1
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The first derivative of the function $\left(\cos ^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right)+x^x\right)$ with respect to $x$ at $x=1$ is

A
$\frac{1}{4}$
B
$\frac{5}{4}$
C
$\frac{-1}{2}$
D
$\frac{3}{4}$
2
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $x^{\frac{2}{5}}+y^{\frac{2}{5}}=\mathrm{a}^{\frac{2}{5}}$ then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

A
$\sqrt[5]{\left(\frac{y}{x}\right)^3}$
B
$\quad-\sqrt[5]{\left(\frac{x}{y}\right)^3}$
C
$\sqrt[5]{\left(\frac{x}{y}\right)^3}$
D
$\quad-\sqrt[5]{\left(\frac{y}{x}\right)^3}$
3
MHT CET 2025 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
For $\mathrm{n} \in \mathbb{N}$ if $y=\mathrm{a} x^{\mathrm{n}+1}+\mathrm{b} x^{-\mathrm{n}}$, then $x^2 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}=$
A
$\mathrm{n}(\mathrm{n}-1) y$
B
$(\mathrm{n}-1) y$
C
$\mathrm{n}(\mathrm{n}+1) y$
D
$(\mathrm{n}+1) y$
4
MHT CET 2025 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The derivative of $\tan ^{-1}\left(\sqrt{1+x^2}-1\right)$ is
A
$\frac{x}{\sqrt{1+x^2}\left(x^2-2 \sqrt{x+1}+1\right)}$
B
$\frac{x}{\sqrt{1+x^2}\left(x^2-2 \sqrt{1+x^2}+3\right)}$
C
$\frac{x}{\sqrt{1+x^2}\left(x^2-2 \sqrt{x^2+1}+2\right)}$
D
$\frac{x}{\sqrt{1+x^2}\left(x^2+2 \sqrt{1+x^2}-3\right)}$
MHT CET Subjects
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