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

The position vectors of the points $A, B, C$ are $\hat{i}+2 \hat{j}-\hat{k}, \hat{i}+\hat{j}+\hat{k}, 2 \hat{i}+3 \hat{j}+2 \hat{k}$ respectively. If $A$ is chosen as the origin, then the cross product of position vectors of $B$ and $C$ are

A
$-5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
B
$-\hat{\mathrm{i}}+0 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
C
$\hat{\mathrm{i}}-\hat{\mathrm{k}}$
D
$5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
2
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the area of a parallelogram whose diagonals are represented by vectors $3 \hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ is $\frac{\sqrt{117}}{2}$ sq. units, then $\lambda=$

A
-1
B
-2
C
-3
D
-4
3
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

$y=\mathrm{e}^x(\mathrm{~A} \cos x+\mathrm{B} \sin x)$ is the solution of the differential equation

A
$x^2 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+\left(1+y^2\right)=0$
B
$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-\frac{\mathrm{d} y}{\mathrm{~d} x}+y=0$
C
$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y=0$
D
$x \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y=0$
4
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

A family has 3 children. The probability that all the three children are girls, given that at least one of them is a girl is

A
$\frac{7}{8}$
B
$\frac{1}{8}$
C
$\frac{1}{7}$
D
$\frac{2}{7}$
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