1
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

$\tan \left(\cos ^{-1} \frac{1}{\sqrt{2}}+\tan ^{-1} \frac{1}{2}\right)=$

A
1
B
2
C
3
D
4
2
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $x^2 y^2=\sin ^{-1} x+\cos ^{-1} x$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ and $y=2$ is

A
$\frac{1}{2}$
B
$2$
C
$-\frac{1}{2}$
D
$-2$
3
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\overline{\mathrm{a}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$, then the vector $\overline{\mathrm{b}}$ satisfying $\overline{\mathrm{a}} \times \overline{\mathrm{b}}+\overline{\mathrm{c}}=\overline{0}$ and $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=3$ is

A
$-\hat{i}+\hat{j}-2 \hat{k}$
B
$-\hat{i}+\hat{j}-\hat{k}$
C
$-\hat{i}-\hat{j}+\hat{k}$
D
$\hat{i}+\hat{j}+\hat{k}$
4
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The expression $((p \wedge q) \vee(p \vee \sim q)) \wedge(\sim p \wedge \sim q)$ is equivalent to

A
$\mathrm{p} \wedge \mathrm{q}$
B
$\mathrm{p} \vee \sim \mathrm{q}$
C
$\mathrm{p} \wedge \sim \mathrm{q}$
D
$(\sim p) \wedge(\sim q)$
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