1
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$. Let $\overline{\mathrm{c}}$ be a vector such that $|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=3$ and $|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=3$ and the angle between $\bar{c}$ and $\bar{a} \times \bar{b}$ is $30^{\circ}$, then $\bar{a} \cdot \bar{c}$ is equal to

A
$\frac{2 \sqrt{2}}{3}$
B
5
C
$-\frac{1}{8}$
D
2
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\int\left(\frac{4 e^x-25}{2 e^x-5}\right) d x=A x+B \log \left(2 e^x-5\right)+c \quad$ (where c is a constant of integration) then

A
$\mathrm{A}=5, \mathrm{~B}=3$
B
$\mathrm{A}=5, \mathrm{~B}=-3$
C
$\mathrm{A}=-5, \mathrm{~B}=3$
D
$\mathrm{A}=-5, \mathrm{~B}=-3$
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The converse of $[p \wedge(\sim q)] \rightarrow r$ is

A
$\sim \mathrm{r} \rightarrow(\sim \mathrm{p} \vee \mathrm{q})$
B
$\mathrm{r} \rightarrow(\sim \mathrm{p} \wedge \sim \mathrm{q})$
C
$(\sim p \vee q) \rightarrow \sim r$
D
$\mathrm{r} \rightarrow(\mathrm{p} \wedge \mathrm{q})$
4
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation obtained by eliminating arbitrary constant from the equation $y^2=(x+c)^3$ is

A
$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3=27 y$
B
$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3=-27 y$
C
$8\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3=27 y$
D
$ 8\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3+27 y=0$
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