1
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
The solution of $\dfrac{\text{d}y}{\text{d}x} = \sin(x + y) + \cos(x + y)$ is
A
$\log\left[1 + \tan\left(\dfrac{x+y}{2}\right)\right] = y + c$, where c is the constant of integration
B
$\log\left[1 - \tan\left(\dfrac{x+y}{2}\right)\right] = y + c$, where c is the constant of integration
C
$\log\left[1 + \tan\left(\dfrac{x+y}{2}\right)\right] = x + c$, where c is the constant of integration
D
$\log\left[1 - \tan\left(\dfrac{x+y}{2}\right)\right] = x + c$, where c is the constant of integration
2
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
The order and degree of the differential equation $3 - \left(\dfrac{\text{d}^3 y}{\text{d}x^3}\right)^{\frac{7}{3}} = \left(\dfrac{\text{d}y}{\text{d}x}\right)^5$ are respectively
A
$3, 7$
B
$7, 3$
C
$5, 7$
D
$3, 5$
3
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
The solution of the differential equation $e^{-x}(y+1)\text{d}y + (\cos^2 x - \sin 2x)y\, \text{d}x = 0$, given that $y = 1$ when $x = 0$ is
A
$\log y + \dfrac{1}{y} + e^x \cos^2 x = 1$
B
$\log y + y + e^x\cos^2 x = 2$
C
$(y+1) + e^x\cos^2 x = 2$
D
$\log\left(y + \dfrac{1}{y}\right) + e^x\cos^2 x = 1$
4
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
A unit vector coplanar with $\hat{i} + \hat{j} + 2\hat{k}$ and $\hat{i} + 2\hat{j} + \hat{k}$ and perpendicular to $\hat{i} + \hat{j} + \hat{k}$ is
A
$\pm\dfrac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$
B
$\pm\dfrac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$
C
$\pm\dfrac{1}{\sqrt{2}}(\hat{j} - \hat{k})$
D
$\pm\dfrac{1}{\sqrt{2}}(\hat{j} + \hat{k})$

MHT CET Papers

All year-wise previous year question papers