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

If $x^2 y^2=\sin ^{-1} \sqrt{x^2+y^2}+\cos ^{-1} \sqrt{x^2+y^2}$ then $\frac{d y}{d x}=$

A
$\frac{-x}{y}$
B
$\frac{x}{y}$
C
$\frac{-y}{x}$
D
$\frac{y}{x}$
3
MHT CET 2020 16th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$f(x)=\sin ^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$$, then $$f^{\prime}(x)=$$

A
$$\frac{1}{2 \sqrt{1+x^2}}$$
B
$$\frac{-1}{2 \sqrt{1+x^2}}$$
C
$$\frac{1}{\sqrt{1-x^2}}$$
D
$$\frac{-1}{2 \sqrt{1-x^2}}$$
4
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x \neq y$$, then $$(1+x)^2 \frac{d y}{d x}=$$

A
1
B
$$\frac{1}{2}$$
C
$$-$$1
D
0
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