question_id stringlengths 8 35 | subject stringclasses 3
values | chapter stringclasses 90
values | topic stringclasses 459
values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
values | answer stringclasses 460
values | explanation stringlengths 1 10.6k | question_type stringclasses 3
values | paper_id stringclasses 154
values | __index_level_0__ int64 2 13.4k |
|---|---|---|---|---|---|---|---|---|---|---|---|
rymMYDE5Xap5FUPpvJ1kls58c1v | maths | indefinite-integrals | integration-by-substitution | The value of the integral <br/>$$\int {{{\sin \theta .\sin 2\theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {1 - \cos 2\theta }}} \,d\theta $$ is : | [{"identifier": "A", "content": "$${1 \\over {18}}{\\left[ {9 - 2{{\\cos }^6}\\theta - 3{{\\cos }^4}\\theta - 6{{\\cos }^2}\\theta } \\right]^{{3 \\over 2}}} + c$$"}, {"identifier": "B", "content": "$${1 \\over {18}}{\\left[ {11 - 18{{\\sin }^2}\\theta + 9{{\\sin }^4}\\theta - 2{{\\sin }^6}\\theta } \\right]^{{3 \\... | ["C"] | null | $$\int {{{2{{\sin }^2}\theta \cos \theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {2{{\sin }^2}\theta }}d\theta } $$<br><br>Let sin$$\theta$$ = t, cos$$\theta$$ d$$\theta$$ = dt<br><br>$$ = \int {({t^6} + {t^4} + {t^2})\sqrt {2{t^4} + 3{... | mcq | jee-main-2021-online-25th-february-morning-slot | 6,457 |
qaDgVipCZFCuXu1Pxz1klt7lrht | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}} dx$$, x > 0, is equal to : (where c is a constant of integration) | [{"identifier": "A", "content": "$${\\log _e}\\sqrt {{x^2} + 5x - 7} + c$$"}, {"identifier": "B", "content": "$$4{\\log _e}|{x^2} + 5x - 7| + c$$"}, {"identifier": "C", "content": "$${1 \\over 4}{\\log _e}|{x^2} + 5x - 7| + c$$"}, {"identifier": "D", "content": "$${\\log _e}|{x^2} + 5x - 7| + c$$"}] | ["B"] | null | $$\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}dx} $$<br><br>$$ = \int {{{8{x^3} + 5(4{x^2})} \over {{x^4} + 5{x^3} - 7{x^2}}}} $$<br><br>$$ = \int {{{8{x^3} + 20{x^2}} \over {{x^4} + 5{x^3} - 7{x^2}}}} $$<br><br>$$ = \int {{{8x + 20} \o... | mcq | jee-main-2021-online-25th-february-evening-slot | 6,458 |
OlbUl68mvnNzXdDFhr1kmizm1hi | maths | indefinite-integrals | integration-by-substitution | For real numbers $$\alpha$$, $$\beta$$, $$\gamma$$ and $$\delta $$, if <br/>$$\int {{{({x^2} - 1) + {{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx} $$<br/><br/> $$ = \alpha {\log _e}\left( {{{\tan }^{ - 1}}\left( {{{{x^2} + 1... | [] | null | 6 | $$\int {{{({x^2} - 1) + {{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx} $$
<br><br>= $$\int {{{{x^2} - 1} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx + \int {{1 \over {{x^4} + 3{x^2} ... | integer | jee-main-2021-online-16th-march-evening-shift | 6,459 |
0zdXCvMp55v13ujE0o1kmlicnc5 | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {4{x^2} - 4x + 6} }}} dx$$ is equal to (where c is a constant of integration) | [{"identifier": "A", "content": "$${1 \\over 2}\\sin \\sqrt {{{(2x - 1)}^2} + 5} + c$$"}, {"identifier": "B", "content": "$${1 \\over 2}\\cos \\sqrt {{{(2x + 1)}^2} + 5} + c$$"}, {"identifier": "C", "content": "$${1 \\over 2}\\cos \\sqrt {{{(2x - 1)}^2} + 5} + c$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\s... | ["A"] | null | $$\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {{{(2x - 1)}^2} + 5} }}} dx$$<br><br>$${(2x - 1)^2} + 5 = {t^2}$$<br><br>$$2(2x - 1)2dx = 2t\,dt$$<br><br>$$2\sqrt {{t^2} - 5} dx = t\,dt$$<br><br>So, $$\int {{{\sqrt {{t^2} - 5} \cos t} \over {2\sqrt {{t^2} - 5} }}dt = {1 \over 2}\sin t + c} $$<br><br>$$... | mcq | jee-main-2021-online-18th-march-morning-shift | 6,460 |
2PEg8x92SALhxOO63W1kmlm3mz5 | maths | indefinite-integrals | integration-by-substitution | If $$f(x) = \int {{{5{x^8} + 7{x^6}} \over {{{({x^2} + 1 + 2{x^7})}^2}}}dx,(x \ge 0),f(0) = 0} $$ and $$f(1) = {1 \over K}$$, then the value of K is | [] | null | 4 | $$\int {{{5{x^8} + 7{x^6}} \over {{{(2{x^7} + {x^2} + 1)}^2}}}dx = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}} \right)}^2}}}dx} } $$<br><br>$$\int {{{{5 \over {{x^6}}} + {7 \over {{x^8}}}} \over {{{\left( {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}} \right)}^2}}}dx} $$<br... | integer | jee-main-2021-online-18th-march-morning-shift | 6,461 |
1ktepadjh | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{dx} \over {{{({x^2} + x + 1)}^2}}} = a{{\tan }^{ - 1}}\left( {{{2x + 1} \over {\sqrt 3 }}} \right) + b\left( {{{2x + 1} \over {{x^2} + x + 1}}} \right) + C} $$, x > 0 where C is the constant of integration, then the value of $$9\left( {\sqrt 3 a + b} \right)$$ is equal to _____________. | [] | null | 15 | $\int \frac{d x}{\left(x^2+x+1\right)^2}$
<br/><br/>$=\int \frac{d x}{\left[\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right]^2}$
<br/><br/>Let $x+\frac{1}{2}=\frac{\sqrt{3}}{2} \tan \theta$
<br/><br/>$$
\Rightarrow d x=\frac{\sqrt{3}}{2} \sec ^2 \theta d \theta
$$
<br/><br/>$\therefore \int \frac{... | integer | jee-main-2021-online-27th-august-morning-shift | 6,462 |
1ktiptozs | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{1 \over {\root 4 \of {{{(x - 1)}^3}{{(x + 2)}^5}} }}} \,dx$$ is equal to : (where C is a constant of integration) | [{"identifier": "A", "content": "$${3 \\over 4}{\\left( {{{x + 2} \\over {x - 1}}} \\right)^{{1 \\over 4}}} + C$$"}, {"identifier": "B", "content": "$${3 \\over 4}{\\left( {{{x + 2} \\over {x - 1}}} \\right)^{{5 \\over 4}}} + C$$"}, {"identifier": "C", "content": "$${4 \\over 3}{\\left( {{{x - 1} \\over {x + 2}}} \\rig... | ["C"] | null | $$\int {{{dx} \over {{{(x - 1)}^{3/4}}{{(x + 2)}^{5/4}}}}} $$<br><br>$$ = \int {{{dx} \over {{{\left( {{{x + 2} \over {x - 1}}} \right)}^{5/4}}.\,{{(x - 1)}^2}}}} $$<br><br>put $${{x + 2} \over {x - 1}} = t$$<br><br>$$ = - {1 \over 3}\int {{{dt} \over {{t^{5/4}}}}} $$<br><br>$$ = {4 \over 3}.{1 \over {{t^{1/4}}}} + C$... | mcq | jee-main-2021-online-31st-august-morning-shift | 6,463 |
1l58fbnfg | maths | indefinite-integrals | integration-by-substitution | <p>If $$\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c} $$, $$g(1) = 0$$, then $$g\left( {{1 \over 2}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$${\\log _e}\\left( {{{\\sqrt 3 - 1} \\over {\\sqrt 3 + 1}}} \\right) + {\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\log _e}\\left( {{{\\sqrt 3 + 1} \\over {\\sqrt 3 - 1}}} \\right) + {\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{{\\sqrt 3... | ["A"] | null | Given, $$\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c} $$, $$g(1) = 0$$
<br/><br/>Let I = $\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx}$
<br/><br/>= $\int {{1 \over x}\sqrt {{{\left( {1 - x} \right)\left( {1 + x} \right)} \over {\left( {1 + x} \right)\left( {1 + x} \right)}}} } dx$
<br/><br/>... | mcq | jee-main-2022-online-26th-june-evening-shift | 6,464 |
1ldv1n42c | maths | indefinite-integrals | integration-by-substitution | <p>Let $$f(x) = \int {{{2x} \over {({x^2} + 1)({x^2} + 3)}}dx} $$. If $$f(3) = {1 \over 2}({\log _e}5 - {\log _e}6)$$, then $$f(4)$$ is equal to</p> | [{"identifier": "A", "content": "$${\\log _e}19 - {\\log _e}20$$"}, {"identifier": "B", "content": "$${\\log _e}17 - {\\log _e}18$$"}, {"identifier": "C", "content": "$${1 \\over 2}({\\log _e}19 - {\\log _e}17)$$"}, {"identifier": "D", "content": "$${1 \\over 2}({\\log _e}17 - {\\log _e}19)$$"}] | ["D"] | null | $f(x)=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x$
<br/><br/>
$$
\begin{aligned}
& \text { Put } x^{2}=t \Rightarrow 2 x d x=d t \\\\
& f(x)=\int \frac{d t}{(t+1)(t+3)}=\int \frac{d t}{(t+2)^{2}-1} \\\\
& =\frac{1}{2} \log _{e}\left|\frac{t+1}{t+3}\right|+C \\\\
& f(x)=\frac{1}{2} \log _{e}\left(\frac... | mcq | jee-main-2023-online-25th-january-morning-shift | 6,466 |
1lgrgryfu | maths | indefinite-integrals | integration-by-substitution | <p>Let $$I(x)=\int \sqrt{\frac{x+7}{x}} \mathrm{~d} x$$ and $$I(9)=12+7 \log _{e} 7$$. If $$I(1)=\alpha+7 \log _{e}(1+2 \sqrt{2})$$, then $$\alpha^{4}$$ is equal to _________.</p> | [] | null | 64 | Given integral: $$\int \sqrt{\frac{x+7}{x}} \, dx$$
<br/><br/>Let's make the substitution $$x = t^2$$. Then, $$dx = 2t \, dt$$.
<br/><br/>Substituting these values, the integral becomes :
<br/><br/>$$\int 2 \sqrt{t^2 + 7} \, dt$$
<br/><br/>Now, let's evaluate this integral :
<br/><br/>$$I(t) = 2\left(\frac{t}{2} \... | integer | jee-main-2023-online-12th-april-morning-shift | 6,468 |
1lgzzmzjh | maths | indefinite-integrals | integration-by-substitution | <p>Let $$I(x)=\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x, x > 0$$. If $$\lim_\limits{x \rightarrow \infty} I(x)=0$$, then $$I(1)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{e+1}{e+2}-\\log _{e}(e+1)$$"}, {"identifier": "B", "content": "$$\\frac{e+1}{e+2}+\\log _{e}(e+1)$$"}, {"identifier": "C", "content": "$$\\frac{e+2}{e+1}-\\log _{e}(e+1)$$"}, {"identifier": "D", "content": "$$\\frac{e+2}{e+1}+\\log _{e}(e+1)$$"}] | ["C"] | null | $$
\begin{aligned}
& \mathrm{I}=\int \frac{x+1}{x\left(1+x e^x\right)^2} d x \\\\
& \text { Put } 1+x e^x=t \Rightarrow x e^x=t-1 \\\\
& \Rightarrow\left(x e^x+e^x\right) d x=d t \\\\
& \Rightarrow e^x(x+1) d x=d t
\end{aligned}
$$
<br/><br/>$$
\therefore I=\int \frac{d t}{e^x \cdot x t^2}=\int \frac{d t}{(t-1) t^2}
$$... | mcq | jee-main-2023-online-8th-april-morning-shift | 6,469 |
jaoe38c1lscnrzev | maths | indefinite-integrals | integration-by-substitution | <p>$$\text { The integral } \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \text { is equal to : }$$</p> | [{"identifier": "A", "content": "$$\\log _{\\mathrm{e}}\\left(\\left|\\tan ^{-1}\\left(x^3+\\frac{1}{x^3}\\right)\\right|\\right)^{1 / 3}+\\mathrm{C}$$\n"}, {"identifier": "B", "content": "$$\\log _{\\mathrm{e}}\\left(\\left|\\tan ^{-1}\\left(x^3+\\frac{1}{x^3}\\right)\\right|\\right)+\\mathrm{C}$$\n"}, {"identifier": ... | ["A"] | null | <p>$$I=\int \frac{x^8-x^2}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} d x$$</p>
<p>$$\begin{aligned}
& \text { Let } \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)=t \\
& \Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) d x=d t \\
& \Rightarrow \frac{x^... | mcq | jee-main-2024-online-27th-january-evening-shift | 6,470 |
jaoe38c1lseyv2z0 | maths | indefinite-integrals | integration-by-substitution | <p>For $$x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, if $$y(x)=\int \frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin ^2 x} d x$$, and $$\lim _\limits{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} y(x)=0$$ then $$y\left(\frac{\pi}{4}\right)$$ is equal to</p> | [{"identifier": "A", "content": "$$-\\frac{1}{\\sqrt{2}} \\tan ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)$$\n"}, {"identifier": "B", "content": "$$\\tan ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{2} \\tan ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)$$\n"}, {"identifier": "D", ... | ["D"] | null | <p>$$\begin{aligned}
& y(x)=\int \frac{\left(1+\sin ^2 x\right) \cos x}{1+\sin ^4 x} d x \\
& \text { Put } \sin x=t \\
& =\int \frac{1+t^2}{t^4+1} d t=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+C \\
& x=\frac{\pi}{2}, t=1 \quad \therefore C=0 \\
& y\left(\frac{\pi}{4}\right)=\frac{1}{\sq... | mcq | jee-main-2024-online-29th-january-morning-shift | 6,471 |
luy6z4yq | maths | indefinite-integrals | integration-by-substitution | <p>Let $$\int \frac{2-\tan x}{3+\tan x} \mathrm{~d} x=\frac{1}{2}\left(\alpha x+\log _e|\beta \sin x+\gamma \cos x|\right)+C$$, where $$C$$ is the constant of integration. Then $$\alpha+\frac{\gamma}{\beta}$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <p>$$I=\int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x$$</p>
<p>Put, $$2 \cos x-\sin x=a(-3 \sin x+\cos x)+ b(3 \cos x+\sin x)$$</p>
<p>$$\begin{aligned}
& a+3 b=2 \quad \text{.... (i)}\\
& -3 a+b=-1 \quad \text{.... (ii)}
\end{aligned}$$</p>
<p>From equation (i) and (ii) $$a=b=\frac... | mcq | jee-main-2024-online-9th-april-morning-shift | 6,473 |
lv5gs8hd | maths | indefinite-integrals | integration-by-substitution | <p>Let $$I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$$. If $$I(0)=3$$, then $$I\left(\frac{\pi}{12}\right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\sqrt3$$"}, {"identifier": "B", "content": "$$2\\sqrt3$$"}, {"identifier": "C", "content": "$$6\\sqrt3$$"}, {"identifier": "D", "content": "$$3\\sqrt3$$"}] | ["D"] | null | <p>$$\begin{aligned}
& I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x \\
& I(x)=\int \frac{6}{(\sin x-\cos x)^2} d x \\
& =\int \frac{6 \sec ^2 x}{(\tan x-1)^2} d x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Let } \tan x=t \Rightarrow \sec ^2 x d x=d t \\
& =\int \frac{6 d t}{(t-1)^2} \\
& =-\frac{6}{(t-1)}+c \\... | mcq | jee-main-2024-online-8th-april-morning-shift | 6,475 |
i7dl4gKHTGzDLWWS | maths | indefinite-integrals | standard-integral | $$\int {{{dx} \over {\cos x - \sin x}}} $$ is equal to | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}\\log \\left| {\\tan \\left( {{x \\over 2} + {{3\\pi } \\over 8}} \\right)} \\right| + C$$ "}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 2 }}\\log \\left| {\\cot \\left( {{x \\over 2}} \\right)} \\right| + C$$ "}, {"identifier": "C", "content": "$${1 \\... | ["A"] | null | $$\int {{{dx} \over {\cos x - \sin x}}} $$
<br><br>$$ = \int {{{dx} \over {\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)}}} $$
<br><br>$$ = {1 \over {\sqrt 2 }}\int {\sec \left( {x + {\pi \over 4}} \right)dx} $$
<br><br>$$ = {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{\pi \over 4} + {x \over 2} + {\pi \over 8}... | mcq | aieee-2004 | 6,476 |
0TSlPppNziu0Gv6m | maths | indefinite-integrals | standard-integral | $$\int {{{dx} \over {\cos x + \sqrt 3 \sin x}}} $$ equals | [{"identifier": "A", "content": "$$\\log \\,\\tan \\,\\left( {{x \\over 2} + {\\pi \\over {12}}} \\right) + C$$ "}, {"identifier": "B", "content": "$$\\log \\,\\tan \\,\\left( {{x \\over 2} - {\\pi \\over {12}}} \\right) + C$$"}, {"identifier": "C", "content": "$$\\,{1 \\over 2}\\,\\log \\,\\tan \\,\\left( {{x \\over... | ["C"] | null | $$I = \int {{{dx} \over {\cos x + \sqrt 3 \sin x}}} $$
<br><br>$$ \Rightarrow I = \int {{{dx} \over {2\left[ {{1 \over 2}\cos x + {{\sqrt 3 } \over 2}\sin x} \right]}}} $$
<br><br>$$ = {1 \over 2}\int {{{dx} \over {\left[ {\sin {\pi \over 6}\cos x + \cos {\pi \over 6}\sin x} \right]}}} $$
<br><br>$$ = {1 \over 2}.\in... | mcq | aieee-2007 | 6,478 |
iKxtxDJH7ZA06wso35mnL | maths | indefinite-integrals | standard-integral | The integral
<br/><br/>$$\int {\sqrt {1 + 2\cot x(\cos ecx + \cot x)\,} \,\,dx} $$
<br/><br/>$$\left( {0 < x < {\pi \over 2}} \right)$$ is equal to :
<br/><br/>(where C is a constant of integration) | [{"identifier": "A", "content": "4 log(sin $${x \\over 2}$$ ) + C"}, {"identifier": "B", "content": "2 log(sin $${x \\over 2}$$ ) + C"}, {"identifier": "C", "content": "2 log(cos $${x \\over 2}$$ ) + C"}, {"identifier": "D", "content": "4 log(cos $${x \\over 2}$$) + C"}] | ["B"] | null | Let, I = $$\int {\sqrt {1 + 2\cot x\cos ec + 2{{\cot }^2}x} .dx} $$<br><br>
$$ \Rightarrow $$ I = $$\int {\sqrt {{{{{\sin }^2}x + 2\cos x + 2{{\cos }^2}x} \over {{{\sin }^2}x}}} .dx} $$<br><br>
$$ \Rightarrow $$ I = $$\int {\sqrt {{{1 + 2\cos x + {{\cos }^2}x} \over {\sin x}}} .dx} $$<br><br>
$$ \Rightarrow $$ I = $$\... | mcq | jee-main-2017-online-8th-april-morning-slot | 6,479 |
u7bfYkLqQ1Ri6xiLX6tgK | maths | indefinite-integrals | standard-integral | If $$f\left( {{{x - 4} \over {x + 2}}} \right) = 2x + 1,$$ (x $$ \in $$ <b>R</b> $$-$${1, $$-$$ 2}), then $$\int f \left( x \right)dx$$ is equal to :
<br/>(where C is a constant of integration) | [{"identifier": "A", "content": "12 log<sub>e</sub> | 1 $$-$$ x | + 3x + C"}, {"identifier": "B", "content": "$$-$$ 12 log<sub>e</sub> | 1 $$-$$ x | $$-$$ 3x + C"}, {"identifier": "C", "content": "12 log<sub>e</sub> | 1 $$-$$ x | $$-$$ 3x + C"}, {"identifier": "D", "content": "$$-$$ 12 log<sub>e</sub> | 1 $$-$$ x | + 3... | ["B"] | null | Let, $${{{x - 4} \over {x + 2}}}$$ = t<br><br>
$$ \Rightarrow $$ x - 4 = t(x+2)<br><br>
$$ \Rightarrow $$ x (1 -t) = 2(t+2)<br><br>
$$ \Rightarrow $$ x = $${{2(t + 2)} \over {1 - t}}$$<br><br>
$$ \therefore $$ f(t) = 2($${{2(t + 2)} \over {1 - t}}$$) + 1<br><br>
= $${{4t + 8} \over {1 - t}} + 1$$<br><br>
= $${{3t + 9} ... | mcq | jee-main-2018-online-15th-april-morning-slot | 6,481 |
1YrtLQqQIm1HgxI1ExfLd | maths | indefinite-integrals | standard-integral | If $$\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} \,\,dx = A\sqrt {7 - 6x - {x^2}} + B{\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right) + C$$
<br/>(where C is a constant of integration), then the ordered pair (A, B) is equal to : | [{"identifier": "A", "content": "(2, 1)"}, {"identifier": "B", "content": "($$-$$ 2, $$-$$1)"}, {"identifier": "C", "content": "($$-$$ 2, 1)"}, {"identifier": "D", "content": "(2, $$-$$1)"}] | ["B"] | null | We can write,
<br><br>7 - 6x - x<sup>2</sup> = 16 - (x + 3)<sup>2</sup>
<br><br>and $${d \over {dx}}\left( {7 - 6x - {x^2}} \right) = - (2x + 6)$$
<br><br>So, $$\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$
<br><br>= $$\int {{{2x + 6 - 1} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$
<br><br>= $$\int {{{2x + 6} \ove... | mcq | jee-main-2018-online-15th-april-evening-slot | 6,482 |
PYnBJilIrz8XFwr7gR18hoxe66ijvwvro3u | maths | indefinite-integrals | standard-integral | $$\int {{e^{\sec x}}}$$ $$(\sec x\tan xf(x) + \sec x\tan x + se{x^2}x)dx$$<br/>
= e<sup>secx</sup>f(x) + C then a possible choice of f(x) is :- | [{"identifier": "A", "content": "x sec x + tan x + 1/2"}, {"identifier": "B", "content": "sec x + xtan x - 1/2"}, {"identifier": "C", "content": "sec x - tan x - 1/2"}, {"identifier": "D", "content": "sec x + tan x + 1/2"}] | ["D"] | null | Given
<br><br>$$\int {{e^{\sec x}}\left( {\sec x\tan x\,f(x) + (sec\,x\,tan\,x\, + se{c^2}x)} \right)} $$<br><br>
$$ = {e^{\sec x}}f(x) + C$$
<br><br>Differentiating both sides with respect to x,
<br><br>$${e^{\sec x}}.\sec x\tan x\,f\left( x \right)$$ + $${e^{\sec x}}\left( {\sec x\tan x + {{\sec }^2}x} \right)$$
<br>... | mcq | jee-main-2019-online-9th-april-evening-slot | 6,483 |
dsxm8kHFIFngSyvhIT3rsa0w2w9jx5brt6e | maths | indefinite-integrals | standard-integral | The integral $$\int {{{2{x^3} - 1} \over {{x^4} + x}}} dx$$ is equal to :
<br/>(Here C is a constant of integration) | [{"identifier": "A", "content": "$${\\log _e}{{\\left| {{x^3} + 1} \\right|} \\over {{x^2}}} + C$$"}, {"identifier": "B", "content": "$${1 \\over 2}{\\log _e}{{\\left| {{x^3} + 1} \\right|} \\over {{x^2}}} + C$$"}, {"identifier": "C", "content": "$${\\log _e}\\left| {{{{x^3} + 1} \\over x}} \\right| + C$$"}, {"identifi... | ["C"] | null | $$\int {{{2{x^3} - 1} \over {{x^4} + x}}dx = \int {{{2x - {x^{ - 2}}} \over {{x^2} + {x^{ - 1}}}}dx = \ln ({x^2} + {x^{ - 1}}} } ) + c$$<br><br>
$$ \Rightarrow \ln ({x^3} + 1) - \ln x + c$$ | mcq | jee-main-2019-online-12th-april-morning-slot | 6,484 |
p6chnBVIYFmsOS0wrDfOu | maths | indefinite-integrals | standard-integral | $$\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx} $$ is equal to<br/>
(where c is a constant of integration) | [{"identifier": "A", "content": "2x + sinx + 2sin2x + c"}, {"identifier": "B", "content": "x + 2sinx + sin2x + c"}, {"identifier": "C", "content": "x + 2sinx + 2sin2x + c"}, {"identifier": "D", "content": "2x + sinx + sin2x + c"}] | ["B"] | null | $$\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx} $$
<br><br>= $$\int {{{2\cos {x \over 2}.\sin {{5x} \over 2}} \over {2\cos {x \over 2}.\sin {x \over 2}}}} dx $$
<br><br>= $$\int {{{\sin \left( {{{5x} \over 2} + {x \over 2}} \right) + \sin \left( {{{5x} \over 2} - {x \over 2}} \right)} \over {\sin x}}} dx$$
... | mcq | jee-main-2019-online-8th-april-morning-slot | 6,485 |
RognVI7rIVMeBEOJCg7k9k2k5iqewxy | maths | indefinite-integrals | standard-integral | If ƒ'(x) = tan<sup>–1</sup>(secx + tanx), $$ - {\pi \over 2} < x < {\pi \over 2}$$,
<br/>and
ƒ(0) = 0, then ƒ(1) is equal to : | [{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "$${{\\pi - 1} \\over 4}$$"}, {"identifier": "C", "content": "$${{\\pi + 1} \\over 4}$$"}, {"identifier": "D", "content": "$${{\\pi + 2} \\over 4}$$"}] | ["C"] | null | ƒ'(x) = tan<sup>–1</sup>(secx + tanx)
<br><br>$$ \Rightarrow $$ ƒ'(x) = $${\tan ^{ - 1}}\left( {{{1 + \sin x} \over {\cos x}}} \right)$$ = $${\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)$$
<br><br>= $${\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right... | mcq | jee-main-2020-online-9th-january-morning-slot | 6,486 |
1l6hzevej | maths | indefinite-integrals | standard-integral | <p>$$
\text { The integral } \int \frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}} \sin 2 x\right)} d x \text { is equal to }
$$</p> | [{"identifier": "A", "content": "$$\\frac{1}{2} \\log _{e}\\left|\\frac{\\tan \\left(\\frac{x}{2}+\\frac{\\pi}{12}\\right)}{\\tan \\left(\\frac{x}{2}+\\frac{\\pi}{6}\\right)}\\right|+C$$"}, {"identifier": "B", "content": "$$\\frac{1}{2} \\log _{e}\\left|\\frac{\\tan \\left(\\frac{x}{2}+\\frac{\\pi}{6}\\right)}{\\tan \\... | ["A"] | null | <p>$$ = \int {{{\left( {1 - {1 \over {\sqrt 3 }}} \right)(\cos x - \sin x)} \over {\left( {1 + {2 \over {\sqrt 3 }}\sin 2x} \right)}}dx} $$</p>
<p>$$ = \int {{{\left( {{{\sqrt 3 - 1} \over {\sqrt 3 }}} \right)\sqrt 2 \sin \left( {{\pi \over 4} - x} \right)} \over {\left( {{2 \over {\sqrt 3 }}} \right)\left( {\sin {\p... | mcq | jee-main-2022-online-26th-july-evening-shift | 6,489 |
1lgvpqqkm | maths | indefinite-integrals | standard-integral | <p>For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$
, where $$e=\sum_\limits{n=0}^{\infty} \frac{1}{n !}$$ ... | [{"identifier": "A", "content": "$$-8$$"}, {"identifier": "B", "content": "$$-4$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4<br/><br/>"}] | ["D"] | null | We have,
<br/><br/>$$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$
<br/><br/>$$
\begin{aligned}
\left(\frac{x}{e}\right)^{2 x} & =\left(\frac{e^{\log _e x}}{e}\rig... | mcq | jee-main-2023-online-10th-april-evening-shift | 6,490 |
1lgyl5enm | maths | indefinite-integrals | standard-integral | <p>The integral $$
\int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) d x
$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\left(\\frac{x}{2}\\right)^{x}+\\left(\\frac{2}{x}\\right)^{x}+C$$"}, {"identifier": "B", "content": "$$\\left(\\frac{x}{2}\\right)^{x}-\\left(\\frac{2}{x}\\right)^{x}+C$$"}, {"identifier": "C", "content": "$$\\left(\\frac{x}{2}\\right)^{x} \\log _{2}\\left(\\frac{2}{x}\\right)+C$$"}... | ["B"] | null | <p>To solve the integral:</p>
<p>$ I = \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) \, dx $</p>
<p>we start by simplifying the logarithmic term:</p>
<p>$ \ln\left(\frac{e x}{2}\right) = \ln(e) + \ln\left(\frac{x}{2}\right) = 1 + \ln\left(\frac{x}{2}\right) $</p>
... | mcq | jee-main-2023-online-8th-april-evening-shift | 6,491 |
lvb294vs | maths | indefinite-integrals | standard-integral | <p>If $$\int \frac{1}{\mathrm{a}^2 \sin ^2 x+\mathrm{b}^2 \cos ^2 x} \mathrm{~d} x=\frac{1}{12} \tan ^{-1}(3 \tan x)+$$ constant, then the maximum value of $$\mathrm{a} \sin x+\mathrm{b} \cos x$$, is :</p> | [{"identifier": "A", "content": "$$\\sqrt{41}$$\n"}, {"identifier": "B", "content": "$$\\sqrt{39}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{40}$$\n"}, {"identifier": "D", "content": "$$\\sqrt{42}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x=\frac{1}{12} \tan ^{-1}(3 \tan x)+c \\
& I=\int \frac{\sec ^2 x}{b^2+a^2 \tan ^2 x} d x \\
& \tan x=t \\
& \Rightarrow \sec ^2 x d x=d t \\
& I=\int \frac{d t}{b^2+a^2 t^2} \\
& =\frac{1}{b a} \tan ^{-1}\left(\frac{a t}{b}\right)+c \\
& I=\frac{1}{a ... | mcq | jee-main-2024-online-6th-april-evening-shift | 6,492 |
uudnlgNnuCb2c1ku | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | The trigonometric equation $${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$ has a solution for : | [{"identifier": "A", "content": "$$\\left| a \\right| \\ge {1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$${1 \\over 2} < \\left| a \\right| < {1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "C", "content": "all real values of $$a$$ "}, {"identifier": "D", "content": "$$\\left| a \\right| \\le {1 \\ove... | ["D"] | null | Given that,
<br><br>$${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$
<br><br>We know,
<br><br>$$ - {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}$$
<br><br>$$\therefore$$ $$\,\,\,$$ $$ - {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}$$
<br><br>$$ \Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over... | mcq | aieee-2003 | 6,493 |
nLj5VndTCR1FFqA1CBjgy2xukewsg59s | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | The domain of the function
<br/>f(x) = $${\sin ^{ - 1}}\left( {{{\left| x \right| + 5} \over {{x^2} + 1}}} \right)$$ is (–
$$\infty $$, -a]$$ \cup $$[a, $$\infty $$). Then a is equal to :
| [{"identifier": "A", "content": "$${{\\sqrt {17} - 1} \\over 2}$$"}, {"identifier": "B", "content": "$${{1 + \\sqrt {17} } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt {17} } \\over 2} + 1$$"}, {"identifier": "D", "content": "$${{\\sqrt {17} } \\over 2}$$"}] | ["B"] | null | f(x) = $${\sin ^{ - 1}}\left( {{{\left| x \right| + 5} \over {{x^2} + 1}}} \right)$$
<br><br>$$ \therefore $$ $$ - 1 \le {{\left| x \right| + 5} \over {{x^2} + 1}} \le 1$$
<br><br>Since |x| + 5 & x<sup>2</sup>
+ 1 is always positive
<br><br>So $${{\left| x \right| + 5} \over {{x^2} + 1}} \ge 0$$
<br><br>That means... | mcq | jee-main-2020-online-2nd-september-morning-slot | 6,494 |
y6Fc34VqoMktCyXZqP1kmkm3flo | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | The number of solutions of the equation <br/><br/>$${\sin ^{ - 1}}\left[ {{x^2} + {1 \over 3}} \right] + {\cos ^{ - 1}}\left[ {{x^2} - {2 \over 3}} \right] = {x^2}$$, for x$$\in$$[$$-$$1, 1], and [x] denotes the greatest integer less than or equal to x, is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "Infinite"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["A"] | null | There are three cases possible for $$x \in [ - 1,1]$$<br><br>Case I : $$x \in \left[ { - 1, - \sqrt {{2 \over 3}} } \right)$$<br><br>$$ \therefore $$ $${\sin ^{ - 1}}(1) + {\cos ^{ - 1}}(0) = {x^2}$$<br><br>$$ \Rightarrow {x^2} = {\pi \over 2} + {\pi \over 2} = \pi $$<br><br>$$ \Rightarrow x = \pm \sqrt \pi $$ $$ \... | mcq | jee-main-2021-online-17th-march-evening-shift | 6,495 |
1krpwpsl7 | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | The number of real roots of the equation $${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$ is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "0"}] | ["D"] | null | $${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$<br><br>For equation to be defined,<br><br>x<sup>2</sup> + x $$\ge$$ 0<br><br>$$\Rightarrow$$ x<sup>2</sup> + x + 1 $$\ge$$ 1<br><br>$$\therefore$$ Only possibility that the equation is defined <br><br>x<sup>2</sup> + x = 0 $$\Ri... | mcq | jee-main-2021-online-20th-july-morning-shift | 6,496 |
1ktczv71w | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | The domain of the function $${{\mathop{\rm cosec}\nolimits} ^{ - 1}}\left( {{{1 + x} \over x}} \right)$$ is : | [{"identifier": "A", "content": "$$\\left( { - 1, - {1 \\over 2}} \\right] \\cup (0,\\infty )$$"}, {"identifier": "B", "content": "$$\\left[ { - {1 \\over 2},0} \\right) \\cup [1,\\infty )$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 2},\\infty } \\right) - \\{ 0\\} $$"}, {"identifier": "D", "content": ... | ["D"] | null | $${{1 + x} \over x} \in ( - \infty , - 1] \cup [1,\infty )$$<br><br>$${1 \over x} \in ( - \infty , - 2] \cup [0,\infty )$$<br><br>$$x \in \left[ { - {1 \over 2},0} \right) \cup (0,\infty )$$<br><br>$$x \in \left[ { - {1 \over 2},0} \right) \cup \{ 0\} $$ | mcq | jee-main-2021-online-26th-august-evening-shift | 6,498 |
1ktk4rt98 | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | The domain of the function<br/><br/>$$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$$ is : | [{"identifier": "A", "content": "$$\\left[ {0,{1 \\over 4}} \\right]$$"}, {"identifier": "B", "content": "$$[ - 2,0] \\cup \\left[ {{1 \\over 4},{1 \\over 2}} \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {{1 \\over 4},{1 \\over 2}} \\right] \\cup \\{ 0\\} $$"}, {"identifier": "D", "content": "$$\\left[ {0,{1... | ["C"] | null | $$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$$<br><br>$$ - 1 \le {{x - 1} \over {x + 1}} \le 1 \Rightarrow 0 \le x < \infty $$ .... (1)<br><br>$$ - 1 \le {{3{x^2} + x - 1} \over {{{(x - 1)}^2}}} \le 1 \Rightarrow x \in \left... | mcq | jee-main-2021-online-31st-august-evening-shift | 6,499 |
1l5c1qzrq | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | <p>The domain of the function <br/><br/>$$f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}}$$ is :</p> | [{"identifier": "A", "content": "$$( - \\infty ,1) \\cup (2,\\infty )$$"}, {"identifier": "B", "content": "$$(2,\\infty )$$"}, {"identifier": "C", "content": "$$\\left[ { - {1 \\over 2},1} \\right) \\cup (2,\\infty )$$"}, {"identifier": "D", "content": "$$\\left[ { - {1 \\over 2},1} \\right) \\cup (2,\\infty ) - \\left... | ["D"] | null | $-1 \leq \frac{x^{2}-5 x+6}{x^{2}-9} \leq 1$ and $x^{2}-3 x+2>0, \neq 1$
<br/><br/>
$$
\frac{(x-3)(2 x+1)}{x^{2}-9} \geq 0 \mid \frac{5(x-3)}{x^{2}-9} \geq 0
$$
<br/><br/>
The solution to this inequality is
<br/><br/>
$x \in\left[\frac{-1}{2}, \infty\right)-\{3\}$
<br/><br/>
for $x^{2}-3 x+2>0$ and $\neq 1$
<br/><br/>
... | mcq | jee-main-2022-online-24th-june-morning-shift | 6,501 |
1l6kib0jf | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | <p>The domain of the function $$f(x)=\sin ^{-1}\left[2 x^{2}-3\right]+\log _{2}\left(\log _{\frac{1}{2}}\left(x^{2}-5 x+5\right)\right)$$, where [t] is the greatest integer function, is :</p> | [{"identifier": "A", "content": "$$\n\\left(-\\sqrt{\\frac{5}{2}}, \\frac{5-\\sqrt{5}}{2}\\right)\n$$"}, {"identifier": "B", "content": "$$\n\\left(\\frac{5-\\sqrt{5}}{2}, \\frac{5+\\sqrt{5}}{2}\\right)\n$$"}, {"identifier": "C", "content": "$$\n\\left(1, \\frac{5-\\sqrt{5}}{2}\\right)\n$$"}, {"identifier": "D", "conte... | ["C"] | null | <p>$$ - 1 \le 2{x^2} - 3 < 2$$</p>
<p>or $$2 \le 2{x^2} < 5$$</p>
<p>or $$1 \le {x^2} < {5 \over 2}$$</p>
<p>$$x \in \left( { - \sqrt {{5 \over 2}} , - 1} \right] \cup \left[ {1,\sqrt {{5 \over 2}} } \right)$$</p>
<p>$${\log _{{1 \over 2}}}({x^2} - 5x + 5) > 0$$</p>
<p>$$0 < {x^2} - 5x + 5 < 1$$</p>
<p>$${x^2} - 5x + 5... | mcq | jee-main-2022-online-27th-july-evening-shift | 6,502 |
1l6rfcn42 | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | <p>The domain of the function $$f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)$$ is :</p> | [{"identifier": "A", "content": "$$[1, \\infty)$$"}, {"identifier": "B", "content": "$$[-1,2]$$"}, {"identifier": "C", "content": "$$[-1, \\infty)$$"}, {"identifier": "D", "content": "$$(-\\infty, 2]$$"}] | ["C"] | null | $f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)$
<br/><br/>$$
-1 \leq \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1
$$
<br/><br/>$$
\begin{aligned}
& \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1 \\\\
& x^{2}-3 x+2 \leq x^{2}+2 x+7 \\\\
& 5 x \geq-5 \\\\
& x \geq-1
\end{aligned}
$$
<br/><br/>And $$
\frac{x^{2}-3 x+2}... | mcq | jee-main-2022-online-29th-july-evening-shift | 6,503 |
lgnwv1sg | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | If the domain of the function
<br/><br/>$f(x)=\log _{e}\left(4 x^{2}+11 x+6\right)+\sin ^{-1}(4 x+3)+\cos ^{-1}\left(\frac{10 x+6}{3}\right)$ is $(\alpha, \beta]$, then
<br/><br/>$36|\alpha+\beta|$ is equal to : | [{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "45"}, {"identifier": "D", "content": "63"}] | ["C"] | null | <p>To find the domain of the function, we need to consider the individual functions and their respective domains. We have:</p>
<p><ol>
<li>$f_1(x) = \ln(4x^2 + 11x + 6)$</li>
<li>$f_2(x) = \sin^{-1}(4x + 3)$</li>
<li>$f_3(x) = \cos^{-1}\left(\frac{10x + 6}{3}\right)$</li>
</ol></p>
<p><ol>
<li>For $f_1(x)$:</li>
</ol><... | mcq | jee-main-2023-online-15th-april-morning-shift | 6,504 |
1lgvqqyhz | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | <p>If the domain of the function $$f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$$ is $$[\alpha, \beta) \mathrm{U}(\gamma, \delta]$$, then $$|3 \alpha+10(\beta+\gamma)+21 \delta|$$ is equal to _________.</p> | [] | null | 24 | Given that $f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$
<br/><br/>Since, the domain for $\sec ^{-1} x$ is $|x| \geq 1$
<br/><br/>Therefore $\left|\frac{2 x}{5 x+3}\right| \geq 1$
<br/><br/>$$
\begin{aligned}
& \frac{2 x}{5 x+3} \leq-1 \text { or } \frac{2 x}{5 x+3} \geq 1 \\\\
& \frac{2 x+5 x+3}{5 x+3} \leq 0 \tex... | integer | jee-main-2023-online-10th-april-evening-shift | 6,505 |
lv0vxckz | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | <p>If the domain of the function $$\sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _{\mathrm{e}}\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right)$$ is $$(\alpha, \beta]$$, then $$3 \alpha+10 \beta$$ is equal to:</p> | [{"identifier": "A", "content": "95"}, {"identifier": "B", "content": "100"}, {"identifier": "C", "content": "97"}, {"identifier": "D", "content": "98"}] | ["C"] | null | <p>$$\begin{aligned}
& \sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _e\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right) \\
& -1 \leq \frac{3 x-22}{2 x-19} \leq 1 \\
& \frac{3 x-22}{2 x-19}+1 \geq 0 \text { and } \frac{3 x-22}{2 x-19}-1 \leq 0 \\
& \frac{3 x-22+2 x-19}{2 x-19} \geq 0 \text { and } \frac{3 x-22-2 x+19}{2 ... | mcq | jee-main-2024-online-4th-april-morning-shift | 6,506 |
r1lgCyi2I37gkOBHCY1kmix729h | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy <br/><br/>$${\sin ^{ - 1}}\left( {{{3x} \over 5}} \right) + {\sin ^{ - 1}}\left( {{{4x} \over 5}} \right) = {\sin ^{ - 1}}x$$ is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $${\sin ^{ - 1}}{{3x} \over 5} + {\sin ^{ - 1}}{{4x} \over 5} = {\sin ^{ - 1}}x$$<br><br>$${\sin ^{ - 1}}\left( {{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} } \right) = {\sin ^{ - 1}}x$$<br><br>$${{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\s... | mcq | jee-main-2021-online-16th-march-evening-shift | 6,508 |
1ktnznemb | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | $${\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))$$ is equal to :<br/><br/>(The inverse trigonometric functions take the principal values) | [{"identifier": "A", "content": "3$$\\pi$$ $$-$$ 11"}, {"identifier": "B", "content": "4$$\\pi$$ $$-$$ 9"}, {"identifier": "C", "content": "4$$\\pi$$ $$-$$ 11"}, {"identifier": "D", "content": "3$$\\pi$$ + 1"}] | ["C"] | null | $${\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))$$<br><br>$$ = (2\pi - 5) + (6 - 2\pi ) - (12 - 4\pi )$$<br><br>$$ = 4\pi - 11$$. | mcq | jee-main-2021-online-1st-september-evening-shift | 6,509 |
1l57ou1of | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>$${\sin ^1}\left( {\sin {{2\pi } \over 3}} \right) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} \right) + {\tan ^{ - 1}}\left( {\tan {{3\pi } \over 4}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$${{11\\pi } \\over {12}}$$"}, {"identifier": "B", "content": "$${{17\\pi } \\over {12}}$$"}, {"identifier": "C", "content": "$${{31\\pi } \\over {12}}$$"}, {"identifier": "D", "content": "$$-$$$${{3\\pi } \\over {4}}$$"}] | ["A"] | null | <p>$${\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) + {\cos ^{ - 1}}\left( {{{ - \sqrt 3 } \over 2}} \right) + {\tan ^{ - 1}}\left( { - 1} \right)$$</p>
<p>$$ = {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4}$$</p>
<p>$$ = {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift | 6,510 |
1l58gom2l | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>If the inverse trigonometric functions take principal values then <br/><br/>$${\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${\\pi \\over 6}$$"}] | ["C"] | null | <p>$${\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right)$$</p>
<p>$$ = {\cos ^{ - 1}}\left( {{3 \over {10}}\,.\,{3 \over 5} + {2 \over 5}\,.\,{4 \over 5}} \right)$$</p>
<p>$$ = {\c... | mcq | jee-main-2022-online-26th-june-evening-shift | 6,511 |
1l5w0n374 | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Let $$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$ and $$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$$ where the inverse trigonometric functions take pr... | [{"identifier": "A", "content": "$$15{x^2} - 8x - 7 = 0$$"}, {"identifier": "B", "content": "$$5{x^2} - 12x + 7 = 0$$"}, {"identifier": "C", "content": "$$25{x^2} - 18x - 7 = 0$$"}, {"identifier": "D", "content": "$$25{x^2} - 32x + 7 = 0$$"}] | ["C"] | null | <p>Given,</p>
<p>$$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$</p>
<p>We know, $$2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1)$$</p>
<p>$$\therefore$$ $$2{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) = {\cos ^1}\left( {2 \times... | mcq | jee-main-2022-online-30th-june-morning-shift | 6,512 |
1l6jduff0 | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>For $$k \in \mathbb{R}$$, let the solutions of the equation $$\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k, 0<|x|<\frac{1}{\sqrt{2}}$$ be $$\alpha$$ and $$\beta$$, where the inverse trigonometric functions take only principal values. If th... | [] | null | 12 | <p>$$\cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}} \right)} \right)} \right)} \right)} \right) = k$$</p>
<p>$$ \Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\sqrt {1 - {x^2}} } \right)} \right)} \right) = k$$</p>
<p>$$ \Rightar... | integer | jee-main-2022-online-27th-july-morning-shift | 6,513 |
1l6m56q48 | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Considering only the principal values of the inverse trigonometric functions, the domain of the function $$f(x)=\cos ^{-1}\left(\frac{x^{2}-4 x+2}{x^{2}+3}\right)$$ is :</p> | [{"identifier": "A", "content": "$$\\left(-\\infty, \\frac{1}{4}\\right]$$"}, {"identifier": "B", "content": "$$\\left[-\\frac{1}{4}, \\infty\\right)$$"}, {"identifier": "C", "content": "$$(-1 / 3, \\infty)$$"}, {"identifier": "D", "content": "$$\\left(-\\infty, \\frac{1}{3}\\right]$$"}] | ["B"] | null | <p>$$ - 1 \le {{{x^2} - 4x + 2} \over {{x^2} + 3}} \le 1$$</p>
<p>$$ \Rightarrow - {x^2} - 3 \le {x^2} - 4x + 2 \le {x^2} + 3$$</p>
<p>$$ \Rightarrow 2{x^2} - 4x + 5 \ge 0$$ & $$ - 4x \le 1$$</p>
<p>$$x \in R$$ & $$x \ge - {1 \over 4}$$</p>
<p>So domain is $$\left[ { - {1 \over 4},\infty } \right)$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift | 6,514 |
1ldyay4yl | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$${\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["B"] | null | <p>$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$</p>
<p>$$= {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{16 + 8\sqrt 3 } \over {12 + 6\sqrt 3 }}} \right)^{{1 \o... | mcq | jee-main-2023-online-24th-january-morning-shift | 6,516 |
jaoe38c1lscmv27h | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Considering only the principal values of inverse trigonometric functions, the number of positive real values of $$x$$ satisfying $$\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}$$ is :</p> | [{"identifier": "A", "content": "more than 2"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <p>$$\begin{aligned}
& \tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4} ; x>0 \\
& \Rightarrow \tan ^{-1} 2 x=\frac{\pi}{4}-\tan ^{-1} x
\end{aligned}$$</p>
<p>Taking tan both sides</p>
<p>$$\begin{aligned}
& \Rightarrow 2 \mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}} \\
& \Rightarrow 2 \mathrm{x}^2+3 \mathrm{x}-1=0 \\
& \math... | mcq | jee-main-2024-online-27th-january-evening-shift | 6,517 |
luxwcnzi | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $$2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5}$$, is __________.</p> | [] | null | 0 | <p>$$\begin{aligned}
& 2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5} \\
& \frac{\pi}{2}+\cos ^{-1} x=\frac{2 \pi}{5} \\
& \cos ^{-1} x=\frac{2 \pi}{5}-\frac{\pi}{2} \\
& \cos ^{-1} x=\frac{-\pi}{10}
\end{aligned}$$</p>
<p>Which is not possible as $$\cos ^{-1} x \in[0, \pi]$$</p>
<p>$$\therefore \quad$$ No solution</p> | integer | jee-main-2024-online-9th-april-evening-shift | 6,519 |
lv2erz4o | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Given that the inverse trigonometric function assumes principal values only. Let $$x, y$$ be any two real numbers in $$[-1,1]$$ such that $$\cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi$$.
Then, the minimum value of $$x^2+y^2+2 x y \sin \alpha$$ is</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "$$\\frac{-1}{2}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \cos ^{-1} x-\frac{\pi}{2}+\cos ^{-1} y=\alpha \\
& \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\
& \because \quad \alpha \in\left[\frac{-\pi}{2}, \pi\right] \\
& \text { then } \frac{\pi}{2}+\alpha \in\left(0, \frac{3 \pi}{2}\right) \\
& \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\... | mcq | jee-main-2024-online-4th-april-evening-shift | 6,520 |
1TDQHlhEobjkoJLP | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ + cosec<sup>-1</sup>$$\left( {{5 \over 4}} \right)$$ = $${\pi \over 2}$$, then the value of x is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Given sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ + cosec<sup>-1</sup>$$\left( {{5 \over 4}} \right)$$ = $${\pi \over 2}$$
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ + sin<sup>-1</sup>$$\left( {{4 \over 5}} \right)$$ = $${\pi \over 2}$$
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup>$$... | mcq | aieee-2007 | 6,523 |
IgvCQgf8GDVXARHa | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of $$cot\left( {\cos e{c^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$ is : | [{"identifier": "A", "content": "$${{6 \\over 17}}$$"}, {"identifier": "B", "content": "$${{3 \\over 17}}$$"}, {"identifier": "C", "content": "$${{4 \\over 17}}$$"}, {"identifier": "D", "content": "$${{5 \\over 17}}$$"}] | ["A"] | null | Given,
<br><br>$$Cot\left( {so{{\sec }^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$
<br><br>$$ = \cot \left( {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$
<br><br>$$ = cot\left( {{{\tan }^{ - 1}}{{{3 \over 4} + {2 \over 3}} \over {1 - {3 \over 4} - {2 \over 3}}}} \right)$$
<br>... | mcq | aieee-2008 | 6,524 |
uhYKiIhxgTCzlkLr | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $$x, y, z$$ are in A.P. and $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$$ and $${\tan ^{ - 1}}z$$ are also in A.P., then : | [{"identifier": "A", "content": "$$x=y=z$$ "}, {"identifier": "B", "content": "$$2x=3y=6z$$ "}, {"identifier": "C", "content": "$$6x=3y=2z$$ "}, {"identifier": "D", "content": "$$6x=4y=3z$$"}] | ["A"] | null | Given that, $$x,y,z\,\,$$ are in $$AP$$
<br><br>So, $$\,\,\,$$ $$2y = x + y$$
<br><br>Also given that,
<br><br> $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y\,\,$$ and $$\,\,\,{\tan ^{ - 1}}z\,\,$$ are in $$AP$$
<br><br>So, $$2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z$$
<br><br>$$ \Rightarrow {\tan ^{ - 1}}\left( {{{2y... | mcq | jee-main-2013-offline | 6,525 |
tb7Srq7xtkaCBPWe | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let $${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right),$$
<br/> where $$\left| x \right| < {1 \over {\sqrt 3 }}.$$ Then a value of $$y$$ is : | [{"identifier": "A", "content": "$${{3x - {x^3}} \\over {1 + 3{x^2}}}$$ "}, {"identifier": "B", "content": "$${{3x + {x^3}} \\over {1 + 3{x^2}}}$$"}, {"identifier": "C", "content": "$${{3x - {x^3}} \\over {1 - 3{x^2}}}$$ "}, {"identifier": "D", "content": "$${{3x + {x^3}} \\over {1 - 3{x^2}}}$$ "}] | ["C"] | null | Given,
<br><br>$${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)$$
<br><br>$$ \Rightarrow {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} \right)}}} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,... | mcq | jee-main-2015-offline | 6,526 |
wFBhtX1szbvhwRLii253K | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of tan<sup>-1</sup> $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right],$$ $$\left| x \right| < {1 \over 2},x \ne 0,$$ is equal to : | [{"identifier": "A", "content": "$${\\pi \\over 4} + {1 \\over 2}{\\cos ^{ - 1}}\\,{x^2}$$"}, {"identifier": "B", "content": "$${\\pi \\over 4} + {\\cos ^{ - 1}}\\,{x^2}$$ "}, {"identifier": "C", "content": "$${\\pi \\over 4} - {1 \\over 2}{\\cos ^{ - 1}}\\,{x^2}$$"}, {"identifier": "D", "content": "$${\\pi \\over ... | ["A"] | null | Given,
<br><br>tan<sup>-1</sup> $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$$
<br><br>Let x<sup>2</sup> = cos $$\theta $$
<br><br>= tan<sup>-1</sup> $$\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \... | mcq | jee-main-2017-online-8th-april-morning-slot | 6,527 |
1AVHKdvqPiPIR7eBWvXeh | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | A value of x satisfying the equation sin[cot<sup>−1</sup> (1+ x)] = cos [tan<sup>−1 </sup>x], is : | [{"identifier": "A", "content": "$$ - {1 \\over 2}$$ "}, {"identifier": "B", "content": "$$-$$ 1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$ {1 \\over 2}$$"}] | ["A"] | null | <p>Let, $${\cot ^{ - 1}}(1 + x) = \alpha $$</p>
<p>$$ \Rightarrow 1 + x = \cot \alpha $$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0v2ql/1fa8d111-320f-4de3-8dab-6d23cbe7b0e2/436fe4d0-d65a-11ec-9a06-bd4ec5b93eb4/file-1l3b0v2qm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.... | mcq | jee-main-2017-online-9th-april-morning-slot | 6,528 |
XPZSMXZi1ZlGHmdnVX3rsa0w2w9jx2gxojq | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$,where –1 $$ \le $$ x $$ \le $$ 1, – 2 $$ \le $$ y $$ \le $$ 2, x $$ \le $$ $${y \over 2}$$
, then for all x, y, 4x<sup>2</sup>
– 4xy cos $$\alpha $$ + y<sup>2</sup>
is equal
to : | [{"identifier": "A", "content": "4 sin<sup>2</sup> $$\\alpha $$"}, {"identifier": "B", "content": "2 sin<sup>2</sup> $$\\alpha $$"}, {"identifier": "C", "content": "4 sin<sup>2</sup> $$\\alpha $$ - 2x<sup>2</sup>y<sup>2</sup>"}, {"identifier": "D", "content": "4 cos<sup>2</sup> $$\\alpha $$ + 2x<sup>2</sup>y<sup>2</sup... | ["A"] | null | $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$<br><br>
$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha $$<br><br>
$$ \Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha $$<br><br>
$$\left( {\cos \alpha ... | mcq | jee-main-2019-online-10th-april-evening-slot | 6,529 |
8ggGtIclV7LzC4NVgY3rsa0w2w9jx6ge0vw | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of $${\sin ^{ - 1}}\left( {{{12} \over {13}}} \right) - {\sin ^{ - 1}}\left( {{3 \over 5}} \right)$$ is equal to : | [{"identifier": "A", "content": "$$\\pi - {\\sin ^{ - 1}}\\left( {{{63} \\over {65}}} \\right)$$"}, {"identifier": "B", "content": "$${\\pi \\over 2} - {\\sin ^{ - 1}}\\left( {{{56} \\over {65}}} \\right)$$"}, {"identifier": "C", "content": "$${\\pi \\over 2} - {\\cos ^{ - 1}}\\left( {{9 \\over {65}}} \\right)$$"}, ... | ["B"] | null | $${\sin ^{ - 1}}{{12} \over {13}} - {\sin ^{ - 1}}{3 \over 5} = {\sin ^{ - 1}}\left( {{{12} \over {13}}.{4 \over 5}.{3 \over 5}.{5 \over {13}}} \right)$$<br><br>
$$ \Rightarrow {\sin ^{ - 1}}{{33} \over {65}} = {\pi \over 2} - {\cos ^{ - 1}}{{33} \over {65}}$$<br><br>
$$ \Rightarrow {\pi \over 2} - {\sin ^{ - 1}}{{56... | mcq | jee-main-2019-online-12th-april-morning-slot | 6,530 |
LzROsT3Flxd6M6C9U01L3 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $$\alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$, $$\beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right)$$ where $$0 < \alpha ,\beta < {\pi \over 2}$$ , then $$\alpha $$ - $$\beta $$ is equal to : | [{"identifier": "A", "content": "$${\\tan ^{ - 1}}\\left( {{9 \\over {14 }}} \\right)$$"}, {"identifier": "B", "content": "$${\\sin ^{ - 1}}\\left( {{9 \\over {5\\sqrt {10} }}} \\right)$$"}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{9 \\over {5\\sqrt {10} }}} \\right)$$"}, {"identifier": "D", "content"... | ["B"] | null | Here $$\cos \alpha = {3 \over 5}$$
<br><br>$$ \therefore $$ $$\tan \alpha = {4 \over 3}$$
<br><br>and $$\tan \beta = {1 \over 3}$$
<br><br>We know,
<br><br>$$\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}$$
<br><br>= $${{{4 \over 3} - {1 \over 3}} \over {... | mcq | jee-main-2019-online-8th-april-morning-slot | 6,531 |
6LQjR16XBRAFABZMYrmun | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If x = sin<sup>$$-$$1</sup>(sin10) and y = cos<sup>$$-$$1</sup>(cos10), then y $$-$$ x is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "7$$\\pi $$"}, {"identifier": "D", "content": "$$\\pi $$"}] | ["D"] | null | x = sin<sup>$$-$$1</sup> sin 10 = 3$$\pi $$ $$-$$ 10
<br><br>y = cos<sup>$$-$$1</sup>cos 10 = 4$$\pi $$ $$-$$ 10
<br><br>y $$-$$ x = (4$$\pi $$ $$-$$ 10) $$-$$ (3$$\pi $$ $$-$$ 10) = $$\pi $$ | mcq | jee-main-2019-online-9th-january-evening-slot | 6,533 |
KKoj2g0io0AWLE2BYi1klrm0vh5 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | A possible value of $$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$$ is : | [{"identifier": "A", "content": "$$\\sqrt 7 - 1$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 7 }}$$"}, {"identifier": "C", "content": "$$2\\sqrt 2 - 1$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt 2 }}$$"}] | ["B"] | null | $$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$$<br><br>$${\sin ^{ - 1}}\left( {{{\sqrt {63} } \over 8}} \right) = \theta $$ $$\sin \theta = {{\sqrt {63} } \over 8}$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265142/exam_images/vmbo5oe8fbvyyyiih8ml.webp" style=... | mcq | jee-main-2021-online-24th-february-evening-slot | 6,536 |
cBLZQ3EuvzBZDVfyOI1klt7uc51 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | cosec$$\left[ {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right]$$ is equal to : | [{"identifier": "A", "content": "$${{75} \\over {56}}$$"}, {"identifier": "B", "content": "$${{65} \\over {56}}$$"}, {"identifier": "C", "content": "$${{56} \\over {33}}$$"}, {"identifier": "D", "content": "$${{65} \\over {33}}$$"}] | ["B"] | null | $$\cos ec\left( {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$$<br><br>$$\cos ec\left( {2{{\tan }^{ - 1}}\left( {{1 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$$<br><br>$$ = \cos ec\left( {{{\tan }^{ - 1}}\left( {{{2\left( {{1 \over 5}} \right)} \over {1 - {... | mcq | jee-main-2021-online-25th-february-evening-slot | 6,537 |
pVJ6lPiMpUPawu7uY81klugzvt7 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${{{{\sin }^1}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$; $$0 < x < 1$$, <br/>then the value of $$\cos \left( {{{\pi c} \over {a + b}}} \right)$$ is : | [{"identifier": "A", "content": "$${{1 - {y^2}} \\over {2y}}$$"}, {"identifier": "B", "content": "$${{1 - {y^2}} \\over {y\\sqrt y }}$$"}, {"identifier": "C", "content": "$$1 - {y^2}$$"}, {"identifier": "D", "content": "$${{1 - {y^2}} \\over {1 + {y^2}}}$$"}] | ["D"] | null | $${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$<br><br>$${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}$$<br><br>Now, $${{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + ... | mcq | jee-main-2021-online-26th-february-morning-slot | 6,538 |
mbltkG0zPddqjhSIul1kluw1xwf | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If 0 < a, b < 1, and tan<sup>$$-$$1</sup>a + tan<sup>$$-$$1</sup>b = $${\pi \over 4}$$, then the value of <br/><br/>$$(a + b) - \left( {{{{a^2} + {b^2}} \over 2}} \right) + \left( {{{{a^3} + {b^3}} \over 3}} \right) - \left( {{{{a^4} + {b^4}} \over 4}} \right) + .....$$ is : | [{"identifier": "A", "content": "$${\\log _e}$$2"}, {"identifier": "B", "content": "e"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{e \\over 2}} \\right)$$"}, {"identifier": "D", "content": "e<sup>2</sup> = 1"}] | ["A"] | null | tan<sup>$$-$$1</sup>a + tan<sup>$$-$$1</sup>b = $${\pi \over 4}$$ 0 < a, b < 1<br><br>$$ \Rightarrow {{a + b} \over {1 - ab}} = 1$$<br><br>a + b = 1 $$-$$ ab<br><br>(a + 1)(b + 1) = 2<br><br>Now $$\left[ {a - {{{a^2}} \over 2} + {{{a^3}} \over 3} + ....} \right] + \left[ {b - {{{b^2}} \over 2} + {{{b^3}} \over 3... | mcq | jee-main-2021-online-26th-february-evening-slot | 6,539 |
1krrolzup | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of $$\tan \left( {2{{\tan }^{ - 1}}\left( {{3 \over 5}} \right) + {{\sin }^{ - 1}}\left( {{5 \over {13}}} \right)} \right)$$ is equal to : | [{"identifier": "A", "content": "$${{ - 181} \\over {69}}$$"}, {"identifier": "B", "content": "$${{220} \\over {21}}$$"}, {"identifier": "C", "content": "$${{ - 291} \\over {76}}$$"}, {"identifier": "D", "content": "$${{151} \\over {63}}$$"}] | ["B"] | null | $$2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) = {\tan ^{ - 1}}\left( {{{6/5} \over {1 - {9 \over {{2^5}}}}}} \right) = {\tan ^{ - 1}}\left( {{{{6 \over 5}} \over {{{16} \over {25}}}}} \right) = {\tan ^{ - 1}}{{15} \over 8}$$<br><br>$$\therefore$$ $$2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) + {\sin ^{ - 1}}\left( {{5 ... | mcq | jee-main-2021-online-20th-july-evening-shift | 6,541 |
1ktei3u10 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$$; 0 < x < 1, a $$\ne$$ 0, then the value of 2x<sup>2</sup> $$-$$ 1 is : | [{"identifier": "A", "content": "$$\\cos \\left( {{{4a} \\over \\pi }} \\right)$$"}, {"identifier": "B", "content": "$$\\sin \\left( {{{2a} \\over \\pi }} \\right)$$"}, {"identifier": "C", "content": "$$\\cos \\left( {{{2a} \\over \\pi }} \\right)$$"}, {"identifier": "D", "content": "$$\\sin \\left( {{{4a} \\over \\pi ... | ["B"] | null | Given $$a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}$$<br><br>$$ = ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$$<br><br>$$ = {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)$$<br><br>$$ \Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }$$<br><br>$$ \Right... | mcq | jee-main-2021-online-27th-august-morning-shift | 6,542 |
1ktfwajo8 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let M and m respectively be the maximum and minimum values of the function <br/>f(x) = tan<sup>$$-$$1</sup> (sin x + cos x) in $$\left[ {0,{\pi \over 2}} \right]$$, then the value of tan(M $$-$$ m) is equal to : | [{"identifier": "A", "content": "$$2 + \\sqrt 3 $$"}, {"identifier": "B", "content": "$$2 - \\sqrt 3 $$"}, {"identifier": "C", "content": "$$3 + 2\\sqrt 2 $$"}, {"identifier": "D", "content": "$$3 - 2\\sqrt 2 $$"}] | ["D"] | null | Let g(x) = sin x + cos x = $$\sqrt 2 $$ sin$$\left( {x + {\pi \over 4}} \right)$$<br><br>g(x)$$\in$$ $$\left[ {1,\sqrt 2 } \right]$$ for x$$\in$$ [0, $$\pi$$/2]<br><br>f(x) = tan<sup>$$-$$1</sup> (sin x + cos x) $$\in$$ $$\left[ {{\pi \over 4},{{\tan }^{ - 1}}\sqrt 2 } \right]$$<br><br>tan$$({\tan ^{ - 1}}\sqrt 2 - ... | mcq | jee-main-2021-online-27th-august-evening-shift | 6,543 |
1l5464te0 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>$$50\tan \left( {3{{\tan }^{ - 1}}\left( {{1 \over 2}} \right) + 2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right) + 4\sqrt 2 \tan \left( {{1 \over 2}{{\tan }^{ - 1}}(2\sqrt 2 )} \right)$$ is equal to ____________.</p> | [] | null | 29 | $50 \tan \left(\tan ^{-1} \frac{1}{2}+2 \tan ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(2)\right)$
$$
+4 \sqrt{2} \tan \left(\frac{\tan ^{-1}}{2}(2 \sqrt{2})\right)
$$
<br/><br/>
$\Rightarrow 50 \tan \left(\pi+\tan ^{-1}\left(\frac{1}{2}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$
<br... | integer | jee-main-2022-online-29th-june-morning-shift | 6,544 |
1l56rlph0 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The value of $$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$$ is :</p> | [{"identifier": "A", "content": "$${{26} \\over {25}}$$"}, {"identifier": "B", "content": "$${{25} \\over {26}}$$"}, {"identifier": "C", "content": "$${{50} \\over {51}}$$"}, {"identifier": "D", "content": "$${{52} \\over {51}}$$"}] | ["A"] | null | <p>$$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$$</p>
<p>$$ = \cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(n + 1) - n} \over {1 + (n + 1)n}}} \right)} } \right)$$</p>
<p>$$ = \cot \left( {\sum\limits_{n = 1}^{50} {({{\tan }^{ - 1}}(n... | mcq | jee-main-2022-online-27th-june-evening-shift | 6,545 |
1l59kzd00 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The value of $${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$ - {\\pi \\over 4}$$"}, {"identifier": "B", "content": "$$ - {\\pi \\over 8}$$"}, {"identifier": "C", "content": "$$ - {{5\\pi } \\over {12}}$$"}, {"identifier": "D", "content": "$$ - {{4\\pi } \\over 9}$$"}] | ["B"] | null | <p>$${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin {\pi \over 4}}}} \right)$$</p>
<p>$$ = {\tan ^{ - 1}}\left( {{{{1 \over {\sqrt 2 }} - 1} \over {{1 \over {\sqrt 2 }}}}} \right)$$</p>
<p>$$ = {\tan ^{ - 1}}(1 - \sqrt 2 ) = - {\tan ^{ - 1}}(\sqrt 2 - 1)$$</p>
<p>$$ = - {\pi \ove... | mcq | jee-main-2022-online-25th-june-evening-shift | 6,546 |
1l5b7rjam | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$x * y = {x^2} + {y^3}$$ and $$(x * 1) * 1 = x * (1 * 1)$$.</p>
<p>Then a value of $$2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$$ is :</p> | [{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$"}, {"identifier": "D", "content": "$${\\pi \\over 6}$$"}] | ["B"] | null | The star "*" in this context represents a binary operation, similar to addition (+), subtraction (-), multiplication (×), and division (÷). It is a custom operation defined by the problem statement, and the specific rules of the operation are provided in the problem.
<br/><br/>
In this case, the operation "*" is defin... | mcq | jee-main-2022-online-24th-june-evening-shift | 6,547 |
1l5c13084 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The set of all values of k for which <br/><br/>$${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3},\,x \in R$$, is the interval :</p> | [{"identifier": "A", "content": "$$\\left[ {{1 \\over {32}},{7 \\over 8}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{1 \\over {24}},{{13} \\over {16}}} \\right)$$"}, {"identifier": "C", "content": "$$\\left[ {{1 \\over {48}},{{13} \\over {16}}} \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {{1 ... | ["A"] | null | <p>$${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3}$$</p>
<p>Let $$f(t) = {t^3} + {\left( {{\pi \over 2} - t} \right)^3}$$</p>
<p>Where $$t = {\tan ^{ - 1}}x$$ ; $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$</p>
<p>$$ = {t^3} + {\left( {{\pi \over 2}} \right)^3} - {{3{\pi ^2}t} \over 4} + {{3... | mcq | jee-main-2022-online-24th-june-morning-shift | 6,548 |
1l6f3v5u9 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$x = \sin (2{\tan ^{ - 1}}\alpha )$$ and $$y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right)$$. If $$S = \{ a \in R:{y^2} = 1 - x\} $$, then $$\sum\limits_{\alpha \in S}^{} {16{\alpha ^3}} $$ is equal to _______________.</p> | [] | null | 130 | <p>$$\because$$ $$x = \sin \left( {2{{\tan }^{ - 1}}\alpha } \right) = {{2\alpha } \over {1 + {\alpha ^2}}}$$ ...... (i)</p>
<p>and $$y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right) = \sin \left( {{{\sin }^{ - 1}}{1 \over {\sqrt 5 }}} \right) = {1 \over {\sqrt 5 }}$$</p>
<p>Now, $${y^2} = 1 - x$$</p>
<... | integer | jee-main-2022-online-25th-july-evening-shift | 6,550 |
1l6gj38zp | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>$$\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$\\frac{1}{4}$$"}, {"identifier": "D", "content": "$$\\frac{5}{4}$$"}] | ["B"] | null | <p>$$\tan \left( {2{{\tan }^{ - 1}}{1 \over 5} + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2} + 2{{\tan }^{ - 1}}{1 \over 8}} \right)$$</p>
<p>$$ = \tan \left( {2{{\tan }^{ - 1}}\left( {{{{1 \over 5} + {1 \over 8}} \over {1 - {1 \over 5}\,.\,{1 \over 8}}}} \right) + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2}} \right)$$</p>
<p>$$ = \... | mcq | jee-main-2022-online-26th-july-morning-shift | 6,551 |
1ldo6tynu | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$$.</p>
<p>If $$\mathrm{n(S)}$$ denotes the number of elements in $$\mathrm{S}$$ then :</p> | [{"identifier": "A", "content": "$$\\mathrm{n}(\\mathrm{S})=0$$"}, {"identifier": "B", "content": "$$\\mathrm{n}(\\mathrm{S})=1$$ and only one element in $$\\mathrm{S}$$ is less than $$\\frac{1}{2}$$."}, {"identifier": "C", "content": "$$\\mathrm{n}(\\mathrm{S})=1$$ and the elements in $$\\mathrm{S}$$ is more than $$\\... | ["D"] | null | $$ {\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)}$$
<br/><br/>$\begin{aligned} & \text { Put } x=\tan \theta \quad \theta \in\left(0, \frac{\pi}{4}\right) \\\\ & 2 \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\cos ^{-1}\left... | mcq | jee-main-2023-online-1st-february-evening-shift | 6,554 |
ldo7xp9f | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let (a, b) $\subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)$,
holds. <br/><br/>If $\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to : | [{"identifier": "A", "content": "$\\frac{\\pi}{16}$\n"}, {"identifier": "B", "content": "$\\frac{\\pi}{48}$\n"}, {"identifier": "C", "content": "$\\frac{\\pi}{8}$\n"}, {"identifier": "D", "content": "$\\frac{\\pi}{12}$"}] | ["D"] | null | $\sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0$
<br/><br/>$\Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4}$
<br/><br/>$\Rightarrow \sin \theta>\frac{1}{\sqrt{2}}$
<br/><br/>So, $\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$
<br/><br/>$\theta \in\left(\frac{\pi}{4}, \frac{3 \pi... | mcq | jee-main-2023-online-31st-january-evening-shift | 6,555 |
1ldom652p | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$S$$ be the set of all solutions of the equation $$\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$$. Then $$\sum_\limits{x \in S} 2 \sin ^{-1}\left(x^{2}-1\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\pi-2 \\sin ^{-1}\\left(\\frac{\\sqrt{3}}{4}\\right)$$"}, {"identifier": "B", "content": "$$\\pi-\\sin ^{-1}\\left(\\frac{\\sqrt{3}}{4}\\right)$$"}, {"identifier": "C", "content": "$$\\frac{-2 \\pi}{3}$$"}, {"identifier": "D", "content": "None"}] | ["D"] | null | $$
\begin{aligned}
& \cos ^{-1}(2 \mathrm{x})=\pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2} \\\\
& \text { Since } \cos ^{-1}(2 \mathrm{x}) \in[0, \pi] \\\\
& \text { R.H.S. } \geq \pi \\\\
& \pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2}=\pi \\\\
& \Rightarrow \cos ^{-1} \sqrt{1-\mathrm{x}^2}=0 \\\\
& \Rightarrow \sqrt{1-\mathrm{x}^... | mcq | jee-main-2023-online-1st-february-morning-shift | 6,556 |
1ldprk1i1 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If $${\sin ^{ - 1}}{\alpha \over {17}} + {\cos ^{ - 1}}{4 \over 5} - {\tan ^{ - 1}}{{77} \over {36}} = 0,0 < \alpha < 13$$, then $${\sin ^{ - 1}}(\sin \alpha ) + {\cos ^{ - 1}}(\cos \alpha )$$ is equal to :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "$$\\pi$$"}, {"identifier": "C", "content": "16 $$-$$ 5$$\\pi$$"}, {"identifier": "D", "content": "0"}] | ["B"] | null | $\sin ^{-1}\left(\frac{\alpha}{17}\right)=-\cos ^{4}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{77}{36}\right)$
<br/><br/>Let $\cos ^{-1}\left(\frac{4}{5}\right)=p$ and $\tan ^{-1}\left(\frac{77}{36}\right)=q$
<br/><br/>$\Rightarrow \sin \left(\sin ^{-1} \frac{\alpha}{17}\right)=\sin (q-p)$
<br/><br/>$=\sin q \cd... | mcq | jee-main-2023-online-31st-january-morning-shift | 6,557 |
ldqv4ewv | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let $a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .$. be consecutive natural numbers.
<br/><br/>Then $\tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots . .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$ is equal to : | [{"identifier": "A", "content": "$\\frac{\\pi}{4}-\\cot ^{-1}(2022)$"}, {"identifier": "B", "content": "$\\frac{\\pi}{4}-\\tan ^{-1}(2022)$"}, {"identifier": "C", "content": "$\\cot ^{-1}(2022)-\\frac{\\pi}{4}$"}, {"identifier": "D", "content": "$\\tan ^{-1}(2022)-\\frac{\\pi}{4}$"}] | null | null | $a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .$. be consecutive natural numbers.
<br/><br/>$$
\begin{aligned}
& \therefore \quad a_2=2, a_3=3, \ldots ., a_{2021}=2021, a_{2022}=2022 \\\\
& \tan ^{-1}\left[\frac{1}{1+a_1 a_2}\right]=\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2}\right) \... | mcqm | jee-main-2023-online-30th-january-evening-shift | 6,558 |
1ldv37zch | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If the sum of all the solutions of $${\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) + {\cot ^{ - 1}}\left( {{{1 - {x^2}} \over {2x}}} \right) = {\pi \over 3}, - 1 < x < 1,x \ne 0$$, is $$\alpha - {4 \over {\sqrt 3 }}$$, then $$\alpha$$ is equal to _____________.</p> | [] | null | 2 | <b>Case-I</b>
<br/><br/>
$-1 < x < 0$
<br/><br/>
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\pi+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$
<br/><br/>
$\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{-\pi}{3}$
<br/><br/>
$2 \tan ^{-1} x=\frac{-\pi}{3}$
<br/><br/>
$\tan ^{-1} x=\frac{-\pi}{6}$
<br/><br/>
$x=\frac{-1... | integer | jee-main-2023-online-25th-january-morning-shift | 6,559 |
1lgoy4b0g | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>For $$x \in(-1,1]$$, the number of solutions of the equation $$\sin ^{-1} x=2 \tan ^{-1} x$$ is equal to __________.</p> | [] | null | 2 | <p>We're given the equation $\sin^{-1}x = 2\tan^{-1}x$ for $x$ in the interval $(-1, 1]$. We want to find the number of solutions.</p>
<p>Step 1: Apply the sine and tangent functions to both sides :</p>
<p>We can rewrite the equation by applying the sine function to both sides :</p>
<p>$$\sin(\sin^{-1}x) = \sin(2\t... | integer | jee-main-2023-online-13th-april-evening-shift | 6,560 |
1lgq0x4gc | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If $$S=\left\{x \in \mathbb{R}: \sin ^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}\right\}$$, then $$\sum_\limits{x \in s}\left(\sin \left(\left(x^{2}+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^{2}+x+5\right) \pi\right)\right)$$ is equal to ___... | [] | null | 4 | Given equation is
<br/><br/>$$
\sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}.
$$
<br/><br/>Let's denote:
<br/><br/>$$A = \sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right),$$
<br/><br/>$$B = \sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right).$$
<br/><... | integer | jee-main-2023-online-13th-april-morning-shift | 6,561 |
jaoe38c1lsfkxule | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$x=\frac{m}{n}$$ ($$m, n$$ are co-prime natural numbers) be a solution of the equation $$\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$$ and let $$\alpha, \beta(\alpha >\beta)$$ be the roots of the equation $$m x^2-n x-m+ n=0$$. Then the point $$(\alpha, \beta)$$ lies on the line</p> | [{"identifier": "A", "content": "$$3 x-2 y=-2$$\n"}, {"identifier": "B", "content": "$$3 x+2 y=2$$\n"}, {"identifier": "C", "content": "$$5 x+8 y=9$$\n"}, {"identifier": "D", "content": "$$5 x-8 y=-9$$"}] | ["C"] | null | <p>Assume $$\sin ^{-1} x=\theta$$</p>
<p>$$\begin{aligned}
& \cos (2 \theta)=\frac{1}{9} \\
& \sin \theta= \pm \frac{2}{3}
\end{aligned}$$</p>
<p>as $$\mathrm{m}$$ and $$\mathrm{n}$$ are co-prime natural numbers,</p>
<p>$$\mathrm{x}=\frac{2}{3}$$</p>
<p>i.e. $$m=2, n=3$$</p>
<p>So, the quadratic equation becomes $$2 x^... | mcq | jee-main-2024-online-29th-january-evening-shift | 6,563 |
lvc57nxl | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>For $$n \in \mathrm{N}$$, if $$\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$$, then $$n$$ is equal to ________.</p> | [] | null | 47 | <p>For $ n \in \mathbb{N} $, if $ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4} $, then $ n $ is equal to <strong></strong><strong></strong>.</p>
<p>Given the equation:</p>
<p>$ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1}(n) = \frac{\pi}{4} $</p>
<p>we can use the identity for the sum... | integer | jee-main-2024-online-6th-april-morning-shift | 6,564 |
P82eExl9uoiR298CyCjgy2xukfg71qct | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | If S is the sum of the first 10 terms of the series <br/><br/>
$${\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....$$<br/><br/>
then tan(S) is equal to : | [{"identifier": "A", "content": "$${10 \\over {11}}$$"}, {"identifier": "B", "content": "$${5 \\over {11}}$$"}, {"identifier": "C", "content": "-$${6 \\over {5}}$$"}, {"identifier": "D", "content": "$${5 \\over {6}}$$"}] | ["D"] | null | S = $${\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....$$
<br><br>= $${\tan ^{ - 1}}\left( {{1 \over {1 + 1 \times 2}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {1 + 2 \times 3}}}... | mcq | jee-main-2020-online-5th-september-morning-slot | 6,566 |
j9QMKExBDBcaXztP8N1kmjbau49 | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | If cot<sup>$$-$$1</sup>($$\alpha$$) = cot<sup>$$-$$1</sup> 2 + cot<sup>$$-$$1</sup> 8 + cot<sup>$$-$$1</sup> 18 + cot<sup>$$-$$1</sup> 32 + ...... upto 100 terms, then $$\alpha$$ is : | [{"identifier": "A", "content": "1.02"}, {"identifier": "B", "content": "1.03"}, {"identifier": "C", "content": "1.01"}, {"identifier": "D", "content": "1.00"}] | ["C"] | null | $${\cot ^{ - 1}}(\alpha ) = co{t^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....100$$ terms<br><br>$$ = {\tan ^{ - 1}}{1 \over 2} + {\tan ^{ - 1}}{1 \over 8} + {\tan ^{ - 1}}{1 \over {18}} + {\tan ^{ - 1}}{1 \over {32}} + ....100$$ term<br><br>$$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{1 \... | mcq | jee-main-2021-online-17th-march-morning-shift | 6,567 |
1ktd06pxr | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | If $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}{1 \over {2{r^2}}} = p} $$, then the value of tan p is : | [{"identifier": "A", "content": "$${{101} \\over {102}}$$"}, {"identifier": "B", "content": "$${{50} \\over {51}}$$"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "$${{51} \\over {50}}$$"}] | ["B"] | null | $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{2 \over {4{r^2}}}} \right) = \sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(2r + 1) - (2r - 1)} \over {1 + (2r + 1)(2r - 1)}}} \right)} } $$<br><br>= $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}(2r + 1) - {{\tan }^{ - 1}}(2r - 1)} $$<br><br>= $${\tan ^{ - 1}}(1... | mcq | jee-main-2021-online-26th-august-evening-shift | 6,568 |
qaoJWPlc6NQngtMw | maths | limits-continuity-and-differentiability | continuity | $$f$$ is defined in $$\left[ { - 5,5} \right]$$ as
<br/><br/>$$f\left( x \right) = x$$ if $$x$$ is rational
<br/><br/>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = - x$$ if $$x$$ is irrational. Then | [{"identifier": "A", "content": "$$f(x)$$ is continuous at every x, except $$x = 0$$"}, {"identifier": "B", "content": "$$f(x)$$ is discontinuous at every $$x,$$ except $$x = 0$$"}, {"identifier": "C", "content": "$$f(x)$$ is continuous everywhere"}, {"identifier": "D", "content": "$$f(x)$$ is discontinuous everywhere"... | ["B"] | null | <b>Let a is a rational number</b> other than $$0,$$ in
<br><br>$$\left[ { - 5,5} \right],$$ then
<br><br>$$f\left( a \right) = a$$ and $$\mathop {\lim }\limits_{x \to a} \,\,f\left( x \right) = - a$$
<br><br>[ As in the immediate neighbourhood of a rational -
<br><br>number, we find irrational numbers ]
<br><br>$$\th... | mcq | aieee-2002 | 6,569 |
0qO7zPDbm3sJgFcz | maths | limits-continuity-and-differentiability | continuity | Let $$f(x) = {{1 - \tan x} \over {4x - \pi }}$$, $$x \ne {\pi \over 4}$$, $$x \in \left[ {0,{\pi \over 2}} \right]$$.
<br/><br/>If $$f(x)$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$, then $$f\left( {{\pi \over 4}} \right)$$ is | [{"identifier": "A", "content": "$$-1$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$-{1 \\over 2}$$"}, {"identifier": "D", "content": "$$1$$"}] | ["C"] | null | $$f\left( x \right) = {{1 - \tan x} \over {4x - \pi }}$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$
<br><br>$$\therefore$$ $$f\left( {{\pi \over 4}} \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} f\left( x \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} \, + f\left( x \right) = \mathop {\l... | mcq | aieee-2004 | 6,570 |
jUvwcp4tbFocqlD1 | maths | limits-continuity-and-differentiability | continuity | The function $$f:R/\left\{ 0 \right\} \to R$$ given by
<br/><br/>$$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
<br/><br/>can be made continuous at $$x$$ = 0 by defining $$f$$(0) as | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$-1$$"}] | ["B"] | null | Given, $$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
<br><br>$$ \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{e^{2x}} - 1} \right) - 2x} \over {x\left( {{e^{2x}} - 1} \right)}}$$ $$\left... | mcq | aieee-2007 | 6,571 |
FvET6cmrFqjSbbCw | maths | limits-continuity-and-differentiability | continuity | The value of $$p$$ and $$q$$ for which the function
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr
q & {,x = 0} \cr
{{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{3/2}}}}} & {,x > 0} \cr
} } \right.$$
<br/><br/>is continuous for all... | [{"identifier": "A", "content": "$$p =$$ $${5 \\over 2}$$, $$q = $$ $${1 \\over 2}$$"}, {"identifier": "B", "content": "$$p =$$ $$-{3 \\over 2}$$, $$q = $$ $${1 \\over 2}$$"}, {"identifier": "C", "content": "$$p =$$ $${1 \\over 2}$$, $$q = $$ $${3 \\over 2}$$"}, {"identifier": "D", "content": "$$p =$$ $${1 \\over 2}$$,... | ["B"] | null | $$L.H.L. = \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sin \left\{ {\left( {p + 1} \right)\left( { - h} \right)} \right\} - \sin \left( { - h} \right)} \over { - h}}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{ - \sin \left( {p + 1} \right)h} \over { - h}} + {{\sin \le... | mcq | aieee-2011 | 6,572 |
kuyx43Dm2TedQW6y | maths | limits-continuity-and-differentiability | continuity | If $$f:R \to R$$ is a function defined by
<br/><br/>$$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$,
<br/><br/>where [x] denotes the greatest integer function, then $$f$$ is | [{"identifier": "A", "content": "continuous for every real $$x$$"}, {"identifier": "B", "content": "discontinuous only at $$x=0$$"}, {"identifier": "C", "content": "discontinuous only at non-zero integral values of $$x$$"}, {"identifier": "D", "content": "continuous only at $$x=0$$"}] | ["A"] | null | Let $$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$$
<br><br>Doubtful points are $$x = n,n \in I$$
<br><br>$$L.H.L$$ $$ = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$
<br><br>$$ = \left( {n - 1} \right)\cos \left( {{{2n - 1} \over ... | mcq | aieee-2012 | 6,573 |
ktLYuFU6rv2nIELWJagcv | maths | limits-continuity-and-differentiability | continuity | Let a, b $$ \in $$ <b>R</b>, (a $$ \ne $$ 0). If the function <i>f</i> defined as
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \cr
{a\,\,\,,} & {1 \le x < \sqrt 2 } \cr
{{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \cr
... | [{"identifier": "A", "content": "$$\\left( {\\sqrt 2 ,1 - \\sqrt 3 } \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - \\sqrt 2 ,1 + \\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {\\sqrt 2 , - 1 + \\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - \\sqrt 2 ,1 - \\sqrt... | ["A"] | null | f(x) is continuous at x = 1
<br><br>$$ \therefore $$ $${{2{{\left( 1 \right)}^2}} \over a} = a$$
<br><br>$$ \Rightarrow $$ a<sup>2</sup> = 2
<br><br>$$ \Rightarrow $$ a = $$ \pm $$ $$\sqrt 2 $$
<br><br>Also f(x) is continuous at x = $$\sqrt 2 $$
<br><br>$$ \therefore... | mcq | jee-main-2016-online-10th-april-morning-slot | 6,574 |
oKPTJbDBgTv4QOlEtDpJg | maths | limits-continuity-and-differentiability | continuity | The value of k for which the function
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr
{k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \cr
} } \right.$$
<br/><br/>is continuous at x = $$... | [{"identifier": "A", "content": "$${{17} \\over {20}}$$ "}, {"identifier": "B", "content": "$${{2} \\over {5}}$$"}, {"identifier": "C", "content": "$${{3} \\over {5}}$$"}, {"identifier": "D", "content": "$$-$$ $${{2} \\over {5}}$$"}] | ["C"] | null | $f(x)=\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}$<br/><br/>
$\lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}=k+\frac{2}{5}$<br/><br/>
$\Rightarrow \lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\frac{2}{5}$<br/><br/>
$\Rightarr... | mcq | jee-main-2017-online-9th-april-morning-slot | 6,575 |
RWZ8zTUsm0YzRTvqcemcN | maths | limits-continuity-and-differentiability | continuity | Let f(x) = $$\left\{ {\matrix{
{{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr
{k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr
} } \right.$$
<br/><br/>Thevaue of k for which f s continuous at x = 2 is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "e"}, {"identifier": "C", "content": "e<sup>-1</sup>"}, {"identifier": "D", "content": "e<sup>-2</sup>"}] | ["C"] | null | Since f(x) is continuous at x = 2.
<br><br>$$\therefore\,\,\,$$$$\mathop {\lim }\limits_{x \to 2} $$ f(x) = f(2)
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} $$ $${\left( {x - 1} \right)^{{1 \over {2 - x}}}}$$ = k ($${{1^\infty }}$$ form)
<br><br>$$ \therefore $$&... | mcq | jee-main-2018-online-15th-april-evening-slot | 6,576 |
6hRA7o4FY7sWQFr59ThCp | maths | limits-continuity-and-differentiability | continuity | If the function f defined as
<br/><br/>$$f\left( x \right) = {1 \over x} - {{k - 1} \over {{e^{2x}} - 1}},x \ne 0,$$ is continuous at
<br/><br/> x = 0, then the ordered pair (k, f(0)) is equal to : | [{"identifier": "A", "content": "(3, 2)"}, {"identifier": "B", "content": "(3, 1)"}, {"identifier": "C", "content": "(2, 1)"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 3},\\,2} \\right)$$"}] | ["B"] | null | If the function is continuous at x = 0, then
<br>$$\mathop {\lim }\limits_{x \to 0} $$ f(x) will exist and f(0) = $$\mathop {\lim }\limits_{x \to 0} $$ f(x)
<br><br>Now, $$\mathop {\lim }\limits_{x \to 0} $$ f(x) = $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {{k - 1} \over {{e^{2x}} - 1}}} \right)$$
<br><... | mcq | jee-main-2018-online-16th-april-morning-slot | 6,577 |
m9GFj6XbKauU5vPk0j9WR | maths | limits-continuity-and-differentiability | continuity | Let f : R $$ \to $$ R be a function defined as
<br/>$$f(x) = \left\{ {\matrix{
5 & ; & {x \le 1} \cr
{a + bx} & ; & {1 < x < 3} \cr
{b + 5x} & ; & {3 \le x < 5} \cr
{30} & ; & {x \ge 5} \cr
} } \right.$$
<br/><br/>Then, f is | [{"identifier": "A", "content": "continuous if a = 0 and b = 5"}, {"identifier": "B", "content": "continuous if a = \u20135 and b = 10"}, {"identifier": "C", "content": "continuous if a = 5 and b = 5 "}, {"identifier": "D", "content": "not continuous for any values of a and b"}] | ["D"] | null | Checking
<br><br>if f(x) is continuous at x = 1 :
<br><br>f(1<sup>$$-$$</sup>) = 5
<br><br>f(1) = 5
<br><br>f(1<sup>+</sup>) = a + b
<br><br>if f(x) is continuous at x = 1,
<br><br>then
<br><br>f(1<sup>$$-$$</sup>) = f(1) = f(1<sup>+</sup>)
<br><br>$$ \Rightarrow $$ 5 = 5 = a + b
<br><br>$$ \therefore $$&... | mcq | jee-main-2019-online-9th-january-morning-slot | 6,578 |
yrQAhYb5CYLwaNHm7fLB7 | maths | limits-continuity-and-differentiability | continuity | Let ƒ : [–1,3] $$ \to $$ R be defined as<br/><br/>
$$f(x) = \left\{ {\matrix{
{\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr
{x + \left| x \right|} & , & {1 \le x < 2} \cr
{x + \left[ x \right]} & , & {2 \le x \le 3} \cr
} } \right.$$<br/><br/>
where [t] ... | [{"identifier": "A", "content": "only three points"}, {"identifier": "B", "content": "four or more points"}, {"identifier": "C", "content": "only two points"}, {"identifier": "D", "content": "only one point"}] | ["A"] | null | $$f(x) = \left\{ {\matrix{
{\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr
{x + \left| x \right|} & , & {1 \le x < 2} \cr
{x + \left[ x \right]} & , & {2 \le x \le 3} \cr
} } \right.$$
<br><br>= $$ = \left\{ {\matrix{
{ - x - 1,} & { - 1 \le x < ... | mcq | jee-main-2019-online-8th-april-evening-slot | 6,579 |
8fazrTwsz2FrcJd12a18hoxe66ijvwpqk6h | maths | limits-continuity-and-differentiability | continuity | If the function ƒ defined on , $$\left( {{\pi \over 6},{\pi \over 3}} \right)$$ by
$$$f(x) = \left\{ {\matrix{
{{{\sqrt 2 {\mathop{\rm cosx}\nolimits} - 1} \over {\cot x - 1}},} & {x \ne {\pi \over 4}} \cr
{k,} & {x = {\pi \over 4}} \cr
} } \right.$$$
is continuous, then
k is equal to | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "1 / $$\\sqrt 2$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "2"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$$ = f($${{\pi \over 4}}$$) = k
<br><br>$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$$ ($${0 \over 0}$$ form) = k
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {\pi \over 4... | mcq | jee-main-2019-online-9th-april-morning-slot | 6,580 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.