1
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The particular solution of the differential equation $\cos \left(\frac{d y}{d x}\right)=7, y=1$ at $x=0$ is

A
$\quad \cos \left(\frac{7}{x}\right)=1$
B
$\quad \cos \left(\frac{y}{x-1}\right)=7$
C
$\quad \cos \left(\frac{y-1}{x}\right)=7$
D
None
2
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

In a triangle $A B C$, with usual notations, $3 \mathrm{~b}=\mathrm{a}+\mathrm{c}$, then $\cot \frac{\mathrm{A}}{2} \cdot \cot \frac{\mathrm{C}}{2}=$

A
1
B
2
C
$\frac{1}{2}$
D
4
3
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$ \int_0^{\frac{\pi}{2}} \frac{d x}{1+(\cot x)^{101}}= $$

A
$\frac{\pi}{2}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{8}$
D
$\pi$
4
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The solution of $\left(1+y^2\right)+\left(x-\mathrm{e}^{\tan ^{-1} y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0$ is

A
$2 x \mathrm{e}^{\tan ^{-1} y}=\mathrm{e}^{2 \tan ^{-1} y}+\mathrm{k}$, where k is the constant of integration
B
$x \cdot \mathrm{e}^{\tan ^{-1} y}=\mathrm{e}^{\tan ^{-1} y}+\mathrm{k}$, where k is the constant of integration
C
$x \cdot \mathrm{e}^{2 \tan ^{-1} y}=\mathrm{e}^{\tan ^{-1} y}+\mathrm{k}$, where k is the constant of integration
D
$\quad x=2+\mathrm{k} \cdot \mathrm{e}^{-\tan ^{-1} y}$, where k is the constant of integration
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