1
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $(\mathrm{a}+\mathrm{b} x) \mathrm{e}^{\frac{y}{x}}=x$, then $x^3 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}$ is equal to

A
$\left(y \frac{\mathrm{~d} y}{\mathrm{~d} x}-x\right)^2$
B
$\left(x \frac{\mathrm{~d} y}{\mathrm{~d} x}-y\right)^2$
C
$\left(x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y\right)^2$
D
$\left(y \frac{\mathrm{~d} y}{\mathrm{~d} x}+x\right)^2$
2
MHT CET 2025 21st April Morning Shift
MCQ (More than One Correct Answer)
+2
-0

If $x=\log \mathrm{t}, \mathrm{t}>0$ and $y=\frac{1}{\mathrm{t}}$ then $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=$

A
$\frac{\mathrm{d} y}{\mathrm{~d} x}$
B
$-\frac{\mathrm{d} y}{\mathrm{~d} x}$
C
$y$
D
$\frac{y}{x}$
3
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $y=\frac{\mathrm{K}^{\cos ^{-1} x}}{1+\mathrm{K}^{\cos ^{-1} x}}$ and $\mathrm{t}=\mathrm{K}^{\cos ^{-1} x}$, then $\frac{\mathrm{d} y}{\mathrm{dt}}=$

A
$\frac{1}{1+\mathrm{K}^{\cos ^{-1} x}}$
B
$\frac{-1}{1+\mathrm{K}^{\cos ^{-1} x}}$
C
$\frac{1}{\left(1+\mathrm{K}^{\cos ^{-1} x}\right)^2}$
D
$\frac{-1}{\left(1+\mathrm{K}^{\cos ^{-1} x}\right)^2}$
4
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The complex numbers $\sin x+i \cos 2 x$ and $\cos x$ - $\mathrm{i} \sin 2 x,(\mathrm{i}=\sqrt{-1})$ are conjugate to each other for,

A
$\quad x=\mathrm{n} \pi, \mathrm{n} \in \mathbb{Z}$
B
$x=\left(\mathrm{n}+\frac{1}{2}\right) \pi, \mathrm{n} \in \mathbb{Z}$
C
$x=(3 \mathrm{n}-1) \pi, \mathrm{n} \in \mathbb{Z}$
D
No value of $x$
MHT CET Papers
EXAM MAP