1
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]_{3 \times 3}$, then $A^{-1}=$

A
A
B
$\mathrm{A}^2$
C
$\mathrm{A}^3$
D
$\mathrm{A}^4$
2
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The area bounded by the parabolas $y=9 x^2, y=\frac{x^2}{16}$ and the line $y=1$ is

A
$\frac{22}{9}$ sq. units
B
$\frac{44}{9}$ sq. units
C
$\frac{8}{9}$ sq. units
D
$\frac{26}{9}$ sq. units
3
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

$\int \frac{\mathrm{d} x}{3 \cos 2 x+5}$ equals

A
$\frac{1}{2} \tan ^{-1}(\tan x)+\mathrm{c}$, where c is the constant of integration.
B
$\frac{1}{2} \tan ^{-1}\left(\frac{\tan x}{2}\right)+\mathrm{c}$, where c is the constant of integration.
C
$\frac{1}{4} \tan ^{-1}\left(\frac{1}{2} \tan x\right)+\mathrm{c}$, where c is the constant of integration.
D
$\frac{1}{4} \tan ^{-1}(\tan x)+\mathrm{c}$, where c is the constant of integration.
4
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The solution of the equation $x^2 y-x^3 \frac{\mathrm{~d} y}{\mathrm{~d} x}=y^4 \cos x$, where $y(0)=1$, is

A
$y^3=3 x^2 \sin x$
B
$x^3=3 y^3 \sin x$
C
$x^3=y^3 \sin x$
D
$y^3=4 x^3 \sin x$
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