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|---|---|---|---|---|---|---|---|---|---|---|---|
1l58h6qb5 | maths | definite-integration | properties-of-definite-integration | <p>The integral $${{24} \over \pi }\int_0^{\sqrt 2 } {{{(2 - {x^2})dx} \over {(2 + {x^2})\sqrt {4 + {x^4}} }}} $$ is equal to ____________.</p> | [] | null | 3 | <p>$$I = {{24} \over \pi }\int_0^{\sqrt 2 } {{{2 - {x^2}} \over {(2 + {x^2})\sqrt {4 + {x^4}} }}dx} $$</p>
<p>Let $$x = \sqrt 2 t \Rightarrow dx = \sqrt 2 dt$$</p>
<p>$$I = {{24} \over \pi }\int_0^1 {{{(2 - 2{t^2})\,.\,\sqrt 2 dt} \over {(2 + 2{t^2})\sqrt {4 + 4{t^4}} }}} $$</p>
<p>$$ = {{12\sqrt 2 } \over \pi }\int_0^... | integer | jee-main-2022-online-26th-june-evening-shift | 5,706 |
1l59kj9rb | maths | definite-integration | properties-of-definite-integration | <p>If $${b_n} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx} \over {\sin x}}dx,\,n \in N} $$, then</p> | [{"identifier": "A", "content": "$${b_3} - {b_2},\\,{b_4} - {b_3},\\,{b_5} - {b_4}$$ are in A.P. with common difference $$-$$2"}, {"identifier": "B", "content": "$${1 \\over {{b_3} - {b_2}}},{1 \\over {{b_4} - {b_3}}},{1 \\over {{b_5} - {b_4}}}$$ are in an A.P. with common difference 2 "}, {"identifier": "C", "content"... | ["D"] | null | <p>$${b_n} - {b_{n - 1}} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx - {{\cos }^2}(n - 1)x} \over {\sin x}}dx} $$</p>
<p>$$ = \int_0^{{\pi \over 2}} {{{ - \sin (2n - 1)x\,.\,\sin x} \over {\sin x}}dx} $$</p>
<p>$$ = \left. {{{\cos (2n - 1)x} \over {2n - 1}}} \right|_0^{\pi /2} = - {1 \over {2n - 1}}$$</p>
<p>So, $${b_... | mcq | jee-main-2022-online-25th-june-evening-shift | 5,707 |
1l59l6mpi | maths | definite-integration | properties-of-definite-integration | <p>The value of b > 3 for which $$12\int\limits_3^b {{1 \over {({x^2} - 1)({x^2} - 4)}}dx = {{\log }_e}\left( {{{49} \over {40}}} \right)} $$, is equal to ___________.</p> | [] | null | 6 | <p>$$I = \int {{1 \over {({x^2} - 1)({x^2} - 4)}}dx = {1 \over 3}\int {\left( {{1 \over {{x^2} - 4}} - {1 \over {{x^2} - 1}}} \right)dx} } $$</p>
<p>$$ = {1 \over 3}\left( {{1 \over 4}\ln \left| {{{x - 2} \over {x + 2}}} \right| - {1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} \right|} \right) + C$$</p>
<p>$$12I = \ln... | integer | jee-main-2022-online-25th-june-evening-shift | 5,708 |
1l5ahmkvl | maths | definite-integration | properties-of-definite-integration | <p>The value of $$\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx} $$ is equal to:</p> | [{"identifier": "A", "content": "$${{{\\pi ^2}} \\over 4}$$"}, {"identifier": "B", "content": "$${{{\\pi ^2}} \\over 2}$$"}, {"identifier": "C", "content": "$${\\pi \\over 4}$$"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$"}] | ["C"] | null | <p>$$\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {\left( {1 + {{\cos }^2}x} \right)\left( {{e^{\cos x}} + {e^{ - \cos x}}} \right)}}dx} $$</p>
<p>Let $$\cos x = t$$</p>
<p>$$\sin xdx = dt$$</p>
<p>$$ = \int\limits_1^{ - 1} {{{ - {e^t}dt} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}} $$</p>... | mcq | jee-main-2022-online-25th-june-morning-shift | 5,709 |
1l5b8114c | maths | definite-integration | properties-of-definite-integration | <p>The value of the integral <br/><br/>$$\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} $$ is equal to</p> | [{"identifier": "A", "content": "2$$\\pi$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$\\pi$$"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$"}] | ["C"] | null | <p>$$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} $$ ...... (i)</p>
<p>$$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^{ - x}})(si{n^6}x + {{\cos }^6}x)}}} $$ ..... (ii)</p>
<p>(i) and (ii)</p>
<p>From equation (i) & (ii)... | mcq | jee-main-2022-online-24th-june-evening-shift | 5,710 |
1l5c2bxms | maths | definite-integration | properties-of-definite-integration | <p>Let $$f(\theta ) = \sin \theta + \int\limits_{ - \pi /2}^{\pi /2} {(\sin \theta + t\cos \theta )f(t)dt} $$. Then the value of $$\left| {\int_0^{\pi /2} {f(\theta )d\theta } } \right|$$ is _____________.</p> | [] | null | 1 | $f(\theta)=\sin \theta\left(1+\int_{-\pi / 2}^{\pi / 2} f(t) d t\right)+\cos \theta\left(\int_{-\pi / 2}^{\pi / 2} t f(t) d t\right)$
<br/><br/>
Clearly $f(\theta)=a \sin \theta+b \cos \theta$
<br/><br/>
Where $a=1+\int_{-\pi / 2}^{\pi / 2}(a \sin t+b \cos t) d t \Rightarrow a=1+2 b\quad\quad...(i)$
<br/><br/>
and $b=\... | integer | jee-main-2022-online-24th-june-morning-shift | 5,711 |
1l5c2cwqz | maths | definite-integration | properties-of-definite-integration | <p>Let $$\mathop {Max}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \alpha $$ and $$\mathop {Min}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \beta $$.</p>
<p>If $$\int\limits_{\beta - {8 \over 3}}^{2\alpha - 1} {Max\left\{ {{{9 - {x^2}} \over {5 - x}},x}... | [] | null | 34 | Let $f(x)=\frac{x^{2}-9}{x-5} \Rightarrow f^{\prime}(x)=\frac{(x-1)(x-9)}{(x-5)^{2}}$
<br/><br/>
So, $\alpha=f(1)=2$ and $\beta=\min (f(0), f(2))=\frac{5}{3}$
<br/><br/>
Now, $\int_{-1}^{3} \max \left\{\frac{x^{2}-9}{x-5}, x\right\} d x=\int_{-1}^{9 / 5} \frac{x^{2}-9}{x-5} d x+\int_{9 / 5}^{3} x d x$
<br/><br/>
$$
=\i... | integer | jee-main-2022-online-24th-june-morning-shift | 5,712 |
1l6f157cx | maths | definite-integration | properties-of-definite-integration | <p>Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Then the value of the integral $$\int_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos (2 \pi x)]}\right) d x$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{52(1-e)}{e}$$"}, {"identifier": "B", "content": "$$\\frac{52}{e}$$"}, {"identifier": "C", "content": "$$\\frac{52(2+e)}{e}$$"}, {"identifier": "D", "content": "$$\\frac{104}{e}$$"}] | ["B"] | null | <p>$$I = \int_{ - 3}^{101} {\left( {\left[ {\sin (\pi x)} \right] + {e^{[\cos (2\pi x)]}}} \right)dx} $$</p>
$$[\sin \pi x]$$ is periodic with period 2 and $${{e^{[\cos (2\pi x)]}}}$$ is periodic with period 1.</p>
<p>So,</p>
<p>$$I = 52\int_0^2 {\left( {\left[ {\sin \pi x} \right] + {e^{[\cos 2\pi x]}}} \right)dx} $$<... | mcq | jee-main-2022-online-25th-july-evening-shift | 5,715 |
1l6f3l2tc | maths | definite-integration | properties-of-definite-integration | <p>Let $${a_n} = \int\limits_{ - 1}^n {\left( {1 + {x \over 2} + {{{x^2}} \over 3} + \,\,.....\,\, + \,\,{{{x^{n - 1}}} \over n}} \right)dx} $$ for every n $$\in$$ N. Then the sum of all the elements of the set {n $$\in$$ N : a<sub>n</sub> $$\in$$ (2, 30)} is ____________.</p> | [] | null | 5 | <p>$$\because$$ $${a_n} = \int\limits_{ - 1}^n {\left( {1 + {x \over 2} + {{{x^2}} \over 3}\, + \,....\, + \,{{{x^{n - 1}}} \over n}} \right)dx} $$</p>
<p>$$ = \left[ {x + {{{x^2}} \over {{2^2}}} + {{{x^3}} \over {{3^2}}}\, + \,......\, + \,{{{x^n}} \over {{n^2}}}} \right]_{ - 1}^n$$</p>
<p>$${a_n} = {{n + 1} \over {{1... | integer | jee-main-2022-online-25th-july-evening-shift | 5,716 |
1l6gjh7xh | maths | definite-integration | properties-of-definite-integration | <p>If $$\mathrm{n}(2 \mathrm{n}+1) \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}} \mathrm{d} x=1177 \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}+1} \mathrm{~d} x$$, then $$\mathrm{n} \in \mathbf{N}$$ is equal to ______________.</p> | [] | null | 24 | <p>$$\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = \int_0^1 {1\,.\,{{(1 - {x^n})}^{2n + 1}}dx} } $$</p>
<p>$$ = \left[ {{{(1 - {x^n})}^{2n + 1}}\,.\,x} \right]_0^1 - \int_0^1 {x\,.\,(2n + 1){{(1 - {x^n})}^{2n}}\,.\, - n{x^{n - 1}}dx} $$</p>
<p>$$ = n(2n + 1)\int_0^1 {(1 - (1 - {x^n})){{(1 - {x^n})}^{2n}}dx} $$</p>
<p>$$ = n(2... | integer | jee-main-2022-online-26th-july-morning-shift | 5,717 |
1l6hydgfm | maths | definite-integration | properties-of-definite-integration | <p>$$
\int\limits_{0}^{20 \pi}(|\sin x|+|\cos x|)^{2} d x \text { is equal to }
$$</p> | [{"identifier": "A", "content": "$$10(\\pi+4)$$"}, {"identifier": "B", "content": "$$10(\\pi+2)$$"}, {"identifier": "C", "content": "$$20(\\pi-2)$$"}, {"identifier": "D", "content": "$$20(\\pi+2)$$"}] | ["D"] | null | <p>$$I = \int\limits_0^{20\pi } {{{\left( {|\sin x| + |\cos x|} \right)}^2}\,dx} $$</p>
<p>$$ = 20\int\limits_0^\pi {\left( {1 + |\sin 2x|} \right)\,dx} $$</p>
<p>$$ = 40\int\limits_0^{{\pi \over 2}} {(1 + \sin 2x)\,dx} $$</p>
<p>$$ = \left. {40\left( {x - {{\cos 2x} \over 2}} \right)} \right|_0^{{\pi \over 2}}$$</p... | mcq | jee-main-2022-online-26th-july-evening-shift | 5,718 |
1l6jbku4w | maths | definite-integration | properties-of-definite-integration | <p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a function defined as</p>
<p>$$f(x)=a \sin \left(\frac{\pi[x]}{2}\right)+[2-x], a \in \mathbb{R}$$ where $$[t]$$ is the greatest integer less than or equal to $$t$$. If $$\mathop {\lim }\limits_{x \to -1 } f(x)$$ exists, then the value of $$\int\limits_{0}^{4} f(x) d x... | [{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["B"] | null | <p>$$f(x) = a\sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]\,a \in R$$</p>
<p>Now,</p>
<p>$$\because$$ $$\mathop {\lim }\limits_{x \to - 1} f(x)$$ exist</p>
<p>$$\therefore$$ $$\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)$$</p>
<p>$$ \Rightarrow a\sin \left( {{{ ... | mcq | jee-main-2022-online-27th-july-morning-shift | 5,719 |
1l6jbqfar | maths | definite-integration | properties-of-definite-integration | <p>Let $$
I=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x
$$. Then</p> | [{"identifier": "A", "content": "$${\\pi \\over 2} < I < {{3\\pi } \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 5} < I < {{5\\pi } \\over {12}}$$"}, {"identifier": "C", "content": "$${{5\\pi } \\over {12}} < I < {{\\sqrt 2 } \\over 3}\\pi $$"}, {"identifier": "D", "content": "$${{3\\pi } \\over 4} < ... | ["C"] | null | <p>I comes out around 1.536 which is not satisfied by any given options.</p>
<p>$$\int\limits_{\pi /4}^{\pi /3} {{{8x - 2x} \over x}dx > I > \int\limits_{\pi /4}^{\pi /3} {{{8\sin x - 2x} \over x}dx} } $$</p>
<p>$${\pi \over 2} > I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x} \over x} - 2} \right)dx} $$</p>
<p>... | mcq | jee-main-2022-online-27th-july-morning-shift | 5,720 |
1l6kiq196 | maths | definite-integration | properties-of-definite-integration | <p>Let $$f(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}$$.</p>
<p>Consider</p>
<p>$$(\mathrm{S} 1): f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$$</p>
<p>$$(\mathrm{S} 2): \int\limits_{-2}^{2} f(x) \mathrm{d} x=12$$</p>
<p>Then... | [{"identifier": "A", "content": "both (S1) and (S2) are correct"}, {"identifier": "B", "content": "both (S1) and (S2) are wrong"}, {"identifier": "C", "content": "only (S1) is correct"}, {"identifier": "D", "content": "only (S2) is correct"}] | ["D"] | null | <p>$$f(x) = 2 + |x| - |x - 1| + |x + 1|,\,x \in R$$</p>
<p>$$\therefore$$ $$f(x) = \left\{ {\matrix{
{ - x} & , & {x < - 1} \cr
{x + 2} & , & { - 1 \le x < 0} \cr
{3x + 2} & , & {0 \le x < 1} \cr
{x + 4} & , & {x \ge 1} \cr
} } \right.$$</p>
<p>$$\therefore$$ $$f'\left( { - {3 \over 2}} \right) +... | mcq | jee-main-2022-online-27th-july-evening-shift | 5,722 |
1l6kjxlll | maths | definite-integration | properties-of-definite-integration | <p>$$\int\limits_{0}^{2}\left(\left|2 x^{2}-3 x\right|+\left[x-\frac{1}{2}\right]\right) \mathrm{d} x$$, where [t] is the greatest integer function, is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{7}{6}$$"}, {"identifier": "B", "content": "$$\\frac{19}{12}$$"}, {"identifier": "C", "content": "$$\\frac{31}{12}$$"}, {"identifier": "D", "content": "$$\\frac{3}{2}$$"}] | ["B"] | null | <p>$$\int\limits_0^2 {|2{x^2} - 3x|dx + \int\limits_0^2 {\left[ {x - {1 \over 2}} \right]dx} } $$</p>
<p>$$ = \int\limits_0^{3/2} {(3x - 2{x^2})dx + \int\limits_{3/2}^2 {(2{x^2} - 3x)dx + \int\limits_0^{1/2} { - 1dx + \int\limits_{1/2}^{3/2} {0\,dx + \int\limits_{3/2}^2 {1dx} } } } } $$</p>
<p>$$ = \left. {\left( {{{3{... | mcq | jee-main-2022-online-27th-july-evening-shift | 5,723 |
1l6klps4r | maths | definite-integration | properties-of-definite-integration | <p>Let $$f(x)=\min \{[x-1],[x-2], \ldots,[x-10]\}$$ where [t] denotes the greatest integer $$\leq \mathrm{t}$$. Then $$\int\limits_{0}^{10} f(x) \mathrm{d} x+\int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x+\int\limits_{0}^{10}|f(x)| \mathrm{d} x$$ is equal to ________________.</p> | [] | null | 385 | <p>$$\because$$ $$f(x) = \min \,\{ [x - 1],[x - 2],\,......,\,[x - 10]\} = [x - 10]$$</p>
<p>Also $$|f(x)| = \left\{ {\matrix{
{ - f(x),} & {if\,x \le 10} \cr
{f(x),} & {if\,x \ge 10} \cr
} } \right.$$</p>
<p>$$\therefore$$ $$\int\limits_0^{10} {f(x)dx + \int\limits_0^{10} {{{(f(x))}^2}dx + \int\limits_0^{... | integer | jee-main-2022-online-27th-july-evening-shift | 5,724 |
1l6m6kw3s | maths | definite-integration | properties-of-definite-integration | <p>If $$\int\limits_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{\left(1+x^{2}\right)^{3}}}} \mathrm{~d} x=\alpha \sqrt{2}+\beta \sqrt{3}$$, where $$\alpha, \beta$$ are integers, then $$\alpha+\beta$$ is equal to __________.</p> | [] | null | 10 | <p>Put $$x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta $$</p>
<p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^3}\theta \,.\,{{\sec }^2}\theta \,d\theta } \over {\sqrt {1 + {{\tan }^2}\theta + \sqrt {{{\sec }^6}\theta } } }}} $$</p>
<p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {... | integer | jee-main-2022-online-28th-july-morning-shift | 5,725 |
1l6nm7c4a | maths | definite-integration | properties-of-definite-integration | <p>Let $$I_{n}(x)=\int_{0}^{x} \frac{1}{\left(t^{2}+5\right)^{n}} d t, n=1,2,3, \ldots .$$ Then :</p> | [{"identifier": "A", "content": "$$50 I_{6}-9 I_{5}=x I_{5}^{\\prime}$$"}, {"identifier": "B", "content": "$$50 I_{6}-11 I_{5}=x I_{5}^{\\prime}$$"}, {"identifier": "C", "content": "$$50 I_{6}-9 I_{5}=I_{5}^{\\prime}$$"}, {"identifier": "D", "content": "$$50 I_{6}-11 I_{5}=I_{5}^{\\prime}$$"}] | ["A"] | null | <p>$${I_n}(x) = \int\limits_0^x {{1 \over {{{({t^2} + 5)}^n}}}dt} $$</p>
<p>$$ = \int\limits_0^x {{1 \over {\underbrace {{{({t^2} + 5)}^n}}_I}} \times \mathop I\limits_{II} \,dt} $$</p>
<p>$$ = \left. {{t \over {{{({t^2} + 5)}^n}}}} \right|_0^x - \int\limits_0^x {{{ - 2nt} \over {{{({t^2} + 5)}^{n + 1}}}} \times t\,dt}... | mcq | jee-main-2022-online-28th-july-evening-shift | 5,726 |
1l6npjdpk | maths | definite-integration | properties-of-definite-integration | <p>The value of the integral $$\int\limits_{0}^{\frac{\pi}{2}} 60 \frac{\sin (6 x)}{\sin x} d x$$ is equal to _________.</p> | [] | null | 104 | <p>$$I = \int\limits_0^{{\pi \over 2}} {60\,.\,{{\sin 6x} \over {\sin x}}dx} $$</p>
<p>$$ = 60\,.\,2\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(4{{\cos }^2}x - 3)\cos x\,dx} $$</p>
<p>$$ = 120\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(1 - 4{{\sin }^2}x)\cos x\,dx} $$</p>
<p>Let $$\sin x = t \Rightarro... | integer | jee-main-2022-online-28th-july-evening-shift | 5,727 |
1l6p1b4ew | maths | definite-integration | properties-of-definite-integration | <p>The integral $$\int\limits_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} \mathrm{~d} x$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\tan ^{-1}(2)$$"}, {"identifier": "B", "content": "$$\\tan ^{-1}(2)-\\frac{\\pi}{4}$$"}, {"identifier": "C", "content": "$$\\frac{1}{2} \\tan ^{-1}(2)-\\frac{\\pi}{8}$$"}, {"identifier": "D", "content": "$$\\frac{1}{2}$$"}] | ["B"] | null | <p>$$I = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx} $$</p>
<p>$$ = \int\limits_0^{\pi /2} {{{(1 + {{\tan }^2}x/2)dx} \over {3(1 + {{\tan }^2}x/2) + 2(2\tan x/2) + (1 - {{\tan }^2}x/2)}}} $$</p>
<p>Let $$\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dt$$</p>
<p>$$I = \int\limits_0^1 {{{2dt} \over {4 + 2{t^... | mcq | jee-main-2022-online-29th-july-morning-shift | 5,728 |
1l6p2ihmv | maths | definite-integration | properties-of-definite-integration | <p>If $$f(\alpha)=\int\limits_{1}^{\alpha} \frac{\log _{10} \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}, \alpha>0$$, then $$f\left(\mathrm{e}^{3}\right)+f\left(\mathrm{e}^{-3}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "$$\\frac{9}{2}$$"}, {"identifier": "C", "content": "$$\\frac{9}{\\log _{e}(10)}$$"}, {"identifier": "D", "content": "$$\\frac{9}{2 \\log _{e}(10)}$$"}] | ["D"] | null | <p>$$f(\alpha ) = \int_1^\alpha {{{{{\log }_{10}}t} \over {1 + t}}dt} $$ ...... (i)</p>
<p>$$f\left( {{1 \over \alpha }} \right) = \int_1^{{1 \over \alpha }} {{{{{\log }_{10}}t} \over {1 + t}}dt} $$</p>
<p>Substituting $$t \to {1 \over p}$$</p>
<p>$$f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}... | mcq | jee-main-2022-online-29th-july-morning-shift | 5,729 |
1l6re2c2a | maths | definite-integration | properties-of-definite-integration | <p>If $$[t]$$ denotes the greatest integer $$\leq t$$, then the value of $$\int_{0}^{1}\left[2 x-\left|3 x^{2}-5 x+2\right|+1\right] \mathrm{d} x$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{\\sqrt{37}+\\sqrt{13}-4}{6}$$"}, {"identifier": "B", "content": "$$\\frac{\\sqrt{37}-\\sqrt{13}-4}{6}$$"}, {"identifier": "C", "content": "$$\\frac{-\\sqrt{37}-\\sqrt{13}+4}{6}$$"}, {"identifier": "D", "content": "$$\\frac{-\\sqrt{37}+\\sqrt{13}+4}{6}$$"}] | ["A"] | null | <p>$$\int_0^1 {\left[ {2x - |3{x^2} - 5x + 2| + 1} \right]dx} $$</p>
<p>$$3{x^2} - 5x + 2 = 0$$</p>
<p>$$ \Rightarrow 3{x^2} - 3x - 2x + 2 = 0$$</p>
<p>$$ \Rightarrow 3x(x - 1) - 2(x - 1) = 0$$</p>
<p>$$ \Rightarrow (x - 1)(3x - 2) = 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9s1mi/e... | mcq | jee-main-2022-online-29th-july-evening-shift | 5,730 |
1ldo61zd1 | maths | definite-integration | properties-of-definite-integration | <p>The value of the integral <br/><br/>$$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{x + {\pi \over 4}} \over {2 - \cos 2x}}dx} $$ is :</p> | [{"identifier": "A", "content": "$${{{\\pi ^2}} \\over {6\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${{{\\pi ^2}} \\over 6}$$"}, {"identifier": "C", "content": "$${{{\\pi ^2}} \\over {3\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${{{\\pi ^2}} \\over {12\\sqrt 3 }}$$"}] | ["A"] | null | $$
I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{(2-\cos 2 x)} d x
$$
<br/><br/>Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
<br/><br/>$$
I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{-x+\frac{\pi}{4}}{2-\cos 2 x}\right) d x
$$
<br/><br/>$\begin{aligned} & \therefore 2 I=\int_{-\frac{\pi}{4}}^{... | mcq | jee-main-2023-online-1st-february-evening-shift | 5,731 |
1ldo73jao | maths | definite-integration | properties-of-definite-integration | <p>If $$\int\limits_0^\pi {{{{5^{\cos x}}(1 + \cos x\cos 3x + {{\cos }^2}x + {{\cos }^3}x\cos 3x)dx} \over {1 + {5^{\cos x}}}} = {{k\pi } \over {16}}} $$, then k is equal to _____________.</p> | [] | null | 13 | $$
\begin{aligned}
& \mathrm{I}=\int_0^\pi \frac{5^{\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right)}{1+5^{\cos x}} d x \\\\
& I=\int_0^\pi \frac{5^{-\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right)}{1+5^{-\cos x}} d x \\\\
& 2 \mathrm{I}=\int_0^\pi\left(1+\cos x \cos 3 x+\cos ^2 x+\... | integer | jee-main-2023-online-1st-february-evening-shift | 5,732 |
1ldonnoj6 | maths | definite-integration | properties-of-definite-integration | <p>If $$\int_\limits{0}^{1}\left(x^{21}+x^{14}+x^{7}\right)\left(2 x^{14}+3 x^{7}+6\right)^{1 / 7} d x=\frac{1}{l}(11)^{m / n}$$ where $$l, m, n \in \mathbb{N}, m$$ and $$n$$ are coprime then $$l+m+n$$ is equal to ____________.</p> | [] | null | 63 | $I=\int_{0}^{1}\left(x^{21}+x^{14}+x^{7}\right)\left(2 x^{14}+3 x^{7}+6\right)^{1 / 7} d x$ <br/><br/>$I=\int_{0}^{1}\left(x^{20}+x^{13}+x^{6}\right)\left(2 x^{21}+3 x^{14}+6 x^{7}\right)^{1 / 7} d x$
<br/><br/>Let $2 x^{21}+3 x^{14}+6 x^{7}=t$
<br/><br/>$\Rightarrow 42\left(x^{20}+x^{13}+x^{6}\right) d x=d t$
<br/>... | integer | jee-main-2023-online-1st-february-morning-shift | 5,734 |
1ldoo6cqc | maths | definite-integration | properties-of-definite-integration | <p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a differentiable function such that $$f^{\prime}(x)+f(x)=\int_\limits{0}^{2} f(t) d t$$. If $$f(0)=e^{-2}$$, then $$2 f(0)-f(2)$$ is equal to ____________.</p> | [] | null | 1 | $f^{\prime}(x)+f(x)=k$
<br/><br/>$$
\begin{aligned}
& \Rightarrow e^{x} f(x)=k e^{x}+c \\\\
& f(x)=k+c e^{-x} \\\\
& k=\int_{0}^{2}\left(k+c e^{-t}\right) d t \\\\
& k=2 k+\left.c \cdot \frac{e^{-t}}{-1}\right|_{0} ^{2} \\\\
& k=2 k+c\left(\frac{e^{-2}}{-1}+1\right) \\\\
& -k=c\left(1-\frac{1}{e^{2}}\right) \\\\
& f(x... | integer | jee-main-2023-online-1st-february-morning-shift | 5,735 |
1ldpt0own | maths | definite-integration | properties-of-definite-integration | <p>The value of $$\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{10}{3}-\\sqrt{3}+\\log _{e} \\sqrt{3}$$"}, {"identifier": "B", "content": "$$\\frac{7}{2}-\\sqrt{3}-\\log _{e} \\sqrt{3}$$"}, {"identifier": "C", "content": "$$\\frac{10}{3}-\\sqrt{3}-\\log _{e} \\sqrt{3}$$"}, {"identifier": "D", "content": "$$-2+3\\sqrt{3}+\\log _{e} \\sqrt{3}... | ["A"] | null | Let I = $$\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$$
<br/><br/>$$
=\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin x(1+\cos x)} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x
$$
<br/><br/>$$
\begin{aligned}
=\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin x... | mcq | jee-main-2023-online-31st-january-morning-shift | 5,737 |
1ldr6vx1p | maths | definite-integration | properties-of-definite-integration | <p>If [t] denotes the greatest integer $$\le \mathrm{t}$$, then the value of $${{3(e - 1)} \over e}\int\limits_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $$ is :</p> | [{"identifier": "A", "content": "$$\\mathrm{e^8-e}$$"}, {"identifier": "B", "content": "$$\\mathrm{e^7-1}$$"}, {"identifier": "C", "content": "$$\\mathrm{e^9-e}$$"}, {"identifier": "D", "content": "$$\\mathrm{e^8-1}$$"}] | ["A"] | null | <p>$$I = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $$</p>
<p>$$ = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{1 + [{x^3}]}}dx} $$ ($$\because$$ $$[x] = 1$$ when $$x \in (12)$$)</p>
<p>$$ = 3(e - 1)\int_1^2 {{x^2}{e^{[{x^3}]}}dx} $$</p>
<p>Let $${x^3} = t$$</p>
<p>$$I = (e - 1)\int_1^8 {{e^{[t]}}dt} $$</p>
<p... | mcq | jee-main-2023-online-30th-january-morning-shift | 5,738 |
1ldsfe8t4 | maths | definite-integration | properties-of-definite-integration | <p>The value of the integral $$\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt} $$ is</p> | [{"identifier": "A", "content": "$${\\tan ^{ - 1}}{1 \\over 2} - {1 \\over 3}{\\tan ^{ - 1}}8 + {\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\tan ^{ - 1}}2 - {1 \\over 3}{\\tan ^{ - 1}}8 + {\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}2 + {1 \\over 3}{\\tan ^{ - 1}}8 - {\\pi \\ove... | ["C"] | null | <p>$$\int_1^2 {{{{t^4} + 1} \over {{t^6} + 1}}dt} $$</p>
<p>$$ = \int_1^2 {{{{{({t^2} + 1)}^2}} \over {{t^6} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{t^6} + 1}}dt} } $$</p>
<p>$$ = \int_1^2 {{{{t^2} + 1} \over {{t^4} - {t^2} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{{({t^3})}^2} + 1}}dt} } $$</p>
<p>$$ = \left. {{{\tan }^{ - ... | mcq | jee-main-2023-online-29th-january-evening-shift | 5,739 |
1ldsfpdr8 | maths | definite-integration | properties-of-definite-integration | <p>The value of the integral $$\int\limits_{1/2}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 2}{\\log _e}2$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}{\\log _e}2$$"}, {"identifier": "C", "content": "$${1 \\over 2}{\\log _e}2$$"}, {"identifier": "D", "content": "$$\\pi {\\log _e}2$$"}] | ["A"] | null | <p>$$I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ ..... (i)</p>
<p>$$x \to {1 \over x}$$</p>
<p>$$I = \int\limits_{{1 \over 2}}^2 {{1 \over x}{{\tan }^{ - 1}}{1 \over x}dx} $$ ..... (ii)</p>
<p>$$2I = \int\limits_{{1 \over 2}}^2 {{1 \over x}\,.\,{\pi \over 2}dx} $$</p>
<p>$$ = \left. {{\pi \ov... | mcq | jee-main-2023-online-29th-january-evening-shift | 5,740 |
1ldsuma7b | maths | definite-integration | properties-of-definite-integration | <p>Let $$f(x) = x + {a \over {{\pi ^2} - 4}}\sin x + {b \over {{\pi ^2} - 4}}\cos x,x \in R$$ be a function which<br/><br/> satisfies $$f(x) = x + \int\limits_0^{\pi /2} {\sin (x + y)f(y)dy} $$. then $$(a+b)$$ is equal to</p> | [{"identifier": "A", "content": "$$ - 2\\pi (\\pi + 2)$$"}, {"identifier": "B", "content": "$$ - \\pi (\\pi - 2)$$"}, {"identifier": "C", "content": "$$ - \\pi (\\pi + 2)$$"}, {"identifier": "D", "content": "$$ - 2\\pi (\\pi - 2)$$"}] | ["A"] | null | $f(x)=x+\int\limits_{0}^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$
<br/><br/>
$f(x)=x+\int\limits_{0}^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x)\quad....(1)$
<br/><br/>
On comparing with
<br/><br/>
$f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x, x \in \mathbb{R}$ then
<br/><br/>
$... | mcq | jee-main-2023-online-29th-january-morning-shift | 5,741 |
1ldu5exe2 | maths | definite-integration | properties-of-definite-integration | <p>The integral $$16\int\limits_1^2 {{{dx} \over {{x^3}{{\left( {{x^2} + 2} \right)}^2}}}} $$ is equal to</p> | [{"identifier": "A", "content": "$${{11} \\over {12}} + {\\log _e}4$$"}, {"identifier": "B", "content": "$${{11} \\over 6} + {\\log _e}4$$"}, {"identifier": "C", "content": "$${{11} \\over {12}} - {\\log _e}4$$"}, {"identifier": "D", "content": "$${{11} \\over 6} - {\\log _e}4$$"}] | ["D"] | null | $I=\int \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}$ <br/><br/>$=\frac{1}{4} \int \frac{x}{x^{2}+2} d x+\frac{1}{4} \int \frac{x}{\left(x^{2}+2\right)^{2}}-\frac{1}{4} \int \frac{d x}{x}+\frac{1}{4} \int \frac{d x}{x^{3}}$<br/><br/> $=\frac{1}{8} \ln \left(x^{2}+2\right)-\frac{\ln x}{4}-\frac{1}{8\left(x^{2}+2\right)}-\f... | mcq | jee-main-2023-online-25th-january-evening-shift | 5,742 |
1ldu6241m | maths | definite-integration | properties-of-definite-integration | <p>If $$\int\limits_{{1 \over 3}}^3 {|{{\log }_e}x|dx = {m \over n}{{\log }_e}\left( {{{{n^2}} \over e}} \right)} $$, where m and n are coprime natural numbers, then $${m^2} + {n^2} - 5$$ is equal to _____________.</p> | [] | null | 20 | $I=\int\limits_{\frac{1}{3}}^{3}|\ln x| d x=-\int\limits_{\frac{1}{3}}^{1} \ln x d x+\int\limits_{1}^{3} \ln x d x$
<br/><br/>
$$
\begin{aligned}
& \left.=-[x \ln x-x]_{\frac{1}{3}}^{1}+x \ln x-x\right]_{1}^{3} \\\\
& =-\left[(0-1)-\left(\frac{1}{3} \ln 3-\frac{1}{3}\right)\right]+[(3 \ln 3-3)-(0-1)] \\\\
& =\frac{2}{3... | integer | jee-main-2023-online-25th-january-evening-shift | 5,743 |
1ldv16kbb | maths | definite-integration | properties-of-definite-integration | <p>The minimum value of the function $$f(x) = \int\limits_0^2 {{e^{|x - t|}}dt} $$ is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$2(e-1)$$"}, {"identifier": "C", "content": "$$e(e-1)$$"}, {"identifier": "D", "content": "$$2e-1$$"}] | ["B"] | null | $$
f(x)=\int_0^2 e^{|x-t|} d t
$$<br/><br/>
For $x>2$<br/><br/>
$$
f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right)
$$<br/><br/>
For $x<0$<br/><br/>
$$
f(x)=\int_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right)
$$<br/><br/>
For $x \in[0,2]$<br/><br/>
$$
\begin{aligned}
& f(x)=\int_0^x e^{x-t} d t + \int_x^2 e^{t-x} d t \\\\
... | mcq | jee-main-2023-online-25th-january-morning-shift | 5,744 |
1ldwx9a09 | maths | definite-integration | properties-of-definite-integration | <p>$$\int\limits_{{{3\sqrt 2 } \over 4}}^{{{3\sqrt 3 } \over 4}} {{{48} \over {\sqrt {9 - 4{x^2}} }}dx} $$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$${\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$$2\\pi $$"}] | ["D"] | null | $$
\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x
$$<br/><br/>
We have $\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C$<br/><br/>
Hence $\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x=\frac{48}{2} \times\left[\sin ^{-1} \frac{2 x}{3}\right]_{... | mcq | jee-main-2023-online-24th-january-evening-shift | 5,745 |
1lgoxpnhf | maths | definite-integration | properties-of-definite-integration | <p>The value of $${{{e^{ - {\pi \over 4}}} + \int\limits_0^{{\pi \over 4}} {{e^{ - x}}{{\tan }^{50}}xdx} } \over {\int\limits_0^{{\pi \over 4}} {{e^{ - x}}({{\tan }^{49}}x + {{\tan }^{51}}x)dx} }}$$ is</p> | [{"identifier": "A", "content": "51"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "49"}] | ["B"] | null | <p>We're given the expression:</p>
<p>$$\frac{e^{-\pi/4} + \int_0^{\pi / 4} e^{-x} \tan ^{50} x dx}{\int_0^{\pi / 4} e^{-x}(\tan x)^{49} dx + \int_0^{\pi / 4} e^{-x}(\tan x)^{51} dx}$$</p>
<p>Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x} \tan... | mcq | jee-main-2023-online-13th-april-evening-shift | 5,749 |
1lgoy1rhw | maths | definite-integration | properties-of-definite-integration | <p>Let $$f_{n}=\int_\limits{0}^{\frac{\pi}{2}}\left(\sum_\limits{k=1}^{n} \sin ^{k-1} x\right)\left(\sum_\limits{k=1}^{n}(2 k-1) \sin ^{k-1} x\right) \cos x d x, n \in \mathbb{N}$$. Then $$f_{21}-f_{20}$$ is equal to _________</p> | [] | null | 41 | Given, $$f_{n}=\int_\limits{0}^{\frac{\pi}{2}}\left(\sum_\limits{k=1}^{n} \sin ^{k-1} x\right)\left(\sum_\limits{k=1}^{n}(2 k-1) \sin ^{k-1} x\right) \cos x d x$$
<br/><br/>$$
\begin{aligned}
& \text { Put } \sin x=t \\\\
& \cos x d x=d t \\\\
& f_n=\int_0^1\left(\sum_{k=1}^n(t)^{k-1}\right)\left(\sum_{k=1}^n(2 k-1)(t)... | integer | jee-main-2023-online-13th-april-evening-shift | 5,750 |
1lgpy0x97 | maths | definite-integration | properties-of-definite-integration | <p>$$\int_\limits{0}^{\infty} \frac{6}{e^{3 x}+6 e^{2 x}+11 e^{x}+6} d x=$$</p> | [{"identifier": "A", "content": "$$\\log _{e}\\left(\\frac{256}{81}\\right)$$"}, {"identifier": "B", "content": "$$\\log _{e}\\left(\\frac{64}{27}\\right)$$"}, {"identifier": "C", "content": "$$\\log _{e}\\left(\\frac{32}{27}\\right)$$"}, {"identifier": "D", "content": "$$\\log _{e}\\left(\\frac{512}{81}\\right)$$"}] | ["C"] | null | $$
\begin{aligned}
& \mathrm{l}=\int_0^{\infty} \frac{6}{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+2\right)\left(\mathrm{e}^{\mathrm{x}}+3\right)} \mathrm{dx} \\\\
& =6 \int_0^{\infty}\left(\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+1}+\frac{-1}{\mathrm{e}^{\mathrm{x}}+2}+\frac{\frac{1}{2}}{\m... | mcq | jee-main-2023-online-13th-april-morning-shift | 5,751 |
1lgq0ygkp | maths | definite-integration | properties-of-definite-integration | <p>Let for $$x \in \mathbb{R}, S_{0}(x)=x, S_{k}(x)=C_{k} x+k \int_{0}^{x} S_{k-1}(t) d t$$, where
<br/><br/>$$C_{0}=1, C_{k}=1-\int_{0}^{1} S_{k-1}(x) d x, k=1,2,3, \ldots$$ Then $$S_{2}(3)+6 C_{3}$$ is equal to ____________.</p> | [] | null | 18 | Given,
<br/><br/>$$
S_k(x)=C_k x+k \int_0^x S_{k-1}(t) d t
$$
<br/><br/>Put $\mathrm{k}=2$ and $\mathrm{x}=3$
<br/><br/>$$
\mathrm{S}_2(3)=\mathrm{C}_2(3)+2 \int_0^3 \mathrm{~S}_1(\mathrm{t}) \mathrm{dt}
$$ .........(i)
<br/><br/>Also,
<br/><br/>$$
\begin{aligned}
& S_1(x)=C_1(x)+\int_0^x S_0(t) d t \\\\
& =C_1 x+\frac... | integer | jee-main-2023-online-13th-april-morning-shift | 5,752 |
1lgrgllfd | maths | definite-integration | properties-of-definite-integration | <p>If $$\int_\limits{-0.15}^{0.15}\left|100 x^{2}-1\right| d x=\frac{k}{3000}$$, then $$k$$ is equal to ___________.</p> | [] | null | 575 | $$
\int\limits_{-0.15}^{0.15}\left|100 x^2-1\right| d x=2 \int\limits_0^{0.15}\left|100 x^2-1\right| \mathrm{dx}
$$
<br/><br/>$$
\text { Now } 100 x^2-1=0 \Rightarrow x^2=\frac{1}{100} \Rightarrow x=0.1
$$
<br/><br/>$$
I=2\left[\int_0^{0.1}\left(1-100 x^2\right) d x+\int_{0.1}^{0.15}\left(100 x^2-1\right) d x\right]
$$... | integer | jee-main-2023-online-12th-april-morning-shift | 5,753 |
1lgsvsu3q | maths | definite-integration | properties-of-definite-integration | <p>Let the function $$f:[0,2] \rightarrow \mathbb{R}$$ be defined as</p>
<p>$$f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]\right\},} & x \in[0,1) \\ e^{\left[x-\log _{e} x\right]}, & x \in[1,2]\end{cases}$$</p>
<p>where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then the value of the in... | [{"identifier": "A", "content": "$$2 e-1$$"}, {"identifier": "B", "content": "$$2 e-\\frac{1}{2}$$"}, {"identifier": "C", "content": "$$1+\\frac{3 e}{2}$$"}, {"identifier": "D", "content": "$$(e-1)\\left(e^{2}+\\frac{1}{2}\\right)$$"}] | ["B"] | null | $$
\begin{aligned}
\operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}\right\} & =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\\\
{\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}\right] } & =1 ; \mathrm{x} \in[1,2)
\end{aligned}
$$
<br/><br/>$$
\therefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}
\mathrm{e}^{\mathrm{x... | mcq | jee-main-2023-online-11th-april-evening-shift | 5,755 |
1lguu4z56 | maths | definite-integration | properties-of-definite-integration | <p>The value of the integral $$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\log _{e}\\left(\\frac{(2+\\sqrt{5})^{2}}{\\sqrt{1+\\sqrt{5}}}\\right)+\\frac{\\sqrt{5}}{2}$$"}, {"identifier": "B", "content": "$$\\log _{e}\\left(\\frac{\\sqrt{2}(2+\\sqrt{5})^{2}}{\\sqrt{1+\\sqrt{5}}}\\right)-\\frac{\\sqrt{5}}{2}$$"}, {"identifier": "C", "content": "$$\\log _{e}\\... | ["B"] | null | $$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$$
<br/><br/>Let $e^x=t \Rightarrow e^x d x=d t$
<br/><br/>When, $x \rightarrow-\log _e 2$, then $t \rightarrow \frac{1}{2}$
<br/><br/>When, $x \rightarrow \log _e 2$, then $t \rightarrow 2$
<br/><br/>$$
I=\int_... | mcq | jee-main-2023-online-11th-april-morning-shift | 5,756 |
1lguwr95k | maths | definite-integration | properties-of-definite-integration | <p>For $$m, n > 0$$, let $$\alpha(m, n)=\int_\limits{0}^{2} t^{m}(1+3 t)^{n} d t$$. If $$11 \alpha(10,6)+18 \alpha(11,5)=p(14)^{6}$$, then $$p$$ is equal to ___________.</p> | [] | null | 32 | We have, $\alpha(m, n)=\int\limits_0^2 t^m(1+3 t)^n d t$
<br/><br/>Also, $11 \alpha(10,6)+18 \alpha(11,5)=P(14)^6$
<br/><br/>$\Rightarrow 11 \int\limits_0^2 t^{10}(1+3 t)^6 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6$
<br/><br/>Using integration by part for expression $t^{10}(1+3 t)^6$
<br/><br/>$$
\begin{array}... | integer | jee-main-2023-online-11th-april-morning-shift | 5,757 |
1lgyoktep | maths | definite-integration | properties-of-definite-integration | <p>Let $$[t]$$ denote the greatest integer function. If $$\int_\limits{0}^{2.4}\left[x^{2}\right] d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}$$, then $$\alpha+\beta+\gamma+\delta$$ is equal to __________.</p> | [] | null | 6 | $$
\begin{aligned}
\int\limits_0^{2.4}\left[x^2\right] d x & =\int\limits_0^1\left[x^2\right] d x+\int\limits_1^{\sqrt{2}}\left[x^2\right] d x \\\\
& +\int\limits_{\sqrt{2}}^{\sqrt{3}}\left[x^2\right] d x+\int\limits_{\sqrt{3}}^2\left[x^2\right] d x+\int\limits_2^{\sqrt{5}}\left[x^2\right] d x+\int\limits_{\sqrt{5}}^{2... | integer | jee-main-2023-online-8th-april-evening-shift | 5,758 |
1lh0010vg | maths | definite-integration | properties-of-definite-integration | <p>Let $$[t]$$ denote the greatest integer $$\leq t$$. Then $$\frac{2}{\pi} \int_\limits{\pi / 6}^{5 \pi / 6}(8[\operatorname{cosec} x]-5[\cot x]) d x$$ is equal to __________.</p> | [] | null | 14 | $$
\begin{aligned}
& \text { Let } \mathrm{I}=\frac{2}{\pi} \int\limits_{\frac{\pi}{6}}^{ \frac{5\pi}{6}}\{8[\operatorname{cosec} x]-5[\cot x]\} d x \\\\
& =\frac{2}{\pi}\left[8 \int\limits_{\frac{\pi}{6}}^{\frac{5 \pi}{6}}[\operatorname{cosec} x] d x-5 \int\limits_{\frac{\pi}{6}}^{\frac{5 \pi}{6}}[\cot x] d x\right]
\... | integer | jee-main-2023-online-8th-april-morning-shift | 5,759 |
1lh21dq8w | maths | definite-integration | properties-of-definite-integration | <p>Let $$5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0$$. Then $$18 \int_\limits{1}^{2} f(x) d x$$ is equal to :</p> | [{"identifier": "A", "content": "$$10 \\log _{\\mathrm{e}} 2+6$$"}, {"identifier": "B", "content": "$$5 \\log _{e} 2-3$$"}, {"identifier": "C", "content": "$$10 \\log _{\\mathrm{e}} 2-6$$"}, {"identifier": "D", "content": "$$5 \\log _{\\mathrm{e}} 2+3$$"}] | ["C"] | null | We have,
<br/><br/>$$
5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x>0
$$ ..........(i)
<br/><br/>On replacing $x$ by $\frac{1}{x}$ in (i), we get
<br/><br/>$$
5 f\left(\frac{1}{x}\right)+4 f(x)=x+3
$$ ..........(ii)
<br/><br/>Now, using Eq. (i) $\times 5-$ (ii) $\times 4$, we get
<br/><br/>$$
\begin{aligned}
& 25... | mcq | jee-main-2023-online-6th-april-morning-shift | 5,760 |
1lh2y8vvh | maths | definite-integration | properties-of-definite-integration | <p>Let $$f(x)$$ be a function satisfying $$f(x)+f(\pi-x)=\pi^{2}, \forall x \in \mathbb{R}$$. Then $$\int_\limits{0}^{\pi} f(x) \sin x d x$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\pi^{2}$$"}, {"identifier": "B", "content": "$$\\frac{\\pi^{2}}{2}$$"}, {"identifier": "C", "content": "$$2 \\pi^{2}$$"}, {"identifier": "D", "content": "$$\\frac{\\pi^{2}}{4}$$"}] | ["A"] | null | Let $I=\int\limits_0^\pi f(x) \sin x d x$ ..........(i)
<br/><br/>$$
\begin{aligned}
& =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\\\
& =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii)
\end{aligned}
$$
<br/><br/>On adding Equations (i) and (ii), we get
<br/><br/>$$
\begin{aligned}
& 2 I=\int\limits_0^\pi[f(x)+f(\... | mcq | jee-main-2023-online-6th-april-evening-shift | 5,761 |
lsam93zl | maths | definite-integration | properties-of-definite-integration | If $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x=\mathrm{a} \pi+\mathrm{b} \sqrt{3}$, where $\mathrm{a}$ and $\mathrm{b}$ are rational numbers, then $9 \mathrm{a}+8 \mathrm{b}$ is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$\\frac{3}{2}$"}] | ["A"] | null | <p>To solve the given integral $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x$, we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles. Specifically for $\cos^4 x$, we can write it in terms of double angles as:</p>... | mcq | jee-main-2024-online-1st-february-evening-shift | 5,762 |
lsamq77b | maths | definite-integration | properties-of-definite-integration | The value of $\int\limits_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} \mathrm{~d} x$ is equal to : | [{"identifier": "A", "content": "-1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $\begin{aligned} & I=\int_0^1\left(2 x^3-3 x^2-x+1\right)^{1 / 3} d x \\\\ & I=\int_0^1\left((2 x-1)\left(x^2-x-1\right)\right)^{1 / 3} d x \\\\ & I=\int_0^1\left[(2(1-x)-1)\left((1-x)^2-(1-x)-1\right)\right]^{1 / 3} d x \\\\ & I=\int_0^1\left((1-2 x)\left(x^2-x-1\right)\right)^{1 / 3} d x\end{aligned}$
<br/><br/>$\beg... | mcq | jee-main-2024-online-1st-february-evening-shift | 5,763 |
lsao9n4x | maths | definite-integration | properties-of-definite-integration | The value of the integral $\int\limits_0^{\pi / 4} \frac{x \mathrm{~d} x}{\sin ^4(2 x)+\cos ^4(2 x)}$ equals :
| [{"identifier": "A", "content": "$\\frac{\\sqrt{2} \\pi^2}{8}$"}, {"identifier": "B", "content": "$\\frac{\\sqrt{2} \\pi^2}{16}$"}, {"identifier": "C", "content": "$\\frac{\\sqrt{2} \\pi^2}{32}$"}, {"identifier": "D", "content": "$\\frac{\\sqrt{2} \\pi^2}{64}$"}] | ["C"] | null | Take $I=\int\limits_0^{\pi / 4} \frac{x d x}{\sin ^4(2 x)+\cos ^4(2 x)}$
<br/><br/>Let $2 x=t$
<br/><br/>$2 d x=d t$
<br/><br/>$d x=\frac{d t}{2}$
<br/><br/>$\begin{aligned} & I=\int\limits_0^{\pi / 2} \frac{t / 2 \cdot 1 / 2 d t}{\sin ^4 t+\cos ^4 t} \\\\ & I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{t d t}{\sin ^4 t+... | mcq | jee-main-2024-online-1st-february-morning-shift | 5,764 |
lsbke98l | maths | definite-integration | properties-of-definite-integration | If $\int\limits_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} \mathrm{~d} x=\mathrm{a}+\mathrm{b} \sqrt{2}+\mathrm{c} \sqrt{3}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are rational numbers, then $2 \mathrm{a}+3 \mathrm{~b}-4 \mathrm{c}$ is equal to : | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "8"}] | ["D"] | null | <p>$$\begin{aligned}
& \int_\limits0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int_\limits0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x \\
& \frac{1}{2}\left[\int_\limits0^1 \sqrt{3+x} d x-\int_\limits0^1(\sqrt{1+x}) d x\right]
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}... | mcq | jee-main-2024-online-27th-january-morning-shift | 5,766 |
lsbkjqqv | maths | definite-integration | properties-of-definite-integration | If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$, and $\mathrm{I}_1=\int\limits_{\mathrm{a}}^{\mathrm{b}} x \sin \left(4 x-x^2\right) \mathrm{d} x, \mathrm{I}_2=\int\limits_{\mathrm{a}}^{\mathrm{b}} \sin \left(4 x-x^2\right) \mathrm{d} x$, then $36 \frac{\mathrm{I}_1}{\mathrm{... | [{"identifier": "A", "content": "80"}, {"identifier": "B", "content": "72"}, {"identifier": "C", "content": "66"}, {"identifier": "D", "content": "88"}] | ["B"] | null | <p>Equation of CE</p>
<p>$$\begin{aligned}
& y-1=-(x-3) \\
& x+y=4
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1cekv6/bbd7771e-59aa-40fe-80f6-e5c8fe8d22a7/2f978020-d3c5-11ee-a50b-bb659a2e1d74/file-1lt1cekv7.png?format=png" data-orsrc="https://app-content.cdn.examgoa... | mcq | jee-main-2024-online-27th-january-morning-shift | 5,767 |
jaoe38c1lscoi3t2 | maths | definite-integration | properties-of-definite-integration | <p>Let $$f(x)=\int_\limits0^x g(t) \log _{\mathrm{e}}\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right) \mathrm{dt}$$, where $$g$$ is a continuous odd function.
If $$\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+\mathrm{e}^x}\right) \mathrm{d} x=\left(\frac{\pi}{\alpha}\right)^2-\alpha$$, then $$\alpha$$ is equal... | [] | null | 2 | <p>$$f(x)=\int_\limits0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t$$</p>
<p>$$f(-x)=\int_\limits0^{-x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t$$</p>
<p>$$f(-x)=-\int_\limits0^x g(-y) \ln \left(\frac{1+y}{1-y}\right) d y$$</p>
<p>$$=-\int_\limits0^x g(y) \ln \left(\frac{1-y}{1+y}\right) d y$$ (g is odd)</p>
<p>$$f(-x... | integer | jee-main-2024-online-27th-january-evening-shift | 5,769 |
jaoe38c1lsd4i9ml | maths | definite-integration | properties-of-definite-integration | <p>Let $$f, g:(0, \infty) \rightarrow \mathbb{R}$$ be two functions defined by $$f(x)=\int\limits_{-x}^x\left(|t|-t^2\right) e^{-t^2} d t$$ and $$g(x)=\int\limits_0^{x^2} t^{1 / 2} e^{-t} d t$$. Then, the value of $$9\left(f\left(\sqrt{\log _e 9}\right)+g\left(\sqrt{\log _e 9}\right)\right)$$ is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <p>$$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\int_\limits{-\mathrm{x}}^{\mathrm{x}}\left(|\mathrm{t}|-\mathrm{t}^2\right) \mathrm{e}^{-\mathrm{t}^2} \mathrm{dt} \\
& \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \cdot\left(|\mathrm{x}|-\mathrm{x}^2\right) \mathrm{e}^{-\mathrm{x}^2} \ldots \ldots \ldots . .(1) \\
& \ma... | mcq | jee-main-2024-online-31st-january-evening-shift | 5,770 |
jaoe38c1lsd4ywg3 | maths | definite-integration | properties-of-definite-integration | <p>$$\left|\frac{120}{\pi^3} \int_\limits0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x\right| \text { is equal to }$$ ________.</p> | [] | null | 15 | <p>$$\begin{aligned}
& \int_\limits0^\pi \frac{x^2 \sin x \cdot \cos x}{\sin ^4 x+\cos ^4 x} d x \\
& =\int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x}{\sin ^4 x+\cos ^4 x}\left(x^2-(\pi-x)^2\right) d x \\
& =\int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x\left(2 \pi x-\pi^2\right)}{\sin ^4 x+\cos ^4 x}... | integer | jee-main-2024-online-31st-january-evening-shift | 5,771 |
jaoe38c1lse5tymd | maths | definite-integration | properties-of-definite-integration | <p>If the integral $$525 \int_\limits0^{\frac{\pi}{2}} \sin 2 x \cos ^{\frac{11}{2}} x\left(1+\operatorname{Cos}^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x$$ is equal to $$(n \sqrt{2}-64)$$, then $$n$$ is equal to _________.</p> | [] | null | 176 | <p>$$I=\int_\limits0^{\frac{\pi}{2}} \sin 2 x \cdot(\cos x)^{\frac{11}{2}}\left(1+(\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} d x$$</p>
<p>Put $$\cos x=t^2 \Rightarrow \sin x d x=-2 t d t$$</p>
<p>$$\begin{aligned}
& \therefore \mathrm{I}=4 \int_\limits0^1 \mathrm{t}^2 \cdot \mathrm{t}^{11} \sqrt{\left(1+\mathrm{t}^5\r... | integer | jee-main-2024-online-31st-january-morning-shift | 5,772 |
jaoe38c1lse60n68 | maths | definite-integration | properties-of-definite-integration | <p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a function defined by $$f(x)=\frac{4^x}{4^x+2}$$ and $$M=\int_\limits{f(a)}^{f(1-a)} x \sin ^4(x(1-x)) d x, N=\int_\limits{f(a)}^{f(1-a)} \sin ^4(x(1-x)) d x ; a \neq \frac{1}{2}$$. If $$\alpha M=\beta N, \alpha, \beta \in \mathbb{N}$$, then the least value of $$\alpha... | [] | null | 5 | <p>$$\mathrm{f}(\mathrm{a})+\mathrm{f}(1-\mathrm{a})=1$$</p>
<p>$$M=\int_\limits{f(a)}^{f(1-a)}(1-x) \cdot \sin ^4 x(1-x) d x$$</p>
<p>$$\mathrm{M}=\mathrm{N}-\mathrm{M} \qquad 2 \mathrm{M}=\mathrm{N}$$</p>
<p>$$\alpha=2 ; \beta=1 \text {; }$$</p>
<p>Ans. 5</p> | integer | jee-main-2024-online-31st-january-morning-shift | 5,773 |
jaoe38c1lsf0gf9u | maths | definite-integration | properties-of-definite-integration | <p>If the value of the integral $$\int_\limits{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{2123}}}\right) d x=\frac{\pi}{4}(\pi+a)-2$$, then the value of $$a$$ is</p> | [{"identifier": "A", "content": "$$-\\frac{3}{2}$$\n"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$\n"}, {"identifier": "D", "content": "2"}] | ["B"] | null | <p>$$\begin{aligned}
& I=\int_\limits{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{2023}}}\right) d x \\
& I=\int_\limits{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^{-x}}+\frac{1+\sin ^2 x}{1+e^{\sin (-x)^{2023}}}\right) d x
\end{aligned}$$</p>
<p>On Adding, we get</p>
<p>$$... | mcq | jee-main-2024-online-29th-january-morning-shift | 5,774 |
jaoe38c1lsflco0v | maths | definite-integration | properties-of-definite-integration | <p>If $$\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}$$, where $$\alpha, \beta$$ and $$\gamma$$ are rational numbers, then
$$3 \alpha+4 \beta-\gamma$$ is equal to _________.</p> | [] | null | 6 | <p>$$\begin{aligned}
& =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x \\
& =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}}|\sin x-\cos x| d x \\
& =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{3}}(\sin x-\cos x) d x \\
& =-1+2 \sqrt{2}-\sqrt{3} ... | integer | jee-main-2024-online-29th-january-evening-shift | 5,775 |
1lsg3nvgv | maths | definite-integration | properties-of-definite-integration | <p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a function defined by $$f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}$$, and $$g(x)=f(f(f(f(x))))$$. Then, $$18 \int_0^{\sqrt{2 \sqrt{5}}} x^2 g(x) d x$$ is equal to</p> | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "33"}, {"identifier": "C", "content": "39"}, {"identifier": "D", "content": "42"}] | ["C"] | null | <p>$$\begin{aligned}
&f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}\\
&f \circ f(x)=\frac{f(x)}{\left(1+f(x)^4\right)^{1 / 4}}=\frac{\frac{x}{\left(1+x^4\right)^{1 / 4}}}{\left(1+\frac{x^4}{1+x^4}\right)^{1 / 4}}=\frac{x}{\left(1+2 x^4\right)^{1 / 4}}\\
&f(f(f(f(x))))=\frac{x}{\left(1+4 x^4\right)^{1 / 4}}
\end{aligned}$$<... | mcq | jee-main-2024-online-30th-january-evening-shift | 5,776 |
1lsg3stoy | maths | definite-integration | properties-of-definite-integration | <p>Let $$y=f(x)$$ be a thrice differentiable function in $$(-5,5)$$. Let the tangents to the curve $$y=f(x)$$ at $$(1, f(1))$$ and $$(3, f(3))$$ make angles $$\pi / 6$$ and $$\pi / 4$$, respectively with positive $$x$$-axis. If $$27 \int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\al... | [{"identifier": "A", "content": "26"}, {"identifier": "B", "content": "$$-$$16"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "$$-$$14"}] | ["A"] | null | <p>$$\begin{aligned}
& y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x) \\
& \left.\frac{d y}{d x}\right)_{(1, f(1))}=f^{\prime}(1)=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \Rightarrow f^{\prime}(1)=\frac{1}{\sqrt{3}} \\
& \left.\frac{d y}{d x}\right)_{(3, f(3))}=f^{\prime}(3)=\tan \frac{\pi}{4}=1 \Rightarrow f^{\prime}(3... | mcq | jee-main-2024-online-30th-january-evening-shift | 5,777 |
1lsg420oq | maths | definite-integration | properties-of-definite-integration | <p>Let $$a$$ and $$b$$ be real constants such that the function $$f$$ defined by $$f(x)=\left\{\begin{array}{ll}x^2+3 x+a & , x \leq 1 \\ b x+2 & , x>1\end{array}\right.$$ be differentiable on $$\mathbb{R}$$. Then, the value of $$\int_\limits{-2}^2 f(x) d x$$ equals</p> | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "19/6"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "15/6"}] | ["C"] | null | <p>To determine the integral of the piecewise function <span class="math-container">$$f$$</span> over the interval <span class="math-container">$$[-2, 2]$$</span>, we first ensure that <span class="math-container">$$f$$</span> is differentiable on <span class="math-container">$$\mathbb{R}$$</span>, as given in the prob... | mcq | jee-main-2024-online-30th-january-evening-shift | 5,778 |
1lsga9myw | maths | definite-integration | properties-of-definite-integration | <p>Let $$f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbf{R}$$ be a differentiable function such that $$f(0)=\frac{1}{2}$$. If the $$\lim _\limits{x \rightarrow 0} \frac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{\mathrm{e}^{x^2}-1}=\alpha$$, then $$8 \alpha^2$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "16"}] | ["B"] | null | <p>$$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{\left(\frac{e^{x^2}-1}{x^2}\right) \times x^2} \\\\
& \lim _{x \rightarrow 0} \frac{\int_0^x f(t) d t}{x} \quad\left(\lim _{x \rightarrow 0} \frac{e^{x^2}-1}{x^2}=1\right) \\\\
& =\lim _{x \rightarrow 0} \frac{f(x)}{1} \text { (using L Hospital) ... | mcq | jee-main-2024-online-30th-january-morning-shift | 5,779 |
1lsgchdrl | maths | definite-integration | properties-of-definite-integration | <p>The value of $$9 \int_\limits0^9\left[\sqrt{\frac{10 x}{x+1}}\right] \mathrm{d} x$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$, is</p> | [] | null | 155 | <p>$$\begin{array}{ll}
\frac{10 x}{x+1}=1 & \Rightarrow x=\frac{1}{9} \\
\frac{10 x}{x+1}=4 & \Rightarrow x=\frac{2}{3} \\
\frac{10 x}{x+1}=9 & \Rightarrow x=9
\end{array}$$</p>
<p>$$\begin{aligned}
& \mathrm{I}=9\left(\int_\limits0^{1 / 9} 0 \mathrm{dx}+\int_\limits{1 / 9}^{2 / 3} 1\mathrm{d} x+\int_\limits{2 / 3}^9 2... | integer | jee-main-2024-online-30th-january-morning-shift | 5,780 |
luxwehsa | maths | definite-integration | properties-of-definite-integration | <p>The integral $$\int_\limits{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x$$ is equal to</p> | [{"identifier": "A", "content": "$$-1/2$$"}, {"identifier": "B", "content": "$$-1/4$$"}, {"identifier": "C", "content": "1/4"}, {"identifier": "D", "content": "1/2"}] | ["B"] | null | <p>$$\begin{aligned}
& \int_\limits{\frac{1}{4}}^{\frac{3}{4}} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x \\
& x=\cos 2 \theta \\
& \Rightarrow d x=(-2 \sin 2 \theta \mathrm{d} \theta)
\end{aligned}$$</p>
<p>Take limit as $$\alpha$$ and $$\beta$$</p>
<p>$$\begin{aligned}
& -2 \int_\limits\alpha^\beta \co... | mcq | jee-main-2024-online-9th-april-evening-shift | 5,781 |
lv0vxbpa | maths | definite-integration | properties-of-definite-integration | <p>$$\text { Let } f(x)=\left\{\begin{array}{lr}
-2, & -2 \leq x \leq 0 \\
x-2, & 0< x \leq 2
\end{array} \text { and } \mathrm{h}(x)=f(|x|)+|f(x)| \text {. Then } \int_\limits{-2}^2 \mathrm{~h}(x) \mathrm{d} x\right. \text { is equal to: }$$</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "1"}] | ["A"] | null | <p>$$f(x)=\left\{\begin{array}{cc}
-2 & -2 \leq x \leq 0 \\
x-2 & 0< x \leq 2
\end{array} h(x)=f|x|+|f(x)|\right.$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwk8ygqe/f2db7258-aa84-4e80-8605-2f6185c78c16/3e8a8d60-198f-11ef-b65b-abc5d1349d93/file-1lwk8ygqf.png?format=png" data-or... | mcq | jee-main-2024-online-4th-april-morning-shift | 5,783 |
lv0vxdgv | maths | definite-integration | properties-of-definite-integration | <p>If the shortest distance between the lines $$\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$$ and $$\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$$ is $$\frac{38}{3 \sqrt{5}} \mathrm{k}$$, and $$\int_\limits 0^{\mathrm{k}}\left[x^2\right] \mathrm{d} x=\alpha-\sqrt{\alpha}$$, where $$[x]$$ denotes the greatest integer functi... | [] | null | 48 | <p>$$L_1: \frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$$</p>
<p>$$\begin{aligned}
& \vec{b}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\
& \vec{a}_1=-2 \hat{i}-3 \hat{j}+5 \hat{k}
\end{aligned}$$</p>
<p>$$L_2=\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$$</p>
<p>$$\begin{aligned}
& \vec{a}_2=3 \hat{i}+2 \hat{j}-4 \hat{k} \\
& \vec{b... | integer | jee-main-2024-online-4th-april-morning-shift | 5,784 |
lv0vxdop | maths | definite-integration | properties-of-definite-integration | <p>If $$\int_0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} \mathrm{~d} x=\frac{1}{\mathrm{a}} \log _{\mathrm{e}}\left(\frac{\mathrm{a}}{3}\right)+\frac{\pi}{\mathrm{b} \sqrt{3}}$$, where $$\mathrm{a}, \mathrm{b} \in \mathrm{N}$$, then $$\mathrm{a}+\mathrm{b}$$ is equal to _________.</p> | [] | null | 8 | <p>$$I=\int_\limits0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} d x=\int_\limits0^{\frac{\pi}{4}} \frac{\sin ^2 x}{\sin ^2 x+\cos ^2 x+\sin x \cos x} d x$$
<p>$$I=\int_\limits0^{\frac{\pi}{4}} \frac{\tan ^2 x}{1+\tan x+\tan ^2 x} d x$$</p>
<p>$$=\int_\limits0^{\frac{\pi}{4}} \frac{\tan x \cdot \sec ^2 x d x}{\le... | integer | jee-main-2024-online-4th-april-morning-shift | 5,785 |
lv2ers9h | maths | definite-integration | properties-of-definite-integration | <p>If the value of the integral $$\int\limits_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$$ is $$\frac{2}{\pi}$$.Then, a value of $$\alpha$$ is</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{2}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{4}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\pi}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{6}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \text { Given, } \int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x=\frac{2}{\pi} \\
& \begin{aligned}
I & =\int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x \\
\Rightarrow I & =\int_\limits0^1\left(\frac{\cos \alpha x}{1+3^x}+\frac{\cos \alpha x}{1+3^{-x}}\right) d x \\
& =\int_\limits0^1 \cos ... | mcq | jee-main-2024-online-4th-april-evening-shift | 5,786 |
lv5grw2g | maths | definite-integration | properties-of-definite-integration | <p>The value of $$k \in \mathbb{N}$$ for which the integral $$I_n=\int_0^1\left(1-x^k\right)^n d x, n \in \mathbb{N}$$, satisfies $$147 I_{20}=148 I_{21}$$ is</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "10"}] | ["C"] | null | <p>$$\begin{aligned}
& I(21)=\int_\limits0^1\left(1-x^k\right)^{21} d x \\
& =\int_\limits0^1\left(1-x^k\right)\left(1-x^k\right)^{20} d x \\
& =\int_\limits0^1\left(1-x^k\right)^{20} d x-\int_0 x^k\left(1-x^k\right)^{20} d x \\
& I(21)=I(20)-\int_\limits0^1 x^k\left(1-x^k\right)^{20} d x
\end{aligned}$... | mcq | jee-main-2024-online-8th-april-morning-shift | 5,788 |
lv7v4o3r | maths | definite-integration | properties-of-definite-integration | <p>The integral $$\int_\limits0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} \mathrm{~d} x$$ is equal to :</p> | [{"identifier": "A", "content": "$$3 \\pi-50 \\log _e 2+20 \\log _e 5$$\n"}, {"identifier": "B", "content": "$$3 \\pi-25 \\log _e 2+10 \\log _e 5$$\n"}, {"identifier": "C", "content": "$$3 \\pi-10 \\log _{\\mathrm{e}}(2 \\sqrt{2})+10 \\log _{\\mathrm{e}} 5$$\n"}, {"identifier": "D", "content": "$$3 \\pi-30 \\log _e 2+2... | ["A"] | null | <p>$$\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$$</p>
<p>$$\begin{aligned}
& \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\
& \begin{array}{l}
3 A-5 B=1 \\
5 A+3 B=0
\end{array}>A=\frac{3}{34} \quad B=\frac{-5}{34}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \int_0^{\pi / 4} \frac{136\left[\frac{3}{3... | mcq | jee-main-2024-online-5th-april-morning-shift | 5,789 |
lv7v3ka9 | maths | definite-integration | properties-of-definite-integration | <p>The value of $$\int_\limits{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{2}$$\n"}, {"identifier": "B", "content": "$$\\pi^2$$\n"}, {"identifier": "C", "content": "$$\\frac{\\pi^2}{2}$$\n"}, {"identifier": "D", "content": "$$2 \\pi^2$$"}] | ["B"] | null | <p>$$\begin{aligned}
& I=\int_\limits{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y \\
& =\int_\limits0^\pi\left(\frac{2 y(1+\sin y)}{1+\cos ^2 y}+\frac{-2 y(1-\sin y)}{1+\cos ^2 y}\right) d y \\
& =\int_\limits0^\pi\left(\frac{2 y+2 y \sin y-2 y+2 y \sin y}{1+\cos ^2 y}\right) d y
\end{aligned}$$</p>
<p>$$I=4 \int_0... | mcq | jee-main-2024-online-5th-april-morning-shift | 5,790 |
lv9s205z | maths | definite-integration | properties-of-definite-integration | <p>Let $$\beta(\mathrm{m}, \mathrm{n})=\int_\limits0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m}, \mathrm{n}>0$$. If $$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x=\mathrm{a} \times \beta(\mathrm{b}, \mathrm{c})$$, then $$100(\mathrm{a}+\mathrm{b}+\mathrm{c})$$ equals _________.</p> | [{"identifier": "A", "content": "2012"}, {"identifier": "B", "content": "1021"}, {"identifier": "C", "content": "1120"}, {"identifier": "D", "content": "2120"}] | ["D"] | null | <p>First, let's rewrite the given integral using the given form of the Beta function. The given integral is:</p>
<p>
<p>$$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x$$</p>
</p>
<p>To use the Beta function, let us make a substitution. Let $ x^{10} = t $. Then, $ dx = \frac{1}{10}t^{-\frac{9}{10}} dt $ or... | mcq | jee-main-2024-online-5th-april-evening-shift | 5,791 |
lv9s20mx | maths | definite-integration | properties-of-definite-integration | <p>If $$f(t)=\int_\limits0^\pi \frac{2 x \mathrm{~d} x}{1-\cos ^2 \mathrm{t} \sin ^2 x}, 0<\mathrm{t}<\pi$$, then the value of $$\int_\limits0^{\frac{\pi}{2}} \frac{\pi^2 \mathrm{dt}}{f(\mathrm{t})}$$ equals __________.</p> | [] | null | 1 | <p>$$\begin{aligned}
& f(t)=\int_\limits0^\pi \frac{2 x d x}{1-\cos ^2 t \sin ^2 x} \\
& x \rightarrow \pi-x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& f(t)=\int_\limits0^\pi \frac{2(\pi-x) d x}{1-\cos ^2 t \sin ^2 x}=2 \pi \int_\limits0^\pi \frac{d x}{1-\cos ^2 t \sin ^2 x}-f(t) \\
& \Rightarrow f(t)=\pi \int_\limits0... | integer | jee-main-2024-online-5th-april-evening-shift | 5,792 |
lvc57ayt | maths | definite-integration | properties-of-definite-integration | <p>$$\int_\limits0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \text { is equal to }$$</p> | [{"identifier": "A", "content": "1/9"}, {"identifier": "B", "content": "1/6"}, {"identifier": "C", "content": "1/3"}, {"identifier": "D", "content": "1/12"}] | ["B"] | null | <p>$$\begin{aligned}
& \int_\limits0^{\pi / 4} \frac{\cos ^2 x \cdot \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \\
& =\int_\limits0^{\pi / 4} \frac{\tan ^2 x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x
\end{aligned}$$</p>
<p>Let $$\tan x=t$$</p>
<p>$$\int_\limits0^1 \frac{t^2 d t}{\left(1+t^3\right)^2}$$</... | mcq | jee-main-2024-online-6th-april-morning-shift | 5,794 |
lvc58e3u | maths | definite-integration | properties-of-definite-integration | <p>Let $$r_k=\frac{\int_0^1\left(1-x^7\right)^k d x}{\int_0^1\left(1-x^7\right)^{k+1} d x}, k \in \mathbb{N}$$. Then the value of $$\sum_\limits{k=1}^{10} \frac{1}{7\left(r_k-1\right)}$$ is equal to _________.</p> | [] | null | 65 | <p>$$r_k=\frac{I_a}{I_b} \text {, where } I_a=\int_0^1\left(1-x^7\right)^k d x$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd3wwxo/4611eb8f-9da2-472d-abed-83945d5d0d49/093f3fc0-15a2-11ef-9ea4-3bb319c2f90f/file-1lwd3wwxp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/imag... | integer | jee-main-2024-online-6th-april-morning-shift | 5,795 |
fRLdYJcHGpcbyscR | maths | differential-equations | formation-of-differential-equations | The differential equation for the family of circle $${x^2} + {y^2} - 2ay = 0,$$ where a is an arbitrary constant is : | [{"identifier": "A", "content": "$$\\left( {{x^2} + {y^2}} \\right)y' = 2xy$$ "}, {"identifier": "B", "content": "$$2\\left( {{x^2} + {y^2}} \\right)y' = xy$$"}, {"identifier": "C", "content": "$$\\left( {{x^2} - {y^2}} \\right)y' =2 xy$$"}, {"identifier": "D", "content": "$$2\\left( {{x^2} - {y^2}} \\right)y' = xy$$"}... | ["C"] | null | $${x^2} + {y^2} - 2ay = 0\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>Differentiate,
<br><br>$$2x + 2y{{dy} \over {dx}} - 2a{{dy} \over {dx}} = 0$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = {{x + yy'} \over {y'}}$$
<br><br>Put in $$(1),$$ $${x^2} + {y^2} - 2\left( {{{x + yy'} \over {y'}}} \right) ... | mcq | aieee-2004 | 5,796 |
Po1lw9JbsWomUyZs | maths | differential-equations | formation-of-differential-equations | The differential equation of all circles passing through the origin and having their centres on the $$x$$-axis is : | [{"identifier": "A", "content": "$${y^2} = {x^2} + 2xy{{dy} \\over {dx}}$$ "}, {"identifier": "B", "content": "$${y^2} = {x^2} - 2xy{{dy} \\over {dx}}$$ "}, {"identifier": "C", "content": "$${x^2} = {y^2} + xy{{dy} \\over {dx}}$$"}, {"identifier": "D", "content": "$${x^2} = {y^2} + 3xy{{dy} \\over {dx}}$$"}] | ["A"] | null | General equation of circles passing through origin
<br><br>and having their centres on the $$x$$-axis is
<br><br>$${x^2} + {y^2} + 2gx = 0\,\,\,\,....\left( i \right)$$
<br><br>On differentiating $$w.r.t.x,$$ we get
<br><br>$$2x + 2y.{{dy} \over {dx}} + 2g = 0$$
<br><br>$$ \Rightarrow g = - \left( {x + y{{dy} \over ... | mcq | aieee-2007 | 5,797 |
YmDJDelJwPq7xlxD | maths | differential-equations | formation-of-differential-equations | The differential equation which represents the family of curves $$y = {c_1}{e^{{c_2}x}},$$ where $${c_1}$$ , and $${c_2}$$ are arbitrary constants, is | [{"identifier": "A", "content": "$$y'' = y'y$$ "}, {"identifier": "B", "content": "$$yy'' = y'$$ "}, {"identifier": "C", "content": "$$yy'' = {\\left( {y'} \\right)^2}$$ "}, {"identifier": "D", "content": "$$y' = {y^2}$$ "}] | ["C"] | null | We have $$y = {c_1}{e^{{c_2}x}}$$
<br><br>$$ \Rightarrow y' = {c_1}{c_2}{e^{{c_2}x}} = {c_2}y$$
<br><br>$$ \Rightarrow {{y'} \over y} = {c_2}$$
<br><br>$$ \Rightarrow {{y''y\left( {y'} \right){}^2} \over {{y^2}}} = 0$$
<br><br>$$ \Rightarrow y''y = {\left( {y'} \right)^2}$$ | mcq | aieee-2009 | 5,798 |
9MRXujl4STa4xPl5c4rEH | maths | differential-equations | formation-of-differential-equations | The differential equation representing the family of ellipse having foci eith on the x-axis or on the $$y$$-axis, center at the origin and passing through the point (0, 3) is : | [{"identifier": "A", "content": "xy y'' + x (y')<sup>2</sup> $$-$$ y y' = 0"}, {"identifier": "B", "content": "x + y y'' = 0"}, {"identifier": "C", "content": "xy y'+ y<sup>2</sup> $$-$$ 9 = 0"}, {"identifier": "D", "content": "xy y' $$-$$ y<sup>2</sup> + 9 = 0"}] | ["D"] | null | Equation of ellipse,
<br><br>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
<br><br>As ellipse passes through (0, 3)
<br><br>$$\therefore\,\,\,$$ $${{{0^2}} \over {{a^2}}} + {{{3^2}} \over {{b^2}}} = 1$$
<br><br>$$ \Rightarrow $$ b<sup>2</sup> = 9
<br><br>$$\therefore\,\,\,$$ Equation of ellipse... | mcq | jee-main-2018-online-16th-april-morning-slot | 5,799 |
Wv8Zsmh00NPdsxTb9E7k9k2k5hk6uoo | maths | differential-equations | formation-of-differential-equations | The differential equation of the family of
curves, x<sup>2</sup> = 4b(y + b), b $$ \in $$ R, is : | [{"identifier": "A", "content": "x(y')<sup>2</sup> = x \u2013 2yy'"}, {"identifier": "B", "content": "x(y')<sup>2</sup> = 2yy' \u2013 x"}, {"identifier": "C", "content": "xy\" = y'"}, {"identifier": "D", "content": "x(y')<sup>2</sup> = x + 2yy'"}] | ["D"] | null | x<sup>2</sup> = 4b(y + b)
<br><br>$$ \Rightarrow $$ 2x = 4by'
<br><br>$$ \Rightarrow $$ b = $${x \over {2y'}}$$
<br><br>$$ \therefore $$ differential equation is
<br><br>x<sup>2</sup> = 4.y.$${x \over {2y'}}$$ + 4$${\left( {{x \over {2y'}}} \right)^2}$$
<br><br>$$ \Rightarrow $$ x<sup>2</sup> = $${{2xy} \over {y'}}$$ +... | mcq | jee-main-2020-online-8th-january-evening-slot | 5,800 |
FT0YVlJ8yJJqRUDqM9jgy2xukg0c7il3 | maths | differential-equations | formation-of-differential-equations | If $$y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$$ is the solution of the
differential equation, <br/><br>$${{dy} \over {dx}} + p\left( x \right)y = {2 \over \pi } cosec\,x$$,<br/><br/>
$$0 < x < {\pi \over 2}$$, then the function p(x) is equal to :</br> | [{"identifier": "A", "content": "cot x"}, {"identifier": "B", "content": "sec x"}, {"identifier": "C", "content": "tan x"}, {"identifier": "D", "content": "cosec x"}] | ["A"] | null | $$y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$$
<br>Differentiate w.r.t x
<br><br>$${{dy} \over {dx}} = {2 \over \pi }$$cosec x - $$\left( {{2 \over \pi }x - 1} \right)$$cosec x.cot x
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ + $$\left( {{2 \over \pi }x - 1} \right)$$cosec x.cot x = $${2 \over \pi }$$cosec ... | mcq | jee-main-2020-online-6th-september-evening-slot | 5,801 |
Fh1ueAAgvs7uHquWTM1klrk5zd2 | maths | differential-equations | formation-of-differential-equations | If a curve y = f(x) passes through the point (1, 2) and satisfies $$x {{dy} \over {dx}} + y = b{x^4}$$, then for what value of b, $$\int\limits_1^2 {f(x)dx = {{62} \over 5}} $$? | [{"identifier": "A", "content": "$${{31} \\over 5}$$"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$${{62} \\over 5}$$"}] | ["B"] | null | $${{dy} \over {dx}} + {y \over x} = b{x^3}$$, $$I.F. = {e^{\int {{{dx} \over x}} }} = x$$<br><br>$$ \therefore $$ $$yx = \int {b{x^4}dx} = {{b{x^5}} \over 5} + C$$<br><br>Passes through (1, 2), we get<br><br>$$2 = {b \over 5} + C$$ ........ (i)<br><br>Also, $$\int\limits_1^2 {\left( {{{b{x^4}} \over 5} + {c \over x}} ... | mcq | jee-main-2021-online-24th-february-evening-slot | 5,802 |
9Fv9AhxXoba7dLVrr21kmlhwgmb | maths | differential-equations | formation-of-differential-equations | The differential equation satisfied by the system of parabolas <br/><br/>y<sup>2</sup> = 4a(x + a) is : | [{"identifier": "A", "content": "$$y{\\left( {{{dy} \\over {dx}}} \\right)^2} - 2x\\left( {{{dy} \\over {dx}}} \\right) - y = 0$$"}, {"identifier": "B", "content": "$$y{\\left( {{{dy} \\over {dx}}} \\right)^2} - 2x\\left( {{{dy} \\over {dx}}} \\right) + y = 0$$"}, {"identifier": "C", "content": "$$y{\\left( {{{dy} \\ov... | ["C"] | null | $${y^2} = 4ax + 4{a^2}$$<br><br>differentiate with respect to x<br><br>$$ \Rightarrow 2y{{dy} \over {dx}} = 4a$$<br><br>$$ \Rightarrow a = \left( {{y \over 2}{{dy} \over {dx}}} \right)$$<br><br>So, required differential equation is <br><br>$${y^2} = \left( {4 \times {y \over 2}{{dy} \over {dx}}} \right)x + 4{\left( {{y... | mcq | jee-main-2021-online-18th-march-morning-shift | 5,803 |
1krzrdrnv | maths | differential-equations | formation-of-differential-equations | Let a curve y = f(x) pass through the point (2, (log<sub>e</sub>2)<sup>2</sup>) and have slope $${{2y} \over {x{{\log }_e}x}}$$ for all positive real value of x. Then the value of f(e) is equal to ______________. | [] | null | 1 | $$y' = {{2y} \over {x\ln x}}$$<br><br>$$ \Rightarrow {{dy} \over y} = {{2dx} \over {x\ln x}}$$<br><br>$$ \Rightarrow \ln |y| = 2\ln |\ln x| + C$$<br><br>put x = 2, y = (ln2)<sup>2</sup><br><br>$$\Rightarrow$$ c = 0<br><br>$$\Rightarrow$$ y = (lnx)<sup>2</sup><br><br>$$\Rightarrow$$ f(e) = 1 | integer | jee-main-2021-online-25th-july-evening-shift | 5,804 |
1ktepvi10 | maths | differential-equations | formation-of-differential-equations | If $${y^{1/4}} + {y^{ - 1/4}} = 2x$$, and <br/><br/>$$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + \alpha x{{dy} \over {dx}} + \beta y = 0$$, then | $$\alpha$$ $$-$$ $$\beta$$ | is equal to __________. | [] | null | 17 | $${y^{{1 \over 4}}} + {1 \over {{y^{{1 \over 4}}}}} = 2x$$<br><br>$$ \Rightarrow {\left( {{y^{{1 \over 4}}}} \right)^2} - 2x{y^{\left( {{1 \over 4}} \right)}} + 1 = 0$$<br><br>$$ \Rightarrow {y^{{1 \over 4}}} = x + \sqrt {{x^2} - 1} $$ or $$x - \sqrt {{x^2} - 1} $$<br><br>So, $${1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{... | integer | jee-main-2021-online-27th-august-morning-shift | 5,805 |
1ktfzg9uq | maths | differential-equations | formation-of-differential-equations | A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2, $$-$$3) from the line 3x + 4y = 5, is given by : | [{"identifier": "A", "content": "$$10{{{d^2}y} \\over {d{x^2}}} = 11$$"}, {"identifier": "B", "content": "$$11{{{d^2}x} \\over {d{y^2}}} = 10$$"}, {"identifier": "C", "content": "$$10{{{d^2}x} \\over {d{y^2}}} = 11$$"}, {"identifier": "D", "content": "$$11{{{d^2}y} \\over {d{x^2}}} = 10$$"}] | ["D"] | null | Length of latus rectum<br><br>$$= {{|3(2) + 4( - 3) - 5|} \over 5} = {{11} \over 5}$$<br><br>$${(x - h)^2} = {{11} \over 5}(y - k)$$<br><br>differentiate w.r.t. 'x' :-<br><br>$$2(x - h) = {{11} \over 5}{{dy} \over {dx}}$$<br><br>again differentiate<br><br>$$2 = {{11} \over 5}{{{d^2}y} \over {d{x^2}}}$$<br><br>$${{11{d^... | mcq | jee-main-2021-online-27th-august-evening-shift | 5,806 |
1l57o2hul | maths | differential-equations | formation-of-differential-equations | <p>Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$$, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "5"}] | ["B"] | null | <p>$${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$$</p>
<p>$$ = bx\,dy + cy\,dy + a\,dy = ax\,dx - by\,dx + a\,dx$$</p>
<p>$$ = cy\,dy + a\,dy - ax\,dx - a\,dx + b(x\,dy + y\,dx) = 0$$</p>
<p>$$ = c\int {y\,dy + a\int {x\,dx - a\int {dx + b\int {d(xy) = 0} } } } $$</p>
<p>$$ = {{c{y^2}} \over 2} + ay - {{a{x... | mcq | jee-main-2022-online-27th-june-morning-shift | 5,807 |
1l5vzx4yi | maths | differential-equations | formation-of-differential-equations | <p>Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$$, represents a circle with center ($$\alpha$$, $$\beta$$). Then, $$\alpha$$ + 2$$\beta$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <p>Given,</p>
<p>$${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$$</p>
<p>$$ \Rightarrow bxdy + cydy + ady = axdx - bydx + adx$$</p>
<p>$$ \Rightarrow bxdy + bydx + (cy + a)dy = (ax + a)dx$$</p>
<p>$$ \Rightarrow b(xdy + ydx) + (cy + a)dy = (ax + a)dx$$</p>
<p>$$ \Rightarrow bd(xy) + (cy + a)dy = (ax + a)dx$$... | mcq | jee-main-2022-online-30th-june-morning-shift | 5,808 |
1l6f1ihu7 | maths | differential-equations | formation-of-differential-equations | <p>Let a smooth curve $$y=f(x)$$ be such that the slope of the tangent at any point $$(x, y)$$ on it is directly proportional to $$\left(\frac{-y}{x}\right)$$. If the curve passes through the points $$(1,2)$$ and $$(8,1)$$, then $$\left|y\left(\frac{1}{8}\right)\right|$$ is equal to</p> | [{"identifier": "A", "content": "$$2 \\log _{e} 2$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$4 \\log _{e} 2$$"}] | ["B"] | null | <p>$${{dy} \over {dx}} \propto {{ - y} \over x}$$</p>
<p>$${{dy} \over {dx}} = {{ - ky} \over x} \Rightarrow \int {{{dy} \over y} = - K\int {{{dx} \over x}} } $$</p>
<p>$$\ln |y| = - K\ln |x| + C$$</p>
<p>If the above equation satisfy (1, 2) and (8, 1)</p>
<p>$$\ln 2 = - K \times 0 + C \Rightarrow C = \ln 2$$</p>
<p... | mcq | jee-main-2022-online-25th-july-evening-shift | 5,809 |
1l6gjjaoy | maths | differential-equations | formation-of-differential-equations | <p>Let a curve $$y=y(x)$$ pass through the point $$(3,3)$$ and the area of the region under this curve, above the $$x$$-axis and between the abscissae 3 and $$x(>3)$$ be $$\left(\frac{y}{x}\right)^{3}$$. If this curve also passes through the point $$(\alpha, 6 \sqrt{10})$$ in the first quadrant, then $$\alpha$$ is e... | [] | null | 6 | <p>$$\int\limits_3^x {f(x)dx = {{\left( {{{f(x)} \over x}} \right)}^3}} $$</p>
<p>$${x^3}\,.\,\int\limits_3^x {f(x)dx = {f^3}(x)} $$</p>
<p>Differentiate w.r.t. x</p>
<p>$${x^3}f(x) + 3{x^2}\,.\,{{{f^3}(x)} \over {{x^3}}} = 3{f^2}(x)f'(x)$$</p>
<p>$$ \Rightarrow 3{y^2}{{dy} \over {dx}} = {x^3}y + {{3{y^3}} \over x}$$</... | integer | jee-main-2022-online-26th-july-morning-shift | 5,810 |
1l6nn544i | maths | differential-equations | formation-of-differential-equations | <p>The differential equation of the family of circles passing through the points $$(0,2)$$ and $$(0,-2)$$ is :</p> | [{"identifier": "A", "content": "$$2 x y \\frac{d y}{d x}+\\left(x^{2}-y^{2}+4\\right)=0$$"}, {"identifier": "B", "content": "$$2 x y \\frac{d y}{d x}+\\left(x^{2}+y^{2}-4\\right)=0$$"}, {"identifier": "C", "content": "$$2 x y \\frac{d y}{d x}+\\left(y^{2}-x^{2}+4\\right)=0$$"}, {"identifier": "D", "content": "$$2 x y ... | ["A"] | null | <p>Family of circles passing through the points (0, 2) and (0, $$-$$2)</p>
<p>$${x^2} + (y - 2)(y + 2) + \lambda x = 0,\,\lambda \in R$$</p>
<p>$${x^2} + {y^2} + \lambda x - 4 = 0$$ ...... (1)</p>
<p>Differentiate w.r.t x</p>
<p>$$2x + 2y{{dy} \over {dx}} + \lambda = 0$$ ....... (2)</p>
<p>Using (1) and (2), eliminat... | mcq | jee-main-2022-online-28th-july-evening-shift | 5,811 |
1ldprlg0x | maths | differential-equations | formation-of-differential-equations | <p> Let a differentiable function $$f$$ satisfy $$f(x)+\int_\limits{3}^{x} \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3$$. Then $$12 f(8)$$ is equal to :</p> | [{"identifier": "A", "content": "19"}, {"identifier": "B", "content": "34"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "1"}] | ["C"] | null | Differentiating both sides we get,
<br/><br/>$$
\begin{aligned}
& f^{\prime}(x)+\frac{f(x)}{x}=\frac{1}{2 \sqrt{x+1}} \\\\
& \Rightarrow \frac{d y}{d x}+\frac{y}{x}=\frac{1}{2 \sqrt{x+1}} \\\\
& \Rightarrow \mathrm{IF}=x \\\\
& \Rightarrow y x=\frac{1}{2} \int \frac{x}{\sqrt{x+1}} d x+c \\\\
& \Rightarrow y x=\frac{1}... | mcq | jee-main-2023-online-31st-january-morning-shift | 5,812 |
lv5grwbn | maths | differential-equations | formation-of-differential-equations | <p>Let $$f(x)$$ be a positive function such that the area bounded by $$y=f(x), y=0$$ from $$x=0$$ to $$x=a>0$$ is $$e^{-a}+4 a^2+a-1$$. Then the differential equation, whose general solution is $$y=c_1 f(x)+c_2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants, is</p> | [{"identifier": "A", "content": "$$\\left(8 e^x+1\\right) \\frac{d^2 y}{d x^2}-\\frac{d y}{d x}=0$$\n"}, {"identifier": "B", "content": "$$\\left(8 e^x+1\\right) \\frac{d^2 y}{d x^2}+\\frac{d y}{d x}=0$$\n"}, {"identifier": "C", "content": "$$\\left(8 e^x-1\\right) \\frac{d^2 y}{d x^2}-\\frac{d y}{d x}=0$$\n"}, {"ident... | ["B"] | null | <p>$$\int_\limits0^a f(x) d x=e^{-a}+4 a^2+a-1$$</p>
<p>Differentiate equation w.r.t. 'a'</p>
<p>$$\begin{aligned}
& f(a)=-e^{-a}+8 a+1 \\
& \Rightarrow f(x)=-e^{-x}+8 x+1
\end{aligned}$$</p>
<p>And $$y=c_1 f(x)+c_2$$</p>
<p>$$\begin{aligned}
& y=c_1\left(-e^{-x}+8 x+1\right)+c_2 \\
& y^{\prime}=c_1\left(e^{-x}+8\right... | mcq | jee-main-2024-online-8th-april-morning-shift | 5,813 |
lv9s20hz | maths | differential-equations | formation-of-differential-equations | <p>The differential equation of the family of circles passing through the origin and having centre at the line $$y=x$$ is :</p> | [{"identifier": "A", "content": "$$\\left(x^2-y^2+2 x y\\right) \\mathrm{d} x=\\left(x^2-y^2+2 x y\\right) \\mathrm{d} y$$\n"}, {"identifier": "B", "content": "$$\\left(x^2+y^2-2 x y\\right) \\mathrm{d} x=\\left(x^2+y^2+2 x y\\right) \\mathrm{d} y$$\n"}, {"identifier": "C", "content": "$$\\left(x^2+y^2+2 x y\\right) \\... | ["D"] | null | <p>Equation of circle passing through origin & having centre at the line $$y=x$$ is</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwegsauf/ce0513c1-e394-4ced-87e0-f3bb34e8f595/28467680-1661-11ef-a209-0180f41f1307/file-1lwegsaug.png?format=png" data-orsrc="https://app-content.cdn.examgoal.n... | mcq | jee-main-2024-online-5th-april-evening-shift | 5,814 |
40MGPqbhsQ7TtAf7 | maths | differential-equations | linear-differential-equations | The solution of the differential equation
<br/><br/>$$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$$ is : | [{"identifier": "A", "content": "$$x{e^{2{{\\tan }^{ - 1}}y}} = {e^{{{\\tan }^{ - 1}}y}} + k$$ "}, {"identifier": "B", "content": "$$\\left( {x - 2} \\right) = k{e^{2{{\\tan }^{ - 1}}y}}$$ "}, {"identifier": "C", "content": "$$2x{e^{{{\\tan }^{ - 1}}y}} = {e^{2{{\\tan }^{ - 1}}y}} + k$$ "}, {"identifier": "D", "content... | ["C"] | null | $$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0$$
<br><br>$$ \Rightarrow {{dx} \over {dy}} + {x \over {\left( {1 + {y^2}} \right)}} = {{{e^{{{\tan }^{ - 1}}y}}} \over {\left( {1 + {y^2}} \right)}}$$
<br><br>$$I.F = e{}^{\int {{1 \over {\left( {1 + {y^2}} \right)}}dy} } =... | mcq | aieee-2003 | 5,815 |
66nkD6pi4hLZiTbu | maths | differential-equations | linear-differential-equations | The solution of the differential equation
<br/><br>$${{dy} \over {dx}} = {{x + y} \over x}$$ satisfying the condition $$y(1)=1$$ is :</br> | [{"identifier": "A", "content": "$$y = \\ln x + x$$ "}, {"identifier": "B", "content": "$$y = x\\ln x + {x^2}$$ "}, {"identifier": "C", "content": "$$y = x{e^{\\left( {x - 1} \\right)}}\\,$$ "}, {"identifier": "D", "content": "$$y = x\\,\\ln x + x$$ "}] | ["D"] | null | $${{dy} \over {dx}} = {{x + y} \over x} = 1 + {y \over x}$$
<br><br>Putting $$y=$$ $$vx$$
<br><br>and $${{dy} \over {dx}} = v + x{{dv} \over {dx}}$$
<br><br>we get
<br><br>$$v + x{{dv} \over {dx}} = 1 + v$$
<br><br>$$ \Rightarrow \int {{{dx} \over x}} = \int {dv} $$
<br><br>$$ \Rightarrow v = \ln {\mkern 1mu} x + c$$... | mcq | aieee-2008 | 5,816 |
MRSMdUCqIIT55v0x | maths | differential-equations | linear-differential-equations | If $${{dy} \over {dx}} = y + 3 > 0\,\,$$ and $$y(0)=2,$$ then $$y\left( {\ln 2} \right)$$ is equal to : | [{"identifier": "A", "content": "$$5$$"}, {"identifier": "B", "content": "$$13$$"}, {"identifier": "C", "content": "$$-2$$"}, {"identifier": "D", "content": "$$7$$"}] | ["D"] | null | $${{dy} \over {dx}} = y + 3 \Rightarrow \int {{{dy} \over {y + 3}}} = \int {dx} $$
<br><br>$$ \Rightarrow \ell n\left| {y + 3} \right| = x + c$$
<br><br>Since $$y\left( 0 \right) = 2,\,\,\,$$ $$\,\,\,\,\,\,\,$$ $$\therefore$$ $$\,\,\,\,\,\,\,$$ $$\ell n5 = c$$
<br><br>$$ \Rightarrow \ell n\left| {y + 3 = x + \ell n5} ... | mcq | aieee-2011 | 5,817 |
3IxYLRiMUbRQ20y2 | maths | differential-equations | linear-differential-equations | Let $$y(x)$$ be the solution of the differential equation
<br/><br>$$\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right).$$ Then $$y(e)$$ is equal to :</br> | [{"identifier": "A", "content": "$$2$$ "}, {"identifier": "B", "content": "$$2e$$ "}, {"identifier": "C", "content": "$$e$$ "}, {"identifier": "D", "content": "$$0$$"}] | ["A"] | null | Given, $${{dy} \over {dx}} + \left( {{1 \over {x\,\log \,x}}} \right)y = 2$$
<br><br>$$I.F. = {e^{\int {{1 \over {x\log x}}dx} }} = {e^{\log \left( {\log x} \right)}} = \log x$$
<br><br>$$y.\log x = \int {2\,\log xdx + c} $$
<br><br>$$y\log x = 2\left[ {x\log x - x} \right] + c$$
<br><br>Put $$x=1,y.0=-2+c$$ $$ \Righta... | mcq | jee-main-2015-offline | 5,818 |
rnmxkZT2jG9mNT3hkCdZl | maths | differential-equations | linear-differential-equations | The solution of the differential equation
<br/><br/>$${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$$
<br/><br/>where 0 $$ \le $$ x < $${\pi \over 2}$$, and y (0) = 1, is given by : | [{"identifier": "A", "content": "y = 1 $$-$$ $${x \\over {\\sec x + \\tan x}}$$ "}, {"identifier": "B", "content": "y<sup>2</sup> = 1 + $${x \\over {\\sec x + \\tan x}}$$"}, {"identifier": "C", "content": "y<sup>2</sup> = 1 $$-$$ $${x \\over {\\sec x + \\tan x}}$$"}, {"identifier": "D", "content": "y = 1 + $${x \\over ... | ["C"] | null | Given,
<br><br>$${{dy} \over {dx}} + {y \over 2}\sec x = {{\tan x} \over {2y}}$$
<br><br>$$ \Rightarrow $$ $$2y{{dy} \over {dx}} + {y^2}\sec x = \tan x$$
<br><br>Now, let
<br><br>y<sup>2</sup> $$=$$ t
<br><br>$$ \Rightarrow $$ 2y$${{dy} \over {dx}} = {{dt} \over {dx}}$$
<br><br>$$ \th... | mcq | jee-main-2016-online-10th-april-morning-slot | 5,819 |
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