1
WB JEE 2024
+1
-0.25

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
+1
-0.25

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
+2
-0.5

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
$$I(R) \text { and } \frac{\pi}{2 R}(1-e^{-R}) \text { are not comparable }$$
4
WB JEE 2024
+2
-0.5

$$\lim _\limits{n \rightarrow \infty} \frac{1}{n^{k+1}}[2^k+4^k+6^k+\ldots .+(2 n)^k]=$$

A
$$\frac{2^k}{k}$$
B
$$\frac{2^{k+1}}{k+1}$$
C
$$\frac{2^k}{k+1}$$
D
$$\frac{2^{\mathrm{k}}}{\mathrm{k}-1}$$
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