2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2019 (Online) 11th January Evening Slot
The solution of the differential equation,
2 , when y(1) = 1, is :
A.
$-$ loge $\left| {{{1 + x - y} \over {1 - x + y}}} \right|$ = x + y $-$ 2
B.
loge $\left| {{{2 - x} \over {2 - y}}} \right|$ = x $-$ y
C.
loge $\left| {{{2 - y} \over {2 - x}}} \right|$ = 2(y $-$ 1)
D.
$-$ loge $\left| {{{1 - x + y} \over {1 + x - y}}} \right|$ = 2(x $-$ 1)
Show Answer
Correct Answer: D
Explanation:
x $-$ y = t 2 $ \Rightarrow $ $\int {{{dt} \over {1 - {t^2}}}} $ = $\int {1dx} $
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2019 (Online) 11th January Morning Slot
If y(x) is the solution of the differential equation ${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,$ where $y\left( 1 \right) = {1 \over 2}{e^{ - 2}},$ then
B.
y(x) is decreasing in (0, 1)
C.
y(loge 2) = ${{{{\log }_e}2} \over 4}$
D.
y(x) is decreasing in $\left( {{1 \over 2},1} \right)$
Show Answer
Correct Answer: D
Explanation:
${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}}$
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2019 (Online) 10th January Evening Slot
Let f be a differentiable function such that f '(x) = 7 - ${3 \over 4}{{f\left( x \right)} \over x},$ (x > 0) and f(1) $ \ne $ 4. Then $\mathop {\lim }\limits_{x \to 0'} \,$ xf$\left( {{1 \over x}} \right)$ :
B.
exists and equals ${4 \over 7}$
Show Answer
Correct Answer: C
Explanation:
$f'(x) = 7 - {3 \over 4}{{f\left( x \right)} \over x}\,\,\,\left( {x > 0} \right)$
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2019 (Online) 10th January Evening Slot
The curve amongst the family of curves represented by the differential equation, (x2 – y2 )dx + 2xy dy = 0 which passes through (1, 1) is :
A.
a circle with centre on the y-axis
B.
an ellipse with major axis along the y-axis
C.
a circle with centre on the x-axis
D.
a hyperbola with transverse axis along the x-axis
Show Answer
Correct Answer: C
Explanation:
(x2 $-$ y2 ) dx + 2xy dy = 0
2 + 1) = $-$ ln x + C
2 + x2 ) = Cx
2 + x2 = 2x
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2019 (Online) 10th January Morning Slot
If ${{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)$ and $y\left( {{\pi \over 4}} \right) = {4 \over 3},$ then $y\left( { - {\pi \over 4}} \right)$ equals -
Show Answer
Correct Answer: A
Explanation:
${{dy} \over {dx}} + 3{\sec ^2}x.y = {\sec ^2}x$
3
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2019 (Online) 9th January Evening Slot
Let f : [0,1] $ \to $ R be such that f(xy) = f(x).f(y), for all x, y $ \in $ [0, 1], and f(0) $ \ne $ 0. If y = y(x) satiesfies the differential equation, ${{dy} \over {dx}}$ = f(x) with y(0) = 1, then y$\left( {{1 \over 4}} \right)$ + y$\left( {{3 \over 4}} \right)$ is equal to :
Show Answer
Correct Answer: A
Explanation:
If f(xy) = f(x) f(y) $\forall $ x, y $ \in $ R and f(0) $ \ne $ 0
2
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2019 (Online) 9th January Morning Slot
If y = y(x) is the solution of the differential equation,
2 , satisfying y(1) = 1, then y($1\over2$) is equal
to :
Show Answer
Correct Answer: B
Explanation:
Given,
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2018 (Online) 16th April Morning Slot
The differential equation representing the family of ellipse having foci eith on the x-axis or on the $y$-axis, center at the origin and passing through the point (0, 3) is :
A.
xy y'' + x (y')2 $-$ y y' = 0
Show Answer
Correct Answer: D
Explanation:
Equation of ellipse,
2 = 9
2 = 9
2 + 9 = 0
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2018 (Offline)
Let y = y(x) be the solution of the differential equation
A.
$ - {4 \over 9}{\pi ^2}$
B.
${4 \over {9\sqrt 3 }}{\pi ^2}$
C.
$ - {8 \over {9\sqrt 3 }}{\pi ^2}$
D.
$ - {8 \over 9}{\pi ^2}$
Show Answer
Correct Answer: D
Explanation:
Given,
2 + c . . . . . (1)
2 + c
2 $-$ ${{{\pi ^2}} \over 2}\,\,.....(2)$
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2018 (Online) 15th April Evening Slot
The curve satifying the differeial equation, (x2 $-$ y2 ) dx + 2xydy = 0 and passing through the point (1, 1) is :
A.
a circle of radius one.
D.
a circle of radius two.
Show Answer
Correct Answer: A
Explanation:
(x2 $-$ y2 ) dx + 2xydy = 0
2 + y2 $-$ 2x = 0, which is a circle of radius one.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2018 (Online) 15th April Morning Slot
Let y = y(x) be the solution of the differential equation ${{dy} \over {dx}} + 2y = f\left( x \right),$
A.
${{{e^2} + 1} \over {2{e^4}}}$
C.
${{{e^2} - 1} \over {{e^3}}}$
D.
${{{e^2} - 1} \over {2{e^3}}}$
Show Answer
Correct Answer: D
Explanation:
When x $ \in $ [0, 1], then ${{dy} \over {dx}}$ + 2y = 1
1 e$-$2x
$-$2x
$-$2 = ${{{e^2} - 1} \over {2{e^2}}}$
2 e$-$2x
2 e$-$2 $ \Rightarrow $ C2 = ${{{e^2} - 1} \over 2}$
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2017 (Online) 9th April Morning Slot
If 2x = y${^{{1 \over 5}}}$ + y${^{ - {1 \over 5}}}$ and
2 $-$ 1) ${{{d^2}y} \over {d{x^2}}}$ + $\lambda $x ${{dy} \over {dx}}$ + ky = 0,
Show Answer
Correct Answer: B
Explanation:
It is given that
$2x = {y^{1/5}} + {y^{ - 1/5}}$
$ \Rightarrow 2x = {y^{1/5}} + 1/{y^{1/5}}$
Therefore, $2x = a + {1 \over a} \Rightarrow {a^2} - 2ax + 1 = 0$
$a = {{2x \pm \sqrt {4{x^2} - 4} } \over 2}$
$ \Rightarrow a = {{2x \pm 2\sqrt {{x^2} - 1} } \over 2}$
$ \Rightarrow a = x \pm \sqrt {{x^2} - 1} $
$ \Rightarrow {y^{1/5}} = x \pm \sqrt {{x^2} - 1} $
$ \Rightarrow y = {(x \pm \sqrt {{x^2} - 1} )^5}$
Therefore,
${{dy} \over {dx}} = 5{(x \pm \sqrt {{x^2} - 1} )^4}\left( {1 \pm {{2x} \over {2\sqrt {{x^2} - 1} }}} \right)$
$ = 5{(x + \sqrt {{x^2} - 1} )^4}\left( {{{\sqrt {{x^2} - 1} \pm x} \over {\sqrt {{x^2} - 1} }}} \right)$
$ \Rightarrow {{dy} \over {dx}} = {{ - 5y} \over {\sqrt {{x^2} - 1} }}$ ...... (1)
$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {{\left[ {\sqrt {{x^2} - 1} \left( { - 5{{dy} \over {dx}}} \right) - 5( - 5y){1 \over 2}{{2x} \over {\sqrt {{x^2} - 1} }}} \right]} \over {({x^2} - 1)}}$
Therefore, $({x^2} - 1){{{d^2}y} \over {d{x^2}}} = - 5\sqrt {{x^2} - 1} {{dy} \over {dx}} + 5y{x \over {\sqrt {{x^2} - 1} }}$
$ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 25y - x{{dy} \over {dx}}$
$ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} + 1x{{dy} \over {dx}} - 25y = 0$
Therefore, $\lambda$ = 1, k = $-$25; hence,
$\lambda + k = - 24$
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2017 (Online) 8th April Morning Slot
The curve satisfying the differential equation, ydx $-$(x + 3y2 )dy = 0 and passing through the point (1, 1), also passes through the point :
A.
$\left( {{1 \over 4}, - {1 \over 2}} \right)$
B.
$\left( { - {1 \over 3},{1 \over 3}} \right)$
C.
$\left( {{1 \over 3}, - {1 \over 3}} \right)$
D.
$\left( {{1 \over 4}, {1 \over 2}} \right)$
Show Answer
Correct Answer: B
Explanation:
Given,
2
2 $-$ 2y
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2017 (Offline)
If $\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$ and y(0) = 1,
Show Answer
Correct Answer: D
Explanation:
$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$
2016
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2016 (Online) 10th April Morning Slot
The solution of the differential equation
A.
y = 1 $-$ ${x \over {\sec x + \tan x}}$
B.
y2 = 1 + ${x \over {\sec x + \tan x}}$
C.
y2 = 1 $-$ ${x \over {\sec x + \tan x}}$
D.
y = 1 + ${x \over {\sec x + \tan x}}$
Show Answer
Correct Answer: C
Explanation:
Given,
2 $=$ t
2 $=$ 1 $-$ ${x \over {\sec x + \tan x}} + c$
2 $=$ 1 $-$ ${x \over {\sec x + \tan x}}$
2016
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2016 (Online) 9th April Morning Slot
If f(x) is a differentiable function in the interval (0, $\infty $) such that f (1) = 1 and
Show Answer
Correct Answer: D
Explanation:
$\mathop {\lim }\limits_{t \to x} {{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1$
2 f '(x) = 1
2 + ${1 \over {3x}}$
2016
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2016 (Offline)
If a curve $y=f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation, $y(1+xy) dx=x$ $dy$, then $f\left( { - {1 \over 2}} \right)$ is equal to :
Show Answer
Correct Answer: B
Explanation:
$y\left( {1 + xy} \right)dx = xdy$
2015
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2015 (Offline)
Let $y(x)$ be the solution of the differential equation
Show Answer
Correct Answer: A
Explanation:
Given, ${{dy} \over {dx}} + \left( {{1 \over {x\,\log \,x}}} \right)y = 2$
2014
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2014 (Offline)
Let the population of rabbits surviving at time $t$ be governed by the differential equation ${{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 200.$ If $p(0)=100,$ then $p(t)$ equals:
A.
$600 - 500\,{e^{t/2}}$
B.
$400 - 300\,{e^{-t/2}}$
C.
$400 - 300\,{e^{t/2}}$
D.
$300 - 200\,{e^{-t/2}}$
Show Answer
Correct Answer: C
Explanation:
Given differential equation is
2013
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2013 (Offline)
At present, a firm is manufacturing $2000$ items. It is estimated that the rate of change of production P w.r.t. additional number of workers $x$ is given by ${{dp} \over {dx}} = 100 - 12\sqrt x .$ If the firm employs $25$ more workers, then the new level of production of items is
Show Answer
Correct Answer: C
Explanation:
Given, Rate of change is ${{dp} \over {dx}} = 100 - 12\sqrt x $
2012
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2012
The population $p$ $(t)$ at time $t$ of a certain mouse species satisfies the differential equation ${{dp\left( t \right)} \over {dt}} = 0.5\,p\left( t \right) - 450.\,\,$ If $p(0)=850,$ then the time at which the population becomes zero is :
C.
${1 \over 2}$$ln$ $18$
Show Answer
Correct Answer: A
Explanation:
Given differential equation is
2011
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2011
Let $I$ be the purchase value of an equipment and $V(t)$ be the value after it has been used for $t$ years. The value $V(t)$ depreciates at a rate given by differential equation ${{dv\left( t \right)} \over {dt}} = - k\left( {T - t} \right),$ where $k>0$ is a constant and $T$ is the total life in years of the equipment. Then the scrap value $V(T)$ of the equipment is
A.
$I - {{k{T^2}} \over 2}$
B.
$I - {{k{{\left( {T - t} \right)}^2}} \over 2}$
Show Answer
Correct Answer: A
Explanation:
${{dV\left( t \right)} \over {dt}} = - k\left( {T - t} \right)$
2011
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2011
If ${{dy} \over {dx}} = y + 3 > 0\,\,$ and $y(0)=2,$ then $y\left( {\ln 2} \right)$ is equal to :
Show Answer
Correct Answer: D
Explanation:
${{dy} \over {dx}} = y + 3 \Rightarrow \int {{{dy} \over {y + 3}}} = \int {dx} $
2010
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2010
Solution of the differential equation
A.
$y\sec x = \tan x + c$
B.
$y\tan x = \sec x + c$
C.
$\tan x = \left( {\sec x + c} \right)y$
D.
$\sec x = \left( {\tan x + c} \right)y$
Show Answer
Correct Answer: D
Explanation:
$\cos xdy = y\left( {\sin x - y} \right)dx$
2009
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2009
The differential equation which represents the family of curves $y = {c_1}{e^{{c_2}x}},$ where ${c_1}$ , and ${c_2}$ are arbitrary constants, is
C.
$yy'' = {\left( {y'} \right)^2}$
Show Answer
Correct Answer: C
Explanation:
We have $y = {c_1}{e^{{c_2}x}}$
2008
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2008
The solution of the differential equation
C.
$y = x{e^{\left( {x - 1} \right)}}\,$
Show Answer
Correct Answer: D
Explanation:
${{dy} \over {dx}} = {{x + y} \over x} = 1 + {y \over x}$
2007
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2007
The differential equation of all circles passing through the origin and having their centres on the $x$-axis is :
A.
${y^2} = {x^2} + 2xy{{dy} \over {dx}}$
B.
${y^2} = {x^2} - 2xy{{dy} \over {dx}}$
C.
${x^2} = {y^2} + xy{{dy} \over {dx}}$
D.
${x^2} = {y^2} + 3xy{{dy} \over {dx}}$
Show Answer
Correct Answer: A
Explanation:
General equation of circles passing through origin
2006
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2006
The differential equation whose solution is $A{x^2} + B{y^2} = 1$
A.
second order and second degree
B.
first order and second degree
C.
first order and first degree
D.
second order and first degree
Show Answer
Correct Answer: D
Explanation:
$A{x^2} + B{y^2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$
2005
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2005
The differential equation representing the family of curves ${y^2} = 2c\left( {x + \sqrt c } \right),$ where $c>0,$ is a parameter, is of order and degree as follows:
Show Answer
Correct Answer: C
Explanation:
${y^2} = 2c\left( {x + \sqrt c } \right)\,\,\,\,\,\,\,\,\,\,...\left( i \right)$
2005
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2005
If $x{{dy} \over {dx}} = y\left( {\log y - \log x + 1} \right),$ then the solution of the equation is :
A.
$y\log \left( {{x \over y}} \right) = cx$
B.
$x\log \left( {{y \over x}} \right) = cy$
C.
$\log \left( {{y \over x}} \right) = cx$
D.
$\log \left( {{x \over y}} \right) = cy$
Show Answer
Correct Answer: C
Explanation:
${{xdy} \over {dx}} = y\left( {\log y - \log x + 1} \right)$
2004
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2004
The differential equation for the family of circle ${x^2} + {y^2} - 2ay = 0,$ where a is an arbitrary constant is :
A.
$\left( {{x^2} + {y^2}} \right)y' = 2xy$
B.
$2\left( {{x^2} + {y^2}} \right)y' = xy$
C.
$\left( {{x^2} - {y^2}} \right)y' =2 xy$
D.
$2\left( {{x^2} - {y^2}} \right)y' = xy$
Show Answer
Correct Answer: C
Explanation:
${x^2} + {y^2} - 2ay = 0\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$
2004
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2004
Solution of the differential equation $ydx + \left( {x + {x^2}y} \right)dy = 0$ is
B.
$ - {1 \over {xy}} + \log y = C$
C.
${1 \over {xy}} + \log y = C$
D.
$ - {1 \over {xy}} = C$
Show Answer
Correct Answer: B
Explanation:
$ydx + \left( {x + {x^2}y} \right)dy = 0$
2003
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2003
The degree and order of the differential equation of the family of all parabolas whose axis is $x$-axis, are respectively.
Show Answer
Correct Answer: C
Explanation:
General equation of parabola whose axis is along X-axis
2003
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2003
The solution of the differential equation
A.
$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + k$
B.
$\left( {x - 2} \right) = k{e^{2{{\tan }^{ - 1}}y}}$
C.
$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$
D.
$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + k$
Show Answer
Correct Answer: C
Explanation:
$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0$
2002
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2002
The order and degree of the differential equation
A.
$\left( {1,{2 \over 3}} \right)$
Show Answer
Correct Answer: C
Explanation:
${\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} =$ (4)3 ${\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}$
2002
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
AIEEE 2002
The solution of the equation $\,{{{d^2}y} \over {d{x^2}}} = {e^{ - 2x}}$
A.
${{{e^{ - 2x}}} \over 4}$
B.
${{{e^{ - 2x}}} \over 4} + cx + d$
C.
${1 \over 4}{e^{ - 2x}} + c{x^2} + d$
D.
$\,{1 \over 4}{e^{ - 4x}} + cx + d$
Show Answer
Correct Answer: B
Explanation:
${{{a^2}y} \over {d{x^2}}} = {e^{ - 2x}};{\mkern 1mu} $
2026
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2026 (Online) 23rd January Evening Shift
If the solution curve $y=f(x)$ of the differential equation
$ \left(x^2-4\right) y^{\prime}-2 x y+2 x\left(4-x^2\right)^2=0, x>2, $
passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$
Show Answer
Correct Answer: 16
Explanation:
$ \left(x^2-4\right) y^{\prime}-2 x y+2 x\left(4-x^2\right)^2=0, x>2 $
rearranging terms.
$ \begin{aligned} & \left(x^2-4\right) \frac{d y}{d x}-2 x y=-2 x\left(x^2-4\right)^2 \\ & \frac{d y}{d x}-\frac{2 x}{x^2-4} y=-2 x\left(x^2-4\right) \end{aligned} $
this is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$
$ P=-\frac{2 x}{x^2-4}, Q=-2 x\left(x^2-4\right) $
finding integrating factor (I.F.).
$ \begin{aligned} \text { I.F. } & =e^{\int P d x}=e^{\int-\frac{2 x}{x^2-4} d x} \\ \text { I.F. } & =e^{-\ln \left(x^2-4\right)}=\frac{1}{x^2-4} \end{aligned} $
the solution is $y \cdot(I . F)=.\int Q \cdot(I . F) d x+$.
$ \begin{aligned} & y \cdot \frac{1}{x^2-4}=\int-2 x\left(x^2-4\right) \cdot \frac{1}{x^2-4} d x+C \\ & \frac{y}{x^2-4}=\int-2 x d x+C \\ & \frac{y}{x^2-4}=-x^2+C \Rightarrow y=\left(x^2-4\right)\left(C-x^2\right) \end{aligned} $
substitute point $(3,15)$
$ 15=\left(3^2-4\right)\left(C-3^2\right) $
$ \begin{aligned} & 15=5(C-9) \Rightarrow 3=C-9 \Rightarrow C=12 \\ & f(x)=\left(x^2-4\right)\left(12-x^2\right) \\ & f(x)=-x^4+16 x^2-48 \end{aligned} $
finding critical points for local maximum.
$ \begin{aligned} & f^{\prime}(x)=-4 x^3+32 x \\ & f^{\prime}(x)=0 \Rightarrow-4 x\left(x^2-8\right)=0 \\ & \text { since } x>2, x^2=8 \Rightarrow x=\sqrt{8}=2 \sqrt{2} \end{aligned} $
$f^{\prime}\left((2 \sqrt{2})^{-}\right)=$positive, $f^{\prime}\left((2 \sqrt{2})^{+}\right)=$negative
so sign of $f^{\prime}(x)$ changing positive to negative about $x=2 \sqrt{2}$ so $x=2 \sqrt{2}$ is point of local maxima
calculating local maximum value.
$ \begin{aligned} & f(2 \sqrt{2})=\left((2 \sqrt{2})^2-4\right)\left(12-(2 \sqrt{2})^2\right) \\ & =(8-4)(12-8) \\ & =4 \times 4=16 \end{aligned} $
the local maximum value is 16 .
2026
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2026 (Online) 23rd January Morning Shift
Let $f$ be a twice differentiable non-negative function such that $(f(x))^2=25+\int_0^x\left((f(\mathrm{t}))^2+\left(f^{\prime}(\mathrm{t})\right)^2\right) \mathrm{dt}$. Then the mean of $f\left(\log _{\mathrm{e}}(1)\right), f\left(\log _{\mathrm{e}}(2)\right), \ldots . ., f\left(\log _{\mathrm{e}}(625)\right)$ is equal to $\_\_\_\_$ .
Show Answer
Correct Answer: 1565
Explanation:
Let $y=f(x)$
$ y^2=25+\int_0^x\left(f(t)^2+f^{\prime}(t)^2\right) d t $
Differentiating w.r.t. $x$
$ \begin{aligned} & 2 y \frac{d y}{d x}=y^2+\left(\frac{d y}{d x}\right)^2 \\ & \Rightarrow\left(\frac{d y}{d x}-y\right)^2=0 \Rightarrow \frac{d y}{y}=d x \\ & \Rightarrow \ln |y|=x+c \end{aligned} $
Such that
$ \begin{aligned} & f(0)^2=25+\int_0^0\left(f^2(t)+f^{\prime}(t)^2\right) d t \\ & \Rightarrow f(0)=5 \\ & \Rightarrow \ln 5=c \end{aligned} $
$ \begin{aligned} & \Rightarrow \ln |y|=x+\ln 5 \\ & \Rightarrow \quad y=5 e^x \\ & f(\ln k)=5 e^{\ln k}=5 k \\ & \text { Mean of }\{5,10, \ldots 625 \times 5\} \\ & =\frac{(5+10+\ldots 5 \times 625)}{625}=\frac{5(625)(626)}{625 \times 2} \\ & =5 \times 313 \\ & =1565 \end{aligned} $
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2025 (Online) 2nd April Evening Shift
Let $y=y(x)$ be the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \sec ^2 x=2 \sec ^2 x+3 \tan x \cdot \sec ^2 x$ such that $y(0)=\frac{5}{4}$. Then $12\left(y\left(\frac{\pi}{4}\right)-\mathrm{e}^{-2}\right)$ is equal to_____________________
Show Answer
Correct Answer: 21
Explanation:
$\begin{aligned}
& \frac{d y}{d x}+2 y \sec ^2 x=2 \sec ^2 x+3 \tan x \sec ^2 x \\
& \text { I.F. }=e^{\int 2 \sec ^2 x d x}
\end{aligned}$
$\begin{aligned}
&\begin{aligned}
& \text { I.F. }=e^{2 \tan x} \\
& y \cdot e^{2 \tan x}=\int e^{2 \tan x}(2+3 \tan x) \sec ^2 x d x
\end{aligned}\\
&\text { Put } \tan x=u\\
&\begin{aligned}
& \sec ^2 x d x=d u \\
& y \cdot e^{2 u}=\int e^{2 u}(2+3 u) d u \\
& y \cdot e^{2 u} \Rightarrow \frac{2 e^{2 u}}{2}+3 \int e^{2 u} \cdot u d u \\
& y \cdot e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\int \frac{e^{2 u}}{2}\right] \\
& y e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\frac{e^{2 u}}{4}\right]+C \\
& y e^{2 \tan x}=e^{2 \tan x}+3\left[\frac{\tan x e^{2 \tan x}}{2}-\frac{e^{2 \tan x}}{4}\right]+C
\end{aligned}
\end{aligned}$
$\begin{aligned}
& F(0)=\frac{5}{4} \\
& \frac{5}{4}=1-\frac{3}{4}+C \\
& \frac{5}{4}-\frac{1}{4}=C \\
& 1=C \\
& y=1+3\left(\frac{\tan x}{2}-\frac{1}{4}\right)+1 \cdot e^{-2 \tan x} \\
& y\left(\frac{\pi}{4}\right)=1+3\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{e^2} \\
& y\left(\frac{\pi}{4}\right)=\frac{7}{4}+\frac{1}{e^2} \\
& 12\left(y\left(\frac{x}{4}\right)-\frac{1}{e^2}\right)=12\left(\frac{7}{4}+\frac{1}{e^2}-\frac{1}{e^2}\right)=21
\end{aligned}$
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2025 (Online) 28th January Evening Shift
If $y=y(x)$ is the solution of the differential equation,
$\sqrt{4-x^2} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\left(\left(\sin ^{-1}\left(\frac{x}{2}\right)\right)^2-y\right) \sin ^{-1}\left(\frac{x}{2}\right),-2 \leq x \leq 2, y(2)=\frac{\pi^2-8}{4}$, then $y^2(0)$ is equal to ___________.
Show Answer
Correct Answer: 4
Explanation:
$\begin{aligned}
& \frac{d y}{d x}+\frac{\left(\sin ^{-1} \frac{x}{2}\right)}{\sqrt{4-x^2}} y=\frac{\left(\sin ^{-3} \frac{x}{2}\right)^3}{\sqrt{4-x^2}} \\
& y e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^2}{2}}=\int \frac{\left(\sin ^{-3} \frac{x}{2}\right)^3}{4-x^2} e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^2}{2}} d x \\
& y=\left(\sin ^{-1} \frac{x}{2}\right)^2-2+c \cdot e^{\frac{-\left(\sin ^{-1} \frac{x}{2}\right)^2}{2}} \\
& y(2)=\frac{\pi^2}{4}-2 \Rightarrow c=0 \\
& y(0)=-2
\end{aligned}$
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2025 (Online) 24th January Evening Shift
Let $y=y(x)$ be the solution of the differential equation
Show Answer
Correct Answer: 1
Explanation:
$\begin{aligned}
& \frac{d y}{d x}+2 y \tan x=\sin x \\
& \text { I.F. }=e^{2 \int \tan x d x}=\sec ^2 x \\
& y \sec ^2 x=\int \frac{\sin x}{\cos ^2 x} d x \\
& =\int \tan x \sec x d x \\
& =\sec x+C \\
& C=-2 \\
& y=\cos x-2 \cos ^2 x \\
& y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}-1 \\
& y^{\prime}=-\sin x+4 \cos ^2 x \sin x \\
& y^{\prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}+2 \\
& y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)=1
\end{aligned}$
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2025 (Online) 24th January Morning Shift
Let $f$ be a differentiable function such that $2(x+2)^2 f(x)-3(x+2)^2=10 \int_0^x(t+2) f(t) d t, x \geq 0$. Then $f(2)$ is equal to ________ .
Show Answer
Correct Answer: 19
Explanation:
$\begin{aligned}
&\text { Differentiate both sides }\\
&\begin{aligned}
& 4(x+2) f(x)+2(x+2)^2 f^{\prime}(x)-6(x+2)=10(x+2) f(x) \\
& 2(x+2)^2 f^{\prime}(x)-6(x+2) f(x)=6(x+2) \\
& (x+2) \frac{d y}{d x}-3 y=3 \\
& \int \frac{d y}{d x}=3 \int \frac{d x}{x+2} \\
& \ln (y+1)=3 \ln (x+2)+C \\
& (y+1)=C(x+2)^3 \\
& f(0)=\frac{3}{2} \\
& f(2)=19
\end{aligned}
\end{aligned}$
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2025 (Online) 22nd January Evening Shift
Let $y=f(x)$ be the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{x y}{x^2-1}=\frac{x^6+4 x}{\sqrt{1-x^2}},-1< x<1$ such that $f(0)=0$. If $6 \int_{-1 / 2}^{1 / 2} f(x) \mathrm{d} x=2 \pi-\alpha$ then $\alpha^2$ is equal to _________ .
Show Answer
Correct Answer: 27
Explanation:
I.F. $\mathrm{e}^{-\frac{1}{2} \int \frac{2 \mathrm{x}}{1-\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{-\frac{1}{2} \ln \left(1-\mathrm{x}^2\right)}=\sqrt{1-\mathrm{x}^2}$
$y \times \sqrt{1-x^2}=\int\left(x^6+4 x\right) d x=\frac{x^7}{7}+2 x^2+c$
Given $y(0)=0 \Rightarrow c=0$
$y=\frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}}$
$\begin{aligned}
&\text { Now, } 6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}} d x=6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2 x^2}{\sqrt{1-x^2}} d x\\
&=24 \int_0^{\frac{1}{2}} \frac{\mathrm{x}^2}{\sqrt{1-\mathrm{x}^2}} \mathrm{dx}
\end{aligned}$
$\begin{aligned}
& \text { Put } x=\sin \theta \\
& d x=\cos \theta d \theta \\
& =24 \int_0^{\frac{\pi}{6}} \frac{\sin ^2 \theta}{\cos \theta} \cos \theta d \theta \\
& =24 \int_0^{\frac{\pi}{6}}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta=12\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{6}} \\
& =12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \\
& =2 \pi-3 \sqrt{3} \\
& \alpha^2=(3 \sqrt{3})^2=27
\end{aligned}$
2024
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2024 (Online) 9th April Evening Shift
For a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$, suppose $f^{\prime}(x)=3 f(x)+\alpha$, where $\alpha \in \mathbb{R}, f(0)=1$ and $\lim _\limits{x \rightarrow-\infty} f(x)=7$. Then $9 f\left(-\log _e 3\right)$ is equal to _________.
Show Answer
Correct Answer: 61
Explanation:
$\begin{aligned}
& f^{\prime}(x)=3 f(x)+\alpha \\
& \Rightarrow \frac{d y}{3 y+\alpha}=d x \\
& \Rightarrow \frac{1}{3} \ln (3 y+\alpha)=x+C \\
& y(0)=1 \Rightarrow C=\frac{1}{3} \ln (3+\alpha) \\
& \frac{1}{3} \ln \left(\frac{3 y+\alpha}{3+\alpha}\right)=x \\
& \Rightarrow y=\frac{1}{3}\left((3+\alpha) e^{3 x}-\alpha\right)=f(x) \\
& \lim _{x \rightarrow-\infty} f(x)=7 \Rightarrow \alpha=-21 \\
& \Rightarrow f(x)=7-6 e^{3 x} \\
& 9 f(-\ln 3)=61
\end{aligned}$
2024
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2024 (Online) 8th April Evening Shift
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 ________
Show Answer
Correct Answer: 4
Explanation:
$\begin{aligned}
& \alpha|x|=|y| e^{x y-\beta} \\
& \frac{x d y-y d x}{y^2}+\frac{x y(x d y+y d x)}{y^2}=0 \\
& -d\left(\frac{x}{y}\right)+\frac{x}{y} d(x y)=0
\end{aligned}$
$\begin{aligned}
& \int d(x y)=\int \frac{d\left(\frac{x}{y}\right)}{\frac{x}{y}} \\
& x y=\ln \left|\frac{x}{y}\right|+\ln c \\
& x y=\ln \left(\left|\frac{x}{y}\right| \cdot c\right) \\
& \because y(1)=2 \\
& 2=\ln \left|\frac{1}{2}\right| c \Rightarrow c=2 e^2 \\
& \therefore \quad \operatorname{solution} x y=\ln \left(\left|\frac{x}{y}\right| \cdot 2 e^2\right) \\
& e^{x y}=\frac{|x|}{|y|} \cdot 2 e^2 \\
& 2|x|=|y| e^{x y-2} \\
& \Rightarrow \alpha=2, \beta=2, \alpha+\beta=4
\end{aligned}$
2024
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2024 (Online) 6th April Evening Shift
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 _________.
Show Answer
Correct Answer: 3
Explanation:
Given the differential equation
$\left(e^y + 1\right) \cos x \, dx + e^y \sin x \, dy = 0,$
we aim to find the value of $ e^{y\left(\frac{\pi}{6}\right)} $ given that the solution $ y(x) $ passes through the point $\left(\frac{\pi}{2}, 0\right)$.
First, we recognize that the differential equation can be rearranged as:
$d\left(e^y \sin x\right) + \cos x \, dx = 0.$
Integrating this expression, we obtain:
$ e^y \sin x + \sin x = C,$
where $ C $ is a constant. Given that the solution passes through the point $\left(\frac{\pi}{2}, 0\right)$, we substitute these values into the equation to find $ C $:
$ e^0 \sin\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = C \Rightarrow 1 + 1 = C \Rightarrow C = 2.$
Thus, the equation simplifies to:
$e^y \sin x + \sin x = 2.$
We now need to determine the value of $ e^{y\left(\frac{\pi}{6}\right)} $. Substituting $ x = \frac{\pi}{6} $ into the equation, we get:
$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \sin\left(\frac{\pi}{6}\right) = 2.$
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, the equation becomes:
$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \frac{1}{2} = 2 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} + 1 = 4 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} = 3.$
Therefore, the value of $ e^{y\left(\frac{\pi}{6}\right)} $ is $ 3 $.
2024
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2024 (Online) 5th April Evening Shift
Let $y=y(x)$ be the solution of the differential equation
$\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 x}{\left(1+x^2\right)^2} y=x \mathrm{e}^{\frac{1}{\left(1+x^2\right)}} ; y(0)=0.$
Then the area enclosed by the curve $f(x)=y(x) \mathrm{e}^{-\frac{1}{\left(1+x^2\right)}}$ and the line $y-x=4$ is ________.
Show Answer
Correct Answer: 18
Explanation:
$\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{1+x^2}} ; y(0)=0$
I.F. of linear differential equation,
$\begin{aligned}
& \text { I.F. }=e^{\int \frac{2 x}{\left(1+x^2\right)^2}} d x=e^{\left(\frac{-1}{1+x^2}\right)} \\
& \Rightarrow y\left(e^{\left(\frac{-1}{1+x^2}\right)}\right)=\int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{\left(\frac{-1}{1+x^2}\right)} d x \\
& =\frac{x^2}{2}+c \\
& \Rightarrow y(0)=0 \Rightarrow 0\left(e^{-1}\right)=c \Rightarrow c=0 \\
& \Rightarrow y=\frac{e^{\frac{1}{1+x^2}} \cdot x^2}{2}
\end{aligned}$
Area between curve $y e^{\left(\frac{-1}{1+x^2}\right)}=\frac{x^2}{2}$ and $y-x=4$
$\begin{aligned}
& \Rightarrow 2(x+4)=x^2 \Rightarrow x^2-2 x-8=0 \\
& \Rightarrow(x-4)(x+2)=0 \\
& \int_{-2}^4\left[(x+4)-\frac{x^2}{2}\right] d x=\frac{x^2}{2}+4 x-\left.\frac{x^3}{6}\right|_{-2} ^4 \\
& =\left(8+16-\frac{64}{6}\right)-\left(2-8+\frac{8}{6}\right) \\
& =30-12=18
\end{aligned}$
2024
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2024 (Online) 4th April Evening Shift
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 $(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}, \gamma, \delta \in \mathbb{Z}$, then $\gamma+\delta$ equals _________.
Show Answer
Correct Answer: 31
Explanation:
$\begin{aligned}
& \frac{d y}{d x}=(x+y+z)^2 \\
& \text { Put } x+y+z=t \\
& \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \\
& \text { Given DE } \Rightarrow \frac{d t}{d x}-1=t^2 \\
& \Rightarrow \frac{d t}{1+t^2}=d x \Rightarrow \tan ^{-1} t=x+c \\
& \Rightarrow x+y+z=\tan (x+c) \\
& \Rightarrow y(x)=\tan (x+c)-x-2 \\
& \because y(0)=-2 \Rightarrow-2=\tan c-0-2 \\
& \qquad \Rightarrow c=0 \\
& \Rightarrow y(x)=\tan x-x-2 \\
& \frac{d y}{d x}=\sec ^2 x-1 \geq 0
\end{aligned}$
$\Rightarrow y(x)$ is increasing if $x \in\left(0, \frac{\pi}{3}\right)$
$\begin{aligned}
& \Rightarrow \alpha=y\left(\frac{\pi}{3}\right), \beta=y(0) \\
& \Rightarrow \alpha=-\frac{\pi}{3}-2+\sqrt{3} \text { and } \beta=-2
\end{aligned}$
Now, $(3 \alpha+\pi)^2+\beta^2=(6+3 \sqrt{3})^2+(-2)^2$
$=67-36 \sqrt{3}=y+\delta \sqrt{3}$.
$\Rightarrow \gamma+\delta=31$
2024
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2024 (Online) 4th April Morning Shift
Let the solution $y=y(x)$ of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}-y=1+4 \sin x$ satisfy $y(\pi)=1$. Then $y\left(\frac{\pi}{2}\right)+10$ is equal to __________.
Show Answer
Correct Answer: 7
Explanation:
$\begin{aligned}
& \frac{d y}{d x}-y=1+4 \sin x \\
& \text { Integrating factor }=e^{-\int d x}=e^{-x}
\end{aligned}$
Solution is $y e^{-x}=\int(1+4 \sin x) e^{-x} d x$
$\begin{aligned}
& =-e^{-x}+2 \cdot e^{-x}(-\sin x-\cos x)+C \\
y(\pi) & =1 \Rightarrow C=0
\end{aligned}$
Hence $y(x)=-1-2(\sin x+\cos x)$
$y\left(\frac{\pi}{2}\right)+10=7$
2024
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2024 (Online) 1st February Evening Shift
If $\frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{1+x-y^2}{y}, x(1)=1$, then $5 x(2)$ is equal to __________.
Show Answer
Correct Answer: 5
Explanation:
$\frac{d x}{d y}-\frac{x}{y}=\frac{1-y^2}{y}$