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

The general solution of

$x(x-1) \frac{\mathrm{d} y}{\mathrm{~d} x}=x^3(2 x-1)+(x-2) y$ is

A
$y(x-1)=x^3+\mathrm{c}(x-1)$, where c is the constant of integration.
B
$y=x^3(x-1)+\mathrm{c}$, where c is the constant of integration.
C
$y(x-1)=x^3(x-1)+\mathrm{cx}^2$, where c is the constant of integration.
D
$y(x-1)=x^3(x-1)+\mathrm{c}$, where c is the constant of integration.
2
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The direction cosines of the line $x-y+2 z=5$ and $3 x+y+z=6$ are

A
$\quad \frac{-3}{5 \sqrt{2}}, \frac{5}{5 \sqrt{2}}, \frac{4}{5 \sqrt{2}}$
B
$\quad \frac{3}{5 \sqrt{2}}, \frac{-5}{5 \sqrt{2}}, \frac{4}{5 \sqrt{2}}$
C
$\frac{3}{5 \sqrt{2}}, \frac{5}{5 \sqrt{2}}, \frac{4}{5 \sqrt{2}}$
D
$\quad \frac{3}{5 \sqrt{2}}, \frac{5}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}$
3
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

20 is divided into two parts so that the product of the cube of one part and the square of the other part is maximum, then these two parts are

A
15,5
B
16,4
C
12,8
D
14,6
4
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The acute angle between the diagonals of a parallelogram whose vertices are $\mathrm{A}(2,-1)$, $B(0,2), C(2,3)$ and $D(4,0)$ is

A
$\cot ^{-1} 2$
B
$\cot ^{-1}\left(\frac{1}{3}\right)$
C
$\tan ^{-1} 2$
D
$\tan ^{-1}\left(\frac{2}{3}\right)$

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