1
WB JEE 2024
MCQ (Single Correct Answer)
+1
-0.25
Change Language

If $$\mathrm{f}(x)=\frac{\mathrm{e}^x}{1+\mathrm{e}^x}, \mathrm{I}_1=\int_\limits{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} x \mathrm{~g}(x(1-x)) \mathrm{d} x$$ and $$\mathrm{I}_2=\int_\limits{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} \mathrm{g}(x(1-x)) \mathrm{d} x$$, then the value of $$\frac{I_2}{I_1}$$ is

A
$$-1$$
B
$$-3$$
C
2
D
1
2
WB JEE 2024
MCQ (Single Correct Answer)
+1
-0.25
Change Language

Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a differentiable function and $$f(1)=4$$. Then the value of $$\lim _\limits{x \rightarrow 1} \int_\limits4^{f(x)} \frac{2 t}{x-1} d t$$, if $$f^{\prime}(1)=2$$ is

A
16
B
8
C
4
D
2
3
WB JEE 2024
MCQ (Single Correct Answer)
+1
-0.25
Change Language

$$ \text { If } \int \frac{\log _e\left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \mathrm{~d} x=\mathrm{f}(\mathrm{g}(x))+\mathrm{c} \text { then } $$

A
$$\mathrm{f}(x)=\frac{x^2}{2}, \mathrm{~g}(x)=\log _{\mathrm{e}}\left(x+\sqrt{1+x^2}\right)$$
B
$$\mathrm{f}(x)=\log _{\mathrm{e}}\left(x+\sqrt{1+x^2}\right), \mathrm{g}(x)=\frac{x^2}{2}$$
C
$$\mathrm{f}(x)=x^2, \mathrm{~g}(x)=\log _{\mathrm{e}}\left(x+\sqrt{1+x^2}\right)$$
D
$$\mathrm{f}(x)=\log _{\mathrm{e}}\left(x-\sqrt{1+x^2}\right), \mathrm{g}(x)=x^2$$
4
WB JEE 2024
MCQ (Single Correct Answer)
+2
-0.5
Change Language

Let $$\mathrm{I}(\mathrm{R})=\int_\limits0^{\mathrm{R}} \mathrm{e}^{-\mathrm{R} \sin x} \mathrm{~d} x, \mathrm{R}>0$$. then,

A
$$I(R)>\frac{\pi}{2 R}\left(1-e^{-R}\right)$$
B
$$I(R)<\frac{\pi}{2 R}\left(1-e^{-R}\right)$$
C
$$I(R)=\frac{\pi}{2 R}\left(1-e^{-R}\right)$$
D
$$I(R) \text { and } \frac{\pi}{2 R}(1-e^{-R}) \text { are not comparable }$$
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