UFM Pure

View all 217 questions →

jee-main 2020 Q69 View
Let $y = y(x)$ be the solution of the differential equation $\cos x\frac{dy}{dx} + 2y\sin x = \sin 2x$, $x \in \left(0, \frac{\pi}{2}\right)$. If $y(\pi/3) = 0$, then $y(\pi/4)$ is equal to:
(1) $2 - \sqrt{2}$
(2) $2 + \sqrt{2}$
(3) $\sqrt{2} - 2$
(4) $\frac{1}{\sqrt{2}} - 1$
jee-main 2021 Q75 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 $y \left( \frac { \pi } { 4 } \right)$ is equal to
(1) $\frac { 1 } { 4 } \log _ { e } 2$
(2) $\left( \frac { 1 } { 2 \sqrt { 2 } } \right) \log _ { e } 2$
(3) $\log _ { e } 2$
(4) $\frac { 1 } { 2 } \log _ { e } 2$
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) $2 \log _ { \mathrm { e } } \left( \frac { 3 \sqrt { 3 } - 8 } { 4 } \right)$
Which of the following is true for $y ( x )$ that satisfies the differential equation $\frac { d y } { d x } = x y - 1 + x - y ; y ( 0 ) = 0$
(1) $y ( 1 ) = \mathrm { e } ^ { - \frac { 1 } { 2 } } - 1$
(2) $y ( 1 ) = e ^ { \frac { 1 } { 2 } } - e ^ { - \frac { 1 } { 2 } }$
(3) $y ( 1 ) = 1$
(4) $y ( 1 ) = e^{\frac{1}{2}} - 1$
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) $\left( \frac { 31 } { 3 } - \frac { 8 } { 3 } \log _ { e } 3 \right)$
If $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ^ { x + y } - 2 ^ { x } } { 2 ^ { y } } , y ( 0 ) = 1$, then $y ( 1 )$ is equal to :
(1) $\log _ { 2 } \left( 1 + \mathrm { e } ^ { 2 } \right)$
(2) $\log _ { 2 } ( 2 \mathrm { e } )$
(3) $\log _ { 2 } ( 2 + e )$
(4) $\log _ { 2 } ( 1 + e )$
Let $f ( x ) = \int _ { 0 } ^ { x } e ^ { t } f ( t ) d t + e ^ { x }$ be a differentiable function for all $x \in R$. Then $f ( x )$ equals:
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
Let a curve $y = y ( x )$ be given by the solution of the differential equation $\cos \left( \frac { 1 } { 2 } \cos ^ { - 1 } \left( e ^ { - x } \right) \right) dx = \left( \sqrt { e ^ { 2 x } - 1 } \right) dy$. If it intersects $y$-axis at $y = - 1$, and the intersection point of the curve with $x$-axis is $( \alpha , 0 )$, then $e ^ { \alpha }$ is equal to $\underline{\hspace{1cm}}$.
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) $2\tan^{-1}\left(\frac{1}{k}\right) = \log_e\frac{k^2 + 1}{k^2}$
Let $y = y ( x )$ be the solution curve of the differential equation $\frac { d y } { d x } + \frac { 1 } { x ^ { 2 } - 1 } y = \left( \frac { x - 1 } { x + 1 } \right) ^ { \frac { 1 } { 2 } } , x > 1$ passing through the point $\left( 2 , \sqrt { \frac { 1 } { 3 } } \right)$. Then $\sqrt { 7 } y ( 8 )$ is equal to
(1) $11 + 6 \log _ { e } 3$
(2) $19$
(3) $12 - 2 \log _ { e } 3$
(4) $19 - 6 \log _ { e } 3$
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) $\frac { 3 } { 8 }$
Let $y = y _ { 1 } ( x )$ and $y = y _ { 2 } ( x )$ be two distinct solutions of the differential equation $\frac { d y } { d x } = x + y$, with $y _ { 1 } ( 0 ) = 0$ and $y _ { 2 } ( 0 ) = 1$ respectively. Then, the number of points of intersection of $y = y _ { 1 } ( x )$ and $y = y _ { 2 } ( x )$ is
(1) 0
(2) 1
(3) 2
(4) 3
If $\frac { d y } { d x } + e ^ { x } \left( x ^ { 2 } - 2 \right) y = \left( x ^ { 2 } - 2 x \right) \left( x ^ { 2 } - 2 \right) e ^ { 2 x }$ and $y ( 0 ) = 0$, then the value of $y ( 2 )$ is
(1) $-1$
(2) 1
(3) 0
(4) $e$
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
If $\sin\left(\frac{y}{x}\right) = \log_e|x| + \frac{\alpha}{2}$ is the solution of the differential equation $x\cos\left(\frac{y}{x}\right)\frac{dy}{dx} = y\cos\left(\frac{y}{x}\right) + x$ and $y(1) = \frac{\pi}{3}$, then $\alpha^2$ is equal to
(1) 12
(2) 3
(3) 4
(4) 9
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 $-1$
Let $y = y ( x ) , y > 0$, be a solution curve of the differential equation $\left( 1 + x ^ { 2 } \right) d y = y ( x - y ) d x$. If $y ( 0 ) = 1$ and $y ( 2 \sqrt { 2 } ) = \beta$, then
(1) $e ^ { 3 \beta - 1 } = e ( 3 + 2 \sqrt { 2 } )$
(2) $e ^ { 3 \beta - 1 } = e ( 5 + \sqrt { 2 } )$
(3) $e ^ { \beta - 1 } = e ^ { - 2 } ( 3 + 2 \sqrt { 2 } )$
(4) $e ^ { \beta - 1 } = e ^ { - 2 } ( 5 + \sqrt { 2 } )$
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) $\frac { x ^ { 2 } } { 7 - 3 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 2 } \right) }$
Let $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$ be the solution of the differential equation $y ( x + 1 ) d x - x ^ { 2 } d y = 0 , y ( 1 ) = e$. Then $\lim _ { x \rightarrow 0 ^ { + } } f ( x )$ is equal to
(1) 0
(2) $\frac { 1 } { e }$
(3) $e ^ { 2 }$
(4) $\frac { 1 } { e ^ { 2 } }$