1
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $y=\sin ^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right]$, then $\frac{d y}{d x}=$

A
$\left(\frac{1}{4}\right) \frac{1}{\sqrt{x^2-1}}$
B
$\left(-\frac{1}{2}\right) \frac{1}{\sqrt{x^2-1}}$
C
$\left(-\frac{1}{2}\right) \frac{1}{\sqrt{1-x^2}}$
D
$\left(\frac{1}{4}\right) \frac{1}{\sqrt{1-x^2}}$
2
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

The area of the $\triangle A B C$ is $10 \sqrt{3} \mathrm{~cm}^2$, angle $B$ is $60^{\circ}$ and its perimeter is 20 cm , then $\ell(A C)=$

A
5 cm
B
10 cm
C
7 cm
D
8 cm
3
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation of the normal to the curve $2 x^2+y^2=12$ at the point $(2,2)$ is

A
$2 x+y-6=0$
B
$x-2 y+2=0$
C
$2 x-y+6=0$
D
$x+2 y+2=0$
4
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta}=$$

A
$-\cot \frac{\theta}{2}$
B
$\cot \frac{\theta}{2}$
C
$\tan \frac{\theta}{2}$
D
$-\tan \frac{\theta}{2}$
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