1
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The value of $\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}+\frac{1}{2} \cos ^{-1} x$ is

A
$\frac{\pi}{2}$
B
$\frac{\pi}{4}$
C
0
D
$\frac{\pi}{3}$
2
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ be three non-zero vectors such that no two of them are collinear and $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{~b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$. If ' $\theta$ ' is the angle between the vectors $\bar{b}$ and $\bar{c}$, then value of $\sin \theta$ is

A
$\frac{2}{3}$
B
$\frac{-\sqrt{2}}{3}$
C
$-\frac{1}{3}$
D
$\frac{2 \sqrt{2}}{3}$
3
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

In an experiment with 15 observations for $x$, the following results were available $\sum x^2=2830, \sum x=170$. One observation 20 was found to be wrong and was replaced by the correct value 30 . Then the corrected variance is

A
78
B
210
C
225
D
88
4
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The area of the region lying in the first quadrant by $y=4 x^2, x=0, y=2, y=4$ is

A
$\frac{1}{6}[8-2 \sqrt{2}]$ sq.units
B
$\frac{1}{3}[8-2 \sqrt{2}]$ sq.units
C
$[8-2 \sqrt{2}]$ sq.units
D
$[8+2 \sqrt{2}]$ sq.units
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