1
MHT CET 2023 9th May Morning Shift
+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
$$\mathrm{e}+\frac{\mathrm{e}^2}{2}+\frac{5}{2}$$
2
MHT CET 2022 11th August Evening Shift
+2
-0

$$\int_\limits{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\mathrm{d} x}{1+\cos x}$$ is equal to

A
$$-2 \sqrt{2}$$
B
$$2$$
C
$$-2-2 \sqrt{2}$$
D
$$-2$$
3
MHT CET 2022 11th August Evening Shift
+2
-0

The value of the integral $$\int_\limits0^1 \sqrt{\frac{1-x}{1+x}} \mathrm{~d} x$$ is

A
$$\left(\frac{\pi}{2}\right)+1$$
B
$$\left(\frac{\pi}{2}\right)-1$$
C
1
D
$$-$$1
4
MHT CET 2022 11th August Evening Shift
+2
-0

$$\int_\limits0^2[x] \mathrm{d} x+\int_\limits0^2|x-1| \mathrm{d} x=$$

(where $$[x]$$ denotes the greatest integer function.)

A
4
B
3
C
1
D
2
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