1
JEE Main 2023 (Online) 25th January Morning Shift
MCQ (Single Correct Answer)
+4
-1

Let $$f(x) = \int {{{2x} \over {({x^2} + 1)({x^2} + 3)}}dx}$$. If $$f(3) = {1 \over 2}({\log _e}5 - {\log _e}6)$$, then $$f(4)$$ is equal to

A
$${\log _e}19 - {\log _e}20$$
B
$${\log _e}17 - {\log _e}18$$
C
$${1 \over 2}({\log _e}19 - {\log _e}17)$$
D
$${1 \over 2}({\log _e}17 - {\log _e}19)$$
2
JEE Main 2022 (Online) 29th July Evening Shift
MCQ (Single Correct Answer)
+4
-1

For $$I(x)=\int \frac{\sec ^{2} x-2022}{\sin ^{2022} x} d x$$, if $$I\left(\frac{\pi}{4}\right)=2^{1011}$$, then

A
$$3^{1010} I\left(\frac{\pi}{3}\right)-I\left(\frac{\pi}{6}\right)=0$$
B
$$3^{1010} I\left(\frac{\pi}{6}\right)-I\left(\frac{\pi}{3}\right)=0$$
C
$$3^{1011} I\left(\frac{\pi}{3}\right)-I\left(\frac{\pi}{6}\right)=0$$
D
$$3^{1011} I\left(\frac{\pi}{6}\right)-I\left(\frac{\pi}{3}\right)=0$$
3
JEE Main 2022 (Online) 26th July Evening Shift
MCQ (Single Correct Answer)
+4
-1

$$\text { The integral } \int \frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}} \sin 2 x\right)} d x \text { is equal to }$$

A
$$\frac{1}{2} \log _{e}\left|\frac{\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)}{\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)}\right|+C$$
B
$$\frac{1}{2} \log _{e}\left|\frac{\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)}{\tan \left(\frac{x}{2}+\frac{\pi}{3}\right)}\right|+C$$
C
$$\log _{e}\left|\frac{\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)}{\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)}\right|+C$$
D
$$\frac{1}{2} \log _{e}\left|\frac{\tan \left(\frac{x}{2}-\frac{\pi}{12}\right)}{\tan \left(\frac{x}{2}-\frac{\pi}{6}\right)}\right|+C$$
4
JEE Main 2022 (Online) 27th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
If $$\int {{{({x^2} + 1){e^x}} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + C}$$, where C is a constant, then $${{{d^3}f} \over {d{x^3}}}$$ at x = 1 is equal to :
A
$$- {3 \over 4}$$
B
$${3 \over 4}$$
C
$$- {3 \over 2}$$
D
$${3 \over 2}$$
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