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1l6dxf7ak | maths | hyperbola | tangent-to-hyperbola | <p>Let the equation of two diameters of a circle $$x^{2}+y^{2}-2 x+2 f y+1=0$$ be $$2 p x-y=1$$ and $$2 x+p y=4 p$$. Then the slope m $$ \in $$ $$(0, \infty)$$ of the tangent to the hyperbola $$3 x^{2}-y^{2}=3$$ passing through the centre of the circle is equal to _______________.</p> | [] | null | 2 | $$
\begin{aligned}
&2 p+f-1=0 \quad\dots(1)\\\\
&2-p f-4 p=0 \quad\dots(2)\\\\
&2=p(f+4) \\\\
&p=\frac{2}{f+4} \\\\
&2 p=1-f \\\\
&\frac{4}{f+4}=1-f \\\\
&f^2+3 f=0 \\\\
&f=0 \text { or }-3
\end{aligned}
$$<br/><br/>
Hyperbola $3 x^2-y^2=3, x^2-\frac{y^2}{3}=1$<br/><br/>
$$
\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{m}^2... | integer | jee-main-2022-online-25th-july-morning-shift |
1ldo4xfyk | maths | hyperbola | tangent-to-hyperbola | <p>Let $$\mathrm{P}\left(x_{0}, y_{0}\right)$$ be the point on the hyperbola $$3 x^{2}-4 y^{2}=36$$, which is nearest to the line $$3 x+2 y=1$$. Then $$\sqrt{2}\left(y_{0}-x_{0}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$-$$9"}, {"identifier": "C", "content": "$$-$$3"}, {"identifier": "D", "content": "9"}] | ["B"] | null | If $\left(x_0, y_0\right)$ is point on hyperbola then tangent at $\left(x_0, y_0\right)$ is parallel to $3 x+2 y=1$
<br/><br/>Equation of tangent $= \frac{x x_0}{12}-\frac{y y_0}{9}=2$
<br/><br/>Slope of tangent $=\frac{-3}{2}$
<br/><br/>Equation of tangent in slope form
<br/><br/>$y=\frac{-3}{2} x \pm \sqrt{12 \cdot \... | mcq | jee-main-2023-online-1st-february-evening-shift |
1lgoy09kt | maths | hyperbola | tangent-to-hyperbola | <p>The foci of a hyperbola are $$( \pm 2,0)$$ and its eccentricity is $$\frac{3}{2}$$. A tangent, perpendicular to the line $$2 x+3 y=6$$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $$\mathrm{x}$$ - and $$\mathrm{y}$$-axes are $$\mathrm{a}$$ and $$\mathrm{b}... | [] | null | 12 | <p>Given that the foci are at $(\pm 2, 0)$, the distance between the foci is $2ae = 2\times2 = 4$. <br/><br/>So, $a = \frac{4}{3}$. </p>
<p>The eccentricity of the hyperbola is given as $\frac{3}{2}$. Thus, $e = \frac{3}{2}$. </p>
<p>For a hyperbola, $e^2 = 1 + \frac{b^2}{a^2}$, which gives $b^2 = a^2(e^2 - 1)$.<br/><b... | integer | jee-main-2023-online-13th-april-evening-shift |
1lgq0o0y8 | maths | hyperbola | tangent-to-hyperbola | <p>Let $$m_{1}$$ and $$m_{2}$$ be the slopes of the tangents drawn from the point $$\mathrm{P}(4,1)$$ to the hyperbola $$H: \frac{y^{2}}{25}-\frac{x^{2}}{16}=1$$. If $$\mathrm{Q}$$ is the point from which the tangents drawn to $$\mathrm{H}$$ have slopes $$\left|m_{1}\right|$$ and $$\left|m_{2}\right|$$ and they make po... | [] | null | 8 | Equation of tangent to the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
<br/><br/>$$
y=m x \pm \sqrt{a^2-b^2 m^2}
$$
<br/><br/>Given the hyperbola $H: \frac{y^2}{25} - \frac{x^2}{16} = 1$, the equation of the tangent to this hyperbola can be written as :
<br/><br/>$$y=mx \pm \sqrt{25 - 16m^2}$$
<br/><br/>We know tha... | integer | jee-main-2023-online-13th-april-morning-shift |
lv2er3wn | maths | hyperbola | tangent-to-hyperbola | <p>Consider a hyperbola $$\mathrm{H}$$ having centre at the origin and foci on the $$\mathrm{x}$$-axis. Let $$\mathrm{C}_1$$ be the circle touching the hyperbola $$\mathrm{H}$$ and having the centre at the origin. Let $$\mathrm{C}_2$$ be the circle touching the hyperbola $$\mathrm{H}$$ at its vertex and having the cent... | [{"identifier": "A", "content": "$$\\frac{28}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{11}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{14}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{10}{3}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhfif9k/1a3e3d85-d2e7-4099-8558-58b863f1c6ee/866ac480-1802-11ef-b156-f754785ad3ce/file-1lwhfif9l.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhfif9k/1a3e3d85-d2e7-4099-8558-58b863f1c6ee/866ac480-1802-11ef-b156-f754785ad3ce... | mcq | jee-main-2024-online-4th-april-evening-shift |
lvb2951p | maths | hyperbola | tangent-to-hyperbola | <p>The length of the latus rectum and directrices of hyperbola with eccentricity e are 9 and $$x= \pm \frac{4}{\sqrt{3}}$$, respectively. Let the line $$y-\sqrt{3} x+\sqrt{3}=0$$ touch this hyperbola at $$\left(x_0, y_0\right)$$. If $$\mathrm{m}$$ is the product of the focal distances of the point $$\left(x_0, y_0\righ... | [] | null | 61 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwaskftu/d526b7cc-575f-4620-8047-7c036e1ade54/189a8020-145c-11ef-a76a-45e5478b32e1/file-1lwaskftv.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwaskftu/d526b7cc-575f-4620-8047-7c036e1ade54/189a8020-145c-11ef-a76a-45e5478b32e1... | integer | jee-main-2024-online-6th-april-evening-shift |
1ktgqhz9h | maths | indefinite-integrals | integration-by-partial-fraction | If $$\int {{{2{e^x} + 3{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}dx = {1 \over {14}}(ux + v{{\log }_e}(4{e^x} + 7{e^{ - x}})) + C} $$, where C is a constant of integration, then u + v is equal to _____________. | [] | null | 7 | $$2{e^x} + 3{e^{ - x}} = A(4{e^x} + 7{e^{ - x}}) + B(4{e^x} - 7{e^{ - x}}) + \lambda $$<br><br>2 = 4A + 4B ; 3 = 7A $$-$$ 7B ; $$\lambda$$ = 0<br><br>$$A + B = {1 \over 2}$$<br><br>$$A - B = {3 \over 7}$$<br><br>$$A = {1 \over 2}\left( {{1 \over 2} + {3 \over 7}} \right) = {{7 + 6} \over {28}} = {{13} \over {28}}$$<br>... | integer | jee-main-2021-online-27th-august-evening-shift |
1ktke5n0m | maths | indefinite-integrals | integration-by-partial-fraction | If $$\int {{{\sin x} \over {{{\sin }^3}x + {{\cos }^3}x}}dx = } $$
<br/><br/>$$\alpha {\log _e}|1 + \tan x| + \beta {\log _e}|1 - \tan x + {\tan ^2}x| + \gamma {\tan ^{ - 1}}\left( {{{2\tan x - 1} \over {\sqrt 3 }}} \right) + C$$, when C is constant of integration, then the value of $$18(\alpha + \beta + {\gamma ^2}... | [] | null | 3 | $$ = \int {{{{{\sin x} \over {{{\cos }^3}x}}} \over {1 + {{\tan }^3}x}}dx = \int {{{\tan x.{{\sec }^2}x} \over {(\tan x + 1)(1 + {{\tan }^2}x - \tan x)}}dx} } $$<br><br>Let $$\tan x = t \Rightarrow {\sec ^2}x.\,dx = dt$$<br><br>$$ = \int {{t \over {(t + 1)({t^2} - t + 1)}}dt} $$<br><br>$$ = \int {\left( {{A \over {t + ... | integer | jee-main-2021-online-31st-august-evening-shift |
Uw47dOEvkjyfJj7r | maths | indefinite-integrals | integration-by-parts | The integral $$\int {\left( {1 + x - {1 \over x}} \right){e^{x + {1 \over x}}}dx} $$ is equal to | [{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264772/exam_images/je7jltiyhhqoek98uddy.webp\" loading=\"lazy\" alt=\"JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 68 English Option 1\"> "}, {"identifier": "B", "content": ... | ["D"] | null | Let $$I = \int {\left( {1 + x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$
<br><br>$$ = \int {{e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}} dx + \int {\left( {x - {1 ... | mcq | jee-main-2014-offline |
k0vrHx9KC2v0TCKPU1nFZ | maths | indefinite-integrals | integration-by-parts | The integral $$\int \, $$cos(log<sub>e</sub> x) dx is equal to : (where C is a constant of integration) | [{"identifier": "A", "content": "$${x \\over 2}$$[sin(log<sub>e</sub> x) $$-$$ cos(log<sub>e</sub> x)] + C"}, {"identifier": "B", "content": "x[cos(log<sub>e</sub> x) + sin(log<sub>e</sub> x)] + C"}, {"identifier": "C", "content": "$${x \\over 2}$$[cos(log<sub>e</sub> x) + sin(log<sub>e</sub> x)] + C"}, {"identifier": ... | ["C"] | null | $${\rm I} = \int {\cos \left( {\ell nx} \right)} dx$$
<br><br>$${\rm I} = \cos (\ln x).x + \int {\sin \left( {\ell nx} \right)dx} $$
<br><br>$${\rm I} = \cos \left( {\ell nx} \right)x + \left[ {\sin \left( {\ell nx} \right).x - \int {\cos \left( {\ell nx} \right)dx} } \right]$$
<br><br>$${\rm I} = {x \over 2}\left[ {\s... | mcq | jee-main-2019-online-12th-january-morning-slot |
1l6re7z1v | maths | indefinite-integrals | integration-by-parts | <p>For $$I(x)=\int \frac{\sec ^{2} x-2022}{\sin ^{2022} x} d x$$, if $$I\left(\frac{\pi}{4}\right)=2^{1011}$$, then</p> | [{"identifier": "A", "content": "$$3^{1010} I\\left(\\frac{\\pi}{3}\\right)-I\\left(\\frac{\\pi}{6}\\right)=0$$"}, {"identifier": "B", "content": "$$3^{1010} I\\left(\\frac{\\pi}{6}\\right)-I\\left(\\frac{\\pi}{3}\\right)=0$$"}, {"identifier": "C", "content": "$$3^{1011} I\\left(\\frac{\\pi}{3}\\right)-I\\left(\\frac{\... | ["A"] | null | <p>Given,</p>
<p>$$I(x) = \int {{{{{\sec }^2}x - 2022} \over {{{\sin }^{2022}}x}}dx} $$</p>
<p>$$ = \int {{{{{\sec }^2}x} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $$</p>
<p>$$ = \int {{1 \over {{{\sin }^{2022}}x}}\,.\,{{\sec }^2}x\,dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $... | mcq | jee-main-2022-online-29th-july-evening-shift |
1lgxgz9mo | maths | indefinite-integrals | integration-by-parts | <p>If $$I(x) = \int {{e^{{{\sin }^2}x}}(\cos x\sin 2x - \sin x)dx} $$ and $$I(0) = 1$$, then $$I\left( {{\pi \over 3}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$ - {e^{{3 \\over 4}}}$$"}, {"identifier": "B", "content": "$$ - {1 \\over 2}{e^{{3 \\over 4}}}$$"}, {"identifier": "C", "content": "$${e^{{3 \\over 4}}}$$"}, {"identifier": "D", "content": "$${1 \\over 2}{e^{{3 \\over 4}}}$$"}] | ["D"] | null | $$
\begin{aligned}
& \text { Given, } I(x)=\int e^{\sin ^2 x}(\cos x \sin 2 x-\sin x) d x \\\\
& =\int e^{\sin ^2 x} \cdot \cos x \cdot \sin 2 x d x-\int \sin x e^{\sin ^2 x} d x \\\\
& =\int \frac{\cos x}{\mathrm{I}} \cdot \frac{e^{\sin ^2 x} \cdot \sin 2 x}{\mathrm{II}} d x-\int \sin x \cdot e^{\sin ^2 x} d x \\\\
& ... | mcq | jee-main-2023-online-10th-april-morning-shift |
1lh21fmey | maths | indefinite-integrals | integration-by-parts | <p>Let $$I(x)=\int \frac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d x$$. If $$I(0)=0$$, then $$I\left(\frac{\pi}{4}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{32}-\\frac{\\pi^{2}}{4(\\pi+4)}$$"}, {"identifier": "B", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{16}-\\frac{\\pi^{2}}{4(\\pi+4)}$$"}, {"identifier": "C", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{16}+\\frac{\\pi^{2}}{4(\\pi+4)}$$"}, {"identifi... | ["A"] | null | We have,
<br/><br/>$$
\begin{aligned}
I(x)= & \int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x \\\\
= & x^2 \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x \\\\
& \quad-\int\left\{\frac{d}{d x}\left(x^2\right) \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x\right\} d x \text { (integration by pa... | mcq | jee-main-2023-online-6th-april-morning-shift |
lv2eqxr8 | maths | indefinite-integrals | integration-by-parts | <p>If $$\int \operatorname{cosec}^5 x d x=\alpha \cot x \operatorname{cosec} x\left(\operatorname{cosec}^2 x+\frac{3}{2}\right)+\beta \log _x\left|\tan \frac{x}{2}\right|+\mathrm{C}$$
where $$\alpha, \beta \in \mathbb{R}$$ and $$\mathrm{C}$$ is the constant of integration, then the value of $$8(\alpha+\beta)$$ equals _... | [] | null | 1 | <p>$$\begin{aligned}
& I=\int(\operatorname{cosec} x)^5 d x=\int(\operatorname{cosec} x)^3(\operatorname{cosec} x)^2 d x \\
& =(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x- \\
& \int\left(\frac{d}{d x}(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x\right) d x
\end{aligned}$$</p>
<p>$$\begin... | integer | jee-main-2024-online-4th-april-evening-shift |
hKe3lNduHiApUmZ8 | maths | indefinite-integrals | integration-by-substitution | $$\int {{{\left\{ {{{\left( {\log x - 1} \right)} \over {1 + {{\left( {\log x} \right)}^2}}}} \right\}}^2}\,\,dx} $$ is equal to | [{"identifier": "A", "content": "$${{\\log x} \\over {{{\\left( {\\log x} \\right)}^2} + 1}} + C$$ "}, {"identifier": "B", "content": "$${x \\over {{x^2} + 1}} + C$$ "}, {"identifier": "C", "content": "$${{x{e^x}} \\over {1 + {x^2}}} + C$$ "}, {"identifier": "D", "content": "$${x \\over {{{\\left( {\\log x} \\right)}^2... | ["D"] | null | $$\int {{{{{\left( {\log x - 1} \right)}^2}} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} dx$$
<br><br>$$ = \int {{{1 + {{\left( {\log x} \right)}^2} - 2\log x} \over {{{\left[ {1 + {{\left( {\log x} \right)}^2}} \right]}^2}}}} $$
<br><br>$$ = \int {\left[ {{1 \over {\left( {1 + {{\left( {\log x} ... | mcq | aieee-2005 |
zX0bJHNzNUnkDwG7 | maths | indefinite-integrals | integration-by-substitution | The value of $$\sqrt 2 \int {{{\sin xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}} $$ is | [{"identifier": "A", "content": "$$\\,x + \\log \\,\\left| {\\,\\cos \\left( {x - {\\pi \\over 4}} \\right)\\,} \\right| + c$$"}, {"identifier": "B", "content": "$$\\,x - \\log \\,\\left| {\\,\\sin \\left( {x - {\\pi \\over 4}} \\right)\\,} \\right| + c$$"}, {"identifier": "C", "content": "$$\\,x + \\log \\,\\left| {... | ["C"] | null | Let $$I = \sqrt 2 \int {{{\sin \,xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}} $$
<br><br>Put $$x - {\pi \over 4} = t$$
<br><br>$$ \Rightarrow dx = dt$$
<br><br>$$ \Rightarrow I = \sqrt 2 \int {{{\sin \left( {t + {\pi \over 4}} \right)} \over {\sin \,t}}} dt$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\, = {{\sqrt 2 ... | mcq | aieee-2008 |
2ACvJSdqk8b5slgg | maths | indefinite-integrals | integration-by-substitution | If the $$\int {{{5\tan x} \over {\tan x - 2}}dx = x + a\,\ln \,\left| {\sin x - 2\cos x} \right| + k,} $$ then $$a$$ is
<br/>equal to : | [{"identifier": "A", "content": "$$-1$$ "}, {"identifier": "B", "content": "$$-2$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["D"] | null | $$\int {{{5\tan x} \over {\tan x - 2}}} dx$$
<br><br>$$ = \int {{{5{{\sin x} \over {\cos x}}} \over {{{\sin x} \over {\cos x}} - 2}}} \,dx$$
<br><br>$$ = \int {\left( {{{5\sin x} \over {\cos x}} \times {{\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$
<br><br>$$ = \int {{{5\,\sin \,x\,dx} \over {\sin x - 2\,\cos x}}... | mcq | aieee-2012 |
7Kbr9AwkQApiexVy | maths | indefinite-integrals | integration-by-substitution | If $$\int {f\left( x \right)dx = \psi \left( x \right),} $$ then $$\int {{x^5}f\left( {{x^3}} \right)dx} $$ is equal to | [{"identifier": "A", "content": "$${1 \\over 3}\\left[ {{x^3}\\psi \\left( {{x^3}} \\right) - \\int {{x^2}\\psi \\left( {{x^3}} \\right)dx} } \\right] + C$$ "}, {"identifier": "B", "content": "$${1 \\over 3}{x^3}\\psi \\left( {{x^3}} \\right) - 3\\int {{x^3}\\psi \\left( {{x^3}} \\right)dx} + C$$ "}, {"identifier": "C... | ["C"] | null | Let $$\int {f\left( x \right)dx = \psi \left( x \right)} $$
<br><br>Let $$I = \int {{x^5}} f\left( {{x^3}} \right)dx$$
<br><br>put $${x^3} = t \Rightarrow 3{x^2}dx = dt$$
<br><br>$$I = {1 \over 3}\int {3.{x^2}} .{x^3}.f\left( {{x^3}} \right).dx$$
<br><br>$$ = {1 \over 3}\int {tf} \left( t \right)dt$$
<br><br>$$ = {1 \o... | mcq | jee-main-2013-offline |
nZ3cEdvpgAwa9myY | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} $$ equals : | [{"identifier": "A", "content": "$$ - {\\left( {{x^4} + 1} \\right)^{{1 \\over 4}}} + c$$"}, {"identifier": "B", "content": "$$ - {\\left( {{{{x^4} + 1} \\over {{x^4}}}} \\right)^{{1 \\over 4}}} + c$$ "}, {"identifier": "C", "content": "$$ {\\left( {{{{x^4} + 1} \\over {{x^4}}}} \\right)^{{1 \\over 4}}} + c$$ "}, {"id... | ["B"] | null | $$1 = \int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} $$
<br><br>$$ = \int {{{dx} \over {{x^3}{{\left( {1 + {x^{ - 4}}} \right)}^{3/4}}}}} $$
<br><br>Let $${x^{ - 4}} = y$$
<br><br>$$ \Rightarrow - 4{x^{ - 3}}\,dx = dy$$
<br><br>$$ \Rightarrow dx = {{ - 1} \over 4}{x^3}dy$$
<br><br>$$\therefore$$ $$I ... | mcq | jee-main-2015-offline |
kUOFTa2sNbMr1KkF | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx$$ is equal to : | [{"identifier": "A", "content": "$${{{x^5}} \\over {2{{\\left( {{x^5} + {x^3} + 1} \\right)}^2}}} + C$$ "}, {"identifier": "B", "content": "$${{ - {x^{10}}} \\over {2{{\\left( {{x^5} + {x^3} + 1} \\right)}^2}}} + C$$ "}, {"identifier": "C", "content": "$${{{-x^5}} \\over {{{\\left( {{x^5} + {x^3} + 1} \\right)}^2}}} + ... | ["D"] | null | $$\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx$$
<br><br>Dividing by $${x^{15}}$$ in numerator and denominator
<br><br>$$\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over 5}} \right)}^3}}}} $$
<br><br>Substitute $$1 + {1 \over {{x^2}... | mcq | jee-main-2016-offline |
LQe1HRHuBYA8dRTJJfiHH | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} $$ is equal to :
<br/><br/>(where C is a constant of integration.) | [{"identifier": "A", "content": "$$ - 2\\sqrt {{{1 + \\sqrt x } \\over {1 - \\sqrt x }}} + C$$ "}, {"identifier": "B", "content": "$$ - 2\\sqrt {{{1 - \\sqrt x } \\over {1 + \\sqrt x }}} + C$$"}, {"identifier": "C", "content": "$$ - \\sqrt {{{1 - \\sqrt x } \\over {1 + \\sqrt x }}} + C$$ "}, {"identifier": "D", "con... | ["B"] | null | I = $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} $$
<br><br>= $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt x \sqrt {1 - x} }}} $$
<br><br>Let 1 + $$\sqrt x $$ = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$$${1 \over {2\sqrt x }}\,dx$$ = dt
<br... | mcq | jee-main-2016-online-10th-april-morning-slot |
7Gnzjnba9fRxVw93NEa9p | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k,$$
<br/><br/>where k is a constant of integration, then A + B +C equals : | [{"identifier": "A", "content": "$${{21} \\over 5}$$ "}, {"identifier": "B", "content": "$${{16} \\over 5}$$"}, {"identifier": "C", "content": "$${{7} \\over 10}$$"}, {"identifier": "D", "content": "$${{27} \\over 10}$$"}] | ["B"] | null | $$\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} $$
<br><br>= $$\int {{{dx} \over {{{\cos }^3}x\sqrt {4\sin x\cos x} }}} $$
<br><br>= $$\int {{{dx} \over {2{{\cos }^4}x\sqrt {\tan x} }}} $$
<br><br>Let tan x = t<sup>2</sup>
<br><br>$$ \Rightarrow $$$$\,\,\,$$ sec<sup>2... | mcq | jee-main-2016-online-9th-april-morning-slot |
5rgUmlOIq0Q0L7jm | maths | indefinite-integrals | integration-by-substitution | Let $${I_n} = \int {{{\tan }^n}x\,dx} ,\,\left( {n > 1} \right).$$
<br/><br/>If $${I_4} + {I_6}$$ = $$a{\tan ^5}x + b{x^5} + C$$, where C is a constant of integration,
<br/><br/>then the ordered pair $$\left( {a,b} \right)$$ is equal to | [{"identifier": "A", "content": "$$\\left( {{1 \\over 5},0} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{1 \\over 5}, - 1} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 5},0} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - {1 \\over 5},1} \\right)$$"}] | ["A"] | null | Given,
<br><br>In = $$\int {{{\tan }^n}x\,dx,\,\,\,n > 1} $$
<br><br>$$\therefore\,\,\,$$ I<sub>4</sub> = $$\int {{{\tan }^4}x\,dx} $$
<br><br>and I<sub>6</sub> = $$\int {{{\tan }^6}} x\,dx$$
<br><br>$$\therefore\,\,\,$$ I = I<sub>4</sub> + I<sub>6</sub>
<br><br>= $$\int {\left( {{{\tan }^4}x + {{\tan }^6}x} \right... | mcq | jee-main-2017-offline |
4FTEIQa2Co5vKRg6Ns20r | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}} $$ $$\left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,$$ is a constant of integration) then the ordered pair (K, A) is equal to : | [{"identifier": "A", "content": "(2, 1)"}, {"identifier": "B", "content": "($$-$$2, 3)"}, {"identifier": "C", "content": "(2, 3)"}, {"identifier": "D", "content": "($$-$$2, 1)"}] | ["C"] | null | Given,
<br><br>$$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx$$
<br><br>Let tanx = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$ sec<sup>2</sup>x dx = dt
<br><br>$$ \Rightarrow $$$$\,\,\,$$ dx = $${{dt} \over {{{\sec }^2}x}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$dx = {{dt} \over {1 + {{\tan }^2}x}}$$
<br><br>$... | mcq | jee-main-2018-online-16th-april-morning-slot |
sWW8X3j3VBEBvcDd | maths | indefinite-integrals | integration-by-substitution | The integral
<br/><br/>$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}} dx$$
<br/><br/>is equal to | [{"identifier": "A", "content": "$${{ - 1} \\over {1 + {{\\cot }^3}x}} + C$$"}, {"identifier": "B", "content": "$${1 \\over {3\\left( {1 + {{\\tan }^3}x} \\right)}} + C$$"}, {"identifier": "C", "content": "$${{ - 1} \\over {3\\left( {1 + {{\\tan }^3}x} \\right)}} + C$$"}, {"identifier": "D", "content": "$${1 \\over {1 ... | ["C"] | null | Given,
<br><br>$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}}\,dx $$
<br><br>$$ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left[ {{{\sin }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\cos }^3}x\left( {... | mcq | jee-main-2018-offline |
Qj3TsBFFhWo0Z64Jt54cd | maths | indefinite-integrals | integration-by-substitution | Let n $$ \ge $$ 2 be a natural number and $$0 < \theta < {\pi \over 2}.$$ Then $$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$ is equal to - (where C is a constant of integration) | [{"identifier": "A", "content": "$${n \\over {{n^2} - 1}}{\\left( {1 + {1 \\over {{{\\sin }^{n - 1}}\\theta }}} \\right)^{{{n + 1} \\over n}}} + C$$"}, {"identifier": "B", "content": "$${n \\over {{n^2} - 1}}{\\left( {1 - {1 \\over {{{\\sin }^{n + 1}}\\theta }}} \\right)^{{{n + 1} \\over n}}} + C$$"}, {"identifier": "C... | ["C"] | null | $$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$
<br><br>$$ = \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}} \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$
<br><br>Put $$1 - {1 \over {{{\sin }^{... | mcq | jee-main-2019-online-10th-january-morning-slot |
H8ln4uLN3lq8lQeMmZ3rsa0w2w9jxacphio | maths | indefinite-integrals | integration-by-substitution | Let $$a \in \left( {0,{\pi \over 2}} \right)$$ be fixed. If the integral
<br/><br>$$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx$$ = A(x) cos 2$$\alpha $$ + B(x) sin 2$$\alpha $$ + C, where C is a
<br/><br/>constant of integration, then the functions A(x) and B(x) are respectively : </br> | [{"identifier": "A", "content": "$$x - \\alpha $$ and $${\\log _e}\\left| {\\cos \\left( {x - \\alpha } \\right)} \\right|$$"}, {"identifier": "B", "content": "$$x + \\alpha $$ and $${\\log _e}\\left| {\\sin \\left( {x - \\alpha } \\right)} \\right|$$"}, {"identifier": "C", "content": "$$x + \\alpha $$ and $${\\log _e}... | ["D"] | null | <p>To solve the given integral, first we simplify the expression in the integral as follows :</p>
<p>$$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx = \int {{{\sin x \over \cos x} + {\sin \alpha \over \cos \alpha }} \over {{\sin x \over \cos x} - {\sin \alpha \over \cos \alpha }}} dx$$ </p>
<p>Simpl... | mcq | jee-main-2019-online-12th-april-evening-slot |
2TA2SAVPabICEfCIXG3rsa0w2w9jx2b6x3x | maths | indefinite-integrals | integration-by-substitution | If $$\int {{x^5}} {e^{ - {x^2}}}dx = g\left( x \right){e^{ - {x^2}}} + c$$, where c is a constant of integration, then $$g$$(–1) is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "- 1"}, {"identifier": "C", "content": "$$ - {5 \\over 2}$$"}, {"identifier": "D", "content": "$$ - {1 \\over 2}$$"}] | ["C"] | null | Let x<sup>2</sup> = t<br><br>
$$ \Rightarrow {1 \over 2}\int {{t^2}{e^{ - t}}dt} $$<br><br>
$$ \Rightarrow {1 \over 2}\left[ { - {t^2}{e^{ - t}} + \int {2t{e^{ - t}}dt} } \right]$$<br><br>
$$ \Rightarrow {{ - {t^2}{e^{ - t}}} \over 2} - t{e^{ - t}} - {e^{ - t}}$$<br><br>
$$ \Rightarrow \left( { - {{{x^4}} \over 2} - {x... | mcq | jee-main-2019-online-10th-april-evening-slot |
NGp83ukLX0F1lubeJ33rsa0w2w9jwxvclk0 | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{dx} \over {{{\left( {{x^2} - 2x + 10} \right)}^2}}}} = A\left( {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{f\left( x \right)} \over {{x^2} - 2x + 10}}} \right) + C$$
<br/><br/> where C is a constant of integration then : | [{"identifier": "A", "content": "A =$${1 \\over {54}}$$ and f(x) = 9(x\u20131)<sup>2</sup>"}, {"identifier": "B", "content": "A =$${1 \\over {54}}$$ and f(x) = 3(x\u20131)"}, {"identifier": "C", "content": "A =$${1 \\over {81}}$$ and f(x) = 3(x\u20131)"}, {"identifier": "D", "content": "A =$${1 \\over {27}}$$ and f(x) ... | ["B"] | null | $$\int {{{dx} \over {{{({x^2} - 2x + 10)}^2}}} = \int {{{dx} \over {{{({{(x - 1)}^2} + 9)}^2}}}} } $$<br><br>
$$Let{\rm{ }}{\left( {x{\rm{ }}-{\rm{ }}1} \right)^2}{\rm{ }} = {\rm{ }}9ta{n^2}\theta \,\,\,\,...\left( i \right)$$<br><br>
$$ \Rightarrow \tan \theta = {{x - 1} \over 3}$$<br><br>
On Differentiating ...(i)<b... | mcq | jee-main-2019-online-10th-april-morning-slot |
QtWndUPN5ICUBfhiUWRfe | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{\rm{se}}{{\rm{c}}^{{\rm{2/ 3}}}}\,{\rm{x }}\,{\rm{cose}}{{\rm{c}}^{{\rm{4 / 3}}}}{\rm{x \,dx}}} $$ is equal to
(Hence C is a constant of integration) | [{"identifier": "A", "content": "-3/4 tan <sup>- 4 / 3</sup> x + C"}, {"identifier": "B", "content": "3tan<sup>\u20131/3</sup>x + C"}, {"identifier": "C", "content": "\u20133cot<sup>\u20131/3</sup>x+ C"}, {"identifier": "D", "content": "- 3tan<sup>\u20131/3</sup>x + C"}] | ["D"] | null | $$\int {{{\sec }^{{2 \over 3}}}} x\cos e{c^{{4 \over 3}}}xdx$$
<br><br>= $$\int {{{{{\sec }^{{2 \over 3}}}x} \over {\cos e{c^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
<br><br>= $$\int {{1 \over {{{\cot }^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
<br><br>Let cot x = t<sup>3</sup>
<br><br>$$ \Rightarrow $$ - cosec<sup>2</sup>x dx = ... | mcq | jee-main-2019-online-9th-april-morning-slot |
AQn2qq06q2fjAXy7QknRR | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C} $$ <br/>
where C is a constant of integration, then the
function ƒ(x) is equal to | [{"identifier": "A", "content": "$${3 \\over {{x^2}}}$$"}, {"identifier": "B", "content": "$$ - {1 \\over {6{x^3}}}$$"}, {"identifier": "C", "content": "$$ - {1 \\over {2{x^3}}}$$"}, {"identifier": "D", "content": "$$ - {1 \\over {2{x^2}}}$$"}] | ["C"] | null | I = $$\int {{{dx} \over {{x^3}{{\left( {1 + {x^6}} \right)}^{{2 \over 3}}}}}} $$
<br><br>= $$\int {{{dx} \over {{x^7}{{\left( {{1 \over {{x^6}}} + 1} \right)}^{{2 \over 3}}}}}} $$
<br><br>Let $${{1 \over {{x^6}}} + 1}$$ = t
<br><br>$$ \Rightarrow $$ $${{ - 6} \over {{x^7}}}dx = dt$$
<br><br>$$ \Rightarrow $$ $${{dx} \o... | mcq | jee-main-2019-online-8th-april-evening-slot |
ye4tbM04B4WyItNxoFvXn | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} \,dx$$ is equal to : (where C is a constant of integration)
| [{"identifier": "A", "content": "$${{{x^{12}}} \\over {6{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^3}}}$$ + $$C$$"}, {"identifier": "B", "content": "$${{{x^4}} \\over {6{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^3}}} + C$$"}, {"identifier": "C", "content": "$${{{x^{12}}} \\over {{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^... | ["A"] | null | $$\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} dx$$
<br><br>$$\int {{{\left( {{3 \over {{x^3}}} + {2 \over {{x^5}}}} \right)dx} \over {{{\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right)}^4}}}} $$
<br><br>Let $$\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}}... | mcq | jee-main-2019-online-12th-january-evening-slot |
xn6ue9zHp1nBwlYZ056NP | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx$$ = f(x) $$\sqrt {2x - 1} $$ + C, where C is a constant of integration, then f(x) is equal to : | [{"identifier": "A", "content": "$${2 \\over 3}$$ (x $$-$$ 4)"}, {"identifier": "B", "content": "$${1 \\over 3}$$ (x + 4)"}, {"identifier": "C", "content": "$${1 \\over 3}$$ (x + 1)"}, {"identifier": "D", "content": "$${2 \\over 3}$$ (x + 2)"}] | ["B"] | null | $$\sqrt {2x - 1} = t \Rightarrow 2x - 1 = {t^2} \Rightarrow 2dx = 2t.dt$$
<br><br>$$\int {{{x + 1} \over {\sqrt {2x - 1} }}dx = \int {{{{{{t^2} + 1} \over 2} + 1} \over t}tdt = \int {{{{t^2} + 3} \over 2}dt} } } $$
<br><br>$$ = {1 \over 2}\left( {{{{t^3}} \over 3} + 3t} \right) = {t \over 6}\left( {{t^2} + 9} \right) ... | mcq | jee-main-2019-online-11th-january-evening-slot |
wH0XiFnWxUutIMcwlOwvx | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))<sup>m</sup> equals : | [{"identifier": "A", "content": "$${1 \\over {27{x^6}}}$$"}, {"identifier": "B", "content": "$${{ - 1} \\over {27{x^9}}}$$"}, {"identifier": "C", "content": "$${1 \\over {9{x^4}}}$$"}, {"identifier": "D", "content": "$${1 \\over {3{x^3}}}$$"}] | ["B"] | null | $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C
<br><br>$$\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,$$
<br><br>Put $${1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}$$
<br><br><b>Case-... | mcq | jee-main-2019-online-11th-january-morning-slot |
GNwlKDvLfwVDVMJ3J4cOC | maths | indefinite-integrals | integration-by-substitution | If $$\int \, $$x<sup>5</sup>.e<sup>$$-$$4x<sup>3</sup></sup> dx = $${1 \over {48}}$$e<sup>$$-$$4x<sup>3</sup></sup> f(x) + C, where C is a constant of inegration, then f(x) is equal to - | [{"identifier": "A", "content": "$$-$$2x<sup>3</sup> $$-$$ 1"}, {"identifier": "B", "content": "$$-$$ 2x<sup>3</sup> + 1 "}, {"identifier": "C", "content": "4x<sup>3</sup> + 1"}, {"identifier": "D", "content": "$$-$$4x<sup>3</sup> $$-$$ 1"}] | ["D"] | null | $$\int {{x^5}} .{e^{ - 4{x^3}}}\,dx = {1 \over {48}}{e^{ - 4{x^3}}}f\left( x \right) + c$$
<br><br>Put $${x^3} = t$$
<br><br>$$3{x^2}\,dx = dt$$
<br><br>$$\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx$$
<br><br>$${1 \over 3}\int {t.{e^{ - 4t}}dt} $$
<br><br>$${1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \in... | mcq | jee-main-2019-online-10th-january-evening-slot |
XbsoEUqQUgLdlc6T8Qlx5 | maths | indefinite-integrals | integration-by-substitution | If $$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx,\,\left( {x \ge 0} \right),$$
<br/><br/>$$f\left( 0 \right) = 0,$$ then the value of $$f(1)$$ is : | [{"identifier": "A", "content": "$$ - $$ $${1 \\over 2}$$"}, {"identifier": "B", "content": "$$ - $$ $${1 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["D"] | null | $$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx$$
<br><br>$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$
<br><br>$$f\left( x \right) = \int {{{5{x^{ - 6}} + 7{x^{ - 8}}} \over {{{\left( {{... | mcq | jee-main-2019-online-9th-january-evening-slot |
Hj0hWZ3LNJm6lC7Dh5vpM | maths | indefinite-integrals | integration-by-substitution | For x<sup>2</sup> $$ \ne $$ n$$\pi $$ + 1, n $$ \in $$ N (the set of natural numbers), the integral <br/><br/>$$\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx} $$ is equal to :
<br/><br/>(where c is a constant of integration) | [{"identifier": "A", "content": "$${\\log _e}\\left| {{1 \\over 2}{{\\sec }^2}\\left( {{x^2} - 1} \\right)} \\right| + c$$"}, {"identifier": "B", "content": "$${1 \\over 2}{\\log _e}\\left| {\\sec \\left( {{x^2} - 1} \\right)} \\right| + c$$"}, {"identifier": "C", "content": "$${1 \\over 2}{\\log _e}\\left| {{{\\sec }^... | ["D"] | null | $$\int {x\sqrt {{{2\sin \left( {{x^2} - } \right) - \sin 2\left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } \,\,dx$$
<br><br>$$ = \int {x\sqrt {{{2\sin \left( {{x^2} - 1} \right) - 2sin\left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)} \over {2\sin \left... | mcq | jee-main-2019-online-9th-january-morning-slot |
7CoM1UAK7oFjOeAfUm7k9k2k5khl5cw | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C$$<br/><br/>
where C is a constant of integration, then the
ordered pair ($$\lambda $$, ƒ($$\theta $$)) is equal to : | [{"identifier": "A", "content": "(\u20131, 1 \u2013 tan$$\\theta $$)"}, {"identifier": "B", "content": "(1, 1 + tan$$\\theta $$)"}, {"identifier": "C", "content": "(\u20131, 1 + tan$$\\theta $$)"}, {"identifier": "D", "content": "(1, 1 \u2013 tan$$\\theta $$)"}] | ["C"] | null | I = $$\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}}$$
<br><br>= $$\int {{{{{\sec }^2}\theta d\theta } \over {{{2\tan \theta } \over {1 - {{\tan }^2}\theta }} + {{1 + {{\tan }^2}\theta } \over {1 - {{\tan }^2}\theta }}}}} $$
<br><br>= $$\int {{{\left( {1 - {{\tan }^2}\theta... | mcq | jee-main-2020-online-9th-january-evening-slot |
H0FHSffmm2l7UtVzAnjgy2xukfqemvyt | maths | indefinite-integrals | integration-by-substitution | If
<br/>$$\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $$ = A$${\log _e}\left| {B\left( \theta \right)} \right| + C$$,
<br/><br/>where C is a constant of integration, then $${{{B\left( \theta \right)} \over A}}$$
<br/>can be : | [{"identifier": "A", "content": "$${{2\\sin \\theta + 1} \\over {5\\left( {\\sin \\theta + 3} \\right)}}$$"}, {"identifier": "B", "content": "$${{2\\sin \\theta + 1} \\over {\\sin \\theta + 3}}$$"}, {"identifier": "C", "content": "$${{5\\left( {2\\sin \\theta + 1} \\right)} \\over {\\sin \\theta + 3}}$$"}, {"iden... | ["C"] | null | $$\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $$
<br><br>= $$\int {{{\cos \theta d\theta } \over {5 + 7\sin \theta - 2\left( {1 - {{\sin }^2}\theta } \right)}}} $$
<br><br>= $$\int {{{\cos \theta d\theta } \over {3 + 7\sin \theta + 2{{\sin }^2}\theta }}} $$
<br><br>Let sin $$\theta... | mcq | jee-main-2020-online-5th-september-evening-slot |
aEHNBpLgZlUpf8cadcjgy2xukfjjgzep | maths | indefinite-integrals | integration-by-substitution | If <br/>$$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$ = $$g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$$<br/><br/>
where c is a constant of integration,
then g(0) is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "e"}, {"identifier": "D", "content": "e<sup>2</sup>"}] | ["B"] | null | I = $$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$
<br><br>= $$\int {\left( {\left( {{e^{2x}} + {e^x} - 1} \right) + \left( {{e^x} - {e^{ - x}}} \right)} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx$$
<br><br>= $$\int {\left( {{e^{2x}} + {e^x} - 1} \ri... | mcq | jee-main-2020-online-5th-september-morning-slot |
E6ZDVN1s3vLFiRj1o5jgy2xukf7g2hr5 | maths | indefinite-integrals | integration-by-substitution | Let $$f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} $$. Then f(3) – f(1) is eqaul to : | [{"identifier": "A", "content": "$$ - {\\pi \\over {12}} + {1 \\over 2} + {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "B", "content": "$$ {\\pi \\over {12}} + {1 \\over 2} - {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "C", "content": "$$ - {\\pi \\over 6} + {1 \\over 2} + {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "D"... | ["B"] | null | $$\int {{{\sqrt x } \over {{{(1 + x)}^2}}}} dx(x > 0)$$<br><br>Put x = tan<sup>2</sup>$$\theta $$ $$ \Rightarrow $$ 2xdx = 2tan$$\theta $$sec<sup>2</sup>$$\theta $$d$$\theta $$<br><br>$$I = \int {{{2{{\tan }^2}\theta .{{\sec }^2}\theta } \over 2}} d\theta = \int {2{{\sin }^2}\theta d\theta = \int {(1 - \cos 2\thet... | mcq | jee-main-2020-online-4th-september-morning-slot |
m2nLubDEJo1qVTMWIO7k9k2k5ith9ii | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}} $$ is equal to :<br/>
(where C is a constant of integration) | [{"identifier": "A", "content": "$${1 \\over 2}{\\left( {{{x - 3} \\over {x + 4}}} \\right)^{{3 \\over 7}}} + C$$"}, {"identifier": "B", "content": "$${\\left( {{{x - 3} \\over {x + 4}}} \\right)^{{1 \\over 7}}} + C$$"}, {"identifier": "C", "content": "$$ - {1 \\over {13}}{\\left( {{{x - 3} \\over {x + 4}}} \\right)^{{... | ["B"] | null | $$\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}} $$
<br><br>= $$\int {{{dx} \over {{{\left( {x + 4} \right)}^2}{{\left( {{{x - 3} \over {x + 4}}} \right)}^{{6 \over 7}}}}}} $$
<br><br>Put $${{{x - 3} \over {x + 4}}}$$ = t
<br><br>$$ \Rightarrow $$ $$\left\{ {{{\left( {x + 4} \right) - \left( {... | mcq | jee-main-2020-online-9th-january-morning-slot |
V06zxcvduCHTTdCKQv7k9k2k5gzjkvq | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}} = f\left( x \right){\left( {1 + {{\sin }^6}x} \right)^{1/\lambda }} + c$$
<br/><br/>where c is a constant of integration, then $$\lambda f\left( {{\pi \over 3}} \right)$$ is equal to | [{"identifier": "A", "content": "$${9 \\over 8}$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "-2"}, {"identifier": "D", "content": "$$-{9 \\over 8}$$"}] | ["C"] | null | Given I = $$\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}}$$
<br><br>Let sin x = t
<br><br>$$ \Rightarrow $$ cos xdx = dt
<br><br>$$ \therefore $$ I = $$\int {{{dt} \over {{t^3}{{\left( {1 + {t^6}} \right)}^{2/3}}}}} $$
<br><br>I = $$\int {{{dt} \over {{t^7}{{\left( {{1 \over {{t^6... | mcq | jee-main-2020-online-8th-january-morning-slot |
jgUAE7YDRC8UIu9j6Gjgy2xukf45aw9x | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx$$ = A(x)$${\tan ^{ - 1}}\left( {\sqrt x } \right)$$ + B(x) + C,
<br/>where C is a constant of integration, then the
ordered pair (A(x), B(x)) can be : | [{"identifier": "A", "content": "(x + 1, -$${\\sqrt x }$$)"}, {"identifier": "B", "content": "(x + 1, $${\\sqrt x }$$)"}, {"identifier": "C", "content": "(x - 1, -$${\\sqrt x }$$)"}, {"identifier": "D", "content": "(x - 1, $${\\sqrt x }$$)"}] | ["A"] | null | Given, I = $$\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx$$
<br><br>Let $${\sin ^{ - 1}}\left( {{{\sqrt x } \over {\sqrt {1 + x} }}} \right)$$ = $$\theta $$
<br><br>$$ \Rightarrow $$ $${{{\sqrt x } \over {\sqrt {1 + x} }} = \sin \theta }$$
<br><br>$$ \Rightarrow $$ tan $$\theta $$ = $${{{\sqrt ... | mcq | jee-main-2020-online-3rd-september-evening-slot |
nGPDGf6uHUWzEnzMjQ1klreafmf | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx = a{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over b}} \right) + c$$, where c is a constant of integration, then
the ordered pair (a, b) is equal to : | [{"identifier": "A", "content": "(-1, 3)"}, {"identifier": "B", "content": "(1, 3)"}, {"identifier": "C", "content": "(1, -3)"}, {"identifier": "D", "content": "(3, 1)"}] | ["B"] | null | Given $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx$$
<br><br>Write sin2x = 1 + sin2x - 1
<br><br>= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {1 + \sin 2x - 1} \right]} }}} dx$$
<br><br>= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x - 1} \righ... | mcq | jee-main-2021-online-24th-february-morning-slot |
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 |
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>$$ = \int {{{8x + 20} \o... | mcq | jee-main-2021-online-25th-february-evening-slot |
OlbUl68mvnNzXdDFhr1kmizm1hi | maths | indefinite-integrals | integration-by-substitution | For real numbers $$\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>= $$\int {{{{x^2} - 1} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx + \int {{1 \over {{x^4} + 3{x^2} ... | integer | jee-main-2021-online-16th-march-evening-shift |
0zdXCvMp55v13ujE0o1kmlicnc5 | maths | indefinite-integrals | integration-by-substitution | The integral $$\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>$$... | mcq | jee-main-2021-online-18th-march-morning-shift |
2PEg8x92SALhxOO63W1kmlm3mz5 | maths | indefinite-integrals | integration-by-substitution | If $$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} $$<br... | integer | jee-main-2021-online-18th-march-morning-shift |
1ktepadjh | maths | indefinite-integrals | integration-by-substitution | If $$\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
$$
<br/><br/>$\therefore \int \frac{... | integer | jee-main-2021-online-27th-august-morning-shift |
1ktiptozs | maths | indefinite-integrals | integration-by-substitution | The integral $$\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>$$ = {4 \over 3}.{1 \over {{t^{1/4}}}} + C$... | mcq | jee-main-2021-online-31st-august-morning-shift |
1l58fbnfg | maths | indefinite-integrals | integration-by-substitution | <p>If $$\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 |
ldr0pnrl | maths | indefinite-integrals | integration-by-substitution | If $\int \sqrt{\sec 2 x-1} d x=\alpha \log _e\left|\cos 2 x+\beta+\sqrt{\cos 2 x\left(1+\cos \frac{1}{\beta} x\right)}\right|+$ constant, then $\beta-\alpha$ is equal to ____________. | [] | null | 1 | <p>$$I = \int {\sqrt {\sec 2x - 1} dx=\int {\sqrt {{{1 - \cos 2x} \over {\cos 2x}}} dx} } $$</p>
<p>$$ = \int {\sqrt {{{2{{\sin }^2}x} \over {2{{\cos }^2}x - 1}}} dx} $$</p>
<p>Let $$\sqrt {2\cos } x = t - \sqrt 2 \sin xdx = dt$$</p>
<p>$$I = \int { - {{dt} \over {\sqrt {{t^2} - 1} }} = - \ln \left| {t + \sqrt {{t^2} ... | integer | jee-main-2023-online-30th-january-evening-shift |
1ldv1n42c | maths | indefinite-integrals | integration-by-substitution | <p>Let $$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 |
lgnz42oa | maths | indefinite-integrals | integration-by-substitution | Let $f(x)=\int \frac{d x}{\left(3+4 x^{2}\right) \sqrt{4-3 x^{2}}},|x|<\frac{2}{\sqrt{3}}$. If $f(0)=0$<br/><br/> and $f(1)=\frac{1}{\alpha \beta} \tan ^{-1}\left(\frac{\alpha}{\beta}\right)$,
$\alpha, \beta>0$, then $\alpha^{2}+\beta^{2}$ is equal to ____________. | [] | null | 28 | $$
\begin{aligned}
& f(x)=\int \frac{d x}{\left(3+4 x^2\right) \sqrt{4-3 x^2}} \\\\
& x=\frac{1}{t} \\\\
& =\int \frac{\frac{-1}{t^2} d t}{\frac{\left(3 t^2+4\right)}{t^2} \frac{\sqrt{4 t^2-3}}{t}} \\\\
& =\int \frac{-d t \cdot t}{\left(3 t^2+4\right) \sqrt{4 t^2-3}}: \text { Put } 4 t^2-3=z^2 \\\\
& =-\frac{1}{4} \int... | integer | jee-main-2023-online-15th-april-morning-shift |
1lgrgryfu | maths | indefinite-integrals | integration-by-substitution | <p>Let $$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 |
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 |
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>$$\begin{aligned}
& \text { Let } \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)=t \\
& \Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) d x=d t \\
& \Rightarrow \frac{x^... | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lseyv2z0 | maths | indefinite-integrals | integration-by-substitution | <p>For $$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 |
jaoe38c1lsfkcmax | maths | indefinite-integrals | integration-by-substitution | <p>If $$\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\sin \theta \cot x}+C$$, where $$C$$ is the integration constant, then $$A B$$ is equal to</p> | [{"identifier": "A", "content": "$$2 \\sec \\theta$$\n"}, {"identifier": "B", "content": "$$8 \\operatorname{cosec}(2 \\theta)$$\n"}, {"identifier": "C", "content": "$$4 \\operatorname{cosec}(2 \\theta)$$\n"}, {"identifier": "D", "content": "$$4 \\sec \\theta$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \text {} \int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x \\
& I=\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x(\sin x \cos \theta-\cos x \sin \theta)}} d x \\
& =\int \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\fr... | mcq | jee-main-2024-online-29th-january-evening-shift |
luy6z4yq | maths | indefinite-integrals | integration-by-substitution | <p>Let $$\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 |
lv3vefpq | maths | indefinite-integrals | integration-by-substitution | <p>If $$\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} \mathrm{~d} x=\mathrm{A}\left(\frac{\alpha x-1}{\beta x+3}\right)^B+\mathrm{C}$$, where $$\mathrm{C}$$ is the constant of integration, then the value of $$\alpha+\beta+20 \mathrm{AB}$$ is _________.</p> | [] | null | 7 | <p>$$\begin{aligned}
& I=\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} d x \\
& \frac{x+3}{x-1}=t \Rightarrow d x=\frac{-4}{(t-1)^2} d t \Rightarrow x=\left(\frac{3+t}{t-1}\right) \\
& \Rightarrow(x-1)^4(x+3)^6=(x-1)^5(x+3)^5\left(\frac{x+3}{x-1}\right) \\
& I=\int \frac{\frac{-4}{(t-1)^2} d t}{t^{1 / 5}\left(\frac{3+t}{t-1}... | integer | jee-main-2024-online-8th-april-evening-shift |
lv5gs8hd | maths | indefinite-integrals | integration-by-substitution | <p>Let $$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 |
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 |
MCOERemigEhmIwRq | maths | indefinite-integrals | standard-integral | If $$\int {{{\sin x} \over {\sin \left( {x - \alpha } \right)}}dx = Ax + B\log \sin \left( {x - \alpha } \right), + C,} $$ then value of
<br/>$$(A, B)$$ is | [{"identifier": "A", "content": "$$\\left( { - \\cos \\alpha ,\\sin \\alpha } \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { \\cos \\alpha ,\\sin \\alpha } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - \\sin \\alpha ,\\cos \\alpha } \\right)$$ "}, {"identifier": "D", "content": "$$\\left( { ... | ["B"] | null | $$\int {{{\sin x} \over {\sin \left( {x - \alpha } \right)}}} dx$$
<br><br>$$ = \int {{{\sin \left( {x - \alpha + \alpha } \right)} \over {\sin \left( {x - \alpha } \right)}}} dx$$
<br><br>$$ = \int {{{\sin \left( {x - \alpha } \right)\cos \alpha + \cos \left( {x - \alpha } \right)\sin \alpha } \over {\sin \left( {x ... | mcq | aieee-2004 |
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 |
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 = $$\... | mcq | jee-main-2017-online-8th-april-morning-slot |
fBGcqpp56sjJxOFIirviI | maths | indefinite-integrals | standard-integral | If $$\,\,\,$$ f$$\left( {{{3x - 4} \over {3x + 4}}} \right)$$ = x + 2, x $$ \ne $$ $$-$$ $${4 \over 3}$$, and
<br/><br/>$$\int {} $$f(x) dx = A log$$\left| {} \right.$$1 $$-$$ x $$\left| {} \right.$$ + Bx + C,
<br/><br/>then the ordered pair (A, B) is equal to :
<br/><br/>(where C is a constant of integration) | [{"identifier": "A", "content": "$$\\left( {{8 \\over 3},{2 \\over 3}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - {8 \\over 3},{2 \\over 3}} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { - {8 \\over 3}, - {2 \\over 3}} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( { {8 \\over ... | ["B"] | null | Given,
<br><br>f$$\left( {{{3x - 4} \over {3x + 4}}} \right)$$ = x + 2, x $$ \ne $$ $$-$$ $${4 \over 3}$$
<br><br>Let, $${{3x - 4} \over {3x + 4}}$$ = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 3x $$-$$ 4 = 3tx + 4t
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 3x $$-$$ 3tx = 4t + 4
<br><br>$$ \Rightarr... | mcq | jee-main-2017-online-9th-april-morning-slot |
u7bfYkLqQ1Ri6xiLX6tgK | maths | indefinite-integrals | standard-integral | If $$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 |
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>= $$\int {{{2x + 6} \ove... | mcq | jee-main-2018-online-15th-april-evening-slot |
PYnBJilIrz8XFwr7gR18hoxe66ijvwvro3u | maths | indefinite-integrals | standard-integral | $$\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)$$
<br>... | mcq | jee-main-2019-online-9th-april-evening-slot |
dsxm8kHFIFngSyvhIT3rsa0w2w9jx5brt6e | maths | indefinite-integrals | standard-integral | The integral $$\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$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
p6chnBVIYFmsOS0wrDfOu | maths | indefinite-integrals | standard-integral | $$\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$$
... | mcq | jee-main-2019-online-8th-april-morning-slot |
RognVI7rIVMeBEOJCg7k9k2k5iqewxy | maths | indefinite-integrals | standard-integral | If ƒ'(x) = tan<sup>–1</sup>(secx + tanx), $$ - {\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 |
GbIwcD7jCGMKEpGbhmjgy2xukf7gh1qh | maths | indefinite-integrals | standard-integral | The integral $$\int {{{\left( {{x \over {x\sin x + \cos x}}} \right)}^2}dx} $$ is equal to <br/>
(where C is a constant of integration): | [{"identifier": "A", "content": "$$\\sec x - {{x\\tan x} \\over {x\\sin x + \\cos x}} + C$$"}, {"identifier": "B", "content": "$$\\sec x + {{x\\tan x} \\over {x\\sin x + \\cos x}} + C$$"}, {"identifier": "C", "content": "$$\\tan x - {{x\\sec x} \\over {x\\sin x + \\cos x}} + C$$"}, {"identifier": "D", "content": "$$\\t... | ["C"] | null | $${\int {\left( {{x \over {x\sin x + \cos x}}} \right)} ^2}dx $$
<br><br>$$= \int {\left( {{x \over {\cos x}}} \right).{{x\cos x\,dx} \over {{{(x\sin x + \cos x)}^2}}}} $$<br><br>= $${x \over {\cos x}}\left( { - {1 \over {x\sin x + \cos x}}} \right) + \int {\left( {{{\cos x + x\sin x} \over {{{\cos }^2}x}}} \right)} \l... | mcq | jee-main-2020-online-4th-september-morning-slot |
1l57o9ptf | maths | indefinite-integrals | standard-integral | If $$\int {{{({x^2} + 1){e^x}} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + C} $$, where C is a constant, then $${{{d^3}f} \over {d{x^3}}}$$ at x = 1 is equal to : | [{"identifier": "A", "content": "$$ - {3 \\over 4}$$"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "$$ - {3 \\over 2}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["B"] | null | <p>$$I = \int {{{{e^x}({x^2} + 1)} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + c} $$</p>
<p>$$ = \int {{{{e^x}({x^2} - 1 + 1 + 1)} \over {{{(x + 1)}^2}}}dx} $$</p>
<p>$$ = \int {{e^x}\left[ {{{x - 1} \over {x + 1}} + {2 \over {{{(x + 1)}^2}}}} \right]dx} $$</p>
<p>$$ = {e^x}\left( {{{x - 1} \over {x + 1}}} \right) + c$$</p>... | mcq | jee-main-2022-online-27th-june-morning-shift |
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 |
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/>$$
\begin{aligned}
\left(\frac{x}{e}\right)^{2 x} & =\left(\frac{e^{\log _e x}}{e}\rig... | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgyl5enm | maths | indefinite-integrals | standard-integral | <p>The integral $$
\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$$"}... | ["B"] | null | <p>To solve the integral:</p>
<p>$ I = \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) \, dx $</p>
<p>we start by simplifying the logarithmic term:</p>
<p>$ \ln\left(\frac{e x}{2}\right) = \ln(e) + \ln\left(\frac{x}{2}\right) = 1 + \ln\left(\frac{x}{2}\right) $</p>
... | mcq | jee-main-2023-online-8th-april-evening-shift |
lvb294vs | maths | indefinite-integrals | standard-integral | <p>If $$\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 |
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 |
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 |
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>$$ \Rightarrow x = \pm \sqrt \pi $$ $$ \... | mcq | jee-main-2021-online-17th-march-evening-shift |
1krpwpsl7 | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | The number of real roots of the equation $${\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 |
1kru9uhcb | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | If the domain of the function $$f(x) = {{{{\cos }^{ - 1}}\sqrt {{x^2} - x + 1} } \over {\sqrt {{{\sin }^{ - 1}}\left( {{{2x - 1} \over 2}} \right)} }}$$ is the interval ($$\alpha$$, $$\beta$$], then $$\alpha$$ + $$\beta$$ is equal to : | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $$O \le {x^2} - x + 1 \le 1$$<br><br>$$ \Rightarrow {x^2} - x \le 0$$<br><br>$$ \Rightarrow x \in [0,1]$$<br><br>Also, $$0 < {\sin ^{ - 1}}\left( {{{2x - 1} \over 2}} \right) \le {\pi \over 2}$$<br><br>$$ \Rightarrow 0 < {{2x - 1} \over 2} \le 1$$<br><br>$$ \Rightarrow 0 < 2x - 1 \le 2$$<br><br>$$1 < 2x \l... | mcq | jee-main-2021-online-22th-july-evening-shift |
1ktczv71w | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | The domain of the function $${{\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\} $$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktk4rt98 | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | The domain of the function<br/><br/>$$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>$$ - 1 \le {{3{x^2} + x - 1} \over {{{(x - 1)}^2}}} \le 1 \Rightarrow x \in \left... | mcq | jee-main-2021-online-31st-august-evening-shift |
1l545i5vo | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | <p>The domain of the function $${\cos ^{ - 1}}\left( {{{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi }} \right)$$ is :</p> | [{"identifier": "A", "content": "$$R - \\left\\{ { - {1 \\over 2},{1 \\over 2}} \\right\\}$$"}, {"identifier": "B", "content": "$$( - \\infty , - 1] \\cup [1,\\infty ) \\cup \\{ 0\\} $$"}, {"identifier": "C", "content": "$$\\left( { - \\infty ,{{ - 1} \\over 2}} \\right) \\cup \\left( {{1 \\over 2},\\infty } \\right) \... | ["D"] | null | <p>$$ - 1 \le {{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi } \le 1$$</p>
<p>$$ \Rightarrow - {\pi \over 2} \le {\sin ^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right) \le {\pi \over 2}$$</p>
<p>$$ \Rightarrow - 1 \le {1 \over {4{x^2} - 1}} \le 1$$</p>
<p>$$\therefore$$ $${1 \over {4{x^2} - 1}... | mcq | jee-main-2022-online-29th-june-morning-shift |
1l5c1qzrq | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | <p>The domain of the function <br/><br/>$$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 |
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>$${x^2} - 5x + 5... | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6rfcn42 | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | <p>The domain of the function $$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 |
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 |
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/>$$
\begin{aligned}
& \frac{2 x}{5 x+3} \leq-1 \text { or } \frac{2 x}{5 x+3} \geq 1 \\\\
& \frac{2 x+5 x+3}{5 x+3} \leq 0 \tex... | integer | jee-main-2023-online-10th-april-evening-shift |
lv0vxckz | maths | inverse-trigonometric-functions | domain-and-range-of-inverse-trigonometric-functions | <p>If the domain of the function $$\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 |
59dUaVXasYJWZEiRdguhw | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | Considering only the principal values of inverse functions, the set
<br/>A = { x $$ \ge $$ 0: tan<sup>$$-$$1</sup>(2x) + tan<sup>$$-$$1</sup>(3x) = $${\pi \over 4}$$} | [{"identifier": "A", "content": "contains two elements "}, {"identifier": "B", "content": "contains more than two elements "}, {"identifier": "C", "content": "is an empty set "}, {"identifier": "D", "content": "is a singleton "}] | ["D"] | null | tan<sup>$$-$$1</sup>(2x) + tan<sup>$$-$$1</sup>(3x) = $$\pi $$/4
<br><br>$$ \Rightarrow \,\,{{5x} \over {1 - 6{x^2}}}$$ = 1
<br><br>$$ \Rightarrow $$ 6x<sup>2</sup> + 5x $$-$$ 1 = 0
<br><br>x = $$-$$1 or x = $${1 \over 6}$$
<br><br>x = $${1 \over 6}$$
<br><br>$$ \because $$ x > 0 | mcq | jee-main-2019-online-12th-january-morning-slot |
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>$${{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\s... | mcq | jee-main-2021-online-16th-march-evening-shift |
1ktnznemb | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | $${\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$$. | mcq | jee-main-2021-online-1st-september-evening-shift |
1l57ou1of | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>$${\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 |
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 |
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