1
WB JEE 2024
MCQ (Single Correct Answer)
+2
-0.5
Change Language

Angle between two diagonals of a cube will be

A
$$\cos ^{-1}\left(\frac{1}{3}\right)$$
B
$$\sin ^{-1}\left(\frac{1}{3}\right)$$
C
$$\frac{\pi}{2}-\cos ^{-1}\left(\frac{1}{3}\right)$$
D
$$\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{3}\right)$$
2
WB JEE 2024
MCQ (Single Correct Answer)
+2
-0.5
Change Language

$$ \text { If } y=\tan ^{-1}\left[\frac{\log _e\left(\frac{e}{x^2}\right)}{\log _e\left(e x^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log _e x}{1-6 \cdot \log _e x}\right] \text {, then } \frac{d^2 y}{d x^2}= $$

A
2
B
1
C
0
D
$$-$$1
3
WB JEE 2024
MCQ (Single Correct Answer)
+2
-0.5
Change Language

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

A
$$\frac{2^k}{k}$$
B
$$\frac{2^{k+1}}{k+1}$$
C
$$\frac{2^k}{k+1}$$
D
$$\frac{2^{\mathrm{k}}}{\mathrm{k}-1}$$
4
WB JEE 2024
MCQ (More than One Correct Answer)
+2
-0
Change Language

The acceleration f $$\mathrm{ft} / \mathrm{sec}^2$$ of a particle after a time $$\mathrm{t}$$ sec starting from rest is given by $$\mathrm{f}=6-\sqrt{1.2 \mathrm{t}}$$. Then the maximum velocity $$\mathrm{v}$$ and time $$\mathrm{T}$$ to attend this velocity are

A
$$\mathrm{T}=20 \mathrm{~sec}$$
B
$$\mathrm{v}=60 \mathrm{~ft} / \mathrm{sec}$$
C
$$\mathrm{T}=30 \mathrm{~sec}$$
D
$$\mathrm{v}=40 \mathrm{~tt} / \mathrm{sec}$$
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