1
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the vectors $\overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\mathrm{pi}+\hat{\mathrm{j}}+\mathrm{q} \hat{\mathrm{k}}$ are mutually orthogonal, then $(p, q)$ is equal to

A
$(3,-2)$
B
$(-2,3)$
C
$(-3,2)$
D
$(2,-3)$
2
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\bar{u}, \bar{v}$ and $\bar{w}$ are three non-coplanar vectors, then $(\bar{u}+\bar{v}-\bar{w}) \cdot[(\bar{u}-\bar{v}) \times(\bar{v}-\bar{w})]$ is equal to

A
$\overline{\mathrm{u}} \cdot(\overline{\mathrm{v}} \times \overline{\mathrm{w}})$
B
$\overline{\mathrm{u}} \cdot(\overline{\mathrm{w}} \times \overline{\mathrm{v}})$
C
$3 \bar{u} \cdot(\bar{v} \times \bar{w})$
D
$0$
3
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The value of $k$, if the slope of one of the lines given by $4 x^2+k x y+y^2=0$ is four times that of the other, is given by

A
4
B
2.5
C
5
D
1
4
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation of $y=\mathrm{e}^x\left(\mathrm{a}+\mathrm{bx}+x^2\right)$ is

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