MHT CET 2023 9th May Morning Shift | MHT CET Year Wise Previous Years Questions

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

$$\left(\tan ^{-1} x\right)^2 \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$\left(\tan ^{-1} x\right) \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$\left(\tan ^{-1} x\right) \mathrm{e}^{2 \tan ^{-1} x}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$\left(\tan ^{-1} x\right)^2 \mathrm{e}^{2 \tan ^{-1} x}+c$$, where $$\mathrm{c}$$ is a constant of integration.

MHT CET 2023 9th May Morning Shift

MCQ (Single Correct Answer)

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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

$$e-\frac{e^2}{2}-\frac{5}{2}$$

$$\mathrm{e}+\frac{\mathrm{e}^2}{2}-\frac{3}{2}$$

$$\mathrm{e}-\frac{\mathrm{e}^2}{2}-\frac{3}{2}$$

$$\mathrm{e}+\frac{\mathrm{e}^2}{2}+\frac{5}{2}$$

MHT CET 2023 9th May Morning Shift

MCQ (Single Correct Answer)

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

$$4 x-2 y+5 z=45$$

$$-4 x+2 y+5 z=45$$

$$4 x-2 y+5 z+45=0$$

$$4 x+2 y-5 z+45=0$$