1
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Derivative of $\sin ^2 x$ with respect to $e^{\cos x}$

A
$2 \sin x \cos ^2 x e^{\cos x}$
B
$\frac{2 \cos x}{\mathrm{e}^{\cos x}}$
C
$\frac{2 \sin x}{\mathrm{e}^{\cos x}}$
D
$\frac{-2 \cos x}{e^{\cos x}}$
2
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

On which of the following lines lies the point of intersection of the line, $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ ?

A
$\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$
B
$\frac{x-4}{1}=\frac{y-5}{1}=\frac{z-5}{-1}$
C
$\frac{x-2}{2}=\frac{y-3}{2}=\frac{z+3}{3}$
D
$\frac{x+3}{3}=\frac{4-y}{3}=\frac{z+1}{-2}$
3
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$\sim[(\mathrm{p} \vee \sim \mathrm{q}) \rightarrow(\mathrm{p} \wedge \sim \mathrm{q})] \equiv$$

A
$(p \wedge \sim q) \wedge(\sim p \vee q)$
B
$(p \wedge \sim q) \wedge(\sim p \wedge q)$
C
$(p \vee \sim q) \wedge(\sim p \vee q)$
D
$(p \vee \sim q) \vee(\sim p \vee q)$
4
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $z^2+z+1=0$ then $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2=$ where $z=w=$ complex cube root of unity

A
4
B
1
C
5
D
2
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