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

With usual notations, in a triangle $A B C$, if $\theta$ is any real number, then $a \cos (B-\theta)+b \cos (A+\theta)$ is

A
$a \cos \theta$
B
$\mathrm{b} \cos \theta$
C
$\cos \theta$
D
$c \cos \theta$
2
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $A=\left[\begin{array}{cc}1 & \tan x \\ -\tan x & 1\end{array}\right]$, then $A^T A^{-1}=$

A
$\left[\begin{array}{cc}\cos 2 x & -\sin 2 x \\ -\sin 2 x & \cos 2 x\end{array}\right]$
B
$\left[\begin{array}{ll}\cos 2 x & -\sin 2 x \\ \sin 2 x & \cos 2 x\end{array}\right]$
C
$\left[\begin{array}{cc}-\cos 2 x & \sin 2 x \\ \sin 2 x & \cos 2 x\end{array}\right]$
D
$\left[\begin{array}{ll}-\cos 2 x & \sin 2 x \\ -\sin 2 x & \cos 2 x\end{array}\right]$
3
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The sum of the degree and order of the differential equation $\sqrt{\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}}=\sqrt[5]{\frac{\mathrm{d} y}{\mathrm{~d} x}-5}$ is

A
1
B
3
C
5
D
7
4
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation whose solution represents the family $x^2 y=4 \mathrm{e}^x+\mathrm{c}$, where c is an arbitrary constant, is

A
$\quad x \frac{\mathrm{~d} y}{\mathrm{~d} x}+x y=0$
B
$\quad x^2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+(2 x-x y)=0$
C
$x \frac{\mathrm{~d} y}{\mathrm{~d} x}+(x-2) y=0$
D
$x^2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 x y-4 \mathrm{e}^x=0$

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