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1ktcz25fq | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y(x) be the solution of the differential equation <br/><br/>2x<sup>2</sup> dy + (e<sup>y</sup> $$-$$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "log<sub>e</sub> 2"}, {"identifier": "D", "content": "log<sub>e</sub> (2e)"}] | ["C"] | null | $$2{x^2}dy + ({e^y} - 2x)dx = 0$$<br><br>$${{dy} \over {dx}} + {{{e^y} - 2x} \over {2{x^2}}} = 0 \Rightarrow {{dy} \over {dx}} + {{{e^y}} \over {2{x^2}}} - {1 \over x} = 0$$<br><br>$${e^{ - y}}{{dy} \over {dx}} - {{{e^{ - y}}} \over x} = - {1 \over {2{x^2}}} \Rightarrow $$ Put $${e^{ - y}} = z$$<br><br>$${{ - dz} \ove... | mcq | jee-main-2021-online-26th-august-evening-shift | 5,951 |
1kteisxmb | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation<br/><br/> $${{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x$$ such that y(0) = 7. Then y($$\pi$$) is equal to : | [{"identifier": "A", "content": "$$2{e^{{\\pi ^2}}} + 5$$"}, {"identifier": "B", "content": "$${e^{{\\pi ^2}}} + 5$$"}, {"identifier": "C", "content": "$$3{e^{{\\pi ^2}}} + 5$$"}, {"identifier": "D", "content": "$$7{e^{{\\pi ^2}}} + 5$$"}] | ["A"] | null | $${{dy} \over {dx}} - 2xy = 2(2\sin x - 5)x - 2\cos x$$<br><br>IF = $${e^{ - {x^2}}}$$<br><br>So, $$y.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx} $$<br><br>$$ \Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c$$<br><br>$$ \Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}$$<br><br>Given at ... | mcq | jee-main-2021-online-27th-august-morning-shift | 5,952 |
1ktiqwm6u | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If $${{dy} \over {dx}} = {{{2^{x + y}} - {2^x}} \over {{2^y}}}$$, y(0) = 1, then y(1) is equal to : | [{"identifier": "A", "content": "log<sub>2</sub>(2 + e)"}, {"identifier": "B", "content": "log<sub>2</sub>(1 + e)"}, {"identifier": "C", "content": "log<sub>2</sub>(2e)"}, {"identifier": "D", "content": "log<sub>2</sub>(1 + e<sup>2</sup>)"}] | ["B"] | null | $${{dy} \over {dx}} = {{{2^{x + y}} - {2^x}} \over {{2^y}}}$$<br><br>$${2^y}{{dy} \over {dx}} = {2^x}({2^y} - 1)$$<br><br>$$\int {{{{2^y}} \over {{2^y} - 1}}dy = \int {{2^x}\,dx} } $$<br><br>$${{\ln ({2^y} - 1)} \over {\ln 2}} = {{{2^x}} \over {\ln 2}} + C$$<br><br>$$ \Rightarrow {\log _2}({2^y} - 1) = {2^x}{\log _2}e ... | mcq | jee-main-2021-online-31st-august-morning-shift | 5,953 |
1ktk8vcgu | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If $$y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$$, x > 0, $$\phi$$ > 0, and y(1) = $$-$$1, then $$\phi \left( {{{{y^2}} \over 4}} \right)$$ is equal to : | [{"identifier": "A", "content": "4 $$\\phi$$ (2)"}, {"identifier": "B", "content": "4$$\\phi$$ (1)"}, {"identifier": "C", "content": "2 $$\\phi$$ (1)"}, {"identifier": "D", "content": "$$\\phi$$ (1)"}] | ["B"] | null | Let, $$y = tx$$<br><br>$${{dy} \over {dx}} = t + x{{dt} \over {dx}}$$<br><br>$$\therefore$$ $$tx\left( {t + x{{dt} \over {dx}}} \right) = x\left( {{t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}} \right)$$<br><br>$${t^2} + xt{{dt} \over {dx}} = {t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}$$<br><br>$$\int {... | mcq | jee-main-2021-online-31st-august-evening-shift | 5,954 |
1l544jesq | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let the solution curve of the differential equation</p>
<p>$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}} $$, $$y(1) = 3$$ be $$y = y(x)$$. Then y(2) is equal to:</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "17"}] | ["A"] | null | <p>Given,</p>
<p>$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16x} $$</p>
<p>$$ \Rightarrow x{{dy} \over {dx}} = y + \sqrt {{y^2} + 16x} $$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = {y \over x} + \sqrt {{{\left( {{y \over x}} \right)}^2} + 16} $$</p>
<p>This is a homogenous different equation.</p>
<p>Let $${y \over x} = v... | mcq | jee-main-2022-online-29th-june-morning-shift | 5,955 |
1l54b68mh | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If y = y(x) is the solution of the differential equation $$\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$$ and y (0) = 0, then $$6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "$$-$$1"}] | ["C"] | null | <p>Given,</p>
<p>$$(1 + {e^{2x}}){{dy} \over {dx}} + 2(1 + {y^2}){e^x} = 0$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = {{ - 2(1 + {y^2}){e^x}} \over {1 + {e^{2x}}}}$$</p>
<p>$$ \Rightarrow \int {{{dy} \over {1 + {y^2}}} = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}} } $$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}(y) = \int ... | mcq | jee-main-2022-online-29th-june-evening-shift | 5,956 |
1l55hl138 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let x = x(y) be the solution of the differential equation <br/><br/>$$2y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 0$$ such that x(1) = 0. Then, x(e) is equal to :</p> | [{"identifier": "A", "content": "$$e{\\log _e}(2)$$"}, {"identifier": "B", "content": "$$ - e{\\log _e}(2)$$"}, {"identifier": "C", "content": "$${e^2}{\\log _e}(2)$$"}, {"identifier": "D", "content": "$$ - {e^2}{\\log _e}(2)$$"}] | ["D"] | null | <p>Given differential equation</p>
<p>$$2y{e^{{x \over {{y^2}}}}}dx + \left( {{y^2} - 4x{e^{{x \over {{y^2}}}}}} \right)dy = 0,\,x(1) = 0$$</p>
<p>$$ \Rightarrow {e^{{x \over {{y^2}}}}}[2ydx - 4xdy] = - {y^2}dy$$</p>
<p>$$ \Rightarrow {e^{{x \over {{y^2}}}}}\left[ {{{2{y^2}dx - 4xydy} \over {{y^4}}}} \right] = {{ - 1}... | mcq | jee-main-2022-online-28th-june-evening-shift | 5,957 |
1l566lm5l | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let the solution curve $$y = y(x)$$ of the differential equation</p>
<p>$$\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y$$</p>
<p>pass through the points (1, 0) and (2$$\alpha$$, $$\alpha$$), $$\... | [{"identifier": "A", "content": "$${1 \\over 2}\\exp \\left( {{\\pi \\over 6} + \\sqrt e - 1} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over 2}\\exp \\left( {{\\pi \\over 6} + e - 1} \\right)$$"}, {"identifier": "C", "content": "$$\\exp \\left( {{\\pi \\over 6} + \\sqrt e + 1} \\right)$$"}, {"identifier... | ["A"] | null | <p>$$\left( {{1 \over {\sqrt {1 - {{{y^2}} \over {{x^2}}}} }} + {e^{{y \over x}}}} \right){{dy} \over {dx}} = 1 + \left( {{1 \over {\sqrt {1 - {{{y^2}} \over {{x^2}}}} }} + {e^{{y \over x}}}} \right){y \over x}$$</p>
<p>Putting y = tx</p>
<p>$$\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)\left( {t + x{{dt} \o... | mcq | jee-main-2022-online-28th-june-morning-shift | 5,958 |
1l58aj791 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let the solution curve y = y(x) of the differential equation <br/><br/>$$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$$ pass through the origin. Then y(2) is equal to _____________.</p> | [] | null | 12 | <p>$$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = \left( {{{6x} \over {{x^2} + 4}}} \right)y + 2x$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} - \left( {{{6x} \over {{x^2} + 4}}} \right)y = 2x$$</p>
<p>$$I.F. = {e^{ - 3\ln ({x^2} + 4)}} = {1 \over {{{({x^2} + 4)}^3}}}$$</p>
<p>So $$... | integer | jee-main-2022-online-26th-june-morning-shift | 5,960 |
1l59kklkn | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If $$y = y(x)$$ is the solution of the differential equation <br/><br/>$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$ such that $$y(e) = {e \over 3}$$, then y(1) is equal to :</p> | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <p>$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$</p>
<p>$$ \Rightarrow 2x(xdy - ydx) + 3{y^3}dx = 0$$</p>
<p>$$ \Rightarrow 2\left( {{{xdy - ydx} \over {{y^2}}}} \right) + 3{{dx} \over x} = 0$$</p>
<p>$$ \Rightarrow - {{2x} \over y} + 3\ln x = C$$</p>
<p>$$\because$$ $$y(e) = {e \over 3} \Rightarrow - 6 + 3 = C \Rig... | mcq | jee-main-2022-online-25th-june-evening-shift | 5,961 |
1l5aihggo | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$g:(0,\infty ) \to R$$ be a differentiable function such that <br/><br/>$$\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c} $$, for all x > 0, where c is an arbitrary constant. Then :<... | [{"identifier": "A", "content": "g is decreasing in $$\\left( {0,{\\pi \\over 4}} \\right)$$"}, {"identifier": "B", "content": "g' is increasing in $$\\left( {0,{\\pi \\over 4}} \\right)$$"}, {"identifier": "C", "content": "g + g' is increasing in $$\\left( {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "D", "cont... | ["D"] | null | $$
\int\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right)}{\left(e^x+1\right)^2}\right) d x=\frac{x g(x)}{e^x+1}+c
$$<br/><br/>
On differentiating both sides w.r.t. $\mathrm{x}$, we get<br/><br/>
$$
\begin{aligned}
&\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right.}{\left(e^x+... | mcq | jee-main-2022-online-25th-june-morning-shift | 5,962 |
1l5aj3huu | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve $$y = y(x)$$ of the differential equation $${y^2}dx + ({x^2} - xy + {y^2})dy = 0$$, which passes through the point (1, 1) and intersects the line $$y = \sqrt 3 x$$ at the point $$(\alpha ,\sqrt 3 \alpha )$$, then value of $${\log _e}(\sqrt 3 \alpha )$$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "$${\\pi \\over 12}$$"}, {"identifier": "D", "content": "$${\\pi \\over 6}$$"}] | ["C"] | null | <p>$${{dy} \over {dx}} = {{{y^2}} \over {xy - {x^2} - {y^2}}}$$</p>
<p>Put $$y = vx$$ we get</p>
<p>$$v + x{{dv} \over {dx}} = {{{v^2}} \over {v - 1 - {v^2}}}$$</p>
<p>$$ \Rightarrow x{{dv} \over {dx}} = {{{v^2} - {v^2} + v + {v^3}} \over {v - 1 - {v^2}}}$$</p>
<p>$$ \Rightarrow \int {{{v - 1 - {v^2}} \over {v(1 + {v^2... | mcq | jee-main-2022-online-25th-june-morning-shift | 5,963 |
1l6dvn1zt | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>The general solution of the differential equation $$\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0$$ is :</p> | [{"identifier": "A", "content": "$$\\left(y^{2}+x\\right)^{4}=\\mathrm{C}\\left|\\left(y^{2}+2 x\\right)^{3}\\right|$$"}, {"identifier": "B", "content": "$$\\left(y^{2}+2 x\\right)^{4}=C\\left|\\left(y^{2}+x\\right)^{3}\\right|$$"}, {"identifier": "C", "content": "$$\\left|\\left(y^{2}+x\\right)^{3}\\right|=\\mathrm{C}... | ["A"] | null | $\left(x-y^{2}\right) d x+y\left(5 x+y^{2}\right) d y=0$
<br/><br/>
$$
y \frac{d y}{d x}=\frac{y^{2}-x}{5 x+y^{2}}
$$<br/><br/>
Let $y^{2}=t$
$$
\frac{1}{2} \cdot \frac{d t}{d x}=\frac{t-x}{5 x+t}
$$
<br/><br/>
Now substitute, $t=v x$
<br/><br/>
$$
\begin{aligned}
& \frac{d t}{d x}=v+x \frac{d v}{d x} \\\\
& \frac{1}{... | mcq | jee-main-2022-online-25th-july-morning-shift | 5,964 |
1l6f3egjv | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation</p>
<p>$$\frac{d y}{d x}=\frac{4 y^{3}+2 y x^{2}}{3 x y^{2}+x^{3}}, y(1)=1$$.</p>
<p>If for some $$n \in \mathbb{N}, y(2) \in[n-1, n)$$, then $$n$$ is equal to _____________.</p> | [] | null | 3 | <p>$${{dy} \over {dx}} = {y \over x}{{(4{y^2} + 2{x^2})} \over {(3{y^2} + {x^2})}}$$</p>
<p>Put $$y = vx$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$$</p>
<p>$$ \Rightarrow v + x{{dv} \over {dx}} = {{v(4{v^2} + 2)} \over {(3{v^2} + 1)}}$$</p>
<p>$$ \Rightarrow x{{dv} \over {dx}} = v\left( {{{(4{v... | integer | jee-main-2022-online-25th-july-evening-shift | 5,965 |
1l6m5qs9v | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If $$y=y(x), x \in(0, \pi / 2)$$ be the solution curve of the differential equation <br/><br/>$$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$$, <br/><br/>with $$y(\pi / 4)=\mathrm{e}^{-\pi}$$, then $$y(\pi / 6)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{2}{\\sqrt{3}} e^{-2 \\pi / 3}$$"}, {"identifier": "B", "content": "$$\\frac{2}{\\sqrt{3}} \\mathrm{e}^{2 \\pi / 3}$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{3}} e^{-2 \\pi / 3}$$"}, {"identifier": "D", "content": "$$\\frac{1}{\\sqrt{3}} e^{2 \\pi / 3}$$"}] | ["A"] | null | <p>$$({\sin ^2}2x){{dy} \over {dx}} + (8{\sin ^2}2x + 2\sin 4x)y$$</p>
<p>$$ = 2{e^{ - 4x}}(2\sin 2x + \cos 2x)$$</p>
<p>$${{dy} \over {dx}} + (8 + 4\cot 2x)y = 2{e^{ - 4x}}\left( {{{2\sin 2x + \cos 2x} \over {{{\sin }^2}2x}}} \right)$$</p>
<p>Integrating factor</p>
<p>$$(I.F.) = {e^{\int {(8 + 4\cot 2x)dx} }}$$</p>
<p... | mcq | jee-main-2022-online-28th-july-morning-shift | 5,967 |
1l6reae56 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve of the differential equation $$\frac{d y}{d x}=\frac{x+y-2}{x-y}$$ passes through the points $$(2,1)$$ and $$(\mathrm{k}+1,2), \mathrm{k}>0$$, then</p> | [{"identifier": "A", "content": "$$2 \\tan ^{-1}\\left(\\frac{1}{k}\\right)=\\log _{e}\\left(k^{2}+1\\right)$$"}, {"identifier": "B", "content": "$$\\tan ^{-1}\\left(\\frac{1}{k}\\right)=\\log _{e}\\left(k^{2}+1\\right)$$"}, {"identifier": "C", "content": "$$2 \\tan ^{-1}\\left(\\frac{1}{k+1}\\right)=\\log _{e}\\left(k... | ["A"] | null | $\frac{d y}{d x}=\frac{x+y-2}{x-y}=\frac{(x-1)+(y-1)}{(x-1)-(y-1)}$
<br/><br/>Let $x-1=X, y-1=Y$
<br/><br/>$$
\frac{d Y}{d X}=\frac{X+Y}{X-Y}
$$
<br/><br/>Let $Y=t X \Rightarrow \frac{d Y}{d X}=t+X \frac{d t}{d X}$
<br/><br/>$t+X \frac{d t}{d X}=\frac{1+t}{1-t}$
<br/><br/>$X \frac{d t}{d X}=\frac{1+t}{1-t}-t=\frac... | mcq | jee-main-2022-online-29th-july-evening-shift | 5,968 |
1l6reeqh6 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution curve of the differential equation $$ \frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$$, which passes through the point $$(0,1)$$. Then $$y(1)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "$$\\frac{3}{2}$$"}, {"identifier": "C", "content": "$$\\frac{5}{2}$$"}, {"identifier": "D", "content": "$$\\frac{7}{2}$$"}] | ["B"] | null | $\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$,
<br/><br/>Integrating factor I.F. $=e^{\int \frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6} d x}$
<br/><br/>$$
\begin{aligned}
& \text { Let } \frac{2 x^{2}+11 x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C... | mcq | jee-main-2022-online-29th-july-evening-shift | 5,969 |
ldoa5dpt | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let $y=y(x)$ be the solution of the differential equation
<br/><br/>$\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0$
<br/><br/>such that $y(1)=1$. Then $\left|(y(2))^{3}-12 y(2)\right|$ is equal to : | [{"identifier": "A", "content": "64"}, {"identifier": "B", "content": "$16 \\sqrt{2}$"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "$32 \\sqrt{2}$"}] | ["D"] | null | $\left(3 y^{2}-5 x^{2}\right) y \cdot d x+2 x\left(x^{2}-y^{2}\right) d y=0$
<br/><br/>$$
\Rightarrow \frac{d y}{d x}=\frac{y\left(5 x^{2}-3 y^{2}\right)}{2 x\left(x^{2}-y^{2}\right)}
$$
<br/><br/>Put $\mathrm{y}=\mathrm{mx}$
<br/><br/>$$
\Rightarrow m+x \cdot \frac{d m}{d x}=\frac{m\left(5-3 m^{2}\right)}{2\left(1-... | mcq | jee-main-2023-online-31st-january-evening-shift | 5,970 |
ldqy3vn1 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The solution of the differential equation
<br/><br/>$\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is : | [{"identifier": "A", "content": "$\\log _e|x+y|+\\frac{x y}{(x+y)^2}=0$"}, {"identifier": "B", "content": "$\\log _e|x+y|-\\frac{x y}{(x+y)^2}=0$"}, {"identifier": "C", "content": "$\\log _e|x+y|+\\frac{2 x y}{(x+y)^2}=0$"}, {"identifier": "D", "content": "$\\log _e|x+y|-\\frac{2 x y}{(x+y)^2}=0$"}] | ["C"] | null | <p>$$y = vx$$</p>
<p>$$v + x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}}} \right)$$</p>
<p>$$x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}} + v} \right)$$</p>
<p>$${{dv} \over {dx}} = - \left( {{{{{(1 + v)}^3}} \over {3 + {v^2}}}} \right)$$</p>
<p>$$ \Rightarrow {{3 + {v^2}} \over ... | mcq | jee-main-2023-online-30th-january-evening-shift | 5,972 |
1ldsuyode | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=f(x)$$ be the solution of the differential equation $$y(x+1)dx-x^2dy=0,y(1)=e$$. Then $$\mathop {\lim }\limits_{x \to {0^ + }} f(x)$$ is equal to</p> | [{"identifier": "A", "content": "$${e^2}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$${1 \\over {{e^2}}}$$"}, {"identifier": "D", "content": "$${1 \\over e}$$"}] | ["B"] | null | <p>Given,</p>
<p>$$y(x + 1)dx - {x^2}dy = 0$$</p>
<p>$$ \Rightarrow \left( {{{x + 1} \over {{x^2}}}} \right)dx = {{dy} \over y}$$</p>
<p>$$ \Rightarrow {1 \over x}dx + {{dx} \over {{x^2}}} = {{dy} \over y}$$</p>
<p>Integrating both sides, we get</p>
<p>$$\int {{{dx} \over x} + \int {{{dx} \over {{x^2}}} = \int {{{dy} \... | mcq | jee-main-2023-online-29th-january-morning-shift | 5,973 |
1lgpxupwa | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ be the solution curves of the differential equation $$\frac{d y}{d x}=y+7$$ with initial conditions $$y_{1}(0)=0$$ and $$y_{2}(0)=1$$ respectively. Then the curves $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ intersect at</p> | [{"identifier": "A", "content": "no point"}, {"identifier": "B", "content": "two points"}, {"identifier": "C", "content": "infinite number of points"}, {"identifier": "D", "content": "one point"}] | ["A"] | null | <p>The given differential equation is</p>
<p>$$\frac{d y}{d x} = y + 7$$</p>
<p>This is a first order linear differential equation and can be solved using an integrating factor. </p>
<p>Rearrange the equation to the standard form of a linear differential equation :</p>
<p>$$\frac{d y}{d x} - y = 7$$</p>
<p>The integrat... | mcq | jee-main-2023-online-13th-april-morning-shift | 5,975 |
1lgsvopsl | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$\frac{d y}{d x}+\frac{5}{x\left(x^{5}+1\right)} y=\frac{\left(x^{5}+1\right)^{2}}{x^{7}}, x > 0$$. If $$y(1)=2$$, then $$y(2)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{693}{128}$$"}, {"identifier": "B", "content": "$$\\frac{697}{128}$$"}, {"identifier": "C", "content": "$$\\frac{637}{128}$$"}, {"identifier": "D", "content": "$$\\frac{679}{128}$$"}] | ["A"] | null | I.F $=\mathrm{e}^{\int \frac{5 \mathrm{dx}}{\mathrm{x}\left(\mathrm{x}^5+1\right)}}=\mathrm{e}^{\int \frac{5 \mathrm{x}^{-6} \mathrm{dx}}{\left(\mathrm{x}^{-5}+1\right)}}$
<br/><br/>Put, $1+\mathrm{x}^{-5}=\mathrm{t} \Rightarrow-5 \mathrm{x}^{-6} \mathrm{dx}=\mathrm{dt}$
<br/><br/>$$ \therefore $$ $$
e^{\int-\frac{d t}... | mcq | jee-main-2023-online-11th-april-evening-shift | 5,976 |
1lguu2d04 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be a solution curve of the differential equation.</p>
<p>$$\left(1-x^{2} y^{2}\right) d x=y d x+x d y$$.</p>
<p>If the line $$x=1$$ intersects the curve $$y=y(x)$$ at $$y=2$$ and the line $$x=2$$ intersects the curve $$y=y(x)$$ at $$y=\alpha$$, then a value of $$\alpha$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{1+3 e^{2}}{2\\left(3 e^{2}-1\\right)}$$"}, {"identifier": "B", "content": "$$\\frac{3 e^{2}}{2\\left(3 e^{2}-1\\right)}$$"}, {"identifier": "C", "content": "$$\\frac{1-3 e^{2}}{2\\left(3 e^{2}+1\\right)}$$"}, {"identifier": "D", "content": "$$\\frac{3 e^{2}}{2\\left(3 e^{2}+1\\... | ["A"] | null | We have,
<br/><br/>$$
\begin{aligned}
& \left(1-x^2 y^2\right) d x=y d x+x d y, y(1)=2 \\\\
& d x=\frac{y d x+x d y}{1-(x y)^2}
\end{aligned}
$$
<br/><br/>On integrating both sides, we get
<br/><br/>$$
\begin{aligned}
\int d x & =\int \frac{d(x y)}{1-(x y)^2} \\\\
x & =\frac{1}{2} \log \left|\frac{1+x y}{1-x y}\right|+... | mcq | jee-main-2023-online-11th-april-morning-shift | 5,977 |
1lgvqhawp | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let the tangent at any point P on a curve passing through the points (1, 1) and $$\left(\frac{1}{10}, 100\right)$$, intersect positive $$x$$-axis and $$y$$-axis at the points A and B respectively. If $$\mathrm{PA}: \mathrm{PB}=1: k$$ and $$y=y(x)$$ is the solution of the differential equation $$e^{\frac{d y}{d x}}=k... | [] | null | 4 | Let the equation of tangent to the curve at $(x, y)$.
<br><br>Whose slope is $\frac{d y}{d x}$ is
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkjbalj/b11eb6ae-312f-49ae-887d-c92b77fa1799/cc33a060-6789-11ee-b7ba-e3cd7ea3cf86/file-6y3zli1lnkjbalk.png?format=png" data-orsrc="https://ap... | integer | jee-main-2023-online-10th-april-evening-shift | 5,978 |
1lh2y4an3 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve $$f(x, y)=0$$ of the differential equation
<br/><br/>$$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$$,
<br/><br/>passes through the points $$(1,0)$$ and $$(\alpha, 2)$$, then $$\alpha^{\alpha}$$ is equal to :</p> | [{"identifier": "A", "content": "$$e^{\\sqrt{2} e^{2}}$$"}, {"identifier": "B", "content": "$$e^{2 e^{\\sqrt{2}}}$$"}, {"identifier": "C", "content": "$$e^{e^{2}}$$"}, {"identifier": "D", "content": "$$e^{2 e^{2}}$$"}] | ["D"] | null | We have, $\left(1+\log _e x\right) \frac{d x}{d y}-x \log _e x=e^{y}, x>0$
<br/><br/>Put $x \log _e x=t$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \left(x \cdot \frac{1}{x}+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\
& \Rightarrow \left(1+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\
& \theref... | mcq | jee-main-2023-online-6th-april-evening-shift | 5,979 |
lsbld6tf | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If the solution of the differential equation <br/><br/>$(2 x+3 y-2) \mathrm{d} x+(4 x+6 y-7) \mathrm{d} y=0, y(0)=3$, is
<br/><br/>$\alpha x+\beta y+3 \log _e|2 x+3 y-\gamma|=6$, then $\alpha+2 \beta+3 \gamma$ is equal to ____________. | [] | null | 29 | <p>$$\begin{array}{ll}
2 x+3 y-2=t & 4 x+6 y-4=2 t \\
2+3 \frac{d y}{d x}=\frac{d t}{d x} & 4 x+6 y-7=2 t-3
\end{array}$$</p>
<p>$$\begin{aligned}
& \frac{d y}{d x}=\frac{-(2 x+3 y-2)}{4 x+6 y-7} \\
& \frac{d t}{d x}=\frac{-3 t+4 t-6}{2 t-3}=\frac{t-6}{2 t-3} \\
& \int \frac{2 t-3}{t-6} d t=\int d x \\
& \int\left(\fra... | integer | jee-main-2024-online-27th-january-morning-shift | 5,982 |
jaoe38c1lscnd6h3 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If $$y=y(x)$$ is the solution curve of the differential equation $$\left(x^2-4\right) \mathrm{d} y-\left(y^2-3 y\right) \mathrm{d} x=0, x>2, y(4)=\frac{3}{2}$$ and the slope of the curve is never zero, then the value of $$y(10)$$ equals :</p> | [{"identifier": "A", "content": "$$\\frac{3}{1+(8)^{1 / 4}}$$\n"}, {"identifier": "B", "content": "$$\\frac{3}{1-(8)^{1 / 4}}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{1-2 \\sqrt{2}}$$\n"}, {"identifier": "D", "content": "$$\\frac{3}{1+2 \\sqrt{2}}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0 \\
& \Rightarrow \int \frac{d y}{y^2-3 y}=\int \frac{d x}{x^2-4} \\
& \Rightarrow \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^2-4} \\
& \Rightarrow \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \\
... | mcq | jee-main-2024-online-27th-january-evening-shift | 5,983 |
jaoe38c1lscot4sr | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve, of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y-2}{x-y}$$ passing through the point $$(2,1)$$ is $$\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta} \log _{\mathrm{e}}\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\log _{\mathrm{e}}|x-1|$$, then $$5 \beta+\... | [] | null | 11 | <p>$$\begin{aligned}
& \frac{d y}{d x}=\frac{x+y-2}{x-y} \\
& \mathrm{x}=\mathrm{X}+\mathrm{h}, \mathrm{y}=\mathrm{Y}+\mathrm{k} \\
& \frac{d Y}{d X}=\frac{X+Y}{X-Y} \\
& \left.\begin{array}{l}
\mathrm{h}+\mathrm{k}-2=0 \\
\mathrm{~h}-\mathrm{k}=0
\end{array}\right\} \mathrm{h}=\mathrm{k}=1 \\
& \mathrm{Y}=\mathrm{vX} ... | integer | jee-main-2024-online-27th-january-evening-shift | 5,984 |
jaoe38c1lsd4y5kd | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation
<br/><br/>$$\sec ^2 x d x+\left(e^{2 y} \tan ^2 x+\tan x\right) d y=0,0< x<\frac{\pi}{2}, y(\pi / 4)=0$$.
<br/><br/>If $$y(\pi / 6)=\alpha$$, then $$e^{8 \alpha}$$ is equal to ____________.</p> | [] | null | 9 | <p>$$\begin{aligned}
& \sec ^2 x \frac{d x}{d y}+e^{2 y} \tan ^2 x+\tan x=0 \\
& \left(\text { Put } \tan x=t \Rightarrow \sec ^2 x \frac{d x}{d y}=\frac{d t}{d y}\right) \\
& \frac{d t}{d y}+e^{2 y} \times t^2+t=0 \\
& \frac{d t}{d y}+t=-t^2 \cdot e^{2 y} \\
& \frac{1}{t^2} \frac{d t}{d y}+\frac{1}{t}=-e^{2 y} \\
& \l... | integer | jee-main-2024-online-31st-january-evening-shift | 5,985 |
jaoe38c1lse53x0f | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$\frac{d y}{d x}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}, x \in\left(0, \frac{\pi}{2}\right)$$ satisfying the condition $$y\left(\frac{\pi}{4}\right)=2$$. Then, $$y\left(\frac{\pi}{3}\right)$$ is</p> | [{"identifier": "A", "content": "$$\\sqrt{3}\\left(2+\\log _e 3\\right)$$\n"}, {"identifier": "B", "content": "$$\\sqrt{3}\\left(1+2 \\log _e 3\\right)$$\n"}, {"identifier": "C", "content": "$$\\sqrt{3}\\left(2+\\log _e \\sqrt{3}\\right)$$\n"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{3}}{2}\\left(2+\\log _e 3\\... | ["C"] | null | <p>$$\begin{aligned}
& \frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)} \\
& =\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^2 x\right)} \\
& \frac{d y}{d x}=\sec ^2 x+y \cdot 2(\operatorname{cosec} 2 x) \\
& \frac{d y}{d x}-2 \operatorname{cosec}(2... | mcq | jee-main-2024-online-31st-january-morning-shift | 5,986 |
jaoe38c1lse5genw | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>The solution curve of the differential equation
$$y \frac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0$$ passing through the point $$(e, 1)$$ is</p> | [{"identifier": "A", "content": "$$\\left|\\log _e \\frac{y}{x}\\right|=y^2$$\n"}, {"identifier": "B", "content": "$$\\left|\\log _e \\frac{y}{x}\\right|=x$$\n"}, {"identifier": "C", "content": "$$\\left|\\log _e \\frac{x}{y}\\right|=y$$\n"}, {"identifier": "D", "content": "$$2\\left|\\log _e \\frac{x}{y}\\right|=y+1$$... | ["C"] | null | <p>$$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}}{\mathrm{y}}\left(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1\right)$$</p>
<p>Let $$\frac{x}{y}=t \Rightarrow x=t y$$</p>
<p>$$\begin{aligned}
& \frac{d x}{d y}=t+y \frac{d t}{d y} \\
& t+y \frac{d t}{d y}=t(\ln (t)+1)
\end{aligned}$$</p>
<p>$$\mathrm{y} \frac{... | mcq | jee-main-2024-online-31st-january-morning-shift | 5,987 |
jaoe38c1lsf0v366 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve $$y=y(x)$$ of the differential equation $$\left(1+y^2\right)\left(1+\log _{\mathrm{e}} x\right) d x+x d y=0, x > 0$$ passes through the point $$(1,1)$$ and $$y(e)=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}$$, then $$\alpha+2 \beta$$ is _________.</p> | [] | null | 3 | <p>$$\begin{aligned}
& \int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^2}=0 \\
& \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=C
\end{aligned}$$</p>
<p>Put $$x=y=1$$</p>
<p>$$\begin{aligned}
& \therefore C=\frac{\pi}{4} \\
& \Rightarrow \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=\frac{\pi}{4}
\end{aligned}$$<... | integer | jee-main-2024-online-29th-january-morning-shift | 5,988 |
jaoe38c1lsfknrcy | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If $$\sin \left(\frac{y}{x}\right)=\log _e|x|+\frac{\alpha}{2}$$ is the solution of the differential equation $$x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$$ and $$y(1)=\frac{\pi}{3}$$, then $$\alpha^2$$ is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["D"] | null | <p>Differential equation :-</p>
<p>$$\begin{aligned}
& x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x \\
& \cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x
\end{aligned}$$</p>
<p>Divide both sides by $$\mathrm{x}^2$$</p>
<p>$$\cos \frac{y}{x}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{x}$$</p>
<p>L... | mcq | jee-main-2024-online-29th-january-evening-shift | 5,989 |
1lsg51qnf | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$Y=Y(X)$$ be a curve lying in the first quadrant such that the area enclosed by the line $$Y-y=Y^{\prime}(x)(X-x)$$ and the co-ordinate axes, where $$(x, y)$$ is any point on the curve, is always $$\frac{-y^2}{2 Y^{\prime}(x)}+1, Y^{\prime}(x) \neq 0$$. If $$Y(1)=1$$, then $$12 Y(2)$$ equals __________.</p> | [] | null | 20 | <p>$$\mathrm{A}=\frac{1}{2}\left(\frac{-\mathrm{y}}{\mathrm{Y}^{\prime}(\mathrm{x})}+\mathrm{x}\right)(\mathrm{y}-\mathrm{xY} / \mathrm{x})=\frac{-\mathrm{y}^2}{2 \mathrm{Y}^{\prime}(\mathrm{x})}+1$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxkuuf/8206ce56-06e1-4b71-9f96-f1cdb4f19... | integer | jee-main-2024-online-30th-january-evening-shift | 5,990 |
1lsgaiiyb | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$\sec x \mathrm{~d} y+\{2(1-x) \tan x+x(2-x)\} \mathrm{d} x=0$$ such that $$y(0)=2$$. Then $$y(2)$$ is equal to:</p> | [{"identifier": "A", "content": "$$2\\{\\sin (2)+1\\}$$\n"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$2\\{1-\\sin (2)\\}$$"}] | ["B"] | null | <p>$$\frac{d y}{d x}=2(x-1) \sin x+\left(x^2-2 x\right) \cos x$$</p>
<p>Now both side integrate</p>
<p>$$\begin{aligned}
& y(x)=\int 2(x-1) \sin x d x+\left[\left(x^2-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right] \\
& y(x)=\left(x^2-2 x\right) \sin x+\lambda \\
& y(0)=0+\lambda \Rightarrow 2=\lambda \\
& y(x)=\left(... | mcq | jee-main-2024-online-30th-january-morning-shift | 5,991 |
lv0vxcw0 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution $$y=y(x)$$ of the differential equation $$(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y-(2 x^2+2 x+3) \mathrm{d} x=0$$ satisfies $$y(-1)=-\frac{\pi}{4}$$, then $$y(0)$$ is equal to :</p> | [{"identifier": "A", "content": "$$-\\frac{\\pi}{12}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{2}$$\n"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$\\frac{\\pi}{4}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0 \\
& \int d y=\int\left(\frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} d x\right. \\
& \int d y=\int \frac{1}{x^2+1} d x+\int \frac{}{x^2+2 x+2} d x \\
& y=\tan ^{-1}(x)+\tan ^{-1}(1+x)+C \\
& y(-1)=\tan ^{-1}(-1)+\tan ^{-1}(1-1)+C ... | mcq | jee-main-2024-online-4th-april-morning-shift | 5,992 |
lv2erury | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$(x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0$$. If $$y(0)=0$$, then $$y(2)$$ is equal to</p> | [{"identifier": "A", "content": "$$2 \\pi$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{8}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\pi}{16}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{32}$$"}] | ["D"] | null | <p>$$\frac{d y}{d x}+\frac{y\left(2 x^3+8 x\right)}{\left(x^2+4\right)^2}=\frac{2}{\left(x^2+4\right)^2}$$</p>
<p>$$\mathrm{IF}=e^{\int \frac{2 x^3+8 x}{\left(x^2+4\right)^2} d x}$$</p>
<p>$$\text { Let }\left(x^2+4\right)^2=t \quad \Rightarrow 2\left(x^2+4\right)(2 x) d x=d t$$</p>
<p>$$\begin{gathered}
=e^{\int \frac... | mcq | jee-main-2024-online-4th-april-evening-shift | 5,993 |
lv2erh30 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$(x+y+2)^2 d x=d y, y(0)=-2$$. Let the maximum and minimum values of the function $$y=y(x)$$ in $$\left[0, \frac{\pi}{3}\right]$$ be $$\alpha$$ and $$\beta$$, respectively. If $$(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}, \gamma, \delta \in \mathbb{Z}... | [] | null | 31 | <p>$$\begin{aligned}
& \frac{d y}{d x}=(x+y+z)^2 \\
& \text { Put } x+y+z=t \\
& \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \\
& \text { Given DE } \Rightarrow \frac{d t}{d x}-1=t^2 \\
& \Rightarrow \frac{d t}{1+t^2}=d x \Rightarrow \tan ^{-1} t=x+c \\
& \Rightarrow x+y+z=\tan (x+c) \\
& \Rightarrow y(x)=\tan (x+c)-... | integer | jee-main-2024-online-4th-april-evening-shift | 5,994 |
lv3ve63m | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$\alpha|x|=|y| \mathrm{e}^{x y-\beta}, \alpha, \beta \in \mathbf{N}$$ be the solution of the differential equation $$x \mathrm{~d} y-y \mathrm{~d} x+x y(x \mathrm{~d} y+y \mathrm{~d} x)=0,y(1)=2$$. Then $$\alpha+\beta$$ is equal to ________</p> | [] | null | 4 | <p>$$\begin{aligned}
& \alpha|x|=|y| e^{x y-\beta} \\
& \frac{x d y-y d x}{y^2}+\frac{x y(x d y+y d x)}{y^2}=0 \\
& -d\left(\frac{x}{y}\right)+\frac{x}{y} d(x y)=0
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \int d(x y)=\int \frac{d\left(\frac{x}{y}\right)}{\frac{x}{y}} \\
& x y=\ln \left|\frac{x}{y}\right|+\ln c \\
& x... | integer | jee-main-2024-online-8th-april-evening-shift | 5,996 |
lv5grwhw | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$(1+y^2) e^{\tan x} d x+\cos ^2 x(1+e^{2 \tan x}) d y=0, y(0)=1$$. Then $$y\left(\frac{\pi}{4}\right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{1}{e^2}$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{e^2}$$\n"}, {"identifier": "C", "content": "$$\\frac{2}{e}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{e}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0 \\
& \frac{d y}{1+y^2}=-\frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}} \\
& \int \frac{d y}{1+y^2}=-\int \frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}}
\end{aligned}$$</p>
<p>Let $$e^{\tan x}=t$$</p>
<p>$$\... | mcq | jee-main-2024-online-8th-april-morning-shift | 5,997 |
lvb294oh | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Suppose the solution of the differential equation $$\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)}$$ represents a circle passing through origin. Then the radius of this circle is :</p> | [{"identifier": "A", "content": "$$\\sqrt{17}$$\n"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{17}}{2}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{2}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)} \\
& \beta x d y-2 \alpha y d y-(\beta \gamma-4 \alpha) d y \\
& =2 x d x+\alpha x d x-\beta y d x+2 d x \\
& \beta \int(x d y+y d y)-\alpha y^2-(\beta \gamma-4 x) y=x^2+\frac{\alpha x^2}{2}+2 x \\
& \beta x... | mcq | jee-main-2024-online-6th-april-evening-shift | 5,998 |
lvb29503 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution $$y(x)$$ of the given differential equation $$\left(e^y+1\right) \cos x \mathrm{~d} x+\mathrm{e}^y \sin x \mathrm{~d} y=0$$ passes through the point $$\left(\frac{\pi}{2}, 0\right)$$, then the value of $$e^{y\left(\frac{\pi}{6}\right)}$$ is equal to _________.</p> | [] | null | 3 | <p>Given the differential equation</p>
<p>$$\left(e^y + 1\right) \cos x \, dx + e^y \sin x \, dy = 0,$$</p>
<p>we aim to find the value of $ e^{y\left(\frac{\pi}{6}\right)} $ given that the solution $ y(x) $ passes through the point $\left(\frac{\pi}{2}, 0\right)$.</p>
<p>First, we recognize that the differential eq... | integer | jee-main-2024-online-6th-april-evening-shift | 5,999 |
FFUHPxJb4rDcgpkbqKevN | maths | differentiation | differentiation-of-a-function-with-respect-to-another-function | If $$x = \sqrt {{2^{\cos e{c^{ - 1}}}}} $$ and $$y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),$$ then $${{dy} \over {dx}}$$ is equal to : | [{"identifier": "A", "content": "$${y \\over x}$$ "}, {"identifier": "B", "content": "$${x \\over y}$$"}, {"identifier": "C", "content": "$$-$$ $${y \\over x}$$"}, {"identifier": "D", "content": "$$-$$ $${x \\over y}$$"}] | ["C"] | null | x = $$\sqrt {{2^{\cos e{c^{ - 1}}t}}} $$
<br><br>$$\therefore\,\,\,\,$$ $${{dx} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ ($${2^{\cos e{c^{ - 1}}t}}\,.\,\log 2$$) $$ \times $$ $${{ - 1} \over {t\sqrt {{t^2} - 1} }}$$
<br><br>$${{dy} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{{{\sec }^... | mcq | jee-main-2018-online-16th-april-morning-slot | 6,000 |
s3JCuNr7c41ktbzUE9jgy2xukfqf5mcc | maths | differentiation | differentiation-of-a-function-with-respect-to-another-function | The derivative of
<br/>$${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$$ with<br/> respect to $${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$$ at x = $${1 \over 2}$$ is : | [{"identifier": "A", "content": "$${{2\\sqrt 3 } \\over 3}$$"}, {"identifier": "B", "content": "$${{2\\sqrt 3 } \\over 5}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over {10}}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over {12}}$$"}] | ["C"] | null | Let f = $${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$$
<br><br>Put x = tan $$\theta $$ $$ \Rightarrow $$ $$\theta $$ = tan<sup>–1</sup> x
<br><br>f = $${\tan ^{ - 1}}\left( {{{\sec \theta - 1} \over {\tan \theta }}} \right)$$
<br><br>$$ \Rightarrow $$ f = $${\tan ^{ - 1}}\left( {{{1 - \cos \thet... | mcq | jee-main-2020-online-5th-september-evening-slot | 6,002 |
h3iEjCw8iLxihq9W | maths | differentiation | differentiation-of-composite-function | Let $$f:\left( { - 1,1} \right) \to R$$ be a differentiable function with $$f\left( 0 \right) = - 1$$ and $$f'\left( 0 \right) = 1$$. Let $$g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}$$. Then $$g'\left( 0 \right) = $$ | [{"identifier": "A", "content": "$$-4$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$-2$$ "}, {"identifier": "D", "content": "$$4$$ "}] | ["A"] | null | $$g'\left( x \right) = 2\left( {f\left( {2f\left( x \right) + 2} \right)} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{d \over {dx}}\left( {f\left( {2f\left( x \right) + 2} \right)} \right)} \right)$$
<br><br>$$ = 2f\left( {2f\left( x \right) + 2} \right)f'\left( {2f\left( x \right... | mcq | aieee-2010 | 6,003 |
mgG2EkWdjrRPzc2r | maths | differentiation | differentiation-of-composite-function | If $$g$$ is the inverse of a function $$f$$ and $$f'\left( x \right) = {1 \over {1 + {x^5}}},$$ then $$g'\left( x \right)$$ is equal to: | [{"identifier": "A", "content": "$${1 \\over {1 + {{\\left\\{ {g\\left( x \\right)} \\right\\}}^5}}}$$ "}, {"identifier": "B", "content": "$$1 + {\\left\\{ {g\\left( x \\right)} \\right\\}^5}$$ "}, {"identifier": "C", "content": "$$1 + {x^5}$$ "}, {"identifier": "D", "content": "$$5{x^4}$$ "}] | ["B"] | null | Since $$f(x)$$ and $$g(x)$$ are inverse of each other
<br><br>$$\therefore$$ $$g'\left( {f\left( x \right)} \right) = {1 \over {f'\left( x \right)}}$$
<br><br>$$ \Rightarrow g'\left( {f\left( x \right)} \right) = 1 + {x^5}$$
<br><br>$$\left( \, \right.$$ As $$\,f'\left( x \right) = {1 \over {1 + {x^5}}}$$ $$\left. \, \... | mcq | jee-main-2014-offline | 6,004 |
nJGKt5gcnsbeag2W6jFE6 | maths | differentiation | differentiation-of-composite-function | If ƒ(1) = 1, ƒ'(1) = 3, then the derivative of
ƒ(ƒ(ƒ(x))) + (ƒ(x))<sup>2
</sup> at x = 1 is : | [{"identifier": "A", "content": "33"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "15"}] | ["A"] | null | Given ƒ(1) = 1, ƒ'(1) = 3
<br><br>Let y = ƒ(ƒ(ƒ(x))) + (ƒ(x))<sup>2
</sup>
<br><br>On differentiating both sides with respect to x we get,
<br><br>$${{dy} \over {dx}}$$ = ƒ'(ƒ(ƒ(x))).ƒ'(ƒ(x)).ƒ'(x) + 2ƒ(x).ƒ'(x)
<br><br>Now at x = 1,
<br><br>$${{dy} \over {dx}}$$ = ƒ'(ƒ(ƒ(1))).ƒ'(ƒ(1)).ƒ'(1) + 2ƒ(1).ƒ'(1)
<br><br>= ƒ'(... | mcq | jee-main-2019-online-8th-april-evening-slot | 6,005 |
HrO6IQwQJN5Eo4N4zJ3rsa0w2w9jx220fw8 | maths | differentiation | differentiation-of-composite-function | Let f(x) = log<sub>e</sub>(sin x), (0 < x < $$\pi $$) and g(x) = sin<sup>–1</sup>
(e<sup>–x</sup>
), (x $$ \ge $$ 0). If $$\alpha $$ is a positive real number such that
a = (fog)'($$\alpha $$) and b = (fog)($$\alpha $$), then : | [{"identifier": "A", "content": "a$$\\alpha $$<sup>2</sup> + b$$\\alpha $$ - a = -2$$\\alpha $$<sup>2</sup>"}, {"identifier": "B", "content": "a$$\\alpha $$<sup>2</sup> + b$$\\alpha $$ + a = 0"}, {"identifier": "C", "content": "a$$\\alpha $$<sup>2</sup> - b$$\\alpha $$ - a = 0"}, {"identifier": "D", "content": "a$$\\al... | ["D"] | null | f(x) = ln(sin x), g(x) = sin<sup>–1</sup> (e<sup>–x</sup>)<br><br>
f(g(x)) = ln(sin(sin<sup>–1</sup> e<sup>–x</sup>)) = -x<br><br>
f(g($$\alpha $$)) = – $$\alpha $$ = b<br><br>
As f(g(x)) = – x <br><br>
$$ \therefore $$ (f(g(x)))' = – 1<br><br>
$$ \Rightarrow $$ (f(g($$\alpha $$)))' = – 1 = a<br><br>
$$ \therefore $$ ... | mcq | jee-main-2019-online-10th-april-evening-slot | 6,006 |
lv0vxcdq | maths | differentiation | differentiation-of-composite-function | <p>Let $$f(x)=x^5+2 \mathrm{e}^{x / 4}$$ for all $$x \in \mathbf{R}$$. Consider a function $$g(x)$$ such that $$(g \circ f)(x)=x$$ for all $$x \in \mathbf{R}$$. Then the value of $$8 g^{\prime}(2)$$ is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "8"}] | ["C"] | null | <p>Given that $$(g \circ f)(x) = x$$ for all $$x \in \mathbf{R}$$. This means $$g(f(x)) = x$$ for all $$x \in \mathbf{R}$$. Differentiating both sides with respect to $$x$$, we get:
<p>$$g'(f(x)) \cdot f'(x) = 1$$</p>
</p>
<p>Now, we want to find the value of $$8g'(2)$$. To do this, we need to find a value of $$x$... | mcq | jee-main-2024-online-4th-april-morning-shift | 6,009 |
lvb294td | maths | differentiation | differentiation-of-composite-function | <p>Suppose for a differentiable function $$h, h(0)=0, h(1)=1$$ and $$h^{\prime}(0)=h^{\prime}(1)=2$$. If $$g(x)=h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$$, then $$g^{\prime}(0)$$ is equal to:</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <p>To determine $$g^{\prime}(0)$$, we start by applying the chain rule and product rule to find the derivative of the given function $$g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$$.</p>
<p>The product rule states that if we have two functions $$u(x)$$ and $$v(x)$$, then the derivative of their product is given... | mcq | jee-main-2024-online-6th-april-evening-shift | 6,010 |
W9kJiVVFBaQQJEim | maths | differentiation | differentiation-of-implicit-function | If $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},$$ then $$\left( {1 + {x^2}} \right){{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is | [{"identifier": "A", "content": "$${n^2}y$$ "}, {"identifier": "B", "content": "$$-{n^2}y$$"}, {"identifier": "C", "content": "$$-y$$ "}, {"identifier": "D", "content": "$$2{x^2}y$$ "}] | ["A"] | null | $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n}$$
<br><br>$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {1 \over 2}{{\left( {1 + {x^2}} \right)}^{ - 1/2}}.2x} \right);$$
<br><br>$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}}... | mcq | aieee-2002 | 6,011 |
lYKxaNHOCAaR6Yxj | maths | differentiation | differentiation-of-implicit-function | Let $$y$$ be an implicit function of $$x$$ defined by $${x^{2x}} - 2{x^x}\cot \,y - 1 = 0$$. Then $$y'(1)$$ equals | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$\\log \\,2$$"}, {"identifier": "C", "content": "$$-\\log \\,2$$ "}, {"identifier": "D", "content": "$$-1$$"}] | ["D"] | null | $${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0$$
<br><br>$$ \Rightarrow 2\,\cot \,y = {x^x} - {x^{ - x}}$$
<br><br>$$ \Rightarrow 2\,\cot \,y\, = u - {1 \over u}$$
<br><br>where $$u = {x^x}$$
<br><br>Differentiating both sides with respect to $$x,$$
<br><br>we get $$ \Rightarrow - 2\cos e{c^2}y{{dy} \over {dx}}$$
<br><br>$$ ... | mcq | aieee-2009 | 6,013 |
DhiE82brjsITwGqGoJmYl | maths | differentiation | differentiation-of-implicit-function | If $$f\left( x \right) = \left| {\matrix{
{\cos x} & x & 1 \cr
{2\sin x} & {{x^2}} & {2x} \cr
{\tan x} & x & 1 \cr
} } \right|,$$ then $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$ | [{"identifier": "A", "content": "does not exist. "}, {"identifier": "B", "content": "exists and is equal to 2. "}, {"identifier": "C", "content": "existsand is equal to 0."}, {"identifier": "D", "content": "exists and is equal to $$-$$ 2."}] | ["D"] | null | Given,<br><br>
$$f\left( x \right) = \left| {\matrix{
{\cos x} & x & 1 \cr
{2\sin x} & {{x^2}} & {2x} \cr
{\tan x} & x & 1 \cr
} } \right|$$<br><br>
= cosx(x<sup>2</sup> - 2x<sup>2</sup>) - x(2 sinx - 2x tanx) + (2x sinx - x<sup>2</sup> tanx)<br>
= x<sup>2</sup> (tanx - cosx)<br... | mcq | jee-main-2018-online-15th-april-morning-slot | 6,015 |
C8sdKeF0ryQ9Msh5eBUCc | maths | differentiation | differentiation-of-implicit-function | If x<sup>2</sup> + y<sup>2</sup> + sin y = 4, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at the point ($$-$$2,0) is : | [{"identifier": "A", "content": "$$-$$ 34"}, {"identifier": "B", "content": "$$-$$ 32"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$-$$ 2"}] | ["A"] | null | Given, x<sup>2</sup> + y<sup>2</sup> + sin y = 4
<br><br>After differentiating the above equation w.r.t.x we get
<br><br>2x + 2y $${{dy} \over {dx}}$$ + cos y $${{dy} \over {dx}}$$ = 0 . . . . (1)
<br><br>$$ \Rightarrow $$ 2x + (2y + cos y) $${{dy... | mcq | jee-main-2018-online-15th-april-morning-slot | 6,016 |
Ne604P2uiWuSdG4XwJ3rsa0w2w9jx5deh3j | maths | differentiation | differentiation-of-implicit-function | If e<sup>y</sup>
+ xy = e, the ordered pair $$\left( {{{dy} \over {dx}},{{{d^2}y} \over {d{x^2}}}} \right)$$ at x = 0 is equal to : | [{"identifier": "A", "content": "$$\\left( {{1 \\over e}, - {1 \\over {{e^2}}}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over e},{1 \\over {{e^2}}}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over e}, - {1 \\over {{e^2}}}} \\right)$$"}, {"identifier": "D", "content": "$$\\l... | ["B"] | null | y = 1 $$ \Rightarrow $$ x = 0<br><br>
$${e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0$$<br><br>
$$ \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}$$<br><br>
$$ \Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}... | mcq | jee-main-2019-online-12th-april-morning-slot | 6,017 |
kPxln5RHIGisQJ1WD17k9k2k5e4fsyg | maths | differentiation | differentiation-of-implicit-function | If $$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} ,\alpha \in \left( {{{3\pi } \over 4},\pi } \right)$$<br/><br/>
$${{dy} \over {d\alpha }}\,\,at\,\alpha = {{5\pi } \over 6}is$$ : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "-4"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "-$${1 \\over 4}$$"}] | ["A"] | null | $$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}$$
<br><br>= $$\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}} $$
<br><br... | mcq | jee-main-2020-online-7th-january-morning-slot | 6,019 |
jKg7va9mIxQ9FIreIH7k9k2k5fnompn | maths | differentiation | differentiation-of-implicit-function | Let y = y(x) be a function of x satisfying
<br/><br>$$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ where k is a constant and
<br/><br>$$y\left( {{1 \over 2}} \right) = - {1 \over 4}$$. Then $${{dy} \over {dx}}$$ at x = $${1 \over 2}$$, is equal to :</br></br> | [{"identifier": "A", "content": "$${2 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$$ - {{\\sqrt 5 } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "D", "content": "$$ - {{\\sqrt 5 } \\over 4}$$"}] | ["B"] | null | $$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ ....(1)
<br><br>On differentiating both side of eq. (1) w.r.t. x we
get,
<br><br>$${{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }}$$
<br><br>= 0 - $$\sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}}$$
<br><br>Put x = $${1 ... | mcq | jee-main-2020-online-7th-january-evening-slot | 6,020 |
1ldpt5swb | maths | differentiation | differentiation-of-implicit-function | <p>Let $$y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right)$$. Then, at x = 1,</p> | [{"identifier": "A", "content": "$$2 y^{\\prime}+\\sqrt{3} \\pi^{2} y=0$$"}, {"identifier": "B", "content": "$$y^{\\prime}+3 \\pi^{2} y=0$$"}, {"identifier": "C", "content": "$$\\sqrt{2} y^{\\prime}-3 \\pi^{2} y=0$$"}, {"identifier": "D", "content": "$$2 y^{\\prime}+3 \\pi^{2} y=0$$"}] | ["D"] | null | $f(x)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right)$
<br/><br/>$$
\begin{aligned}
& f^{\prime}(x)=3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\
& \cos \left(\frac{\pi}{3} \... | mcq | jee-main-2023-online-31st-january-morning-shift | 6,022 |
1lgoy3hxl | maths | differentiation | differentiation-of-implicit-function | <p>Let $$f(x)=\sum_\limits{k=1}^{10} k x^{k}, x \in \mathbb{R}$$. If $$2 f(2)+f^{\prime}(2)=119(2)^{\mathrm{n}}+1$$ then $$\mathrm{n}$$ is equal to ___________</p> | [] | null | 10 | Given, $f(x)=\sum_\limits{k=1}^{10} k x^{k}$
<br/><br/>$$
\begin{aligned}
& f(x)=x+2 x^2+\ldots \ldots \ldots+10 x^{10} \\\\
& f(x) . x=x^2+2 x^3+\ldots \ldots \ldots+9 x^{10}+10 x^{11} \\\\
& f(x)(1-x)=x+x^2+x^3+\ldots \ldots \ldots+x^{10}-10 x^{11} \\\\
& \therefore f(x)=\frac{x\left(1-x^{10}\right)}{(1-x)^2}-\frac{1... | integer | jee-main-2023-online-13th-april-evening-shift | 6,023 |
1lgzxhecr | maths | differentiation | differentiation-of-implicit-function | <p>Let $$f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\}$$. Then $$f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{2}{3 \\sqrt{3}}$$"}, {"identifier": "B", "content": "$$\\frac{2}{9}$$"}, {"identifier": "C", "content": "$$\\frac{-1}{3 \\sqrt{3}}$$"}, {"identifier": "D", "content": "$$\\frac{-2}{3}$$"}] | ["B"] | null | $$f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}$$
<br/><br/>$$
\begin{aligned}
& =\frac{\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x-1}{\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x} \\\\
& =\frac{\cos \left(x-\frac{\pi}{4}\right)-1}{\sin \left(x-\frac{\pi}{4}\right)} \\\\
& =\frac{-2 \sin ^2\left(\fra... | mcq | jee-main-2023-online-8th-april-morning-shift | 6,024 |
cOntCBTD82jRclEC | maths | differentiation | differentiation-of-inverse-trigonometric-function | If $$y = \sec \left( {{{\tan }^{ - 1}}x} \right),$$ then $${{{dy} \over {dx}}}$$ at $$x=1$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$\\sqrt 2 $$ "}] | ["A"] | null | Let $$y = \sec \left( {{{\tan }^{ - 1}}x} \right)$$
<br><br>and $${\tan ^{ - 1}}\,\,x = \theta .$$
<br><br>$$ \Rightarrow x = \tan \theta $$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267182/exam_images/gheykswaxc1dfck1kir0.webp" loading="lazy" alt="JEE Main 2013 (O... | mcq | jee-main-2013-offline | 6,025 |
yy401t5DTBxS3W73 | maths | differentiation | differentiation-of-inverse-trigonometric-function | If for $$x \in \left( {0,{1 \over 4}} \right)$$, the derivatives of
<br/><br/>$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals | [{"identifier": "A", "content": "$${{{3x\\sqrt x } \\over {1 - 9{x^3}}}}$$"}, {"identifier": "B", "content": "$${{{3x} \\over {1 - 9{x^3}}}}$$"}, {"identifier": "C", "content": "$${{3 \\over {1 + 9{x^3}}}}$$"}, {"identifier": "D", "content": "$${{9 \\over {1 + 9{x^3}}}}$$"}] | ["D"] | null | Let y = $${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$
<br><br>= $${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$$
<br><br>= 2$${\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)$$
<br><br>$$ \therefore $$ $${{dy} \ove... | mcq | jee-main-2017-offline | 6,026 |
WaMVJei76O03McRXFgNSd | maths | differentiation | differentiation-of-inverse-trigonometric-function | If f(x) = sin<sup>-1</sup> $$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right),$$ then f'$$\left( { - {1 \over 2}} \right)$$ equals : | [{"identifier": "A", "content": "$$ - \\sqrt 3 {\\log _e}\\sqrt 3 $$"}, {"identifier": "B", "content": "$$ \\sqrt 3 {\\log _e}\\sqrt 3 $$"}, {"identifier": "C", "content": "$$ - \\sqrt 3 {\\log _e}\\, 3 $$"}, {"identifier": "D", "content": "$$ \\sqrt 3 {\\log _e}\\, 3 $$"}] | ["B"] | null | Since f(x) = sin$$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right)$$
<br><br>Suppose 3<sup>x</sup> = tan t
<br><br> $$ \Rightarrow $$ f(x) = sin<sup>$$-$$1</sup> $$\left( {{{2\tan t} \over {1 + {{\tan }^2}t}}} \right)$$ = sin<sup>$$-$$1</sup> (sin2t) = 2t
<br><br> &nb... | mcq | jee-main-2018-online-15th-april-evening-slot | 6,027 |
9onUmZNU2npIbQ1VhNA44 | maths | differentiation | differentiation-of-inverse-trigonometric-function | If $$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$,
<br/><br/>x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ then $$dy \over dx$$ is equal to: | [{"identifier": "A", "content": "$$2x - {\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6} - x$$"}, {"identifier": "C", "content": "$${\\pi \\over 3} - x$$"}, {"identifier": "D", "content": "$$x - {\\pi \\over 6}$$"}] | ["D"] | null | $$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\... | mcq | jee-main-2019-online-8th-april-morning-slot | 6,028 |
9rwVQ5XrboppQiu5tE7k9k2k5gqvdkp | maths | differentiation | differentiation-of-inverse-trigonometric-function | Let ƒ(x) = (sin(tan<sup>–1</sup>x) + sin(cot<sup>–1</sup>x))<sup>2</sup> – 1, |x| > 1.
<br/>If $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$ and $$y\left( {\sqrt 3 } \right) = {\pi \over 6}$$,
then y($${ - \sqrt 3 }$$) is equal to : | [{"identifier": "A", "content": "$${{5\\pi } \\over 6}$$"}, {"identifier": "B", "content": "$$ - {\\pi \\over 6}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${{2\\pi } \\over 3}$$"}] | ["B"] | null | Given ƒ(x) = (sin(tan<sup>–1</sup>x) + sin(cot<sup>–1</sup>x))<sup>2</sup> – 1
<br><br> = (sin(tan<sup>–1</sup>x) + sin($${\pi \over 2}$$ - tan<sup>–1</sup>x))<sup>2</sup> – 1
<br><br> = (sin(tan<sup>–1</sup>x) + cos(tan<sup>–1</sup>x))<sup>2</sup> – 1
<br><br>= sin<sup>2</sup>(tan<sup>–1</sup>x) + cos<sup>2</sup>(tan... | mcq | jee-main-2020-online-8th-january-morning-slot | 6,029 |
o7hlhlRCMBt3fIKQdBjgy2xukezff9hs | maths | differentiation | differentiation-of-inverse-trigonometric-function | If y = $$\sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left\{ {{3 \over 5}\cos kx - {4 \over 5}\sin kx} \right\}} $$,
<br/><br/>then $${{dy} \over {dx}}$$ at x = 0 is _______. | [] | null | 91 | Put, $$\cos \alpha = {3 \over 5},\sin \alpha = {4 \over 5}$$<br><br>
$$ \therefore {3 \over 5}\cos kx - {4 \over 5}\sin \,kx$$<br><br>
$$ = \cos \alpha .\cos kx - \sin \alpha .\sin kx$$<br><br>
$$ = \cos \left( {\alpha + kx} \right)$$<br><br>
So, $$y = \sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left( {\cos \left( {\al... | integer | jee-main-2020-online-2nd-september-evening-slot | 6,030 |
1ktbc8v27 | maths | differentiation | differentiation-of-inverse-trigonometric-function | Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then : | [{"identifier": "A", "content": "$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$"}, {"identifier": "B", "content": "$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$"}, {"identifier": "C", "content": "$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$"}, {"identifier": "D", "content": "$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$"}] | ["C"] | null | $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$<br><br>$${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x $$<br><br>or $$f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$$<br><br>$$ = \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$... | mcq | jee-main-2021-online-26th-august-morning-shift | 6,032 |
1ktg2zxf8 | maths | differentiation | differentiation-of-inverse-trigonometric-function | If $$y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$$, then $${{dy} \over {dx}}$$ at $$x = {{5\pi } \over 6}$$ is : | [{"identifier": "A", "content": "$$ - {1 \\over 2}$$"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "0"}] | ["A"] | null | We have,
<br/><br/>$$
y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)
$$
<br/><br/>$$
=\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}... | mcq | jee-main-2021-online-27th-august-evening-shift | 6,033 |
1l5banuqb | maths | differentiation | differentiation-of-inverse-trigonometric-function | <p>If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$$, then</p> | [{"identifier": "A", "content": "$$xy'' + 2y' = 0$$"}, {"identifier": "B", "content": "$${x^2}y'' - 6y + {{3\\pi } \\over 2} = 0$$"}, {"identifier": "C", "content": "$${x^2}y'' - 6y + 3\\pi = 0$$"}, {"identifier": "D", "content": "$$xy'' - 4y' = 0$$"}] | ["B"] | null | <p>Let $${x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)$$</p>
<p>$$\therefore$$ $$y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )$$</p>
<p>$$ = {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)$$</p>
<p>$$\therefore$$ $$y = {\pi \over 4} - {... | mcq | jee-main-2022-online-24th-june-evening-shift | 6,034 |
luxwdj6m | maths | differentiation | differentiation-of-inverse-trigonometric-function | <p>If $$\log _e y=3 \sin ^{-1} x$$, then $$(1-x^2) y^{\prime \prime}-x y^{\prime}$$ at $$x=\frac{1}{2}$$ is equal to</p> | [{"identifier": "A", "content": "$$9 e^{\\pi / 2}$$\n"}, {"identifier": "B", "content": "$$9 e^{\\pi / 6}$$\n"}, {"identifier": "C", "content": "$$3 e^{\\pi / 2}$$\n"}, {"identifier": "D", "content": "$$3 e^{\\pi / 6}$$"}] | ["A"] | null | <p>$$\begin{aligned}
&\log _e y=3 \sin ^{-1} x\\
&\begin{aligned}
& y=e^{3 \sin ^{-1} x} \\
& \frac{d y}{d x}=e^{3 \sin ^{-1} x} \cdot \frac{3}{\sqrt{1-x^2}}
\end{aligned}
\end{aligned}$$</p>
<p>$$\sqrt{1-x^2} \frac{d y}{d x}=3 y$$</p>
<p>Again differentiate</p>
<p>$$\begin{aligned}
& \sqrt{1-x^2} \cdot y^{\prime \prim... | mcq | jee-main-2024-online-9th-april-evening-shift | 6,035 |
jmcGBPjprXE1bdtL | maths | differentiation | differentiation-of-logarithmic-function | If $${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}},$$ then $${{{dy} \over {dx}}}$$ is | [{"identifier": "A", "content": "$${y \\over x}$$ "}, {"identifier": "B", "content": "$${{x + y} \\over {xy}}$$ "}, {"identifier": "C", "content": "$$xy$$ "}, {"identifier": "D", "content": "$${x \\over y}$$"}] | ["A"] | null | $${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}$$
<br><br>$$ \Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)$$
<br><br>Differentiating both sides.
<br><br>$$\therefore$$ $${m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)$$
<br><br... | mcq | aieee-2006 | 6,038 |
oMWsMZ2BcQMVesDBdN3ng | maths | differentiation | differentiation-of-logarithmic-function | If xlog<sub>e</sub>(log<sub>e</sub>x) $$-$$ x<sup>2</sup> + y<sup>2</sup> = 4(y > 0), then $${{dy} \over {dx}}$$ at x = e is equal to : | [{"identifier": "A", "content": "$${{\\left( {1 + 2e} \\right)} \\over {2\\sqrt {4 + {e^2}} }}$$"}, {"identifier": "B", "content": "$${{\\left( {1 + 2e} \\right)} \\over {\\sqrt {4 + {e^2}} }}$$"}, {"identifier": "C", "content": "$${{\\left( {2e - 1} \\right)} \\over {2\\sqrt {4 + {e^2}} }}$$"}, {"identifier": "D", "co... | ["C"] | null | Differentiating with respect to x,
<br><br>$$x.{1 \over {\ell nx}}.{1 \over x} + \ell n(\ell nx) - 2x + 2y.{{dy} \over {dx}} = 0$$
<br><br>at $$x = e$$ we get
<br><br>$$1 - 2e + 2y{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2y}}$$
<br><br>$$ \Rightarrow {{dy} \over... | mcq | jee-main-2019-online-11th-january-morning-slot | 6,039 |
ltTrASebDziGIvmUP3p9h | maths | differentiation | differentiation-of-logarithmic-function | For x > 1, if (2x)<sup>2y</sup> = 4e<sup>2x$$-$$2y</sup>,
<br/><br/>then (1 + log<sub>e</sub> 2x)<sup>2</sup> $${{dy} \over {dx}}$$ is equal to : | [{"identifier": "A", "content": "$${{x\\,{{\\log }_e}2x - {{\\log }_e}2} \\over x}$$"}, {"identifier": "B", "content": "log<sub>e</sub> 2x"}, {"identifier": "C", "content": "x log<sub>e</sub> 2x"}, {"identifier": "D", "content": "$${{x\\,{{\\log }_e}2x + {{\\log }_e}2} \\over x}$$"}] | ["A"] | null | (2x)<sup>2y</sup> = 4e<sup>2x-2y</sup>
<br><br>2y$$\ell $$n2x = $$\ell $$n4 + 2x $$-$$ 2y
<br><br>y = $${{x + \ell n2} \over {1 + \ell n2x}}$$
<br><br>y ' = $${{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}}$$
<br><br>y '$${\left( {1 + \ell n2x} \ri... | mcq | jee-main-2019-online-12th-january-morning-slot | 6,040 |
1l57o8xuz | maths | differentiation | differentiation-of-logarithmic-function | <p>If $${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5},\,|y| < 2$$, then :</p> | [{"identifier": "A", "content": "$${x^2}y'' + xy' - 25y = 0$$"}, {"identifier": "B", "content": "$${x^2}y'' - xy' - 25y = 0$$"}, {"identifier": "C", "content": "$${x^2}y'' - xy' + 25y = 0$$"}, {"identifier": "D", "content": "$${x^2}y'' + xy' + 25y = 0$$"}] | ["D"] | null | <p>$${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5}\,\,\,\,\,\,\,\,\,|y| < 2$$</p>
<p>Differentiating on both side</p>
<p>$$ - {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}$$</p>
<p>$${{ - xy'} \over... | mcq | jee-main-2022-online-27th-june-morning-shift | 6,042 |
1l58gt0p9 | maths | differentiation | differentiation-of-logarithmic-function | <p>Let f : R $$\to$$ R satisfy $$f(x + y) = {2^x}f(y) + {4^y}f(x)$$, $$\forall$$x, y $$\in$$ R. If f(2) = 3, then $$14.\,{{f'(4)} \over {f'(2)}}$$ is equal to ____________.</p> | [] | null | 248 | <p>$$\because$$ $$f(x + y) = {2^x}f(y) + {4^y}f(x)$$ ....... (1)</p>
<p>Now, $$f(y + x){2^y}f(x) + {4^x}f(y)$$ ...... (2)</p>
<p>$$\therefore$$ $${2^x}f(y) + {4^y}f(x) = {2^y}f(x) + {4^x}f(y)$$</p>
<p>$$({4^y} - {2^y})f(x) = ({4^x} - {2^x})f(y)$$</p>
<p>$${{f(x)} \over {{4^x} - {2^x}}} = {{f(y)} \over {{4^y} - {2^y}}} ... | integer | jee-main-2022-online-26th-june-evening-shift | 6,043 |
1ldo5zk2p | maths | differentiation | differentiation-of-logarithmic-function | <p>If $$y(x)=x^{x},x > 0$$, then $$y''(2)-2y'(2)$$ is equal to</p> | [{"identifier": "A", "content": "$$4(\\log_{e}2)^{2}+2$$"}, {"identifier": "B", "content": "$$8\\log_{e}2-2$$"}, {"identifier": "C", "content": "$$4\\log_{e}2+2$$"}, {"identifier": "D", "content": "$$4(\\log_{e}2)^{2}-2$$"}] | ["D"] | null | $\begin{aligned} & y=x^x \\\\ & y^{\prime}=x^x(1+\ln x) \\\\ & y^{\prime \prime}=x^x(1+\ln x)^2+\frac{x^x}{x} \\\\ & f^{\prime \prime}(2)-2 f^{\prime}(2)=\left(4(1+\ln 2)^2+2\right)-(2)(4(1+\ln 2)) \\\\ & =4\left(1+(\ln 2)^2\right)+2-8 \\\\ & =4(\ln 2)^2-2 \\\\ & \end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift | 6,045 |
1lh23oisg | maths | differentiation | differentiation-of-logarithmic-function | <p>If $$2 x^{y}+3 y^{x}=20$$, then $$\frac{d y}{d x}$$ at $$(2,2)$$ is equal to :</p> | [{"identifier": "A", "content": "$$-\\left(\\frac{3+\\log _{e} 16}{4+\\log _{e} 8}\\right)$$"}, {"identifier": "B", "content": "$$-\\left(\\frac{2+\\log _{e} 8}{3+\\log _{e} 4}\\right)$$"}, {"identifier": "C", "content": "$$-\\left(\\frac{3+\\log _{e} 8}{2+\\log _{e} 4}\\right)$$"}, {"identifier": "D", "content": "$$-\... | ["B"] | null | Given, $2 x^y+3 y^x=20$ ..........(i)
<br/><br/>Let $u=x^y$
<br/><br/>On taking log both sides, we get
<br/><br/>$\log u=y \log x$
<br/><br/>On differentiating both sides with respect to $x$, we get
<br/><br/>$$
\begin{array}{rlrl}
& \frac{1}{u} \frac{d u}{d x} =y \frac{1}{x}+\log x \frac{d y}{d x} \\\\
& \Rightarro... | mcq | jee-main-2023-online-6th-april-morning-shift | 6,046 |
jaoe38c1lsfkl66w | maths | differentiation | differentiation-of-logarithmic-function | <p>$$\text { Let } y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1 < x<1 \text {. Then at } x=\frac{1}{2} \text {, the value of } 225\left(y^{\prime}-y^{\prime \prime}\right) \text { is equal to }$$</p> | [{"identifier": "A", "content": "732"}, {"identifier": "B", "content": "736"}, {"identifier": "C", "content": "742"}, {"identifier": "D", "content": "746"}] | ["B"] | null | <p>$$\begin{aligned}
& y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\
& \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4}
\end{aligned}$$</p>
<p>Again,</p>
<p>$$\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$$</p>
<p>Again</p>
<p>$$y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\fr... | mcq | jee-main-2024-online-29th-january-evening-shift | 6,047 |
ghiwTEg0WvaIpfIxQkmJb | maths | differentiation | differentiation-of-parametric-function | If x $$=$$ 3 tan t and y $$=$$ 3 sec t, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at t $$ = {\pi \over 4},$$ is : | [{"identifier": "A", "content": "$${1 \\over {3\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over {6\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${3 \\over {2\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over 6}$$"}] | ["B"] | null | x = 3 tan t and y = 3 sec t
<br><br>So that $${{dx} \over {dt}}$$ = 3sec<sup>2</sup>t and $${{dy} \over {dt}}$$ = 3 sec t tan t
<br><br>$${{dy} \over {dx}}$$ = $${{dy/dt} \over {dx/dt}}$$ = sin t
<br><br>$${{{d^2}y} \over {d{x^2}}}$$ = (cos t)$$.{{dt} \over {dx}}$$
<br><br>$${{{d^2}y} \over {d{x^2}}} = \left( {\cos t}... | mcq | jee-main-2019-online-9th-january-evening-slot | 6,048 |
JDlJzbuL7enwokKQJU7k9k2k5k6v2n8 | maths | differentiation | differentiation-of-parametric-function | If $$x = 2\sin \theta - \sin 2\theta $$ and $$y = 2\cos \theta - \cos 2\theta $$,<br/>
$$\theta \in \left[ {0,2\pi } \right]$$, then $${{{d^2}y} \over {d{x^2}}}$$ at $$\theta $$ = $$\pi $$ is : | [{"identifier": "A", "content": "$${3 \\over 8}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "-$${3 \\over 4}$$"}] | ["A"] | null | $$x = 2\sin \theta - \sin 2\theta $$
<br><br>$$ \Rightarrow $$ $${{dx} \over {d\theta }}$$ = $$2\cos \theta - 2\cos 2\theta $$
<br><br>$$y = 2\cos \theta - \cos 2\theta $$
<br><br>$$ \Rightarrow $$ $${{dy} \over {d\theta }}$$ = –2sin$$\theta $$ + 2sin2$$\theta $$
<br><br>$${{dy} \over {dx}} = {{{{dy} \over {d\theta ... | mcq | jee-main-2020-online-9th-january-evening-slot | 6,049 |
1l6nm5311 | maths | differentiation | differentiation-of-parametric-function | <p>Let $$x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}$$ and <br/><br/>$$y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \frac{\pi}{2}\right)$$. <br/><br/>Then $$\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$$ at $$t=\frac{\pi}{4}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{-2 \\sqrt{2}}{3}$$"}, {"identifier": "B", "content": "$$\\frac{2}{3}$$"}, {"identifier": "C", "content": "$$\\frac{1}{3}$$"}, {"identifier": "D", "content": "$$ \\frac{-2}{3}$$"}] | ["D"] | null | <p>$$x = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t} $$</p>
<p>$$\therefore$$ $${{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }}$$</p>
<p>$$\therefore$$ $${{dy} \over {dx}} = \tan 3t,\,\left( {\mathrm{at}\,t = ... | mcq | jee-main-2022-online-28th-july-evening-shift | 6,050 |
lmUN4HWdFdPr7roD | maths | differentiation | methods-of-differentiation | $${{{d^2}x} \over {d{y^2}}}$$ equals: | [{"identifier": "A", "content": "$$ - {\\left( {{{{d^2}y} \\over {d{x^2}}}} \\right)^{ - 1}}{\\left( {{{dy} \\over {dx}}} \\right)^{ - 3}}$$ "}, {"identifier": "B", "content": "$${\\left( {{{{d^2}y} \\over {d{x^2}}}} \\right)^{}}{\\left( {{{dy} \\over {dx}}} \\right)^{ - 2}}$$ "}, {"identifier": "C", "content": "$$ - \... | ["C"] | null | $${{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)$$
<br><br>$$ = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}$$
<br><br>$$ = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}$$
<br><br>$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y... | mcq | aieee-2011 | 6,051 |
DdF7R8tTzx4Kv40prc7k9k2k5khz7tx | maths | differentiation | methods-of-differentiation | Let ƒ and g be differentiable functions on R
such that fog is the identity function. If for some
a, b $$ \in $$ R, g'(a) = 5 and g(a) = b, then ƒ'(b) is
equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$${1 \\over 5}$$"}] | ["D"] | null | Given the function composition f(g(x)) is the identity function, it means f(g(x)) = x for all x.
<br><br>$$ \Rightarrow $$ ƒ'(g(x)) g'(x) = 1
<br><br>put x = a
<br><br>$$ \Rightarrow $$ ƒ'(b) g'(a) = 1
<br><br>$$ \Rightarrow $$ ƒ'(b) = $${1 \over 5}$$ | mcq | jee-main-2020-online-9th-january-evening-slot | 6,052 |
1l54ubqt9 | maths | differentiation | methods-of-differentiation | <p>Let f and g be twice differentiable even functions on ($$-$$2, 2) such that $$f\left( {{1 \over 4}} \right) = 0$$, $$f\left( {{1 \over 2}} \right) = 0$$, $$f(1) = 1$$ and $$g\left( {{3 \over 4}} \right) = 0$$, $$g(1) = 2$$. Then, the minimum number of solutions of $$f(x)g''(x) + f'(x)g'(x) = 0$$ in $$( - 2,2)$$ is e... | [] | null | 4 | Let $h(x)=f(x) \cdot g^{\prime}(x)$
<br/><br/>
As $f(x)$ is even $f\left(\frac{1}{2}\right)=\left(\frac{1}{4}\right)=0$
<br/><br/>
$\Rightarrow f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{4}\right)=0$
<br/><br/>
and $g(x)$ is even $\Rightarrow g^{\prime}(x)$ is odd <br/><br/>
and $g(1)=2$ ensures one root of $g^{\prime... | integer | jee-main-2022-online-29th-june-evening-shift | 6,053 |
1ldswyfk5 | maths | differentiation | methods-of-differentiation | <p>Let $$f:\mathbb{R}\to\mathbb{R}$$ be a differentiable function that satisfies the relation $$f(x+y)=f(x)+f(y)-1,\forall x,y\in\mathbb{R}$$. If $$f'(0)=2$$, then $$|f(-2)|$$ is equal to ___________.</p> | [] | null | 3 | $f(x+y)=f(x)+f(y)-1$
<br/><br/>
$$
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\\\
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f^{\prime}(0)=2 \\\\
& f^{\prime}(x)=2 \Rightarrow d y=2 d x \\\\
& y=2 x+C \\\\
& \mathrm{x}=0, \mathrm{y}=1, \mathrm{c}=1 \\\\
& \mathrm{y}=... | integer | jee-main-2023-online-29th-january-morning-shift | 6,054 |
1lgq0jgd2 | maths | differentiation | methods-of-differentiation | <p>For the differentiable function $$f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$$, let $$3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10$$, then $$\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|$$ is equal to</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "$$\\frac{29}{5}$$"}, {"identifier": "C", "content": "$$\\frac{33}{5}$$"}, {"identifier": "D", "content": "7"}] | ["A"] | null | <ol>
<li><p>Given the equation: $$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$$</p>
</li>
<li><p>Replace $$x$$ with $$\frac{1}{x}$$ in the original equation:
<br/>$$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$$</p>
</li>
<li><p>Now, we have two equations:</p>
</li>
</ol>
<p>$$3f(x) + 2f\left(\frac{1}{x}\righ... | mcq | jee-main-2023-online-13th-april-morning-shift | 6,055 |
lsan2rgn | maths | differentiation | methods-of-differentiation | If $y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x$, then $96 y^{\prime}\left(\frac{\pi}{6}\right)$ is equal to : | [] | null | 105 | $\begin{aligned} & y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x \\\\ & y=\frac{(\sqrt{x}+1)(\sqrt{x})\left((\sqrt{x})^3-1\right)}{(\sqrt{x})\left((\sqrt{x})^2+(\sqrt{x})+1\right)}+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x \\\\ & y=(\sqrt{x}+1)(... | integer | jee-main-2024-online-1st-february-evening-shift | 6,056 |
1lsg8vyuz | maths | differentiation | methods-of-differentiation | <p>Let $$g: \mathbf{R} \rightarrow \mathbf{R}$$ be a non constant twice differentiable function such that $$\mathrm{g}^{\prime}\left(\frac{1}{2}\right)=\mathrm{g}^{\prime}\left(\frac{3}{2}\right)$$. If a real valued function $$f$$ is defined as $$f(x)=\frac{1}{2}[g(x)+g(2-x)]$$, then</p> | [{"identifier": "A", "content": "$$f^{\\prime \\prime}(x)=0$$ for atleast two $$x$$ in $$(0,2)$$\n"}, {"identifier": "B", "content": "$$f^{\\prime}\\left(\\frac{3}{2}\\right)+f^{\\prime}\\left(\\frac{1}{2}\\right)=1$$\n"}, {"identifier": "C", "content": "$$f^{\\prime \\prime}(x)=0$$ for no $$x$$ in $$(0,1)$$\n"}, {"ide... | ["A"] | null | <p>$$f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0$$</p>
<p>Also $$\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}... | mcq | jee-main-2024-online-30th-january-morning-shift | 6,059 |
luy6z53a | maths | differentiation | methods-of-differentiation | <p>Let $$f(x)=a x^3+b x^2+c x+41$$ be such that $$f(1)=40, f^{\prime}(1)=2$$ and $$f^{\prime \prime}(1)=4$$. Then $$a^2+b^2+c^2$$ is equal to:</p> | [{"identifier": "A", "content": "54"}, {"identifier": "B", "content": "51"}, {"identifier": "C", "content": "73"}, {"identifier": "D", "content": "62"}] | ["B"] | null | <p>Given the polynomial function:</p>
<p>$$f(x) = ax^3 + bx^2 + cx + 41$$</p>
<p>We are provided the following conditions from the problem:</p>
<p>1. $$f(1) = 40$$</p>
<p>2. $$f^{\prime}(1) = 2$$</p>
<p>3. $$f^{\prime \prime}(1) = 4$$</p>
<p>First, calculate $f(1)$:</p>
<p>$$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = ... | mcq | jee-main-2024-online-9th-april-morning-shift | 6,060 |
aLwt0WnHgsFA2mmz | maths | differentiation | successive-differentiation | If $$f\left( x \right) = {x^n},$$ then the value of
<p>$$f\left( 1 \right) - {{f'\left( 1 \right)} \over {1!}} + {{f''\left( 1 \right)} \over {2!}} - {{f'''\left( 1 \right)} \over {3!}} + ..........{{{{\left( { - 1} \right)}^n}{f^n}\left( 1 \right)} \over {n!}}$$ is</p> | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$${{2^n}}$$ "}, {"identifier": "C", "content": "$${{2^n} - 1}$$ "}, {"identifier": "D", "content": "$$0$$"}] | ["D"] | null | $$f\left( x \right) = {x^n} \Rightarrow f\left( 1 \right) = 1$$
<br><br>$$f'\left( x \right) = n{x^{n - 1}} \Rightarrow f'\left( 1 \right) = n$$
<br><br>$$f''\left( x \right) = n\left( {n - 1} \right){x^{n - 2}}$$
<br><br>$$ \Rightarrow f''\left( 1 \right) = n\left( {n - 1} \right)$$
<br><br>$$\therefore$$ $${f^n}\lef... | mcq | aieee-2003 | 6,061 |
7QKM8xqcNeiQvnGv4PFtQ | maths | differentiation | successive-differentiation | Let f : R $$ \to $$ R be a function such that f(x) = x<sup>3</sup> + x<sup>2</sup>f'(1) + xf''(2) + f'''(3), x $$ \in $$ R. Then f(2) equals - | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "$$-$$ 2"}, {"identifier": "C", "content": "$$-$$ 4"}, {"identifier": "D", "content": "8"}] | ["B"] | null | f(x) = x<sup>3</sup> + x<sup>2</sup>f '(1) + xf ''(2) + f '''(3)
<br><br>$$ \Rightarrow $$ f '(x) = 3x<sup>2</sup> + 2xf '(1) + f ''(x) . . . . . (1)
<br><br>$$ \Rightarrow $$ f ''(x) = 6x + 2f '(1) . . . . . . (2)
<br><br>$$ \Rightarrow $$... | mcq | jee-main-2019-online-10th-january-morning-slot | 6,063 |
xrX2b9GpaeqoQ7utJtjgy2xukf0qq2t5 | maths | differentiation | successive-differentiation | If y<sup>2</sup> + log<sub>e</sub> (cos<sup>2</sup>x) = y, <br/>$$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, then : | [{"identifier": "A", "content": "|y''(0)| = 2"}, {"identifier": "B", "content": "|y'(0)| + |y''(0)| = 3"}, {"identifier": "C", "content": "y''(0) = 0"}, {"identifier": "D", "content": "|y'(0)| + |y\"(0)| = 1"}] | ["A"] | null | Given y<sup>2</sup> + log<sub>e</sub> (cos<sup>2</sup>x) = y .....(1)
<br><br>Put x = 0, we get
<br><br>y<sup>2</sup> + log<sub>e</sub> (1) = y
<br><br>$$ \Rightarrow $$ y<sup>2</sup> = y
<br><br>$$ \Rightarrow $$ y = 0, 1
<br><br>Differentiating (1) we get
<br><br>2yy' + $${1 \over {\cos x}}\left( { - \sin x} \right)... | mcq | jee-main-2020-online-3rd-september-morning-slot | 6,064 |
1l56u56af | maths | differentiation | successive-differentiation | <p>If $$y(x) = {\left( {{x^x}} \right)^x},\,x > 0$$, then $${{{d^2}x} \over {d{y^2}}} + 20$$ at x = 1 is equal to ____________.</p> | [] | null | 16 | <p>$$\because$$ $$y(x) = {\left( {{x^x}} \right)^x}$$</p>
<p>$$\therefore$$ $$y = {x^{{x^2}}}$$</p>
<p>$$\therefore$$ $${{dy} \over {dx}} = {x^2}\,.\,{x^{{x^2} - 1}} + {x^{{x^2}}}\ln x\,.\,2x$$</p>
<p>$$\therefore$$ $${{dx} \over {dy}} = {1 \over {{x^{{x^2} + 1}}(1 + 2\ln x)}}$$ ..... (i)</p>
<p>Now, $${{{d^2}x} \over ... | integer | jee-main-2022-online-27th-june-evening-shift | 6,065 |
1l6klla1m | maths | differentiation | successive-differentiation | <p>For the curve $$C:\left(x^{2}+y^{2}-3\right)+\left(x^{2}-y^{2}-1\right)^{5}=0$$, the value of $$3 y^{\prime}-y^{3} y^{\prime \prime}$$, at the point $$(\alpha, \alpha)$$, $$\alpha>0$$, on C, is equal to ____________.</p> | [] | null | 16 | <p>$$\because$$ $$C:({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$$ for point ($$\alpha$$, $$\alpha$$)</p>
<p>$${\alpha ^2} + {\alpha ^2} - 3 + {({\alpha ^2} - {\alpha ^2} - 1)^5} = 0$$</p>
<p>$$\therefore$$ $$\alpha = \sqrt 2 $$</p>
<p>On differentiating $$({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$$ we ge... | integer | jee-main-2022-online-27th-july-evening-shift | 6,066 |
1ldsfuib3 | maths | differentiation | successive-differentiation | <p>Let $$f$$ and $$g$$ be the twice differentiable functions on $$\mathbb{R}$$ such that</p>
<p>$$f''(x)=g''(x)+6x$$</p>
<p>$$f'(1)=4g'(1)-3=9$$</p>
<p>$$f(2)=3g(2)=12$$.</p>
<p>Then which of the following is NOT true?</p> | [{"identifier": "A", "content": "$$g(-2)-f(-2)=20$$"}, {"identifier": "B", "content": "There exists $$x_0\\in(1,3/2)$$ such that $$f(x_0)=g(x_0)$$"}, {"identifier": "C", "content": "$$|f'(x)-g'(x)| < 6\\Rightarrow -1 < x < 1$$"}, {"identifier": "D", "content": "If $$-1 < x < 2$$, then $$|f(x)-g(x)| < 8$$"}] | ["D"] | null | <p>$$f''(x) = g''(x) + 6x$$</p>
<p>$$ \Rightarrow f'(x) = g'(x) + 3{x^2} + C$$</p>
<p>$$f'(1) = g'(1) + 3 + C$$</p>
<p>$$ \Rightarrow g = 3 + 3 + C \Rightarrow C = 3$$</p>
<p>$$ \Rightarrow f'(x) = g'(x) + 3{x^2} + 3$$</p>
<p>$$ \Rightarrow f(x) = g(x) + {x^2} + 3x + C'$$</p>
<p>$$x = 2$$</p>
<p>$$f(2) = g(2) + 14 + C'... | mcq | jee-main-2023-online-29th-january-evening-shift | 6,069 |
1ldv1t76j | maths | differentiation | successive-differentiation | <p>Let $$y(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})(1 + {x^{16}})$$. Then $$y' - y''$$ at $$x = - 1$$ is equal to</p> | [{"identifier": "A", "content": "496"}, {"identifier": "B", "content": "976"}, {"identifier": "C", "content": "464"}, {"identifier": "D", "content": "944"}] | ["A"] | null | $$
\begin{aligned}
& y=\frac{1-x^{32}}{1-x}=1+x+x^2+x^3+\ldots+x^{31} \\\\
& y^{\prime}=1+2 x+3 x^2+\ldots+31 x^{30} \\\\
& y^{\prime}(-1)=1-2+3-4+\ldots+31=16 \\\\
& y^{\prime \prime}(x)=2+6 x+12 x^2+\ldots+31.30 x^{29} \\\\
& y^{\prime \prime}(-1)=2-6+12 \ldots 31.30=480 \\\\
& y^{\prime \prime}(-1)-y^{\prime}(-1)=-4... | mcq | jee-main-2023-online-25th-january-morning-shift | 6,070 |
1ldwx6r58 | maths | differentiation | successive-differentiation | <p>If $$f(x) = {x^3} - {x^2}f'(1) + xf''(2) - f'''(3),x \in \mathbb{R}$$, then</p> | [{"identifier": "A", "content": "$$2f(0) - f(1) + f(3) = f(2)$$"}, {"identifier": "B", "content": "$$f(1) + f(2) + f(3) = f(0)$$"}, {"identifier": "C", "content": "$$f(3) - f(2) = f(1)$$"}, {"identifier": "D", "content": "$$3f(1) + f(2) = f(3)$$"}] | ["A"] | null | $$
f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R
$$<br/><br/>
Let $\mathrm{f}^{\prime}(1)=\mathrm{a}, \mathrm{f}^{\prime \prime}(2)=\mathrm{b}, \mathrm{f}^{\prime \prime \prime}(3)=\mathrm{c}$<br/><br/>
$$
\begin{aligned}
& f(x)=x^3-a x^2+b x-c \\\\
& f^{\prime}(x)=3 x^2-2 a x+b... | mcq | jee-main-2023-online-24th-january-evening-shift | 6,071 |
lsble0m7 | maths | differentiation | successive-differentiation | Let $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in \mathbf{R}$. Then $f^{\prime}(10)$ is equal to ____________. | [] | null | 202 | <p>$$\begin{aligned}
& f(x)=x^3+x^2 \cdot f^{\prime}(1)+x \cdot f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\
& f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2) \\
& f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \\
& f^{\prime \prime \prime}(x)=6 \\
& f^{\prime}(1)=-5, f^{\prime \prime}(2)=2, f^{\prime \prim... | integer | jee-main-2024-online-27th-january-morning-shift | 6,072 |
1lsg94q54 | maths | differentiation | successive-differentiation | <p>If $$f(x)=\left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\
2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x
\end{array}\right|,$$ then $$\frac{1}{5} f^{\prime}(0)=$$ is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <p>$$\begin{aligned}
& \left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\
2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x
\end{array}\right| \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\
& \left|... | mcq | jee-main-2024-online-30th-january-morning-shift | 6,073 |
lv9s204y | maths | differentiation | successive-differentiation | <p>If $$y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}$$, then at $$\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "2"}] | ["D"] | null | <p>$$\begin{aligned}
& y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2} \\
& =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{4 \cos ^3 \theta+8 \cos ^2 \theta+2 \cos \theta-2} \\
& =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)... | mcq | jee-main-2024-online-5th-april-evening-shift | 6,075 |
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