1
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
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$$\left( {{1 \over x}} \right)$$ :
A
does not exist
B
exists and equals $${4 \over 7}$$
C
exists and equals 4
D
exists and equals 0
2
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
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
$$-$$ $${4 \over 3}$$
3
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
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
4
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
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
$${{13} \over {16}}$$
EXAM MAP
Medical
NEET