $$f(x)=\log _e\left(\frac{2 x+3}{4 x^2+x-3}\right)+\cos ^{-1}\left(\frac{2 x-1}{x+2}\right)$$ is $$(\alpha, \beta]$$, then the value of $$5 \beta-4 \alpha$$ is equal to</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "10"}] | ["B"] | null | <p>$$\begin{aligned}
& \frac{2 x+3}{4 x^2+x-3}>0 \text { and }-1 \leq \frac{2 x-1}{x+2} \leq 1 \\
& \frac{2 x+3}{(4 x-3)(x+1)}>0 \quad \frac{3 x+1}{x+2} \geq 0 \& \frac{x-3}{x+2} \leq 0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxumdb/8342ad24-c8e3-4299-8fcb-c60fcb809b21/dcf834f0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxumdc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxumdb/8342ad24-c8e3-4299-8fcb-c60fcb809b21/dcf834f0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxumdc.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Functions Question 16 English Explanation"></p>
<p>$$(-\infty,-2) \cup\left[\frac{-1}{3}, \infty\right)$$ ..... (1)</p>
<p>$$(-2,3]$$ ..... (2)</p>
<p>$$\left[\frac{-1}{3}, 3\right]$$ ..... (3) $$\quad (1) \cap(2) \cap(3)$$</p>
<p>$$\begin{aligned}
& \left(\frac{3}{4}, 3\right] \\
& \alpha=\frac{3}{4} \beta=3 \\
& 5 \beta-4 \alpha=15-3=12
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
1lsgb11sw | maths | functions | domain | <p>If the domain of the function $$f(x)=\cos ^{-1}\left(\frac{2-|x|}{4}\right)+\left\{\log _e(3-x)\right\}^{-1}$$ is $$[-\alpha, \beta)-\{\gamma\}$$, then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <p>$$\begin{aligned}
& -1 \leq\left|\frac{2-|x|}{4}\right| \leq 1 \\
& \Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1 \\
& -4 \leq 2-|x| \leq 4 \\
& -6 \leq-|x| \leq 2 \\
& -2 \leq|x| \leq 6 \\
& |x| \leq 6
\end{aligned}$$</p>
<p>$$\Rightarrow x \in[-6,6]$$ .... (1)</p>
<p>Now, $$3-x\ne 1$$</p>
<p>And $$x\ne2$$ .... (2)</p>
<p>and $$3-x>0$$</p>
<p>$$x<3$$ .... (3)</p>
<p>$$\begin{aligned}
& \text { From (1), (2) and (3) } \\
& \Rightarrow x \in[-6,3)-\{2\} \\
& \alpha=6 \\
& \beta=3 \\
& \gamma=2 \\
& \alpha+\beta+\gamma=11
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
luy6z50x | maths | functions | domain | <p>If the domain of the function $$f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$$ is $$\mathbf{R}-(\alpha, \beta)$$, then $$12 \alpha \beta$$ is equal to :</p> | [{"identifier": "A", "content": "40"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "32"}] | ["D"] | null | <p>$$
\begin{array}{ll}
f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right) & \\
-1 \leq \frac{x-1}{2 x+3} \leq 1 & \frac{x-1}{2 x+3}+1 \geq 0 \\
\frac{x-1}{2 x+3}-1 \leq 0 & \frac{x-1+2 x+3}{2 x+3} \geq 0 \\
\frac{x-1-2 x-3}{2 x+3} \leq 0 & \frac{3 x+2}{2 x+3} \geq 0
\end{array}
$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lw3c14xb/42504263-4110-4311-9e4d-11eb95a83473/f131dbf0-1041-11ef-9f6c-75804a813f04/file-jaoe38c1lw3c14xc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/jaoe38c1lw3c14xb/42504263-4110-4311-9e4d-11eb95a83473/f131dbf0-1041-11ef-9f6c-75804a813f04/file-jaoe38c1lw3c14xc.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Functions Question 11 English Explanation 1"></p>
<p>$$\begin{aligned}
& \frac{-x-4}{2 x+3} \leq 0 \\
& \frac{x+4}{2 x+3} \geq 0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lw3c32ar/ca8579c0-7ba5-4526-8059-fbf044a6256c/26c8bb30-1042-11ef-9f6c-75804a813f04/file-jaoe38c1lw3c32as.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/jaoe38c1lw3c32ar/ca8579c0-7ba5-4526-8059-fbf044a6256c/26c8bb30-1042-11ef-9f6c-75804a813f04/file-jaoe38c1lw3c32as.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Functions Question 11 English Explanation 2"></p>
<p>$$\begin{aligned}
& x \in(-\infty,-4] \cup\left[\frac{-2}{3}, \infty\right) \\
& \text { Domain : } R-\left(-4, \frac{-2}{3}\right) \\
& \alpha=-4 \\
& \beta=\frac{-2}{3} \\
& 12 \times \alpha \beta=12 \times 4 \times \frac{2}{3}=32
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
eDxALxbmMByHC8s7 | maths | functions | even-and-odd-functions | The function $$f\left( x \right)$$ $$ = \log \left( {x + \sqrt {{x^2} + 1} } \right)$$, is | [{"identifier": "A", "content": "neither an even nor an odd function"}, {"identifier": "B", "content": "an even function"}, {"identifier": "C", "content": "an odd function"}, {"identifier": "D", "content": "a periodic function"}] | ["C"] | null | $$f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right)$$
<br><br>$$f\left( { - x} \right) = \log \left\{ { - x + \sqrt {{x^2} + 1} } \right\}$$
<br><br>$$ = \log \left\{ {{{ - {x^2} + {x^2} + 1} \over {x + \sqrt {{x^2} + 1} }}} \right\}$$
<br><br>$$ = - \log \left( {x + \sqrt {{x^2} + 1} } \right) = - f\left( x \right)$$
<br><br>$$ \Rightarrow f\left( x \right)\,\,\,\,$$ is an odd function. | mcq | aieee-2003 |
7lH4aEIStXIhkbBM | maths | functions | even-and-odd-functions | The graph of the function y = f(x) is symmetrical about the line x = 2, then | [{"identifier": "A", "content": "$$f\\left( x \\right) = - f\\left( { - x} \\right)$$ "}, {"identifier": "B", "content": "$$f\\left( {2 + x} \\right) = f\\left( {2 - x} \\right)$$"}, {"identifier": "C", "content": "$$f\\left( x \\right) = f\\left( { - x} \\right)$$ "}, {"identifier": "D", "content": "$$f\\left( {x + 2} \\right) = f\\left( {x - 2} \\right)$$"}] | ["B"] | null | Let us consider a graph symm. with respect to line $$x=2$$ as shown in the figure.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263707/exam_images/uiwdp9jmte0l8nkstssm.webp" loading="lazy" alt="AIEEE 2004 Mathematics - Functions Question 131 English Explanation">
<br><br>From the figure
<br><br>$$f\left( {{x_1}} \right) = f\left( {{x_2}} \right),$$
<br><br>where $${x_1} = 2 - x$$ and $${x_2} = 2 + x$$
<br><br>$$\therefore$$ $$\,\,\,\,f\left( {2 - x} \right) = f\left( {2 + x} \right)$$ | mcq | aieee-2004 |
7Gxl5AoOoGNHBVk7crwGB | maths | functions | even-and-odd-functions | Let ƒ(x) = a<sup>x</sup>
(a > 0) be written as
<br/>ƒ(x) = ƒ<sub>1</sub>
(x) + ƒ<sub>2</sub>
(x), where ƒ<sub>1</sub>
(x) is an even
function of ƒ<sub>2</sub>
(x) is an odd function. <br/>Then
ƒ<sub>1</sub>
(x + y) + ƒ<sub>1</sub>
(x – y) equals | [{"identifier": "A", "content": "2\u0192<sub>1</sub>\n(x)\u0192<sub>1</sub>\n(y)\n"}, {"identifier": "B", "content": "2\u0192<sub>1</sub>\n(x + y)\u0192<sub>1</sub>\n(x \u2013 y)"}, {"identifier": "C", "content": "2\u0192<sub>1</sub>\n(x)\u0192<sub>2</sub>\n(y)"}, {"identifier": "D", "content": "2\u0192<sub>1</sub>\n(x + y)\u0192<sub>2</sub>\n(x \u2013 y)"}] | ["A"] | null | f(x) = a<sup>x</sup>
<br><br>As f<sub>1</sub>(x) is even function then
<br><br>f<sub>1</sub>(x) = $${{{f\left( x \right) + f\left( { - x} \right)} \over 2}}$$
<br><br>= $${{{a^x} + {a^{ - x}}} \over 2}$$
<br><br>As f<sub>2</sub>(x) is odd function then
<br><br>f<sub>2</sub>(x) = $${{{f\left( x \right) - f\left( { - x} \right)} \over 2}}$$
<br><br>= $${{{a^x} - {a^{ - x}}} \over 2}$$
<br><br>Now,
<br><br>ƒ<sub>1</sub>
(x + y) + ƒ<sub>1</sub>
(x – y)
<br><br>= $${{{a^{x + y}} + {a^{ - \left( {x + y} \right)}} + {a^{x + y}} - {a^{ - \left( {x - y} \right)}}} \over 2}$$
<br><br>Also ƒ<sub>1</sub>
(x)ƒ<sub>1</sub>
(y) = $$\left( {{{{a^x} + {a^{ - x}}} \over 2}} \right)\left( {{{{a^y} + {a^{ - y}}} \over 2}} \right)$$
<br><br>= $${{{a^{x + y}} + {a^{x - y}} + {a^{ - x + y}} + {a^{ - x - y}}} \over 4}$$
<br><br>= $${{{{f_1}\left( {x + y} \right) + {f_2}\left( {x - y} \right)} \over 2}}$$
<br><br>$$ \therefore $$ ƒ<sub>1</sub>
(x + y) + ƒ<sub>1</sub>
(x – y) = 2ƒ<sub>1</sub>
(x)ƒ<sub>1</sub>
(y)
| mcq | jee-main-2019-online-8th-april-evening-slot |
lv3ve421 | maths | functions | even-and-odd-functions | <p>Let $$f(x)=\left\{\begin{array}{ccc}-\mathrm{a} & \text { if } & -\mathrm{a} \leq x \leq 0 \\ x+\mathrm{a} & \text { if } & 0< x \leq \mathrm{a}\end{array}\right.$$ where $$\mathrm{a}> 0$$ and $$\mathrm{g}(x)=(f(|x|)-|f(x)|) / 2$$. Then the function $$g:[-a, a] \rightarrow[-a, a]$$ is</p> | [{"identifier": "A", "content": "neither one-one nor onto.\n"}, {"identifier": "B", "content": "both one-one and onto.\n"}, {"identifier": "C", "content": "one-one.\n"}, {"identifier": "D", "content": "onto"}] | ["A"] | null | <p>$$\begin{aligned}
& f(x)=\left\{\begin{array}{l}
-a \quad \text { if }-a \leq x \leq 0 \\
x+a \quad \text { if } 0< x \leq a
\end{array}\right. \\
& f(|x|)=\left\{\begin{array}{cc}
-a & -a \leq|x| \leq 0 \\
|x|+a & \text { if } 0 < |x| \leq a
\end{array}\right.
\end{aligned}$$</p>
<p>$$|x|<0$$ is not possible, so</p>
<p>$$\begin{aligned}
& f(|x|)= \begin{cases}x+a & -a \leq x \leq 0 \\
-x+a & 0 < x \leq a\end{cases} \\
& |f(x)|= \begin{cases}a & -a \leq x \leq 0 \\
x+a & 0 < x \leq a\end{cases} \\
& h(x)= \begin{cases}-\frac{x}{2} & -a \leq x \leq 0 \\
0 & 0 < x \leq a\end{cases}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4k23pz/525fa5bf-d94c-4b57-ab0f-dd8a7e2507fb/1dc1b670-10ee-11ef-8836-ed8ff380bbe9/file-1lw4k23q0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4k23pz/525fa5bf-d94c-4b57-ab0f-dd8a7e2507fb/1dc1b670-10ee-11ef-8836-ed8ff380bbe9/file-1lw4k23q0.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Evening Shift Mathematics - Functions Question 8 English Explanation"></p>
<p>Neither one-one nor onto.</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
8TmUW15AHe2hgz5a | maths | functions | functional-equations | If $$f:R \to R$$ satisfies $$f$$(x + y) = $$f$$(x) + $$f$$(y), for all x, y $$ \in $$ R and $$f$$(1) = 7, then $$\sum\limits_{r = 1}^n {f\left( r \right)} $$ is | [{"identifier": "A", "content": "$${{7n\\left( {n + 1} \\right)} \\over 2}$$"}, {"identifier": "B", "content": "$${{7n} \\over 2}$$"}, {"identifier": "C", "content": "$${{7\\left( {n + 1} \\right)} \\over 2}$$"}, {"identifier": "D", "content": "$$7n + \\left( {n + 1} \\right)$$"}] | ["A"] | null | $$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right).$$
<br><br>Function should be $$f(x)=mx$$
<br><br>$$f\left( 1 \right) = 7;$$
<br><br>$$\therefore$$ $$m=7,$$ $$f\left( x \right) = 7x$$
<br><br>$$\sum\limits_{r = 1}^n {f\left( r \right)} = 7\sum\limits_1^n {r = {{7n\left( {n + 1} \right)} \over 2}} $$ | mcq | aieee-2003 |
0SD6FFBfLQleiCCW | maths | functions | functional-equations | A real valued function f(x) satisfies the functional equation
<br/><br/>f(x - y) = f(x)f(y) - f(a - x)f(a + y)
<br/><br/>where a is given constant and f(0) = 1, f(2a - x) is equal to | [{"identifier": "A", "content": "- f(x)"}, {"identifier": "B", "content": "f(x)"}, {"identifier": "C", "content": "f(a) + f(a - x)"}, {"identifier": "D", "content": "f(- x)"}] | ["A"] | null | $$f\left( {2a - x} \right) = f\left( {a - \left( {x - a} \right)} \right)$$
<br><br>$$ = f\left( a \right)f\left( {x - a} \right) - f\left( 0 \right)f\left( x \right)$$
<br><br>$$ = f\left( a \right)f\left( {x - a} \right) - f\left( x \right)$$
<br><br>$$ = - f\left( x \right)$$
<br><br>$$\left[ {} \right.$$ as $$x = 0,y = 0,f\left( 0 \right) = {f^2}\left( 0 \right) - {f^2}\left( a \right)$$
<br><br>$$\,\,\,\,\,\,\,\, \Rightarrow {f^2}\left( a \right) = 0 \Rightarrow f\left( a \right) = \left. 0 \right]$$
<br><br>$$ \Rightarrow f\left( {2a - x} \right) = - f\left( x \right)$$ | mcq | aieee-2005 |
l87cb3gu | maths | functions | functional-equations | If $f(x)+2 f\left(\frac{1}{x}\right)=3 x, x \neq 0$, and $\mathrm{S}=\{x \in \mathbf{R}: f(x)=f(-x)\}$; then $\mathrm{S}:$ | [{"identifier": "A", "content": "is an empty set."}, {"identifier": "B", "content": "contains exactly one element."}, {"identifier": "C", "content": "contains exactly two elements."}, {"identifier": "D", "content": "contains more than two elements."}] | ["C"] | null | We have, $f(x)+2 f\left(\frac{1}{x}\right)=3 x, \quad x \neq 0$
$\ldots$ (i)<br/><br/>
On replacing $x$ by $\frac{1}{x}$ in the above equation, we get<br/><br/>
$$
\begin{aligned}
& f\left(\frac{1}{x}\right)+2 f(x) =\frac{3}{x} \\\\
\Rightarrow & \,\, 2 f(x)+f\left(\frac{1}{x}\right) =\frac{3}{x} \,\,\,\,\,...(ii)
\end{aligned}
$$<br/><br/>
On multiplying Eq. (ii) by 2, we get<br/><br/>
$$
4 f(x)+2 f\left(\frac{1}{x}\right)=\frac{6}{x}\quad...(iii)
$$<br/><br/>
and subtracting Eq. (i) from Eq. (iii), we get<br/><br/>
$$[4 f(x)+2 f\left(\frac{1}{x}\right)] - [f(x)+2 f\left(\frac{1}{x}\right)]=\frac{6}{x} - 3x$$<br/><br/>
$\Rightarrow {3 f(x)=\frac{6}{x}-3 x}$<br/><br/>
$\Rightarrow f(x)=\frac{2}{x}-x$<br/><br/>
Now, consider $\quad f(x)=f(-x)$<br/><br/>
$$
\begin{aligned}
&\Rightarrow \frac{2}{x}-x =-\frac{2}{x}+x \\\\
&\Rightarrow \frac{4}{x} =2 x \\\\
&\Rightarrow 2 x^2 =4 \\\\
&\Rightarrow x^2 =2 \\\\
&\Rightarrow x =\pm \sqrt{2}
\end{aligned}
$$<br/><br/>
Hence, $S$ contains exactly two elements. | mcq | jee-main-2016-offline |
2v2CovVtbLHzzXns | maths | functions | functional-equations | Let $$a$$, b, c $$ \in R$$. If $$f$$(x) = ax<sup>2</sup> + bx + c is such that
<br/>$$a$$ + b + c = 3 and $$f$$(x + y) = $$f$$(x) + $$f$$(y) + xy, $$\forall x,y \in R,$$
<br/><br/>then $$\sum\limits_{n = 1}^{10} {f(n)} $$ is equal to | [{"identifier": "A", "content": "165"}, {"identifier": "B", "content": "190"}, {"identifier": "C", "content": "255"}, {"identifier": "D", "content": "330"}] | ["D"] | null | f(x) = ax<sup>2</sup> + bx + c
<br><br>f(1) = a + b + c = 3 $$ \Rightarrow $$ f (1) = 3
<br><br>Now f(x + y) = f(x) + f(y) + xy ...(1)
<br><br>Put x = y = 1 in eqn (1)
<br><br>f(2) = f(1) + f(1) + 1
<br><br>= 2f(1) + 1
<br><br>$$ \Rightarrow $$ f(2) = 7
<br><br>Similarly f(3) = 12
<br><br>f(4) = 18
<br><br>$$\sum\limits_{n = 1}^{10} {f(n)} $$ = 3 + 7 + 12 + 18 + 25 + 33 + 42 + 52 + 63 + 75 = 330 | mcq | jee-main-2017-offline |
7T8qHanHgmWJe870g3Bhs | maths | functions | functional-equations | If $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$, $$\left| x \right| < 1$$ then $$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ is equal to | [{"identifier": "A", "content": "2f(x<sup>2</sup>)"}, {"identifier": "B", "content": "2f(x)"}, {"identifier": "C", "content": "(f(x))<sup>2</sup>"}, {"identifier": "D", "content": "-2f(x)"}] | ["B"] | null | Given, $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$
<br><br>$$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ = $$\ln \left( {{{1 - {{2x} \over {1 + {x^2}}}} \over {1 + {{2x} \over {1 + {x^2}}}}}} \right)$$
<br><br>= $$\ln \left( {{{{x^2} - 2x + 1} \over {{x^2} + 2x + 1}}} \right)$$
<br><br>= $$\ln {\left( {{{1 - x} \over {1 + x}}} \right)^2}$$
<br><br>= $$2\ln \left( {{{1 - x} \over {1 + x}}} \right)$$
<br><br>= 2f(x) | mcq | jee-main-2019-online-8th-april-morning-slot |
vZDInSndsBCwo3mCTX18hoxe66ijvwp4erx | maths | functions | functional-equations | Let $$\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)} $$ where the function
ƒ satisfies
<br/>ƒ(x + y) = ƒ(x)ƒ(y) for all natural
numbers x, y and ƒ(1) = 2. then the natural
number 'a' is | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Given ƒ(1) = 2
<br><br> and ƒ(x + y) = ƒ(x)ƒ(y)
<br><br>When x = 1 and y = 1 then,
<br><br>ƒ(1 + 1) = ƒ(1)ƒ(1)
<br><br>$$ \Rightarrow $$ f(2) = (f(1))<sup>2</sup> = 2<sup>2</sup>
<br><br>Also when x = 2 and y = 1 then,
<br><br>ƒ(2 + 1) = ƒ(2)ƒ(1)
<br><br>$$ \Rightarrow $$ f(3) = 2<sup>3</sup>
<br><br>$$ \therefore $$ Similarly f(4) = 2<sup>4</sup>
<br>.
<br>.
<br>.
<br>.
<br><br>f(x) = 2<sup>x</sup>
<br><br>$$ \therefore $$ f(a + k) = 2<sup>a + k</sup>
<br><br>Now given,
<br><br>$$\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)} $$
<br><br>$$ \Rightarrow $$ $$\sum\limits_{k = 1}^{10} {{2^{a + k}}} $$ = $$16\left( {{2^{10}} - 1} \right)$$
<br><br>$$ \Rightarrow $$ $${2^{a + 1}} + {2^{a + 2}} + .... + {2^{a + 10}}$$ = $$16\left( {{2^{10}} - 1} \right)$$
<br><br>$$ \Rightarrow $$ $${2^a}\left[ {2 + {2^2} + .... + {2^{10}}} \right]$$ = $$16\left( {{2^{10}} - 1} \right)$$
<br><br>$$ \Rightarrow $$ $${2^a} \times {{2\left( {{2^{10}} + 1} \right)} \over {2 - 1}}$$ = $$16\left( {{2^{10}} - 1} \right)$$
<br><br>$$ \Rightarrow $$ $${2^a} = 8$$
<br><br>$$ \Rightarrow $$ $$a$$ = 3 | mcq | jee-main-2019-online-9th-april-morning-slot |
yAWIEwUDWnjs2sVq2ejMn | maths | functions | functional-equations | Let f<sub>k</sub>(x) = $${1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$ for k = 1, 2, 3, ... Then for all x $$ \in $$ R, the value of f<sub>4</sub>(x) $$-$$ f<sub>6</sub>(x) is equal to | [{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$${5 \\over {12}}$$"}, {"identifier": "C", "content": "$${{ - 1} \\over {12}}$$"}, {"identifier": "D", "content": "$${1 \\over {12}}$$"}] | ["D"] | null | f<sub>4</sub>(x) $$-$$ f<sub>6</sub>(x)
<br><br>= $${1 \over 4}$$ (sin<sup>4</sup> x + cos<sup>4</sup> x) $$-$$ $${1 \over 6}$$ (sin<sup>6</sup> x + cos<sup>6</sup> x)
<br><br>= $${1 \over 4}$$ (1$$-$$ $${1 \over 2}$$ sin<sup>2</sup> 2x) $$-$$ $${1 \over 6}$$ (1 $$-$$ $${3 \over 4}$$ sin<sup>2</sup> 2x) = $${1 \over 12}$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
kmmWFv4B3ZGx0zoZFq7k9k2k5khqkpe | maths | functions | functional-equations | Let a – 2b + c = 1.<br/><br/>
If $$f(x)=\left| {\matrix{
{x + a} & {x + 2} & {x + 1} \cr
{x + b} & {x + 3} & {x + 2} \cr
{x + c} & {x + 4} & {x + 3} \cr
} } \right|$$, then: | [{"identifier": "A", "content": "\u0192(50) = 1"}, {"identifier": "B", "content": "\u0192(\u201350) = \u20131"}, {"identifier": "C", "content": "\u0192(50) = \u2013501"}, {"identifier": "D", "content": "\u0192(\u201350) = 501"}] | ["A"] | null | R<sub>1</sub> $$ \to $$ R<sub>1</sub> + R<sub>3</sub> – 2R<sub>2</sub>
<br><br>f(x) = $$\left| {\matrix{
{a + c - 2b} & 0 & 0 \cr
{x + b} & {x + 3} & {x + 2} \cr
{x + c} & {x + 4} & {x + 3} \cr
} } \right|$$
<br><br>= (a + c – 2b) ((x + 3)<sup>2</sup> – (x + 2)(x + 4))
<br><br>= x<sup>2</sup> + 6x + 9 – x<sup>2</sup> – 6x – 8 = 1
<br><br>$$ \therefore $$ f(x) = 1
<br><br>$$ \Rightarrow $$ f(50) = 1 | mcq | jee-main-2020-online-9th-january-evening-slot |
oOLSAdvLWP0U3MIENGjgy2xukez67a9z | maths | functions | functional-equations | Let f : R $$ \to $$ R be a function which satisfies
<br/>f(x + y) = f(x) + f(y) $$\forall $$ x, y $$ \in $$ R. If f(1) = 2 and
<br/>g(n) = $$\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)} $$, n $$ \in $$ N then the value of n, for
which g(n) = 20, is : | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["C"] | null | Given f(1) = 2 ;
<br><br>f(x + y) = f(x) + f(y)
<br><br>When x = y = 1 $$ \Rightarrow $$
f(2) = 2 + 2 = 4
<br><br>When x = 2, y = 1 $$ \Rightarrow $$ f(3) = 4 + 2 = 6
<br><br> g(n) = $$\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)} $$
<br><br>= f(1) + f(2) +.........+ f(n - 1)
<br><br>= 2 + 4 + 6 + ......+ 2(n - 1)
<br><br>= 2 $$ \times $$ $$\frac{\left( n-1\right) \left( n\right) }{2} $$
<br><br>= n<sup>2</sup> - n
<br><br> Given g(n) = 20
<br><br>$$ \Rightarrow $$ n<sup>2</sup>– n = 20
<br><br>$$ \Rightarrow $$ n<sup>2</sup> – n – 20 = 0
<br><br>$$ \Rightarrow $$ n = 5 | mcq | jee-main-2020-online-2nd-september-evening-slot |
o26Zk0Ad7pZ3fy2Cu9jgy2xukfuvv0hp | maths | functions | functional-equations | If f(x + y) = f(x)f(y) and $$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$ , x, y $$ \in $$ N, where N is the set of all natural number, then the
value of
$${{f\left( 4 \right)} \over {f\left( 2 \right)}}$$ is : | [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over 9}$$"}, {"identifier": "C", "content": "$${1 \\over 3}$$"}, {"identifier": "D", "content": "$${4 \\over 9}$$"}] | ["D"] | null | f(x + y) = f(x)f(y)
<br><br>$$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$
<br><br>$$ \Rightarrow $$ f(1) + f(2) + f(3) + ........$$\infty $$ = 2 ....(1)
<br><br>On f(x + y) = f(x) f(y)
<br>* Put x = 1, y = 1
<br>f(2) = (f(1))<sup>2</sup>
<br>* Put x = 2, y = 1
<br>f(3) = f(2). f(1) = f((1))<sup>3</sup>
<br>* Put x = 2, y = 2
<br>f(4) = f((2))<sup>2</sup> = f((1))<sup>4</sup>
<br><br>Now put these values in equation (1)
<br><br> f(1) + f((1))<sup>2</sup> + f((1))<sup>3</sup> + ....... = 2
<br><br>$$ \Rightarrow $$ $${{f\left( 1 \right)} \over {1 - f\left( 1 \right)}}$$ = 2
<br><br>$$ \Rightarrow $$ f(1) = $${2 \over 3}$$
<br><br>Now f(2) = $${\left( {{2 \over 3}} \right)^2}$$
<br><br>and f(4) = $${\left( {{2 \over 3}} \right)^4}$$
<br><br>$$ \therefore $$ $${{f\left( 4 \right)} \over {f\left( 2 \right)}}$$
<br><br>= $${{{{\left( {{2 \over 3}} \right)}^4}} \over {{{\left( {{2 \over 3}} \right)}^2}}}$$ = $${4 \over 9}$$ | mcq | jee-main-2020-online-6th-september-morning-slot |
xRBLs2qkeW9u7OBLNTjgy2xukg4n20of | maths | functions | functional-equations | Suppose that a function f : R $$ \to $$ R satisfies<br/>
f(x + y) = f(x)f(y) for all x, y $$ \in $$ R and f(1) = 3.<br/> If $$\sum\limits_{i = 1}^n {f(i)} = 363$$ then n is equal to ________ . | [] | null | 5 | f(x + y) = f(x) f(y)
<br><br>put x = y = 1
<br>$$ \therefore $$ f(2) = (ƒ(1))<sup>2</sup>
= 3<sup>2</sup>
<br><br>put x = 2, y = 1
<br>$$ \therefore $$ f(3) = (ƒ(1))<sup>3</sup>
= 3<sup>3</sup>
<br><br>Similarly f(x) = 3<sup>x</sup>
<br>$$ \Rightarrow $$ f(i) = 3<sup>i</sup>
<br><br>Given, $$\sum\limits_{i = 1}^n {f(i)} = 363$$
<br><br>$$ \Rightarrow $$ 3 + 3<sup>2</sup>
+ 3<sup>3</sup> +.... + 3<sup>n</sup>
= 363
<br><br>$$ \Rightarrow $$ $${{3\left( {{3^n} - 1} \right)} \over {3 - 1}}$$ = 363
<br><br>$$ \Rightarrow $$ 3<sup>n</sup> - 1 = $${{363 \times 2} \over 3}$$ = 242
<br><br>$$ \Rightarrow $$ 3<sup>n</sup> = 243 = 3<sup>5</sup>
<br><br>$$ \Rightarrow $$ n = 5 | integer | jee-main-2020-online-6th-september-evening-slot |
k8qq4kBTBPpAjRvtRg1klrmmoe0 | maths | functions | functional-equations | If a + $$\alpha$$ = 1, b + $$\beta$$ = 2 and $$af(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x},x \ne 0$$, then the value of the expression $${{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}}$$ is __________. | [] | null | 2 | $$af(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x}$$ ........(i)<br><br>Replace $$x $$ with $$ {1 \over x}$$<br><br>$$af\left( {{1 \over x}} \right) + af(x) = {b \over x} + \beta x$$ ..... (ii)<br><br>(i) + (ii)<br><br>$$(a + \alpha )\left[ {f(x) + f\left( {{1 \over x}} \right)} \right] = \left( {x + {1 \over x}} \right)(b + \beta )$$<br><br>$${{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}} = {{\beta + b} \over {a + \alpha }} = {2 \over 1} = 2$$ | integer | jee-main-2021-online-24th-february-evening-slot |
TBP1H0CdJEYh9UShXc1klt7f0ou | maths | functions | functional-equations | A function f(x) is given by $$f(x) = {{{5^x}} \over {{5^x} + 5}}$$, then the sum of the series $$f\left( {{1 \over {20}}} \right) + f\left( {{2 \over {20}}} \right) + f\left( {{3 \over {20}}} \right) + ....... + f\left( {{{39} \over {20}}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${{{39} \\over 2}}$$"}, {"identifier": "B", "content": "$${{{19} \\over 2}}$$"}, {"identifier": "C", "content": "$${{{49} \\over 2}}$$"}, {"identifier": "D", "content": "$${{{29} \\over 2}}$$"}] | ["A"] | null | $$f(x) = {{{5^x}} \over {{5^x} + 5}}$$ ..... (i)<br><br>$$f(2 - x) = {{{5^{2 - x}}} \over {{5^{2 - x}} + 5}}$$<br><br>$$f(2 - x) = {5 \over {{5^x} + 5}}$$ .... (ii)<br><br>Adding equation (i) and (ii) <br><br>$$f(x) + f(2 - x) = 1$$<br><br>$$f\left( {{1 \over {20}}} \right) + f\left( {{{39} \over {20}}} \right) = 1$$<br><br>$$f\left( {{2 \over {20}}} \right) + f\left( {{{38} \over {20}}} \right) = 1$$<br><br>$$\eqalign{
& : \cr
& : \cr} $$<br><br>$$f\left( {{{19} \over {20}}} \right) + f\left( {{{21} \over {20}}} \right) = 1$$<br><br>and $$f\left( {{{20} \over {20}}} \right) = f(1) = {1 \over 2}$$<br><br>$$ \therefore $$ Sum = $$19 + {1 \over 2} = {{39} \over 2}$$ | mcq | jee-main-2021-online-25th-february-evening-slot |
GZkr2iQcFKkQTUhoWJ1kmm3zbjp | maths | functions | functional-equations | If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x<sup>3</sup>) + x g(x<sup>3</sup>) is divisible by x<sup>2</sup> + x + 1, then P(1) is equal to ___________. | [] | null | 0 | Given, p(x) = f(x<sup>3</sup>) + xg(x<sup>3</sup>)<br><br>We know, x<sup>2</sup> + x + 1 = (x $$-$$ $$\omega$$) (x $$-$$ $$\omega$$<sup>2</sup>)<br><br>Given, p(x) is divisible by x<sup>2</sup> + x + 1. So, roots of p(x) is $$\omega$$ and $$\omega$$<sup>2</sup>.<br><br>As root satisfy the equation,<br><br>So, put x = $$\omega$$<br><br>p($$\omega$$) = f($$\omega$$<sup>3</sup>) + $$\omega$$g($$\omega$$<sup>3</sup>) = 0<br><br>= f(1) + $$\omega$$g(1) = 0 [$$\omega$$<sup>3</sup> = 1]<br><br>= f(1) + $$\left( { - {1 \over 2} + {{i\sqrt 3 } \over 2}} \right)$$ g(1) = 0<br><br>$$ \Rightarrow $$ f(1) $$-$$ $${{g(1)} \over 2} + i\left( {{{\sqrt 3 g(1)} \over 2}} \right)$$ = 0 + i0<br><br>Comparing both sides, we get<br><br>f(1) $$-$$ $${{g(1)} \over 2}$$ = 0<br><br>and $${{{\sqrt 3 } \over 2}g(1) = 0}$$ $$ \Rightarrow $$ g(1) = 0<br><br>So, f(1) = 0<br><br>Now, p(1) = f(1) + 1 . g(1) = 0 + 0 = 0 | integer | jee-main-2021-online-18th-march-evening-shift |
1krxhvhxq | maths | functions | functional-equations | Let f : R $$\to$$ R be defined as $$f(x + y) + f(x - y) = 2f(x)f(y),f\left( {{1 \over 2}} \right) = - 1$$. Then, the value of $$\sum\limits_{k = 1}^{20} {{1 \over {\sin (k)\sin (k + f(k))}}} $$ is equal to : | [{"identifier": "A", "content": "cosec<sup>2</sup>(21) cos(20) cos(2)"}, {"identifier": "B", "content": "sec<sup>2</sup>(1) sec(21) cos(20)"}, {"identifier": "C", "content": "cosec<sup>2</sup>(1) cosec(21) sin(20)"}, {"identifier": "D", "content": "sec<sup>2</sup>(21) sin(20) sin(2)"}] | ["C"] | null | f(x) = cos$$\lambda$$x<br><br>$$\because$$ $$f\left( {{1 \over 2}} \right)$$ = $$-$$1<br><br>So, $$-$$1 = cos$${\lambda \over 2}$$<br><br>$$\Rightarrow$$ $$\lambda$$ = 2$$\pi$$<br><br>Thus f(x) = cos2$$\pi$$x<br><br>Now k is natural number<br><br>Thus f(k) = 1<br><br>$$\sum\limits_{k = 1}^{20} {{1 \over {\sin k\sin (k + 1)}} = {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {\left[ {{{\sin \left( {(k + 1) - k} \right)} \over {\sin k.\sin (k + 1)}}} \right]} } $$<br><br>$$ = {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {(\cot k - \cot (k + 1)} $$)<br><br>$$ = {{\cot 1 - \cot 21} \over {\sin 1}} = \cos e{c^2}1\cos ec(21).\sin 20$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1ks0d47kv | maths | functions | functional-equations | Let S = {1, 2, 3, 4, 5, 6, 7}. Then the number of possible functions f : S $$\to$$ S <br/>such that f(m . n) = f(m) . f(n) for every m, n $$\in$$ S and m . n $$\in$$ S is equal to _____________. | [] | null | 490 | F(mn) = f(m) . f(n)<br><br>Put m = 1 f(n) = f(1) . f(n) $$\Rightarrow$$ f(1) = 1<br><br>Put m = n = 2<br><br>$$f(4) = f(2).f(2)\left\{ \matrix{
f(2) = 1 \Rightarrow f(4) = 1 \hfill \cr
or \hfill \cr
f(2) = 2 \Rightarrow f(4) = 4 \hfill \cr} \right.$$<br><br>Put m = 2, n = 3<br><br>$$f(6) = f(2).f(3)\left\{ \matrix{
when\,f(2) = 1 \hfill \cr
f(3) = 1\,to\,7 \hfill \cr
\hfill \cr
f(2) = 2 \hfill \cr
f(3) = 1\,or\,2\,or\,3 \hfill \cr} \right.$$<br><br>f(5), f(7) can take any value <br><br>Total = (1 $$\times$$ 1 $$\times$$ 7 $$\times$$ 1 $$\times$$ 7 $$\times$$ 1 $$\times$$ 7) + (1 $$\times$$ 1 $$\times$$ 3 $$\times$$ 1 $$\times$$ 7 $$\times$$ 1 $$\times$$ 7)<br><br>= 490 | integer | jee-main-2021-online-27th-july-morning-shift |
1ktk52hwc | maths | functions | functional-equations | Let f : N $$\to$$ N be a function such that f(m + n) = f(m) + f(n) for every m, n$$\in$$N. If f(6) = 18, then f(2) . f(3) is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "36"}] | ["B"] | null | f(m + n) = f(m) + f(n)<br><br>Put m = 1, n = 1<br><br>f(2) = 2f(1)<br><br>Put m = 2, n = 1<br><br>f(3) = f(2) + f(1) = 3f(1)<br><br>Put m = 3, n = 3<br><br>f(6) = 2f(3) $$\Rightarrow$$ f(3) = 9<br><br>$$\Rightarrow$$ f(1) = 3, f(2) = 6<br><br>f(2) . f(3) = 6 $$\times$$ 9 = 54 | mcq | jee-main-2021-online-31st-august-evening-shift |
1l546a0kk | maths | functions | functional-equations | <p>Let c, k $$\in$$ R. If $$f(x) = (c + 1){x^2} + (1 - {c^2})x + 2k$$ and $$f(x + y) = f(x) + f(y) - xy$$, for all x, y $$\in$$ R, then the value of $$|2(f(1) + f(2) + f(3) + \,\,......\,\, + \,\,f(20))|$$ is equal to ____________.</p> | [] | null | 3395 | <p>f(x) is polynomial</p>
<p>Put y = 1/x in given functional equation we get</p>
<p>$$f\left( {x + {1 \over x}} \right) = f(x) + f\left( {{1 \over x}} \right) - 1$$</p>
<p>$$ \Rightarrow (c + 1){\left( {x + {1 \over x}} \right)^2} + (1 - {c^2})\left( {x + {1 \over x}} \right) + 2K$$</p>
<p>$$ = (c + 1){x^2} + (1 - {c^2})x + 2K + (c + 1){1 \over {{x^2}}} + (1 - {c^2}){1 \over x} + 2K - 1$$</p>
<p>$$ \Rightarrow 2(c + 1) = 2K - 1$$ ..... (1)</p>
<p>and put $$x = y = 0$$ we get</p>
<p>$$f(0) = 2 + f(0) - 0 \Rightarrow f(0) = 0 \Rightarrow k = 0$$</p>
<p>$$\therefore$$ $$k = 0$$ and $$2c = - 3 \Rightarrow c = - 3/2$$</p>
<p>$$f(x) = - {{{x^2}} \over 2} - {{5x} \over 4} = {1 \over 4}(5x + 2{x^2})$$</p>
<p>$$\left| {2\sum\limits_{i = 1}^{20} {f(i)} } \right| = \left| {{{ - 2} \over 4}\left( {{{5.20.21} \over 2} + {{2.20.21.41} \over 6}} \right)} \right|$$</p>
<p>$$ = \left| {{{ - 1} \over 2}(2730 + 5740)} \right|$$</p>
<p>$$ = \left| { - {{6790} \over 2}} \right| = 3395$$.</p> | integer | jee-main-2022-online-29th-june-morning-shift |
1l57ovytd | maths | functions | functional-equations | <p>Let f : R $$\to$$ R be a function defined by $$f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$$. Then $$f\left( {{1 \over {100}}} \right) + f\left( {{2 \over {100}}} \right) + f\left( {{3 \over {100}}} \right) + \,\,\,.....\,\,\, + \,\,\,f\left( {{{99} \over {100}}} \right)$$ is equal to ______________.</p> | [] | null | 99 | <p>Given,</p>
<p>$$f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$$</p>
<p>$$\therefore$$ $$f(1 - x) = {{2{e^{2(1 - x)}}} \over {{e^{2(1 - x)}} + e}}$$</p>
<p>$$ = {{2\,.\,{{{e^2}} \over {{e^{2x}}}}} \over {{{{e^2}} \over {{e^{2x}}}} + e}}$$</p>
<p>$$ = {{2{e^2}} \over {{e^2} + {e^{2x}}\,.\,e}}$$</p>
<p>$$ = {{2{e^2}} \over {e(e + {e^{2x}})}}$$</p>
<p>$$ = {{2e} \over {e + {e^{2x}}}}$$</p>
<p>$$\therefore$$ $$f(x) + f(1 - x) = {{2{e^{2x}}} \over {{e^{2x}} + e}} + {{2e} \over {{e^{2x}} + e}}$$</p>
<p>$$ = {{2({e^{2x}} + e)} \over {{e^{2x}} + e}}$$</p>
<p>$$ = 2$$ ...... (1)</p>
<p>Now,</p>
<p>$$f\left( {{1 \over {100}}} \right) + f\left( {{{99} \over {100}}} \right)$$</p>
<p>$$ = f\left( {{1 \over {100}}} \right) + f\left( {1 - {1 \over {100}}} \right)$$</p>
<p>$$ = 2$$ [as $$f(x) + f(1 - x) = 2$$]</p>
<p>Similarly,</p>
<p>$$f\left( {{2 \over {100}}} \right) + f\left( {1 - {2 \over {100}}} \right) = 2$$</p>
<p>$$ \vdots $$</p>
<p>$$f\left( {{{49} \over {100}}} \right) + f\left( {1 - {{49} \over {100}}} \right) = 2$$</p>
<p>$$\therefore$$ Total sum $$ = 49 \times 2$$</p>
<p>Remaining term $$ = f\left( {{{50} \over {100}}} \right) = f\left( {{1 \over 2}} \right)$$</p>
<p>Put $$x = {1 \over 2}$$ in equation (1), we get</p>
<p>$$f\left( {{1 \over 2}} \right) + f\left( {1 - {1 \over 2}} \right) = 2$$</p>
<p>$$ \Rightarrow 2f\left( {{1 \over 2}} \right) = 2$$</p>
<p>$$ \Rightarrow f\left( {{1 \over 2}} \right) = 1$$</p>
<p>$$\therefore$$ Sum $$ = 49 \times 2 + 1 = 99$$</p> | integer | jee-main-2022-online-27th-june-morning-shift |
1l5ahps1l | maths | functions | functional-equations | <p>Let f : N $$\to$$ R be a function such that $$f(x + y) = 2f(x)f(y)$$ for natural numbers x and y. If f(1) = 2, then the value of $$\alpha$$ for which</p>
<p>$$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)} $$</p>
<p>holds, is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <p>Given,</p>
<p>$$f(x + y) = 2f(x)f(y)$$</p>
<p>and $$f(1) = 2$$</p>
<p>For x = 1 and y = 1,</p>
<p>$$f(1 + 1) = 2f(1)f(1)$$</p>
<p>$$ \Rightarrow f(2) = 2{\left( {f(1)} \right)^2} = 2{(2)^2} = {2^3}$$</p>
<p>For x = 1, y = 2,</p>
<p>$$f(1 + 2) = 2f(1)y(2)$$</p>
<p>$$ \Rightarrow f(3) = 2\,.\,2\,.\,{2^3} = {2^5}$$</p>
<p>For x = 1, y = 3,</p>
<p>$$f(1 + 3) = 2f(1)f(3)$$</p>
<p>$$ \Rightarrow f(4) = 2\,.\,2\,.\,{2^5} = {2^7}$$</p>
<p>For x = 1, y = 4,</p>
<p>$$f(1 + 4) = 2f(1)f(4)$$</p>
<p>$$ \Rightarrow f(5) = 2\,.\,2\,.\,{2^7} = {2^9}$$ ..... (1)</p>
<p>Also given</p>
<p>$$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)} $$</p>
<p>$$ \Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,...\,\, + \,\,f(\alpha + 10) = {{512} \over 3}({2^{20}} - 1)$$</p>
<p>$$ \Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,....\,\, + f(\alpha + 10) = {{{2^9}\left( {{{({2^2})}^{10}} - 1} \right)} \over {{2^2} - 1}}$$</p>
<p>This represent a G.P with first term = 2<sup>9</sup> and common ratio = 2<sup>2</sup></p>
<p>$$\therefore$$ First term $$ = f(\alpha + 1) = {2^9}$$ ..... (2)</p>
<p>From equation (1), $$f(5) = {2^9}$$</p>
<p>$$\therefore$$ From (1) and (2), we get</p>
<p>$$f(\alpha + 1) = {2^9} = f(5)$$</p>
<p>$$ \Rightarrow f(\alpha + 1) = f(5)$$</p>
<p>$$ \Rightarrow f(\alpha + 1) = f(4 + 1)$$</p>
<p>Comparing both sides we get,</p>
<p>$$\alpha = 4$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l6jeicx4 | maths | functions | functional-equations | <p> Let $$f(x)=2 x^{2}-x-1$$ and $$\mathrm{S}=\{n \in \mathbb{Z}:|f(n)| \leq 800\}$$. Then, the value of $$\sum\limits_{n \in S} f(n)$$ is equal to ___________.</p> | [] | null | 10620 | <p>$$\because$$ $$\left| {f(n)} \right| \le 800$$</p>
<p>$$ \Rightarrow - 800 \le 2{n^2} - n - 1 \le 800$$</p>
<p>$$ \Rightarrow 2{n^2} - n - 801 \le 0$$</p>
<p>$$\therefore$$ $$n \in \left[ {{{ - \sqrt {6409} + 1} \over 4},{{\sqrt {6409} + 1} \over 4}} \right]$$ and $$n \in z$$</p>
<p>$$\therefore$$ $$n = - 19, - 18, - 17,\,..........,\,19,20.$$</p>
<p>$$\therefore$$ $$\sum {\left( {2{x^2} - x - 1} \right) = 2\sum {{x^2} - \sum {x - \sum 1 } } } $$.</p>
<p>$$ = 2\,.\,2\,.\,\left( {{1^2} + {2^2}\, + \,...\, + \,{{19}^2}} \right) + 2\,.\,{20^2} - 20 - 40$$</p>
<p>$$ = 10620$$</p> | integer | jee-main-2022-online-27th-july-morning-shift |
1l6m6qe90 | maths | functions | functional-equations | <p>For $$\mathrm{p}, \mathrm{q} \in \mathbf{R}$$, consider the real valued function $$f(x)=(x-\mathrm{p})^{2}-\mathrm{q}, x \in \mathbf{R}$$ and $$\mathrm{q}>0$$. Let $$\mathrm{a}_{1}$$, $$\mathrm{a}_{2^{\prime}}$$ $$\mathrm{a}_{3}$$ and $$\mathrm{a}_{4}$$ be in an arithmetic progression with mean $$\mathrm{p}$$ and positive common difference. If $$\left|f\left(\mathrm{a}_{i}\right)\right|=500$$ for all $$i=1,2,3,4$$, then the absolute difference between the roots of $$f(x)=0$$ is ___________.</p> | [] | null | 50 | <p>$$\because$$ $${a_1},{a_2},{a_3},{a_4}$$</p>
<p>$$\therefore$$ $${a_2} = p - 3d,\,{a_2} = p - d,\,{a_3} = p + d$$ and $${a_4} = p + 3d$$</p>
<p>Where $$d > 0$$</p>
<p>$$\because$$ $$\left| {f({a_i})} \right| = 500$$</p>
<p>$$ \Rightarrow |9{d^2} - q| = 500$$</p>
<p>and $$|{d^2} - q| = 500$$ ..... (i)</p>
<p>either $$9{d^2} - q = {d^2} - q$$</p>
<p>$$ \Rightarrow d = 0$$ not acceptable</p>
<p>$$\therefore$$ $$9{d^2} - q = q - {d^2}$$</p>
<p>$$\therefore$$ $$5{d^2} - q = 0$$ ..... (ii)</p>
<p>Roots of $$f(x) = 0$$ are $$p + \sqrt q $$ and $$p - \sqrt q $$</p>
<p>$$\therefore$$ absolute difference between roots $$ = \left| {2\sqrt q } \right| = 50$$</p> | integer | jee-main-2022-online-28th-july-morning-shift |
1l6nld95e | maths | functions | functional-equations | <p>$$
\text { Let } f(x)=a x^{2}+b x+c \text { be such that } f(1)=3, f(-2)=\lambda \text { and } $$ $$f(3)=4$$. If $$f(0)+f(1)+f(-2)+f(3)=14$$, then $$\lambda$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$4"}, {"identifier": "B", "content": "$$\\frac{13}{2}$$"}, {"identifier": "C", "content": "$$\\frac{23}{2}$$"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <p>$$f(1) = a + b + c = 3$$ ..... (i)</p>
<p>$$f(3) = 9a + 3b + c = 4$$ .... (ii)</p>
<p>$$f(0) + f(1) + f( - 2) + f(3) = 14$$</p>
<p>OR $$c + 3 + (4a - 2b + c) + 4 = 14$$</p>
<p>OR $$4a - 2b + 2c = 7$$ ..... (iii)</p>
<p>From (i) and (ii) $$8a + 2b = 1$$ ..... (iv)</p>
<p>From (iii) $$ - (2) \times $$ (i)</p>
<p>$$ \Rightarrow 2a - 4b = 1$$ ..... (v)</p>
<p>From (iv) and (v) $$a = {1 \over 6},\,b = {{ - 1} \over 6}$$ and $$c = 3$$</p>
<p>$$f( - 2) = 4a - 2b + c$$</p>
<p>$$ = {4 \over 6} + {2 \over 6} + 3 = 4$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldo6xo14 | maths | functions | functional-equations | <p>Let $$f:\mathbb{R}-{0,1}\to \mathbb{R}$$ be a function such that $$f(x)+f\left(\frac{1}{1-x}\right)=1+x$$. Then $$f(2)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{9}{4}$$"}, {"identifier": "B", "content": "$$\\frac{7}{4}$$"}, {"identifier": "C", "content": "$$\\frac{7}{3}$$"}, {"identifier": "D", "content": "$$\\frac{9}{2}$$"}] | ["A"] | null | $\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)=1+\mathrm{x} \\\\ & \mathrm{x}=2 \Rightarrow \mathrm{f}(2)+\mathrm{f}(-1)=3 ........(1) \\\\ & \mathrm{x}=-1 \Rightarrow \mathrm{f}(-1)+\mathrm{f}\left(\frac{1}{2}\right)=0 .........(2) \\\\ & \mathrm{x}=\frac{1}{2} \Rightarrow \mathrm{f}\left(\frac{1}{2}\right)+\mathrm{f}(2)=\frac{3}{2} ........(3) \\\\ & (1)+(3)-(2) \Rightarrow 2 \mathrm{f}(2)=\frac{9}{2} \\\\ & \therefore \mathrm{f}(2)=\frac{9}{4}\end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift |
ldo9i6yu | maths | functions | functional-equations | The absolute minimum value, of the function
<br/><br/>$f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right]$,
<br/><br/>where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$, is : | [{"identifier": "A", "content": "$\\frac{3}{4}$"}, {"identifier": "B", "content": "$\\frac{3}{2}$"}, {"identifier": "C", "content": "$\\frac{1}{4}$"}, {"identifier": "D", "content": "$\\frac{5}{4}$"}] | ["A"] | null | $\mathrm{f}(\mathrm{x})=\left|\mathrm{x}^{2}-\mathrm{x}+1\right|+\left[\mathrm{x}^{2}-\mathrm{x}+1\right] ; \mathrm{x} \in[-1,2]$
<br/><br/>Let $g(x)=x^{2}-x+1$
<br/><br/>$$
=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}
$$
<br/><br/>$$
\because\left|\mathrm{x}^{2}-\mathrm{x}+1\right| \text { and }\left[\mathrm{x}^{2}-\mathrm{x}+1\right]
$$
<br/><br/>Both have minimum value at $\mathrm{x}=1 / 2$
<br/><br/>$\Rightarrow$ Minimum $\mathrm{f}(\mathrm{x})=\frac{3}{4}+0$
<br/><br/>$=\frac{3}{4}$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldon97rl | maths | functions | functional-equations | <p>Let $$f(x) = \left| {\matrix{
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr
} } \right|,\,x \in \left[ {{\pi \over 6},{\pi \over 3}} \right]$$. If $$\alpha$$ and $$\beta$$ respectively are the maximum and the minimum values of $$f$$, then</p> | [{"identifier": "A", "content": "$${\\alpha ^2} - {\\beta ^2} = 4\\sqrt 3 $$"}, {"identifier": "B", "content": "$${\\beta ^2} - 2\\sqrt \\alpha = {{19} \\over 4}$$"}, {"identifier": "C", "content": "$${\\beta ^2} + 2\\sqrt \\alpha = {{19} \\over 4}$$"}, {"identifier": "D", "content": "$${\\alpha ^2} + {\\beta ^2} = {9 \\over 2}$$"}] | ["B"] | null | $$f(x) = \left| {\matrix{
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr
} } \right|$$
<br/><br/>$C_{1} \rightarrow C_{1}+C_{2}+C_{3}$
<br/><br/>$$
\begin{aligned}
& = (2+\sin 2 x)\left|\begin{array}{ccc}1 & \cos ^{2} x & \sin 2 x \\1 & 1+\cos ^{2} x & \sin 2 x \\1 & \cos ^{2} x & 1+\sin 2 x\end{array}\right| \\\\
& R_{2} \rightarrow R_{2} \rightarrow R_{1} ; R_{3} \rightarrow R_{3} \rightarrow R_{1} \\\\
& = (2+\sin 2 x)\left|\begin{array}{ccc}1 & \cos ^{2} x & \sin 2 x \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right| \\\\
& f(x)=2+\sin 2 x ; x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right] \\\\
& f(x)_{\max }=2+1=3 \text { for } x=\frac{\pi}{4} \\\\
& f(x)_{\min }=2+\frac{\sqrt{3}}{2} \text { for } x=\frac{\pi}{6}, \frac{\pi}{3} \\\\
& \beta^{2}-2 \sqrt{\alpha}=4+\frac{3}{4}+2 \sqrt{3}-2 \sqrt{3} \\\\
& =\frac{19}{4}
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
ldqyb0zm | maths | functions | functional-equations | Let $A=\{1,2,3,5,8,9\}$. Then the number of possible functions $f: A \rightarrow A$ such that $f(m \cdot n)=f(m) \cdot f(n)$ for every $m, n \in A$ with $m \cdot n \in A$ is equal to ___________. | [] | null | 432 | <p>$$f(1.n)=f(1).f(n)\Rightarrow f(1)=1$$.</p>
<p>$$f(3.3)=(f(3))^2$$</p>
<p>Hence, the possibilities for $$(t(3),(9))$$ are $$(1,1)$$ and $$(3,9)$$.</p>
<p>Other three i.e. $$f(2),f(5),f(8)$$</p>
<p>Can be chosen in 6$$^3$$ ways.</p>
<p>Hence, total number of functions</p>
<p>$$6^3\times2=432$$</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldsf1d3h | maths | functions | functional-equations | <p>Consider a function $$f:\mathbb{N}\to\mathbb{R}$$, satisfying $$f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2$$ with $$f(1)=1$$. Then $$\frac{1}{f(2022)}+\frac{1}{f(2028)}$$ is equal to</p> | [{"identifier": "A", "content": "8000"}, {"identifier": "B", "content": "8400"}, {"identifier": "C", "content": "8100"}, {"identifier": "D", "content": "8200"}] | ["C"] | null | <p>$$f(1) + 2f(2) + 3f(3)\, + \,...\, + \,nf(n) = n(n + 1) + (n)$$ ..... (i)</p>
<p>$$n \to n + 1$$</p>
<p>$$f(1) + 2f(2)\, + \,...\, + \,(n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1)$$ ...... (ii)</p>
<p>(i) and (ii) gives</p>
<p>$$3f(3) - 2f(2) = 0$$</p>
<p>$$4f(4) - 3f(3) = 0$$</p>
<p>$$ \vdots $$</p>
<p>$$(n + 1)f(n + 1) - nf(n) = 0$$</p>
<p>$$ \Rightarrow f(n + 1) = {{2f(2)} \over {n + 1}}$$</p>
<p>$$f(n) = {1 \over {2n}}$$</p>
<p>$${1 \over {f(2022)}} + {1 \over {f(2028)}} = 8100$$</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldswo1xf | maths | functions | functional-equations | <p>Suppose $$f$$ is a function satisfying $$f(x + y) = f(x) + f(y)$$ for all $$x,y \in N$$ and $$f(1) = {1 \over 5}$$. If $$\sum\limits_{n = 1}^m {{{f(n)} \over {n(n + 1)(n + 2)}} = {1 \over {12}}} $$, then $$m$$ is equal to __________.</p> | [] | null | 10 | $\because f(1)=\frac{1}{5} ~\therefore f(2)=f(1)+f(1)=\frac{2}{5}$
<br/><br/>
$f(2)=\frac{2}{5} \quad\quad f(3)=f(2)+f(1)=\frac{3}{5}$
<br/><br/>
$f(3)=\frac{3}{5}$
<br/><br/>
$\therefore \sum\limits_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}$
<br/><br/>
$=\frac{1}{5} \sum\limits_{n=1}^{m}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
<br/><br/>
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$
<br/><br/>
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}$
<br/><br/>
$\therefore m=10$ | integer | jee-main-2023-online-29th-january-morning-shift |
1ldu5gt5g | maths | functions | functional-equations | <p>Let $$f:\mathbb{R}\to\mathbb{R}$$ be a function defined by $$f(x) = {\log _{\sqrt m }}\{ \sqrt 2 (\sin x - \cos x) + m - 2\} $$, for some $$m$$, such that the range of $$f$$ is [0, 2]. Then the value of $$m$$ is _________</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "2"}] | ["C"] | null | We know that $\sin x-\cos x \in[-\sqrt{2}, \sqrt{2}]$
<br/><br/>
$$
\begin{aligned}
& \log _{\sqrt{M}}(\sqrt{2}(\sin x-\cos ) +M-2) \\\\
&\quad\quad\in {\left[\log _{\sqrt{M}}(M-4), \log _{\sqrt{M}} M\right] }
\end{aligned}
$$
<br/><br/>
$\Rightarrow \log _{\sqrt{M}}(M-4)=0 \Rightarrow M=5$
| mcq | jee-main-2023-online-25th-january-evening-shift |
1ldu5xulj | maths | functions | functional-equations | <p>Let $$f(x) = 2{x^n} + \lambda ,\lambda \in R,n \in N$$, and $$f(4) = 133,f(5) = 255$$. Then the sum of all the positive integer divisors of $$(f(3) - f(2))$$ is</p> | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "58"}, {"identifier": "C", "content": "61"}, {"identifier": "D", "content": "59"}] | ["A"] | null | $f(x)=2 x^{n}+\lambda, \lambda \in \mathbb{R}, n \in \mathbb{N}$
<br/><br/>
$f(4)=2 \cdot 4^{n}+\lambda=133, f(5)=2 \cdot 5^{n}+\lambda=255$
<br/><br/>
$f(5)-f(4)=2 \cdot\left(5^{n} \cdot 4^{n}\right)=122 \Rightarrow n=3$
<br/><br/>
$\Rightarrow f(3)-f(2)=2 \cdot\left(3^{n} \cdot 2^{n}\right)=2 \cdot\left(3^{3}-2^{3}\right)=2 \times 19$
<br/><br/>
Required sum $=1+2+19+38=60$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldwvz91z | maths | functions | functional-equations | <p>Let $$f(x)$$ be a function such that $$f(x+y)=f(x).f(y)$$ for all $$x,y\in \mathbb{N}$$. If $$f(1)=3$$ and $$\sum\limits_{k = 1}^n {f(k) = 3279} $$, then the value of n is</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["B"] | null | $$
\begin{aligned}
& \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{N}, \mathrm{f}(1)=3 \\\\
& \mathrm{f}(2)=\mathrm{f}^2(1)=3^2 \\\\
& \mathrm{f}(3)=\mathrm{f}(1) \mathrm{f}(2)=3^3 \\\\
& \mathrm{f}(4)=3^4 \\\\
& \mathrm{f}(\mathrm{k})=3^{\mathrm{k}} \\\\
& \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}(\mathrm{k})=3279 \\\\
& \mathrm{f}(1)+\mathrm{f}(2)+\mathrm{f}(3)+\ldots \ldots \ldots+\mathrm{f}(\mathrm{k})=3279 \\\\
& 3+3^2+3^3+\ldots \ldots \ldots 3^{\mathrm{k}}=3279 \\\\
& \frac{3\left(3^{\mathrm{k}}-1\right)}{3-1}=3279 \\\\
& \frac{3^{\mathrm{k}}-1}{2}=1093 \\\\
& 3^{\mathrm{k}}-1=2186 \\\\
& 3^{\mathrm{k}}=2187 \\\\
& \text{So, } \mathrm{k}=7
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldwwwfha | maths | functions | functional-equations | <p>If $$f(x) = {{{2^{2x}}} \over {{2^{2x}} + 2}},x \in \mathbb{R}$$, then $$f\left( {{1 \over {2023}}} \right) + f\left( {{2 \over {2023}}} \right)\, + \,...\, + \,f\left( {{{2022} \over {2023}}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "2011"}, {"identifier": "B", "content": "2010"}, {"identifier": "C", "content": "1010"}, {"identifier": "D", "content": "1011"}] | ["D"] | null | $$
\begin{aligned}
& f(x)=\frac{4^x}{4^x+2} \\\\
& f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2} \\\\
& =\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)} \\\\
& =\frac{4^x}{4^{\mathrm{x}}+2}+\frac{2}{2+4^{\mathrm{x}}} \\\\
& =1 \\\\
& \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(1-\mathrm{x})=1 \\\\
& \text { Now } \mathrm{f}\left(\frac{1}{2023}\right)+\mathrm{f}\left(\frac{2}{2023}\right)+\mathrm{f}\left(\frac{3}{2023}\right)+\ldots \ldots .+ \\\\
& \ldots \ldots \ldots . .+\mathrm{f}\left(1-\frac{3}{2023}\right)+\mathrm{f}\left(1-\frac{2}{2023}\right)+\mathrm{f}\left(1-\frac{1}{2023}\right)
\end{aligned}
$$<br/><br/>
Now sum of terms equidistant from beginning and end is 1<br/><br/>
$$
\begin{aligned}
& \text { Sum }=1+1+1+\ldots \ldots \ldots+1 \text { (1011 times }) \\\\
& =1011
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-evening-shift |
1lgxszp3h | maths | functions | functional-equations | <p>If $$f(x) = {{(\tan 1^\circ )x + {{\log }_e}(123)} \over {x{{\log }_e}(1234) - (\tan 1^\circ )}},x > 0$$, then the least value of $$f(f(x)) + f\left( {f\left( {{4 \over x}} \right)} \right)$$ is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "8"}] | ["B"] | null | Given that $f(x)=\frac{\left(\tan 1^{\circ}\right) x+\log _e(123)}{x \log _e(1234)-\left(\tan 1^{\circ}\right)}$
<br/><br/>Let us consider a similar function of $(x)$,
<br/><br/>$\therefore f(x)=\frac{A x+B}{C x-A}$
<br/><br/>$\text { Now, } $
<br/><br/>$$
\begin{aligned}
&f(f(x)) =\frac{A\left(\frac{A x+B}{C x-A}\right)+B}{C\left(\frac{A x+B}{C x-A}\right)-A} \\\\
& =\frac{A^2 x+A B+B C x-A B}{A C x+B C-A C x+A^2} \\\\
& =\frac{x\left(A^2+B C\right)}{\left(B C+A^2\right)}=x
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \therefore \quad f(f(x))=x \\\\
& \text { Similarly, } f\left(f\left(\frac{4}{x}\right)\right)=\frac{4}{x} \\\\
& \text { Apply, AM } \geq \text { GM } \\\\
& \left(x+\frac{4}{x}\right) \geq 2 \sqrt{x \cdot \frac{4}{x}}=4(x>0) \\\\
& \Rightarrow f(f(x))+f\left(f\left(\frac{4}{x}\right)\right) \geq 4
\end{aligned}
$$
<br/><br/>Hence, the least value is 4 . | mcq | jee-main-2023-online-10th-april-morning-shift |
luy9clgk | maths | functions | functional-equations | <p>If a function $$f$$ satisfies $$f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})$$ for all $$\mathrm{m}, \mathrm{n} \in \mathbf{N}$$ and $$f(1)=1$$, then the largest natural number $$\lambda$$ such that $$\sum_\limits{\mathrm{k}=1}^{2022} f(\lambda+\mathrm{k}) \leq(2022)^2$$ is equal to _________.</p> | [] | null | 1010 | <p>$$\begin{aligned}
& f(m+n)=f(m)+f(n) \\
& f(x)=k x \\
& \because f(1)=1 \\
& \Rightarrow k=1 \\
& \Rightarrow f(x)=x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \sum_{k=1}^{2022} f(\lambda+k)=\sum_{k=1}^{2022}(\lambda+k)=\underbrace{\lambda+\lambda+\ldots+\lambda}_{2022}+(1+2+\ldots+2022) \\
& =2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^2 \\
& \Rightarrow \lambda \leq \frac{2021}{2} \\
& \text { largest } \lambda=1010
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-morning-shift |
lv7v47tt | maths | functions | functional-equations | <p>If $$S=\{a \in \mathbf{R}:|2 a-1|=3[a]+2\{a \}\}$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$ and $$\{t\}$$ represents the fractional part of $$t$$, then $$72 \sum_\limits{a \in S} a$$ is equal to _________.</p> | [] | null | 18 | <p>$$\begin{aligned}
& S:\{a \in R:|2 a-1|=3[a]+2\{a\}\} \\
& |2 a-1|=3[a]+2(a-[a]) \\
& |2 a-1|=[a]+2 a
\end{aligned}$$</p>
<p>Case I: If $$0 < a < \frac{1}{2}$$</p>
<p>$$\begin{aligned}
& 1-2 a=0+2 a \\
& \Rightarrow a=\frac{1}{4}
\end{aligned}$$</p>
<p>Case II: If $$\frac{1}{2} < a < 1$$</p>
<p>$$2 a-1=0+2 a$$</p>
<p>No solution</p>
<p>Case III: If $$1 \leq a<2$$</p>
<p>$$2 a-1=1+2 a$$</p>
<p>$$\Rightarrow$$ No solution</p>
<p>$$\therefore$$ only solution is $$a=\frac{1}{4}$$</p>
<p>$$72 \sum_\limits{a \in S} a=72 \times \frac{1}{4}=18$$</p> | integer | jee-main-2024-online-5th-april-morning-shift |
NmOxptYb4egNVmQdBRShM | maths | functions | inverse-functions | Let <i> f</i> : A $$ \to $$ B be a function defined as f(x) = $${{x - 1} \over {x - 2}},$$ Where A = <b>R</b> $$-$$ {2} and B = <b>R</b> $$-$$ {1}. Then <i>f</i> is : | [{"identifier": "A", "content": "invertible and $${f^{ - 1}}(y) = $$ $${{3y - 1} \\over {y - 1}}$$ "}, {"identifier": "B", "content": "invertible and $${f^{ - 1}}\\left( y \\right) = {{2y - 1} \\over {y - 1}}$$ "}, {"identifier": "C", "content": "invertible and $${f^{ - 1}}\\left( y \\right) = {{2y + 1} \\over {y - 1}}$$"}, {"identifier": "D", "content": "not invertible"}] | ["B"] | null | Assume,
<br><br>y = f(x)
<br><br>$$ \Rightarrow $$ y = $${{x - 1} \over {x - 2}}$$
<br><br>$$ \Rightarrow $$ yx - 2y = x - 1
<br><br>$$ \Rightarrow $$ (y - 1)x = 2y - 1
<br><br>$$ \Rightarrow $$ x = $${{2y - 1} \over {y - 1}}$$ = f<sup> -1</sup>(y)
<br><br>As on the given domain the function is invertible and its inverse can be computed as shown above. | mcq | jee-main-2018-online-15th-april-evening-slot |
S9ViRTELMsFpCdXCyu7k9k2k5gzd75n | maths | functions | inverse-functions | The inverse function of
<br/><br/>f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$, x $$ \in $$ (-1, 1), is : | [{"identifier": "A", "content": "$${1 \\over 4}{\\log _e}\\left( {{{1 - x} \\over {1 + x}}} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over 4}\\left( {{{\\log }_8}e} \\right){\\log _e}\\left( {{{1 - x} \\over {1 + x}}} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over 4}\\left( {{{\\log }_8}e} \\right){\\log _e}\\left( {{{1 + x} \\over {1 - x}}} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 4}{\\log _e}\\left( {{{1 + x} \\over {1 - x}}} \\right)$$"}] | ["C"] | null | f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$ = y
<br><br>$$ \therefore $$ $${{y + 1} \over {y - 1}} = {{{{2.8}^{2x}}} \over { - {{2.8}^{ - 2x}}}}$$
<br><br>$$ \Rightarrow $$ $${{1 + y} \over {1 - y}}$$ = 8<sup>4x</sup>
<br><br>$$ \Rightarrow $$ $${\log _e}\left( {{{1 + y} \over {1 - y}}} \right)$$ = 4x $${\log _e}8$$
<br><br>$$ \Rightarrow $$ x = $${1 \over {4{{\log }_e}8}}{\log _e}\left( {{{1 + y} \over {1 - y}}} \right)$$
<br><br>$$ \therefore $$ f<sup>-1</sup>(x) = $${1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)$$ | mcq | jee-main-2020-online-8th-january-morning-slot |
3KiVygqDw488GbnoRq1kmjaxjig | maths | functions | inverse-functions | The inverse of $$y = {5^{\log x}}$$ is : | [{"identifier": "A", "content": "$$x = {5^{\\log y}}$$"}, {"identifier": "B", "content": "$$x = {y^{{1 \\over {\\log 5}}}}$$"}, {"identifier": "C", "content": "$$x = {5^{{1 \\over {\\log y}}}}$$"}, {"identifier": "D", "content": "$$x = {y^{\\log 5}}$$"}] | ["B"] | null | $$y = {5^{\log x}}$$<br><br>$$ \Rightarrow \log y = \log x.log5$$<br><br>$$ \Rightarrow \log x = {{\log y} \over {\log 5}} = {\log _5}y$$<br><br>$$ \Rightarrow x = {e^{{{\log }_5}y}}$$<br><br>$$ \Rightarrow x = {y^{{{\log }_5}e}}$$<br><br>$$ \Rightarrow x = {y^{{1 \over {\log 5}}}}$$ | mcq | jee-main-2021-online-17th-march-morning-shift |
LQM2NmZMuUOeUUPSBE1kmm3nlbt | maths | functions | inverse-functions | Let f : R $$-$$ {3} $$ \to $$ R $$-$$ {1} be defined by f(x) = $${{x - 2} \over {x - 3}}$$.<br/><br/>Let g : R $$ \to $$ R be given as g(x) = 2x $$-$$ 3. Then, the sum of all the values of x for which f<sup>$$-$$1</sup>(x) + g<sup>$$-$$1</sup>(x) = $${{13} \over 2}$$ is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "7"}] | ["B"] | null | Finding inverse of f(x)<br><br>$$y = {{x - 2} \over {x - 3}} \Rightarrow xy - 3y = x - 2 \Rightarrow x(y - 1) = 3y - 2$$<br><br>$$ \therefore $$ $${f^{ - 1}}(x) = {{3x - 2} \over {x - 1}}$$<br><br>Similarly for $${g^{ - 1}}(x)$$<br><br>$$y = 2x - 3 \Rightarrow x = {{y + 3} \over 2} \Rightarrow {g^{ - 1}}(x) = {{x + 3} \over 2}$$<br><br>$$ \therefore $$ $${{3x - 2} \over {x - 1}} + {{x + 3} \over 2} = {{13} \over 2}$$<br><br>$$ \Rightarrow 6x - 4 + {x^2} + 2x - 3 = 13x - 13$$<br><br>$$ \Rightarrow {x^2} - 5x + 6 = 0$$<br><br>$$ \Rightarrow (x - 2)(x - 3) = 0$$<br><br>$$ \Rightarrow $$ x = 2 or 3 | mcq | jee-main-2021-online-18th-march-evening-shift |
1lgoxs5ux | maths | functions | inverse-functions | <p>The range of $$f(x)=4 \sin ^{-1}\left(\frac{x^{2}}{x^{2}+1}\right)$$ is</p> | [{"identifier": "A", "content": "$$[0,2 \\pi]$$"}, {"identifier": "B", "content": "$$[0,2 \\pi)$$"}, {"identifier": "C", "content": "$$[0, \\pi)$$"}, {"identifier": "D", "content": "$$[0, \\pi]$$"}] | ["B"] | null | $$
\begin{aligned}
& \frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}<1 \\\\
\therefore & 0 \leq \frac{x^2}{1+x^2}<1 \\\\
\Rightarrow & 0 \leq \sin ^{-1}\left(\frac{x^2}{1+x^2}\right)<\frac{\pi}{2} \\\\
\Rightarrow & 0 \leq 4 \sin ^{-1}\left(\frac{x^2}{1+x^2}\right)<2 \pi
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-evening-shift |
bIuxRNjZnQV19sc9 | maths | functions | periodic-functions | The period of $${\sin ^2}\theta $$ is | [{"identifier": "A", "content": "$${\\pi ^2}$$ "}, {"identifier": "B", "content": "$$\\pi $$ "}, {"identifier": "C", "content": "$$2\\pi $$ "}, {"identifier": "D", "content": "$$\\pi /2$$ "}] | ["B"] | null | The period of $${\sin ^2}\theta $$ is = $$\pi $$
<br><br><b>Note :</b>
<br>(1) When $$n$$ is odd then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}\theta $$, $${\sec ^n}\theta $$ = $$2\pi $$
<br><br>(2) When $$n$$ is even then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}\theta $$, $${\sec ^n}\theta $$ = $$\pi $$
<br><br>(3) When $$n$$ is even/odd then the period of $${\tan ^n}\theta $$, $${\cot ^n}\theta $$ = $$\pi $$
<br><br>(3) When $$n$$ is even/odd then the period of $$\left| {{{\sin }^n}\theta } \right|$$, $$\left| {{{\cos }^n}\theta } \right|$$, $$\left| {{{\csc }^n}\theta } \right|$$, $$\left| {{{\sec }^n}\theta } \right|$$, $$\left| {{{\tan }^n}\theta } \right|$$, $$\left| {{{\cot }^n}\theta } \right|$$ = $$\pi $$ | mcq | aieee-2002 |
rGh90lKpMIAeYj89 | maths | functions | periodic-functions | Which one is not periodic? | [{"identifier": "A", "content": "$$\\left| {\\sin 3x} \\right| + {\\sin ^2}x$$ "}, {"identifier": "B", "content": "$$\\cos \\sqrt x + {\\cos ^2}x$$ "}, {"identifier": "C", "content": "$$\\cos \\,4x + {\\tan ^2}x$$ "}, {"identifier": "D", "content": "$$cos\\,2x + \\sin x$$ "}] | ["B"] | null | $$\sqrt x $$ is non periodic function and $$\cos \left( {something} \right)$$ is a periodic function so here in $$\cos \sqrt x $$ $$ \to $$ inside periodic function there is non periodic function which always produce non periodic function.
<br><br>$${{{\cos }^2}x}$$ is a periodic function with period $$\pi $$
<br><br><b>Note :</b>
<br>(1) When $$n$$ is odd then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}\theta $$, $${\sec ^n}\theta $$ = $$2\pi $$
<br><br>(2) When $$n$$ is even then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}\theta $$, $${\sec ^n}\theta $$ = $$\pi $$
<br><br>(3) When $$n$$ is even/odd then the period of $${\tan ^n}\theta $$, $${\cot ^n}\theta $$ = $$\pi $$
<br><br>(3) When $$n$$ is even/odd then the period of $$\left| {{{\sin }^n}\theta } \right|$$, $$\left| {{{\cos }^n}\theta } \right|$$, $$\left| {{{\csc }^n}\theta } \right|$$, $$\left| {{{\sec }^n}\theta } \right|$$, $$\left| {{{\tan }^n}\theta } \right|$$, $$\left| {{{\cot }^n}\theta } \right|$$ = $$\pi $$
<br><br>$$\cos \sqrt x + {\cos ^2}x$$ = non periodic function + periodic function = non periodic function | mcq | aieee-2002 |
twJaTXmPzxEKgG6b | maths | functions | range | The range of the function f(x) = $${}^{7 - x}{P_{x - 3}}$$ is | [{"identifier": "A", "content": "{1, 2, 3, 4, 5}"}, {"identifier": "B", "content": "{1, 2, 3, 4, 5, 6}"}, {"identifier": "C", "content": "{1, 2, 3, 4}"}, {"identifier": "D", "content": "{1, 2, 3}"}] | ["D"] | null | The range of the function $f(x) = {}^{7-x}P_{x-3}$ can be found by considering the possible values of $f(x)$ as $x$ varies over its domain.
<br/><br/>
The domain of $f(x)$ is the set of all real numbers such that
<br/><br/>(i) $x \geq 3$ (since the permutation function is only defined for non-negative integers)
<br/><br/>(ii) 7 - x > 0 $$ \Rightarrow $$ x < 7
<br/><br/>(iii) x - 3 $$ \le $$ 7 - x $$ \Rightarrow $$ 2x $$ \le $$ 10 $$ \Rightarrow $$ x $$ \le $$ 5.
<br/><br/>
To find the range of $f(x)$, we need to consider what values the expression ${}^{7-x}P_{x-3}$ can take as $x$ varies over its domain.
<br/><br/>
For $x=3$, we have ${}^{7-x}P_{x-3} = {}^{4}P_{0} = 1$.
<br/><br/>
For $x=4$, we have ${}^{7-x}P_{x-3} = {}^{3}P_{1} = 3$.
<br/><br/>
For $x=5$, we have ${}^{7-x}P_{x-3} = {}^{2}P_{2} = 2$.
<br/><br/>
Therefore, the range of $f(x)$ is the set of all non-negative integers less than or equal to 3, i.e., ${1, 2, 3}$. | mcq | aieee-2004 |
m5lbIUG3trWiugfMUNHOx | maths | functions | range | Let f : R $$ \to $$ R be defined by f(x) = $${x \over {1 + {x^2}}},x \in R$$. Then the range of f is : | [{"identifier": "A", "content": "$$\\left[ { - {1 \\over 2},{1 \\over 2}} \\right]$$"}, {"identifier": "B", "content": "$$R - \\left[ { - {1 \\over 2},{1 \\over 2}} \\right]$$"}, {"identifier": "C", "content": "($$-$$ 1, 1) $$-$$ {0}"}, {"identifier": "D", "content": "R $$-$$ [$$-$$1, 1]"}] | ["A"] | null | f(0) = 0 & f(x) is odd
<br><br>Further, if x > 0 then
<br><br>f(x) = $$f(x) = {1 \over {x + {1 \over x}}} \in \left( {0,{1 \over 2}} \right]$$
<br><br>Hence, $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
ohIDlBis3k5AggYqUF7k9k2k5hjmi7l | maths | functions | range | Let ƒ : (1, 3) $$ \to $$ R be a function defined by<br/>
$$f(x) = {{x\left[ x \right]} \over {1 + {x^2}}}$$ , where [x] denotes the greatest
integer $$ \le $$ x. Then the range of ƒ is | [{"identifier": "A", "content": "$$\\left( {{2 \\over 5},{1 \\over 2}} \\right) \\cup \\left( {{3 \\over 4},{4 \\over 5}} \\right]$$"}, {"identifier": "B", "content": "$$\\left( {{3 \\over 5},{4 \\over 5}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{2 \\over 5},{4 \\over 5}} \\right]$$"}, {"identifier": "D", "content": "$$\\left( {{2 \\over 5},{3 \\over 5}} \\right] \\cup \\left( {{3 \\over 4},{4 \\over 5}} \\right)$$"}] | ["A"] | null | f(x) = $$\left\{ {\matrix{
{{x \over {{x^2} + 1}},} & {1 < x < 2} \cr
{{{2x} \over {{x^2} + 1}},} & {2 \le x < 3} \cr
} } \right.$$
<br><br>$$ \therefore $$ f(x) is decreasing function
<br><br>$$ \therefore $$ Range is $$\left( {{2 \over 5},{1 \over 2}} \right) \cup \left( {{3 \over 4},{4 \over 5}} \right]$$. | mcq | jee-main-2020-online-8th-january-evening-slot |
17P9YSWU4u46PPcqJx1kmhxmh7h | maths | functions | range | The range of a$$\in$$R for which the <br/><br/>function f(x) = (4a $$-$$ 3)(x + log<sub>e</sub> 5) + 2(a $$-$$ 7) cot$$\left( {{x \over 2}} \right)$$ sin<sup>2</sup>$$\left( {{x \over 2}} \right)$$, x $$\ne$$ 2n$$\pi$$, n$$\in$$N has critical points, is : | [{"identifier": "A", "content": "[1, $$\\infty $$)"}, {"identifier": "B", "content": "($$-$$3, 1)"}, {"identifier": "C", "content": "$$\\left[ { - {4 \\over 3},2} \\right]$$"}, {"identifier": "D", "content": "($$-$$$$\\infty $$, $$-$$1]"}] | ["C"] | null | $$f(x) = (4a - 3)(x + \ln 5) + 2(a - 7)\left( {{{\cos {x \over 2}} \over {\sin {x \over 2}}}.{{\sin }^2}{x \over 2}} \right)$$<br><br>$$f(x) = (4a - 3)(x + \ln 5) + (a - 7)\sin x$$<br><br>$$f'(x) = (4a - 3) + (a - 7)\cos x = 0$$<br><br>$$\cos x = {{ - (4a - 3)} \over {a - 7}}$$<br><br>$$ - 1 \le - {{4a - 3} \over {a - 7}} \le 1$$<br><br>$$ - 1 \le {{4a - 3} \over {a - 7}} \le 1$$<br><br>$${{4a - 3} \over {a - 7}} - 1 \le 0$$ and $${{4a - 3} \over {a - 7}} + 1 \ge 0$$<br><br>$$ \Rightarrow {{ - 4} \over 3} \le a \le 2$$ | mcq | jee-main-2021-online-16th-march-morning-shift |
1kto99h4t | maths | functions | range | The range of the function, <br/><br/>$$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$$ is : | [{"identifier": "A", "content": "$$\\left( {0,\\sqrt 5 } \\right)$$"}, {"identifier": "B", "content": "[$$-$$2, 2]"}, {"identifier": "C", "content": "$$\\left[ {{1 \\over {\\sqrt 5 }},\\sqrt 5 } \\right]$$"}, {"identifier": "D", "content": "[0, 2]"}] | ["D"] | null | $$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$$<br><br>$$f(x) = {\log _{\sqrt 5 }}\left[ {3 + 2\cos \left( {{\pi \over 4}} \right)\cos (x) - 2\sin \left( {{{3\pi } \over 4}} \right)\sin (x)} \right]$$<br><br>$$f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$$<br><br>Since $$ - \sqrt 2 \le \cos x - \sin x \le \sqrt 2 $$<br><br>$$ \Rightarrow {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 \left( { - \sqrt 2 } \right) \le f(x) \le {{\log }_{\sqrt 5 }}\left[ {3 + \sqrt 2 \left( {\sqrt 2 } \right)} \right]} \right]$$<br><br>$$ \Rightarrow {\log _{\sqrt 5 }}(1) \le f(x) \le {\log _{\sqrt 5 }}(5)$$<br><br>So, Range of f(x) is [0, 2]<br><br>Option (d) | mcq | jee-main-2021-online-1st-september-evening-shift |
ldo7i3kx | maths | functions | range | Let $f: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}$ be real valued function<br/><br/> defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.
<br/><br/>Then range of $f$ is | [{"identifier": "A", "content": "$ \\left(-\\infty,-\\frac{21}{4}\\right] \\cup[1, \\infty) $"}, {"identifier": "B", "content": "$\\left(-\\infty,-\\frac{21}{4}\\right) \\cup(0, \\infty) $"}, {"identifier": "C", "content": "$\\left(-\\infty,-\\frac{21}{4}\\right] \\cup[0, \\infty) $"}, {"identifier": "D", "content": "$\\left(-\\infty,-\\frac{21}{4}\\right] \\cup\\left[\\frac{21}{4}, \\infty\\right)$"}] | ["C"] | null | $y=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$
<br/><br/>$\Rightarrow(y-1) x^{2}-(8 y+2) x+12 y-1=0$
<br/><br/>Let $y \neq 1$, then $D \geq 0$
<br/><br/>$$
4(4 y+1)^{2}-4(y-1)(12 y-1) \geq 0
$$
<br/><br/>$\Rightarrow 16 y^{2}+1+8 y-\left(12 y^{2}-13 y+1\right) \geq 0$
<br/><br/>$\Rightarrow 4 y^{2}+21 y \geq 0$
<br/><br/>$\Rightarrow y \in\left(-\infty,-\frac{21}{4}\right) \cup[0, \infty)-\{1\}$
<br/><br/>$$
\text { for } y=1 \text {, }
$$
<br/><br/>$$
\begin{aligned}
& -8 x+12=2 x+1 \\\\
& x=\frac{11}{10} \quad \therefore I \in R
\end{aligned}
$$
<br/><br/>$\therefore $ Range $=\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldprz63h | maths | functions | range | If the domain of the function $$f(x)=\frac{[x]}{1+x^{2}}$$, where $$[x]$$ is greatest integer $$\leq x$$, is $$[2,6)$$, then its range is | [{"identifier": "A", "content": "$$\\left(\\frac{5}{37}, \\frac{2}{5}\\right]-\\left\\{\\frac{9}{29}, \\frac{27}{109}, \\frac{18}{89}, \\frac{9}{53}\\right\\}$$"}, {"identifier": "B", "content": "$$\\left(\\frac{5}{37}, \\frac{2}{5}\\right]$$"}, {"identifier": "C", "content": "$$\\left(\\frac{5}{26}, \\frac{2}{5}\\right]$$"}, {"identifier": "D", "content": "$$\\left(\\frac{5}{26}, \\frac{2}{5}\\right]-\\left\\{\\frac{9}{29}, \\frac{27}{109}, \\frac{18}{89}, \\frac{9}{53}\\right\\}$$"}] | ["B"] | null | $f(x)=\frac{k}{1+x^{2}}$ is a decreasing function where $k>0$
<br/><br/>$$
\begin{gathered}
\therefore \quad x \in[2,3) \Rightarrow f(x)=\frac{2}{1+x^{2}} \in\left(\frac{2}{10}, \frac{2}{5}\right]=R_{1} \\\\
x \in[3,4) \Rightarrow f(x)=\frac{3}{1+x^{2}} \in\left(\frac{3}{17}, \frac{3}{10}\right]=R_{2} \\\\
x \in[4,5) \Rightarrow f(x)=\frac{4}{1+x^{2}} \in\left(\frac{4}{26}, \frac{4}{17}\right]=R_{3} \\\\
x \in[5,6) \Rightarrow f(x)=\frac{5}{1+x^{2}} \in\left(\frac{5}{37}, \frac{5}{26}\right]=R_{4} \\\\
\text { Range }=R_{1} \cup R_{2} \cup R_{3} \cup R_{4} \\\\
=\left(\frac{5}{37}, \frac{2}{5}\right]
\end{gathered}
$$ | mcq | jee-main-2023-online-31st-january-morning-shift |
ldqy1wog | maths | functions | range | The range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$ is : | [{"identifier": "A", "content": "$[2 \\sqrt{2}, \\sqrt{11}]$"}, {"identifier": "B", "content": "$[\\sqrt{5}, \\sqrt{13}]$"}, {"identifier": "C", "content": "$[\\sqrt{2}, \\sqrt{7}]$"}, {"identifier": "D", "content": "$[\\sqrt{5}, \\sqrt{10}]$"}] | ["D"] | null | <p>$$f(x) = \sqrt {3 - x} + \sqrt {x + 2} $$</p>
<p>$$y' = {{ - 1} \over {2\sqrt 3 - x}} + {1 \over {2\sqrt {x + 2} }} = 0$$</p>
<p>$$ \Rightarrow \sqrt x + 2 = \sqrt 3 - x$$</p>
<p>$$ \Rightarrow x = {1 \over 2}$$</p>
<p>$$y\left( {{1 \over 2}} \right) = \sqrt {{5 \over 2}} + \sqrt {{5 \over 2}} = \sqrt {10} $$</p>
<p>$${y_{\min }}$$ at $$x = - 2$$ or $$x = 3$$, i.e., $$\sqrt 5 $$</p>
<p>$$\therefore$$ $$f(x) \in [\sqrt 5 ,\sqrt {10} ]$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1lh2ylef4 | maths | functions | range | <p>Let the sets A and B denote the domain and range respectively of the function $$f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$$, where $$\lceil x\rceil$$ denotes the smallest integer greater than or equal to $$x$$. Then among the statements</p>
<p>(S1) : $$A \cap B=(1, \infty)-\mathbb{N}$$ and</p>
<p>(S2) : $$A \cup B=(1, \infty)$$</p> | [{"identifier": "A", "content": "only $$(\\mathrm{S} 2)$$ is true"}, {"identifier": "B", "content": "only (S1) is true"}, {"identifier": "C", "content": "neither (S1) nor (S2) is true"}, {"identifier": "D", "content": "both (S1) and (S2) are true"}] | ["B"] | null | $$
f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}
$$
<br/><br/>If $\mathrm{x} \in \mathrm{I},\lceil\mathrm{x}\rceil=[\mathrm{x}]$ (greatest integer function)
<br/><br/>If $x \notin I,\lceil x\rceil=[x]+1$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}
\frac{1}{\sqrt{[\mathrm{x}]-\mathrm{x}}}, \mathrm{x} \in \mathrm{I} \\\\
\frac{1}{\sqrt{[\mathrm{x}]+1-\mathrm{x}}}, \mathrm{x} \notin \mathrm{I}
\end{array}\right. \\\\
& \Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}
\frac{1}{\sqrt{-\{\mathrm{x}\}}}, \mathrm{x} \in \mathrm{I}, \text { (does not exist) } \\\\
\frac{1}{\sqrt{1-\{\mathrm{x}\}}}, \mathrm{x} \notin \mathrm{I}
\end{array}\right. \\\\
& \Rightarrow \text { domain of } \mathrm{f}(\mathrm{x})=\mathrm{R}-\mathrm{I}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Now, } \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{1-\{\mathrm{x}\}}}, \mathrm{x} \notin \mathrm{I} \\\\
& \Rightarrow 0<\{\mathrm{x}\}<1 \\\\
& \Rightarrow 0<\sqrt{1-\{\mathrm{x}\}}<1 \\\\
& \Rightarrow \frac{1}{\sqrt{1-\{\mathrm{x}\}}}>1 \\\\
& \Rightarrow \text { Range }=(1, \infty) \\\\
& \Rightarrow \mathrm{A}=\mathrm{R}-\mathrm{I}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& B=(1, \infty) \\\\
& \text { So, } A \cap B=(1, \infty)-N \\\\
& A \cup B \neq(1, \infty) \\\\
& \Rightarrow S 1 \text { is only correct }
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-evening-shift |
luxwcbm5 | maths | functions | range | <p>Let the range of the function $$f(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R}$$ be $$[a, b]$$. If $$\alpha$$ and $$\beta$$ ar respectively the A.M. and the G.M. of $$a$$ and $$b$$, then $$\frac{\alpha}{\beta}$$ is equal to</p> | [{"identifier": "A", "content": "$$\\pi$$\n"}, {"identifier": "B", "content": "$$\\sqrt{\\pi}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{2}$$"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <p>$$\begin{aligned}
& F(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R} \\
& \sin 3 x+\cos 3 x \in[-\sqrt{2}, \sqrt{2}] \\
& 2+\sin 3 x+\cos 3 x \in[2-\sqrt{2}, 2+\sqrt{2}] \\
& \Rightarrow \frac{1}{2+\sin 3 x+\cos 3 x} \in\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right] \\
& \Rightarrow a=\frac{1}{2+\sqrt{2}}, b=\frac{1}{2-\sqrt{2}} \\
& \alpha=\frac{a+b}{2}=\frac{\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}}}{2} \\
& =\frac{4}{2 \times 2}=1 \\
& \beta=\sqrt{a b}=\sqrt{\left(\frac{1}{2+\sqrt{2}}\right) \times\left(\frac{1}{2-\sqrt{2}}\right)} \\
& =\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \\
& \Rightarrow \frac{\alpha}{\beta}=\sqrt{2} \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
lv5gt1wc | maths | functions | range | <p>If the range of $$f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}, \theta \in \mathbb{R}$$ is $$[\alpha, \beta]$$, then the sum of the infinite G.P., whose first term is 64 and the common ratio is $$\frac{\alpha}{\beta}$$, is equal to __________.</p> | [] | null | 96 | <p>To determine the range of the function $$f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}$$, let's start by simplifying the expression. Let $$\sin^2 \theta = x$$, so $$\cos^2 \theta = 1 - x$$. The function then transforms into:</p>
<p>$$ f(x) = \frac{x^2 + 3(1-x)}{x^2 + (1-x)} $$</p>
<p>Simplify the numerator and denominator separately:</p>
<p>Numerator: $$ x^2 + 3 - 3x $$</p>
<p>Denominator: $$ x^2 + 1 - x $$</p>
<p>Thus, the function becomes:</p>
<p>$$ f(x) = \frac{x^2 + 3 - 3x}{x^2 + 1 - x} = \frac{x^2 - 3x + 3}{x^2 - x + 1} $$</p>
<p>Next, we need to find the range of this function. Let's analyze the function by testing specific values of $$x$$ in the interval $$[0, 1]$$ (since $$\sin^2 \theta$$ ranges from 0 to 1):</p>
<p>When $$x = 0$$:</p>
<p>$$ f(0) = \frac{0^2 - 3(0) + 3}{0^2 - 0 + 1} = \frac{3}{1} = 3 $$</p>
<p>When $$x = 1$$:</p>
<p>$$ f(1) = \frac{1^2 - 3(1) + 3}{1^2 - 1 + 1} = \frac{1 - 3 + 3}{1 - 1 + 1} = \frac{1}{1} = 1 $$</p>
<p>It appears that $$f(x)$$ achieves values within $$[1, 3]$$. To confirm this, we need to solve the quadratic inequality:</p>
<p>$$ 1 \leq \frac{x^2 - 3x + 3}{x^2 - x + 1} \leq 3 $$</p>
<p>By solving the inequalities, it can be confirmed that the function indeed ranges from 1 to 3 on the interval [0,1]. Hence, we have:</p>
<p>$$ \alpha = 1 $$</p>
<p>$$ \beta = 3 $$</p>
<p>The common ratio of the infinite geometric progression is:</p>
<p>$$ \frac{\alpha}{\beta} = \frac{1}{3} $$</p>
<p>Given the first term $a = 64$, the sum $S$ of the infinite geometric progression can be given as:</p>
<p>$$ S = \frac{a}{1 - r} $$</p>
<p>Substituting the values $a = 64$ and $r = \frac{1}{3}$, we get:</p>
<p>$$ S = \frac{64}{1 - \frac{1}{3}} = \frac{64}{\frac{2}{3}} = 64 \times \frac{3}{2} = 96 $$</p>
<p>Therefore, the sum of the infinite geometric progression is 96.</p> | integer | jee-main-2024-online-8th-april-morning-shift |
lvb29471 | maths | functions | range | <p>If the function $$f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0$$ attains the maximum value at $$x=\frac{1}{\mathrm{e}}$$ then :</p> | [{"identifier": "A", "content": "$$\\mathrm{e}^\\pi<\\pi^{\\mathrm{e}}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{e}^{2 \\pi}<(2 \\pi)^{\\mathrm{e}}$$\n"}, {"identifier": "C", "content": "$$(2 e)^\\pi>\\pi^{(2 e)}$$\n"}, {"identifier": "D", "content": "$$\\mathrm{e}^\\pi>\\pi^{\\mathrm{e}}$$"}] | ["D"] | null | <p>$$f\left(\frac{1}{\pi}\right)< f\left(\frac{1}{e}\right) \quad \text { as } \frac{1}{\pi}<\frac{1}{e}$$</p>
<p>$$\begin{aligned}
& \Rightarrow\left(\frac{1}{1}\right)^{\frac{2}{\pi}}<\left(\frac{1}{\frac{1}{e}}\right)^{\frac{2}{e}} \\
& \Rightarrow(\pi)^{\frac{2}{\pi}}<(e)^{\frac{2}{e}} \\
& \Rightarrow \pi^e < e^\pi
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
lvb294cr | maths | functions | range | <p>Let $$f(x)=\frac{1}{7-\sin 5 x}$$ be a function defined on $$\mathbf{R}$$. Then the range of the function $$f(x)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\left[\\frac{1}{8}, \\frac{1}{5}\\right]$$\n"}, {"identifier": "B", "content": "$$\\left[\\frac{1}{7}, \\frac{1}{6}\\right]$$\n"}, {"identifier": "C", "content": "$$\\left[\\frac{1}{7}, \\frac{1}{5}\\right]$$\n"}, {"identifier": "D", "content": "$$\\left[\\frac{1}{8}, \\frac{1}{6}\\right]$$"}] | ["D"] | null | <p>$$\begin{aligned}
& f(x)=\frac{1}{7-\sin 5 x} \\\\
& -1 \leq \sin 5 x \leq 1 \\\\
& -1 \leq-\sin 5 x \leq 1 \\\\
& -1+7 \leq 7-\sin 5 x \leq 1+7 \\\\
& 6 \leq 7-\sin 5 x \leq 8 \\\\
& \frac{1}{8} \leq \frac{1}{7-\sin 5 x} \leq \frac{1}{6} \\\\
& \frac{1}{8} \leq f(x) \leq \frac{1}{6} \\\\
& \text { Range }=\left[\frac{1}{8}, \frac{1}{6}\right]
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
LYcYPZMeePF9jKMj | maths | height-and-distance | height-and-distance | A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is $${60^ \circ }$$ and when he retires $$40$$ meters away from the tree the angle of elevation becomes $${30^ \circ }$$. The breadth of the river is : | [{"identifier": "A", "content": "$$60\\,\\,m$$ "}, {"identifier": "B", "content": "$$30\\,\\,m$$"}, {"identifier": "C", "content": "$$40\\,\\,m$$"}, {"identifier": "D", "content": "$$20\\,\\,m$$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265502/exam_images/egbnkczftvlxditjtcbz.webp" loading="lazy" alt="AIEEE 2004 Mathematics - Height and Distance Question 40 English Explanation">
<br><br>From the figure
<br><br>$$\tan {60^ \circ } = {y \over x}$$
<br><br>$$ \Rightarrow y = \sqrt {3x} .......\left( 1 \right)$$
<br><br>$$\tan {30^ \circ } = {y \over {x + 40}}$$
<br><br>$$ \Rightarrow y = {{x + 40} \over {\sqrt 3 }}........\left( 2 \right)$$
<br><br>From $$(1)$$ and $$(2),$$
<br><br>$$\sqrt 3 x = {{x + 40} \over {\sqrt 3 }} \Rightarrow x = 20m$$ | mcq | aieee-2004 |
tWejUYNwkeIGCAA9 | maths | height-and-distance | height-and-distance | A tower stands at the centre of a circular park. $$A$$ and $$B$$ are two points on the boundary of the park such that $$AB(=a)$$ subtends an angle of $${60^ \circ }$$ at the foot of the tower, and the angle of elevation of the top of the tower from $$A$$ or $$B$$ is $${30^ \circ }$$. The height of the tower is : | [{"identifier": "A", "content": "$$a/\\sqrt 3 $$ "}, {"identifier": "B", "content": "$$a\\sqrt 3 $$"}, {"identifier": "C", "content": "$$2a/\\sqrt 3 $$ "}, {"identifier": "D", "content": "$$2a\\sqrt 3 $$"}] | ["A"] | null | In the $$\Delta AOB,\,\,\angle AOB = {60^ \circ },$$ and
<br><br>$$\angle OBA = \angle OAB$$
<br><br>(since $$OA=OB=AB$$ radius of same circle).
<br><br>$$\therefore$$ $$\Delta AOB$$ is a equilateral triangle.
<br><br>Let the height of tower is $$h$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267731/exam_images/mtwwbj0355rmby4b8ntt.webp" loading="lazy" alt="AIEEE 2007 Mathematics - Height and Distance Question 39 English Explanation">
<br><br>Given distance between two points $$A$$ & $$B$$ lie on boundary of
<br><br>circular park, subtends an angle of $${60^ \circ }$$ at the foot of the tower
<br><br>is $$AB$$ i.e. $$AB$$$$=a.$$ A tower $$OC$$ stands at the center of a circular
<br><br>park. Angle of elevation of the top of the tower from $$A$$ and $$B$$ is $${30^ \circ }$$ .
<br><br>In $$\Delta OAC\,\,\tan {30^ \circ } = {h \over a}$$
<br><br>$$ \Rightarrow {1 \over {\sqrt 3 }} = {h \over a} \Rightarrow h = {a \over {\sqrt 3 }}$$ | mcq | aieee-2007 |
GR58oSFYu58GzaCR | maths | height-and-distance | height-and-distance | $$AB$$ is a vertical pole with $$B$$ at the ground level and $$A$$ at the top. $$A$$ man finds that the angle of elevation of the point $$A$$ from a certain point $$C$$ on the ground is $${60^ \circ }$$. He moves away from the pole along the line $$BC$$ to a point $$D$$ such that $$CD=7$$ m. From $$D$$ the angle of elevation of the point $$A$$ is $${45^ \circ }$$. Then the height of the pole is : | [{"identifier": "A", "content": "$${{7\\sqrt 3 } \\over 2} {1 \\over {\\sqrt {3 - 1} }}m$$ "}, {"identifier": "B", "content": "$${{7\\sqrt 3 } \\over 2}\\left( {\\sqrt {3 } + 1 } \\right)m$$ "}, {"identifier": "C", "content": "$${{7\\sqrt 3 } \\over 2}\\left( {\\sqrt {3 } - 1 } \\right)m$$"}, {"identifier": "D", "content": "$${{7\\sqrt 3 } \\over 2} {1 \\over {\\sqrt {3 + 1} }}m$$ "}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266701/exam_images/izrw5r6ggvq1o2nql6cu.webp" loading="lazy" alt="AIEEE 2008 Mathematics - Height and Distance Question 38 English Explanation">
<br><br>In $$\Delta ABC$$
<br><br>$${h \over x} = \tan {60^ \circ } = \sqrt 3 $$
<br><br>$$ \Rightarrow x = {h \over {\sqrt 3 }}$$
<br><br>In $$\Delta ABD{h \over {x + 7}}$$
<br><br>$$ = \tan {45^ \circ } = 1$$
<br><br>$$ \Rightarrow h = x + 7 \Rightarrow h - {h \over {\sqrt 3 }} = 7$$
<br><br>$$ \Rightarrow h = {{7\sqrt 3 } \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}$$
<br><br>$$ \Rightarrow h = {{7\sqrt 3 } \over 2}\left( {\sqrt 3 + 1\,m} \right)$$ | mcq | aieee-2008 |
XLrrI8GbRHPBpVqu | maths | height-and-distance | height-and-distance | $$ABCD$$ is a trapezium such that $$AB$$ and $$CD$$ are parallel and $$BC \bot CD.$$ If $$\angle ADB = \theta ,\,BC = p$$ and $$CD = q,$$ then AB is equal to: | [{"identifier": "A", "content": "$${{\\left( {{p^2} + {q^2}} \\right)\\sin \\theta } \\over {p\\cos \\theta + q\\sin \\theta }}$$ "}, {"identifier": "B", "content": "$${{{p^2} + {q^2}\\cos \\theta } \\over {p\\cos \\theta + q\\sin \\theta }}$$ "}, {"identifier": "C", "content": "$${{{p^2} + {q^2}} \\over {{p^2}\\cos \\theta + {q^2}\\sin \\theta }}$$ "}, {"identifier": "D", "content": "$${{\\left( {{p^2} + {q^2}} \\right)\\sin \\theta } \\over {{{\\left( {p\\cos \\theta + q\\sin \\theta } \\right)}^2}}}$$ "}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266634/exam_images/yvyztb8wfjakqr2ttuvt.webp" loading="lazy" alt="JEE Main 2013 (Offline) Mathematics - Height and Distance Question 41 English Explanation">
<br><br>From Sine Rule
<br><br>$${{AB} \over {\sin \theta }} = {{\sqrt {{p^2} + {q^2}} } \over {\sin \left( {\pi - \left( {\theta + \alpha } \right)} \right)}}$$
<br><br>$$AB = {{\sqrt {{p^2} + {q^2}} \sin \theta } \over {\sin \theta \cos \alpha + \cos \theta \sin \alpha }}$$
<br><br>$$ = {{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {q\sin \theta + p\cos \theta }}$$
<br><br>(As $$\cos \alpha = {q \over {\sqrt {{p^2} + {q^2}} }}$$ and $$\sin \alpha = {p \over {\sqrt {{p^2} + {q^2}} }}$$ ) | mcq | jee-main-2013-offline |
c3vyLJcH7Y9uv1nY | maths | height-and-distance | height-and-distance | A bird is sitting on the top of a vertical pole $$20$$ m high and its elevation from a point $$O$$ on the ground is $${45^ \circ }$$. It files off horizontally straight away from the point $$O$$. After one second, the elevation of the bird from $$O$$ is reduced to $${30^ \circ }$$. Then the speed (in m/s) of the bird is : | [{"identifier": "A", "content": "$$20\\sqrt 2 $$ "}, {"identifier": "B", "content": "$$20\\left( {\\sqrt 3 - 1} \\right)$$ "}, {"identifier": "C", "content": "$$40\\left( {\\sqrt 2 - 1} \\right)$$"}, {"identifier": "D", "content": "$$40\\left( {\\sqrt 3 - \\sqrt 2 } \\right)$$ "}] | ["B"] | null | Let the speed be $$y$$ $$m/sec$$.
<br><br>Let $$AC$$ be the vertical pole of height $$20$$ $$m.$$
<br><br>Let $$O$$ be the point on the ground such that $$\angle AOC = {45^ \circ }$$
<br><br>Let $$OC = x$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267056/exam_images/ahksb1duj8ufepfas9sz.webp" loading="lazy" alt="JEE Main 2014 (Offline) Mathematics - Height and Distance Question 37 English Explanation">
<br><br>Time $$t=1$$ $$s$$
<br><br>From $$\Delta AOC,\,\,\tan {45^ \circ } = {{20} \over x}\,\,\,\,\,\,\,.....\left( i \right)$$
<br><br>and from $$\Delta BOD,\,\,\tan {30^ \circ } = {{20} \over {x + y}}...\left( {ii} \right)$$
<br><br>From $$(i)$$ and $$(ii),$$ we have $$x=20$$
<br><br>and $${1 \over {\sqrt 3 }} = {{20} \over {x + y}}$$
<br><br>$$ \Rightarrow {1 \over {\sqrt 3 }} = {{20} \over {20 + y}}$$
<br><br>$$ \Rightarrow 20 + y = 20\sqrt 3 $$
<br><br>So, $$y = 20\left( {\sqrt 3 - 1} \right)\,\,i.e.,$$
<br><br>speed $$ = 20\left( {\sqrt 3 - 1} \right)m/s$$ | mcq | jee-main-2014-offline |
xU2eo9pTfAzBTHRg | maths | height-and-distance | height-and-distance | If the angles of elevation of the top of a tower from three collinear points $$A, B$$ and $$C,$$ on a line leading to the foot of the tower, are $${30^ \circ }$$, $${45^ \circ }$$ and $${60^ \circ }$$ respectively, then the ratio, $$AB:BC,$$ is : | [{"identifier": "A", "content": "$$1:\\sqrt 3 $$ "}, {"identifier": "B", "content": "$$2:3$$"}, {"identifier": "C", "content": "$$\\sqrt 3 :1$$ "}, {"identifier": "D", "content": "$$\\sqrt 3 :\\sqrt 2 $$ "}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91njbgq/c332d856-caa2-46a4-8d1d-eda9ad96fe55/6437c2a0-47fc-11ed-8757-0f869593f41f/file-1l91njbgr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91njbgq/c332d856-caa2-46a4-8d1d-eda9ad96fe55/6437c2a0-47fc-11ed-8757-0f869593f41f/file-1l91njbgr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2015 (Offline) Mathematics - Height and Distance Question 36 English Explanation">
<br>As $$PB$$ bisects $$\angle APC,$$ therefore $$AB$$ $$:$$ $$BC$$ $$=PA:PC$$
<br><br>Also in $$\Delta APQ,\sin {30^ \circ } = {h \over {PA}} \Rightarrow PA = 2h$$
<br><br>and in $$\Delta CPQ,$$ $$\sin {60^ \circ } = {h \over {PC}} \Rightarrow PC = {{2h} \over {\sqrt 3 }}$$
<br><br>$$\therefore$$ $$AB:BC = 2h:{{2h} \over {\sqrt 3 }} = \sqrt 3 :1$$ | mcq | jee-main-2015-offline |
dkDTQQj1IIHo6pmfr63Q9 | maths | height-and-distance | height-and-distance | The angle of elevation of the top of a vertical tower from a point A, due east of it is 45<sup>o</sup>. The angle of elevation of the top of the same tower from a point B, due south of A is 30<sup>o</sup>. If the distance between A and B is $$54\sqrt 2 \,m,$$ then the height of the tower (in metres), is : | [{"identifier": "A", "content": "$$36\\sqrt 3 $$"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "$$54\\sqrt 3 $$ "}, {"identifier": "D", "content": "108"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267746/exam_images/sfbzoig20kahtq3la2hf.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Mathematics - Height and Distance Question 30 English Explanation">
<br><br>Let the height of tower = h
<br><br>In triangle PQA,
<br><br>tan45<sup>o</sup> = $${h \over {QA}}$$
<br><br>$$ \Rightarrow $$ h = QA
<br><br>In triangle PQB,
<br><br>tan30<sup>0</sup> = $${h \over {BQ}}$$
<br><br>$$ \Rightarrow $$ BQ = $$\sqrt 3 $$h
<br><br>In triangle BAQ
<br><br>$$\angle $$ QAB = 90<sup>o</sup>.
<br><br>$$ \therefore $$ QA<sup>2</sup> + AB<sup>2</sup> = QB<sup>2</sup>
<br><br>$$ \Rightarrow $$ h<sup>2</sup> + (54$$\sqrt 2 $$)<sup>2</sup> = ($$\sqrt 3 $$h)<sup>2</sup>
<br><br>$$ \Rightarrow $$ 2h<sup>2</sup> = (54$$\sqrt 2 $$)<sup>2</sup>
<br><br>$$ \Rightarrow $$ $$\sqrt 2 h$$ = 54$$\sqrt 2 $$
<br><br>$$ \Rightarrow $$ h = 54 m | mcq | jee-main-2016-online-10th-april-morning-slot |
l87d3ma7 | maths | height-and-distance | height-and-distance | A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30<sup>o</sup>. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60<sup>o</sup>. Then the time taken (in minutes) by him, from B to reach the pillar, is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "5"}] | ["D"] | null | According to given information, we have the following figure<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l87czgsa/9146332a-d177-4510-8b69-14ebf4d88eae/9d5eea90-3753-11ed-9417-1312e45a73c7/file-1l87czgsb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l87czgsa/9146332a-d177-4510-8b69-14ebf4d88eae/9d5eea90-3753-11ed-9417-1312e45a73c7/file-1l87czgsb.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2016 (Offline) Mathematics - Height and Distance Question 3 English Explanation"><br>
Now, from $\triangle A C D$ and $\triangle B C D$, we have<br><br>
$\tan 30^{\circ}=\frac{h}{x+y}$<br><br>
and $\tan 60^{\circ}=\frac{h}{y}$<br><br>
$\Rightarrow h=\frac{x+y}{\sqrt{3}}\quad...(i)$<br><br>
and $h=\sqrt{3} y\quad...(ii)$<br><br>
From Eqs. (i) and (ii),<br><br>
$\frac{x+y}{\sqrt{3}}=\sqrt{3} y \Rightarrow x+y=3 y$<br><br>
$\Rightarrow x-2 y=0 \Rightarrow y=\frac{x}{2}$<br><br> $\because$ Speed is uniform.<br><br>
$\therefore$ Distance $y$ will be cover in $5 \mathrm{~min}$.<br><br>
$\because$ Distance $x$ covered in $10 \mathrm{~min}$.<br><br>
$\therefore$ Distance $\frac{x}{2}$ will be cover in $5 \mathrm{~min}$.
| mcq | jee-main-2016-offline |
DcXLZz7J4dXtDQAC | maths | height-and-distance | height-and-distance | Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point
on the ground such that AP = 2AB. If $$\angle $$BPC = $$\beta $$, then tan$$\beta $$ is equal to: | [{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "$${2 \\over 9}$$"}, {"identifier": "C", "content": "$${4 \\over 9}$$"}, {"identifier": "D", "content": "$${6 \\over 7}$$"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263839/exam_images/lz3a9qs9vzwnjl9bmeiu.webp" loading="lazy" alt="JEE Main 2017 (Offline) Mathematics - Height and Distance Question 34 English Explanation">
<br><br>Let the height of tower $$AB = x$$ and $$LCPA = \propto $$
<br><br>From the diagram you can see,
<br><br>$$\tan \left( { \propto + \beta } \right) = {x \over {2x}} = {1 \over 2}$$
<br><br>we know,
<br><br>$$\tan \left( { \propto + \beta } \right) = {{\tan \propto + \tan \beta } \over {1 - \tan \propto \tan \beta }}$$
<br><br>$$\therefore\,\,\,$$ $${{\tan \propto + \tan \beta } \over {1 - \tan \propto \tan \beta }} = {1 \over 2}....\left( 1 \right)$$
<br><br>From the diagram,
<br><br>$$\tan \widehat \propto = {{x/2} \over {2x}} = {1 \over 4}......\left( 2 \right)$$
<br><br>Putting value of $$\tan \propto $$ in eq$$(1)$$,
<br><br>$${{{1 \over 4} + \tan \beta } \over {1 - {1 \over 4}\tan \beta }} = {1 \over 2}$$
<br><br>$$ \Rightarrow 1 - {1 \over 4}\tan \beta $$ $$ = {1 \over 2} + 2\tan \beta $$
<br><br>$$ \Rightarrow {{9\tan \beta } \over 4} = {1 \over 2}$$
<br><br>$$ \Rightarrow \tan \beta = {2 \over 9}$$ | mcq | jee-main-2017-offline |
u9KCmuptWLfgAMh2 | maths | height-and-distance | height-and-distance | PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles
of elevation of the top of the tower at P, Q and R are respectively 45$$^\circ $$, 30$$^\circ $$ and 30$$^\circ $$, then the height of the tower (in m) is : | [{"identifier": "A", "content": "$$50\\sqrt 2 $$"}, {"identifier": "B", "content": "100"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "$$100\\sqrt 3 $$"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266852/exam_images/nlqfkpsnhv5bqrg6xjyo.webp" style="max-width:100%;height:auto;max-height:300px;margin: 0 auto" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Height and Distance Question 35 English Explanation">
<br><br>Let height of tower $$TM = h$$
<br><br>From triangle $$PMT,$$
<br><br>$$\tan {45^o} = {h \over {PM}}$$
<br><br>$$ \Rightarrow \,\,\,PM = h$$
<br><br>From triangle $$TRM,$$
<br><br>$$\tan {30^o} = {{TM} \over {RM}}$$
<br><br>$$ \Rightarrow \,\,\,RM = \sqrt3 h$$
<br><br>From triangle. $$PMR,$$
<br><br>$$P{M^2} + M{R^2} = {\left( {200} \right)^2}$$
<br><br>$$ \Rightarrow \,\,\,h{}^2 + {\left( {\sqrt 3 h} \right)^2} = {200^2}$$
<br><br>$$ \Rightarrow \,\,\,4{h^2} = {200^2}$$
<br><br>$$ \Rightarrow \,\,\,{h^2} = {{{{200}^2}} \over 4}$$
<br><br>$$ \Rightarrow \,\,\,h = {{200} \over 2} = 100\,m$$
<br><br>$$\therefore\,\,\,$$ height of tower $$=100$$ $$m.$$ | mcq | jee-main-2018-offline |
ZnQ7c1CR8ZKBJ5XWE59SM | maths | height-and-distance | height-and-distance | An aeroplane flying at a constant speed, parallel to the horizontal ground, $$\sqrt 3 $$ kmabove it, is obsered at an elevation of $${60^o}$$ from a point on the ground. If, after five seconds, its elevation from the same point, is $${30^o}$$, then the speed (in km / hr) of the aeroplane, is : | [{"identifier": "A", "content": "1500"}, {"identifier": "B", "content": "1440"}, {"identifier": "C", "content": "750"}, {"identifier": "D", "content": "720"}] | ["B"] | null | For $$\Delta $$OA, A, OA<sub>1</sub> = $${{\sqrt 3 } \over {\tan {{60}^o}}}$$ = 1 km
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266827/exam_images/o576rgrvrso9cx8btgxu.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15th April Morning Slot Mathematics - Height and Distance Question 33 English Explanation">
<br><br>For $$\Delta $$OB<sub>1</sub>, B, OB<sub>1</sub> = $${{\sqrt 3 } \over {\tan {{30}^o}}}$$ = 3km
<br><br>As, a distance of 3 $$-$$ 1 = 2 km is convered in 5 seconds.
<br><br>Therefore the speed of the plane is
<br><br>$${{2 \times 3600} \over 5}$$ = 1440 km/hr | mcq | jee-main-2018-online-15th-april-morning-slot |
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