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jee-main 2021 Q75 View
If $y = y ( x )$ is the solution of the differential equation, $\frac { d y } { d x } + 2 y \tan x = \sin x , y \left( \frac { \pi } { 3 } \right) = 0$, then the maximum value of the function $y ( x )$ over $R$ is equal to :
(1) 8
(2) $\frac { 1 } { 2 }$
(3) $- \frac { 15 } { 4 }$
(4) $\frac { 1 } { 8 }$
jee-main 2021 Q76 View
Let $y = y ( x )$ be the solution of the differential equation $\cos x ( 3 \sin x + \cos x + 3 ) d y = ( 1 + y \sin x ( 3 \sin x + \cos x + 3 ) ) d x , 0 \leq x \leq \frac { \pi } { 2 } , y ( 0 ) = 0$. Then, $y \left( \frac { \pi } { 3 } \right)$ is equal to:
(1) $2 \log _ { \mathrm { e } } \left( \frac { 2 \sqrt { 3 } + 9 } { 6 } \right)$
(2) $2 \log _ { \mathrm { e } } \left( \frac { 2 \sqrt { 3 } + 10 } { 11 } \right)$
(3) $2 \log _ { \mathrm { e } } \left( \frac { \sqrt { 3 } + 7 } { 2 } \right)$
(4) $2 \log _ { \mathrm { e } } \left( \frac { 3 \sqrt { 3 } - 8 } { 4 } \right)$
jee-main 2021 Q76 View
Let $y = y ( x )$ satisfies the equation $\frac { d y } { d x } - | A | = 0$, for all $x > 0$, where $A = \left[ \begin{array} { c c c } y & \sin x & 1 \\ 0 & - 1 & 1 \\ 2 & 0 & \frac { 1 } { x } \end{array} \right]$. If $y ( \pi ) = \pi + 2$, then the value of $y \left( \frac { \pi } { 2 } \right)$ is:
(1) $\frac { \pi } { 2 } + \frac { 4 } { \pi }$
(2) $\frac { \pi } { 2 } - \frac { 1 } { \pi }$
(3) $\frac { 3 \pi } { 2 } - \frac { 1 } { \pi }$
(4) $\frac { \pi } { 2 } - \frac { 4 } { \pi }$
jee-main 2021 Q77 View
If the curve $y = y ( x )$ is the solution of the differential equation $2 \left( x ^ { 2 } + x ^ { 5 / 4 } \right) d y - y \left( x + x ^ { 1 / 4 } \right) d x = 2 x ^ { 9 / 4 } d x , x > 0$ which passes through the point $\left( 1,1 - \frac { 4 } { 3 } \log _ { \mathrm { e } } 2 \right)$, then the value of $y ( 16 )$ is equal to
(1) $4 \left( \frac { 31 } { 3 } + \frac { 8 } { 3 } \log _ { e } 3 \right)$
(2) $\left( \frac { 31 } { 3 } + \frac { 8 } { 3 } \log _ { e } 3 \right)$
(3) $4 \left( \frac { 31 } { 3 } - \frac { 8 } { 3 } \log _ { \mathrm { e } } 3 \right)$
(4) $\left( \frac { 31 } { 3 } - \frac { 8 } { 3 } \log _ { e } 3 \right)$
jee-main 2021 Q82 View
If $y = y ( x )$ is the solution of the differential equation $\frac { d y } { d x } + ( \tan x ) y = \sin x , 0 \leq x \leq \frac { \pi } { 3 }$, with $y ( 0 ) = 0$, then $y \left( \frac { \pi } { 4 } \right)$ equal to
jee-main 2022 Q75 View
If the solution curve of the differential equation $\frac{dy}{dx} = \frac{x + y - 2}{x - y}$ passes through the point $(2, 1)$ and $(k + 1, 2)$, $k > 0$, then
(1) $2\tan^{-1}\left(\frac{1}{k}\right) = \log_e(k^2 + 1)$
(2) $\tan^{-1}\left(\frac{1}{k}\right) = \log_e(k^2 + 1)$
(3) $2\tan^{-1}\left(\frac{1}{k+1}\right) = \log_e(k^2 + 2k + 2)$
(4) $2\tan^{-1}\left(\frac{1}{k}\right) = \log_e\frac{k^2 + 1}{k^2}$
jee-main 2022 Q76 View
Let $y = y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} + \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6} y = \frac{x + 3}{x + 1}$, $x > -1$, which passes through the point $(0, 1)$. Then $y(1)$ is equal to
(1) $\frac{1}{2}$
(2) $\frac{3}{2}$
(3) $\frac{5}{2}$
(4) $\frac{7}{2}$
jee-main 2022 Q76 View
If the solution curve of the differential equation $\left( \left( \tan ^ { - 1 } y \right) - x \right) d y = \left( 1 + y ^ { 2 } \right) d x$ passes through the point $( 1,0 )$ then the abscissa of the point on the curve whose ordinate is $\tan ( 1 )$ is
(1) 2
(2) $\frac { 2 } { e }$
(3) $\frac { 3 } { e }$
(4) $2 e$
jee-main 2022 Q76 View
If $\frac { d y } { d x } + 2 y \tan x = \sin x , 0 < x < \frac { \pi } { 2 }$ and $y \left( \frac { \pi } { 3 } \right) = 0$, then the maximum value of $y ( x )$ is
(1) $\frac { 1 } { 8 }$
(2) $\frac { 3 } { 4 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 3 } { 8 }$
jee-main 2022 Q78 View
Let the solution curve $y = f(x)$ of the differential equation $\frac { d y } { d x } + \frac { x y } { x ^ { 2 } - 1 } = \frac { x ^ { 4 } + 2 x } { \sqrt { 1 - x ^ { 2 } } }$, $x \in ( - 1, 1 )$ pass through the origin. Then $\int _ { - \frac { \sqrt { 3 } } { 2 } } ^ { \frac { \sqrt { 3 } } { 2 } } f(x) \, d x$ is equal to
jee-main 2022 Q89 View
Let $y = y ( x )$ be the solution of the differential equation $\left( 1 - x ^ { 2 } \right) d y = \left( x y + \left( x ^ { 3 } + 2 \right) \sqrt { 1 - x ^ { 2 } } \right) d x , - 1 < x < 1$ and $y ( 0 ) = 0$. If $\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \sqrt { 1 - x ^ { 2 } } y ( x ) d x = k$ then $k ^ { - 1 }$ is equal to
jee-main 2022 Q89 View
Let $y = y ( x )$ be the solution curve of the differential equation $\sin \left( 2 x ^ { 2 } \right) \log _ { e } \left( \tan x ^ { 2 } \right) d y + \left( 4 x y - 4 \sqrt { 2 } x \sin \left( x ^ { 2 } - \frac { \pi } { 4 } \right) \right) d x = 0,0 < x < \sqrt { \frac { \pi } { 2 } }$, which passes through the point $\left( \sqrt { \frac { \pi } { 6 } } , 1 \right)$. Then $\left| y \left( \sqrt { \frac { \pi } { 3 } } \right) \right|$ is equal to $\_\_\_\_$ .
jee-main 2023 Q75 View
Let $x = x ( y )$ be the solution of the differential equation $2 ( y + 2 ) \log _ { e } ( y + 2 ) d x + \left( x + 4 - 2 \log _ { e } ( y + 2 ) \right) d y = 0 , y > - 1$ with $x \left( e ^ { 4 } - 2 \right) = 1$. Then $x \left( e ^ { 9 } - 2 \right)$ is equal to
(1) 3
(2) $\frac { 4 } { 9 }$
(3) $\frac { 32 } { 9 }$
(4) $\frac { 10 } { 3 }$
jee-main 2023 Q77 View
If $y = y(x)$ is the solution curve of the differential equation $\frac{dy}{dx} + y\tan x = x\sec x$, $0 \leq x \leq \frac{\pi}{3}$, $y(0) = 1$, then $y\left(\frac{\pi}{6}\right)$ is equal to
(1) $\frac{\pi}{12} - \frac{\sqrt{3}}{2}\log_e\frac{2}{e\sqrt{3}}$
(2) $\frac{\pi}{12} + \frac{\sqrt{3}}{2}\log_e\frac{2\sqrt{3}}{e}$
(3) $\frac{\pi}{12} - \frac{\sqrt{3}}{2}\log_e\frac{2\sqrt{3}}{e}$
(4) $\frac{\pi}{12} + \frac{\sqrt{3}}{2}\log_e\frac{2}{e\sqrt{3}}$
jee-main 2023 Q77 View
Let a differentiable function $f$ satisfy $f(x) + \int_3^x \frac{f(t)}{t}\,dt = \sqrt{x+1}$, $x \geq 3$. Then $12f(8)$ is equal to:
(1) 34
(2) 19
(3) 17
(4) 1
jee-main 2023 Q83 View
Let $y = y ( t )$ be a solution of the differential equation $\frac { d y } { d t } + \alpha y = \gamma e ^ { - \beta t }$ where $\alpha > 0 , \beta > 0$ and $\gamma > 0$. Then $\lim _ { t \rightarrow \infty } y ( t )$
(1) is 0
(2) does not exist
(3) is 1
(4) is $-1$
jee-main 2023 Q84 View
Let $\mathrm { y } = \mathrm { y } ( \mathrm { x } )$ be the solution curve of the differential equation $\frac { d y } { d x } = \frac { y } { x } \left( 1 + x ^ { 2 } \left( 1 + \log _ { e } x \right) \right) , \mathrm { x } > 0 , \mathrm { y } ( 1 ) = 3$. Then $\frac { \mathrm { y } ^ { 2 } ( \mathrm { x } ) } { 9 }$ is equal to:
(1) $\frac { x ^ { 2 } } { 5 - 2 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 3 } \right) }$
(2) $\frac { x ^ { 2 } } { 2 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 3 } \right) - 3 }$
(3) $\frac { x ^ { 2 } } { 3 x ^ { 3 } \left( 1 + \log _ { e } x ^ { 2 } \right) - 2 }$
(4) $\frac { x ^ { 2 } } { 7 - 3 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 2 } \right) }$
jee-main 2023 Q84 View
Let $y = y ( x )$ be the solution of the differential equation $x \log _ { e } x \frac { d y } { d x } + y = x ^ { 2 } \log _ { e } x , ( x > 1 )$. If $y ( 2 ) = 2$, then $y ( e )$ is equal to (1) $\frac { 4 + e ^ { 2 } } { 4 }$ (2) $\frac { 1 + e ^ { 2 } } { 4 }$ (3) $\frac { 2 + e ^ { 2 } } { 2 }$ (4) $\frac { 1 + e ^ { 2 } } { 2 }$
jee-main 2023 Q88 View
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f'(x) + f(x) = \int_0^2 f(t)\, dt$. If $f(0) = e^{-2}$, then $2f(0) - f(2)$ is equal to $\_\_\_\_$.
jee-main 2024 Q76 View
Let $y = y(x)$ be the solution of the differential equation $\frac { d y } { d x } = \frac { \tan x + y } { \sin x \sec x - \sin x \tan x } , \quad x \in \left(0, \frac { \pi } { 2 }\right)$ satisfying the condition $y\left(\frac { \pi } { 4 }\right) = 2$. Then, $y\left(\frac { \pi } { 3 }\right)$ is
(1) $\sqrt { 32 } + \log _ { e } \sqrt { 3 }$
(2) $\frac { \sqrt { 3 } } { 2} \left(2 + \log _ { e } 3\right)$
(3) $\sqrt { 3 } \left(1 + 2 \log _ { e } 3\right)$
(4) $\sqrt { 3 } \left(2 + \log _ { e } 3\right)$
jee-main 2024 Q76 View
Let $y = y ( x )$ be the solution of the differential equation $\left( 1 + y ^ { 2 } \right) e ^ { \tan x } d x + \cos ^ { 2 } x \left( 1 + e ^ { 2 \tan x } \right) d y = 0 , y ( 0 ) = 1$. Then $y \left( \frac { \pi } { 4 } \right)$ is equal to
(1) $\frac { 2 } { e }$
(2) $\frac { 2 } { e ^ { 2 } }$
(3) $\frac { 1 } { e }$
(4) $\frac { 1 } { e ^ { 2 } }$
jee-main 2024 Q76 View
Let $y = y ( x )$ be the solution curve of the differential equation $\sec y \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 x \sin y = x ^ { 3 } \cos y , y ( 1 ) = 0$. Then $y ( \sqrt { 3 } )$ is equal to : (1) $\frac { \pi } { 3 }$ (2) $\frac { \pi } { 6 }$ (3) $\frac { \pi } { 12 }$ (4) $\frac { \pi } { 4 }$
jee-main 2024 Q77 View
Let $y = y ( x )$ be the solution of the differential equation $\left( 1 + x ^ { 2 } \right) \frac { d y } { d x } + y = e ^ { \tan ^ { - 1 } x } , y ( 1 ) = 0$. Then $y ( 0 )$ is
(1) $\frac { 1 } { 2 } \left( e ^ { \pi / 2 } - 1 \right)$
(2) $\frac { 1 } { 2 } \left( 1 - e ^ { \pi / 2 } \right)$
(3) $\frac { 1 } { 4 } \left( 1 - e ^ { \pi / 2 } \right)$
(4) $\frac { 1 } { 4 } \left( e ^ { \pi / 2 } - 1 \right)$
jee-main 2024 Q77 View
Let $x = x ( t )$ and $y = y ( t )$ be solutions of the differential equations $\frac { \mathrm { dx } } { \mathrm { dt } } + \mathrm { ax } = 0$ and $\frac { \mathrm { dy } } { \mathrm { dt } } + $ by $= 0$ respectively, $\mathrm { a } , \mathrm { b } \in \mathrm { R }$. Given that $x ( 0 ) = 2 ; y ( 0 ) = 1$ and $3 y ( 1 ) = 2 x ( 1 )$, the value of $t$, for which $x ( t ) = y ( t )$, is:
(1) $\log _ { \frac { 2 } { 3 } } 2$
(2) $\log _ { 4 } 3$
(3) $\log _ { 3 } 4$
(4) $\log _ { \frac { 4 } { 3 } } 2$
jee-main 2024 Q78 View
Let $y = y ( x )$ be the solution of the differential equation $\left( 2 x \log _ { e } x \right) \frac { d y } { d x } + 2 y = \frac { 3 } { x } \log _ { e } x , x > 0$ and $y \left( e ^ { - 1 } \right) = 0$. Then, $y ( e )$ is equal to
(1) $- \frac { 3 } { \mathrm { e } }$
(2) $- \frac { 3 } { 2 e }$
(3) $- \frac { 2 } { 3 e }$
(4) $- \frac { 2 } { \mathrm { e } }$

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