1
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)$$, then $$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)$$ at $$x=0$$ is

A
3
B
5
C
8
D
1
2
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

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

A
$$\sim \mathrm{p} \vee \mathrm{q}$$
B
$$p \wedge q$$
C
$$\mathrm{p} \vee \mathrm{q}$$
D
$$p \vee \sim q$$
3
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The raw data $$x_1, x_2, \ldots \ldots, x_{\mathrm{n}}$$ is an A.P. with common difference $$\mathrm{d}$$ and first term $$0, \bar{x}$$ and $$\sigma^2$$ are mean and variance of $$x_{\mathrm{i}}, \mathrm{i}=1,2, \ldots \ldots \mathrm{n}$$, then $$\sigma^2$$ is

A
$$\frac{\left(n^2+1\right) d^2}{24}$$
B
$$\frac{\left(\mathrm{n}^2-1\right) \mathrm{d}^2}{24}$$
C
$$\frac{\left(\mathrm{n}^2+1\right) \mathrm{d}^2}{12}$$
D
$$\frac{\left(\mathrm{n}^2-1\right) \mathrm{d}^2}{12}$$
4
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The particular solution of differential equation $$\mathrm{e}^{\frac{d y}{d x}}=(x+1), y(0)=3$$ is

A
$$y=x \log x-x+2$$
B
$$y=(x+1) \log (x+1)-x+3$$
C
$$y=(x+1) \log (x+1)+x-3$$
D
$$y=x \log x+x-2$$
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