1
JEE Main 2019 (Online) 10th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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 -
A
$${1 \over 3} + {e^6}$$
B
$${1 \over 3}$$
C
$${1 \over 3}$$ + e3
D
$$-$$ $${4 \over 3}$$
2
JEE Main 2019 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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 :
A
3
B
4
C
2
D
5
3
JEE Main 2019 (Online) 9th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If y = y(x) is the solution of the differential equation,

x$$dy \over dx$$ + 2y = x2, satisfying y(1) = 1, then y($$1\over2$$) is equal to :
A
$$ {{7} \over {64}}$$
B
$$ {{49} \over {16}}$$
C
$$ {{1} \over {4}}$$
D
$$ {{13} \over {16}}$$
4
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
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
B
x + y y'' = 0
C
xy y'+ y2 $$-$$ 9 = 0
D
xy y' $$-$$ y2 + 9 = 0
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