1
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$\int \frac{\mathrm{e}^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \mathrm{d} x, x > 0=$$

A
$$\left(\tan ^{-1} x\right)^2 \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
B
$$\left(\tan ^{-1} x\right) \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
C
$$\left(\tan ^{-1} x\right) \mathrm{e}^{2 \tan ^{-1} x}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
D
$$\left(\tan ^{-1} x\right)^2 \mathrm{e}^{2 \tan ^{-1} x}+c$$, where $$\mathrm{c}$$ is a constant of integration.
2
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$\mathrm{f}(x)$$ is a function satisfying $$\mathrm{f}^{\prime}(x)=\mathrm{f}(x)$$ with $$\mathrm{f}(0)=1$$ and $$\mathrm{g}(x)$$ is a function that satisfies $$\mathrm{f}(x)+\mathrm{g}(x)=x^2$$. Then the value of the integral $$\int_\limits0^1 f(x) g(x) d x$$ is

A
$$e-\frac{e^2}{2}-\frac{5}{2}$$
B
$$\mathrm{e}+\frac{\mathrm{e}^2}{2}-\frac{3}{2}$$
C
$$\mathrm{e}-\frac{\mathrm{e}^2}{2}-\frac{3}{2}$$
D
$$\mathrm{e}+\frac{\mathrm{e}^2}{2}+\frac{5}{2}$$
3
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The foot of the perpendicular drawn from the origin to the plane is $$(4,-2,5)$$, then the Cartesian equation of the plane is

A
$$4 x-2 y+5 z=45$$
B
$$-4 x+2 y+5 z=45$$
C
$$4 x-2 y+5 z+45=0$$
D
$$4 x+2 y-5 z+45=0$$
4
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$\lim _\limits{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$$ equals

A
$$\frac{1}{24}$$
B
$$\frac{1}{16}$$
C
$$\frac{1}{8}$$
D
$$\frac{1}{4}$$
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