1
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The function
$$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr } } \right.$$ is :
A
continuous on R–{–1} and differentiable on R–{–1, 1}
B
both continuous and differentiable on R–{1}
C
both continuous and differentiable on R–{–1}
D
continuous on R–{1} and differentiable on R–{–1, 1}
2
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The integral
$$\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $$
is equal to:
A
$$ - {1 \over {9}}$$
B
$$ - {1 \over {18}}$$
C
$$ {7 \over {18}}$$
D
$${9 \over 2}$$
3
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let $$f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$$ be a differentiable function such that f(1) = e and
$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$. If f(x) = 1, then x is equal to :
A
$${1 \over e}$$
B
e
C
$${1 \over 2e}$$
D
2e
4
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The solution of the differential equation

$${{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$$ is:

(where c is a constant of integration)
A
$$x - {1 \over 2}{\left( {{{\log }_e}\left( {y + 3x} \right)} \right)^2} = C$$
B
$$y + 3x - {1 \over 2}{\left( {{{\log }_e}x} \right)^2} = C$$
C
x – loge(y+3x) = C
D
x – 2loge(y+3x) = C

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