JEE Advanced
2026
MCQ
Let $y : (-\infty, \infty) \to (0, \infty)$ be the solution of the differential equation
$\frac{dy}{dx} = \frac{e^{5x} y^3 + y^3}{e^x + e^x y^4},$
satisfying $y(0) = \frac{1}{\sqrt{2}}$. Then the value of $y(\log_e 2)$ is
JEE Advanced
2026
MSQ
Let $y = f(x)$ be the real valued function defined on the interval $(0, \infty)$, satisfying $y(1) = 0$ and the differential equation
$ x \frac{dy}{dx} = y - x^3. $
Then which of the following statements is (are) TRUE?
JEE Advanced
2024
MCQ
Let $f(x)$ be a continuously differentiable function on the interval $(0, \infty)$ such that $f(1)=2$ and
$ \lim\limits_{t \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1 $
for each $x>0$. Then, for all $x>0, f(x)$ is equal to :
JEE Advanced
2023
MCQ
Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f(1)=\frac{1}{3}$ and $3 \int\limits_1^x f(t) d t=x f(x)-\frac{x^3}{3}, x \in[1, \infty)$. Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is :
JEE Advanced
2022
MCQ
For $x \in \mathbb{R}$, let the function $y(x)$ be the solution of the differential equation
$
\frac{d y}{d x}+12 y=\cos \left(\frac{\pi}{12} x\right), \quad y(0)=0
$
Then, which of the following statements is/are TRUE ?
JEE Advanced
2019
MSQ
Let $\Gamma $ denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to I` at a point P intersect the y-axis at YP. If PYP has length 1 for each point P on I`, then which of the following options is/are correct?
JEE Advanced
2017
MCQ
If y = y(x) satisfies the differential equation
${8\sqrt x \left( {\sqrt {9 + \sqrt x } } \right)dy = {{\left( {\sqrt {4 + \sqrt {9 + \sqrt x } } } \right)}^{ - 1}}}$
dx, x > 0 and y(0) = $\sqrt 7 $, then y(256) =
JEE Advanced
2017
MSQ
If $g(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$, then
JEE Advanced
2016
MSQ
A solution curve of the differential equation
$\left( {{x^2} + xy + 4x + 2y + 4} \right){{dy} \over {dx}} - {y^2} = 0,$ $x>0,$ passes through the
point $(1,3)$. Then the solution curve
JEE Advanced
2016
MCQ
Let $f:(0,\infty ) \to R$ be a differentiable function such that $f'(x) = 2 - {{f(x)} \over x}$ for all $x \in (0,\infty )$ and $f(1) \ne 1$. Then
JEE Advanced
2015
MSQ
Let $y(x)$ be a solution of the differential equation
$\left( {1 + {e^x}} \right)y' + y{e^x} = 1.$
If $y(0)=2$, then which of the following statement is (are) true?
JEE Advanced
2015
MSQ
Consider the family of all circles whose centres lie on the straight line $y=x,$ If this family of circle is represented by the differential equation $Py'' + Qy' + 1 = 0,$ where $P, Q$ are functions of $x,y$ and $y'$ $\left( {here\,\,\,y' = {{dy} \over {dx}},y'' = {{{d^2}y} \over {d{x^2}}}} \right)$ then which of the following statements is (are) true?
JEE Advanced
2014
MCQ
The function $y=f(x)$ is the solution of the differential equation
${{dy} \over {dx}} + {{xy} \over {{x^2} - 1}} = {{{x^4} + 2x} \over {\sqrt {1 - {x^2}} }}\,$ in $(-1,1)$ satisfying $f(0)=0$.
Then $\int\limits_{ - {{\sqrt 3 } \over 2}}^{{{\sqrt 3 } \over 2}} {f\left( x \right)} \,d\left( x \right)$ is
JEE Advanced
2013
MCQ
A curve passes through the point $\left( {1,{\pi \over 6}} \right)$. Let the slope of
the curve at each point $(x,y)$ be ${y \over x} + \sec \left( {{y \over x}} \right),x > 0.$
Then the equation of the curve is
JEE Advanced
2012
MSQ
If $y(x)$ satisfies the differential equation $y' - y\,tan\,x = 2x\,secx$ and $y(0)=0,$ then
JEE Advanced
2009
MCQ
Match the statements/expressions in Column I with the values given in Column II:
|
Column I |
|
Column II |
| (A) |
The number of solutions of the equation $x{e^{\sin x}} - \cos x = 0$ in the interval $\left( {0,{\pi \over 2}} \right)$ |
(P) |
1 |
| (B) |
Value(s) of $k$ for which the planes $kx + 4y + z = 0,4x + ky + 2z = 0$ and $2x + 2y + z = 0$ intersect in a straight line |
(Q) |
2 |
| (C) |
Value(s) of $k$ for which $|x - 1| + |x - 2| + |x + 1| + |x + 2| = 4k$ has integer solution(s) |
(R) |
3 |
| (D) |
If $y' = y + 1$ and $y(0) = 1$ then value(s) of $y(\ln 2)$ |
(S) |
4 |
|
|
(T) |
5 |
JEE Advanced
2009
MCQ
Match the statements/expressions in Column I with the open intervals in Column II :
|
Column I |
|
Column II |
| (A) |
Interval contained in the domain of definition of non-zero solutions of the differential equation ${(x - 3)^2}y' + y = 0$ |
(P) |
$\left( { - {\pi \over 2},{\pi \over 2}} \right)$ |
| (B) |
Interval containing the value of the integral $\int\limits_1^5 {(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)dx} $ |
(Q) |
$\left( {0,{\pi \over 2}} \right)$ |
| (C) |
Interval in which at least one of the points of local maximum of ${\cos ^2}x + \sin x$ lies |
(R) |
$\left( {{\pi \over 8},{{5\pi } \over 4}} \right)$ |
| (D) |
Interval in which ${\tan ^{ - 1}}(\sin x + \cos x)$ is increasing |
(S) |
$\left( {0,{\pi \over 8}} \right)$ |
|
|
(T) |
$( - \pi ,\pi )$ |
JEE Advanced
2008
MCQ
Let a solution $y=y(x)$ of the differential equation,
$x\sqrt {{x^2} - 1} \,\,dy - y\sqrt {{y^2} - 1} \,dx = 0$ satify $y\left( 2 \right) = {2 \over {\sqrt 3 }}.$
STATEMENT-1 : $y\left( x \right) = \sec \left( {{{\sec }^{ - 1}}x - {\pi \over 6}} \right)$ and
STATEMENT-2 : $y\left( x \right)$ given by ${1 \over y} = {{2\sqrt 3 } \over x} - \sqrt {1 - {1 \over {{x^2}}}} $
JEE Advanced
2007
MCQ
The differential equation $\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{y}$ determines a family of circles with :
JEE Advanced
2005
MCQ
The differential equation ${{dy} \over {dx}} = {{\sqrt {1 - {y^2}} } \over y}$ determines a family of circles with
JEE Advanced
2005
MCQ
For the primitive integral equation $ydx + {y^2}dy = x\,dy;$
$x \in R,\,\,y > 0,y = y\left( x \right),\,y\left( 1 \right) = 1,$ then $y(-3)$ is
JEE Advanced
2005
MCQ
The solution of primitive integral equation $\left( {{x^2} + {y^2}} \right)dy = xy$
$dx$ is $y=y(x),$ If $y(1)=1$ and $\left( {{x_0}} \right) = e$, then ${{x_0}}$ is equal to
JEE Advanced
2005
MCQ
If $y=y(x)$ and it follows the relation $x\cos \,y + y\,cos\,x = \pi $ then $y''(0)=$
JEE Advanced
2004
MCQ
If $y=y(x)$ and ${{2 + \sin x} \over {y + 1}}\left( {{{dy} \over {dx}}} \right) = - \cos x,y\left( 0 \right) = 1,$
then $y\left( {{\pi \over 2}} \right)$ equals
JEE Advanced
2003
MCQ
If $y(t)$ is a solution of $\left( {1 + t} \right){{dy} \over {dt}} - ty = 1$ and $y\left( 0 \right) = - 1,$ then $y(1)$ is equal to
JEE Advanced
2000
MCQ
If ${x^2} + {y^2} = 1,$ then
JEE Advanced
1999
MCQ
A solution of the differential equation
${\left( {{{dy} \over {dx}}} \right)^2} - x{{dy} \over {dx}} + y = 0$ is
JEE Advanced
1999
MSQ
The differential equation representing the family of curves
${y^2} = 2c\left( {x + \sqrt c } \right),$ where $c$ is a positive parameter, is of
JEE Advanced
1998
MCQ
The order of the differential equation whose general solution is given by
$y = \left( {{C_1} + {C_2}} \right)\cos \left( {x + {C_3}} \right) - {C_4}{e^{x + {C_5}}},$ where
${C_1},{C_2},{C_3},{C_4},{C_5},$ are arbitrary constants, is