1
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let y = y(x) be the solution of the differential equation,

$${({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1$$

such that y(0) = 0. If $$\sqrt ay(1)$$ = $$\pi \over 32$$ , then the value of 'a' is :
A
$${1 \over 2}$$
B
$${1 \over 16}$$
C
1
D
$${1 \over 4}$$
2
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The sum of the solutions of the equation
$$\left| {\sqrt x - 2} \right| + \sqrt x \left( {\sqrt x - 4} \right) + 2 = 0$$
(x > 0) is equal to:
A
9
B
12
C
4
D
10
3
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$, $$\left| x \right| < 1$$ then $$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ is equal to
A
2f(x2)
B
2f(x)
C
(f(x))2
D
-2f(x)
4
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let ƒ : [0, 2] $$ \to $$ R be a twice differentiable function such that ƒ''(x) > 0, for all x $$ \in $$ (0, 2). If $$\phi $$(x) = ƒ(x) + ƒ(2 – x), then $$\phi $$ is :
A
decreasing on (0, 2)
B
decreasing on (0, 1) and increasing on (1, 2)
C
increasing on (0, 2)
D
increasing on (0, 1) and decreasing on (1, 2)

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