1
MHT CET 2023 11th May Evening Shift
+2
-0

Let $$f:[-1,2] \rightarrow[0, \infty)$$ be a continuous function such that $$\mathrm{f}(x)=\mathrm{f}(1-x), \forall x \in[-1,2]$$

Let $$\mathrm{R}_1=\int_{-1}^2 x \mathrm{f}(x) \mathrm{d} x$$ and $$\mathrm{R}_2$$ be the area of the region bounded by $$y=\mathrm{f}(x), x=-1, x=2$$ and the $$\mathrm{X}$$-axis, then $$\mathrm{R}_2$$ is

A
$$\frac{1}{2} R_1$$
B
$$2 R_1$$
C
$$3 R_1$$
D
$$\frac{1}{3} R_1$$
2
MHT CET 2023 11th May Morning Shift
+2
-0

$$\int\limits_0^\pi \frac{d x}{4+3 \cos x}=$$

A
$$\frac{2 \pi}{7}$$
B
$$\frac{\pi}{\sqrt{7}}$$
C
$$\frac{\pi}{2 \sqrt{7}}$$
D
$$\frac{\pi}{7}$$
3
MHT CET 2023 11th May Morning Shift
+2
-0

Let $$\mathrm{f}(x)=\int \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x, x \geq 0$$, then $$\mathrm{f}(3)-\mathrm{f}(1)$$ is equal to

A
$$-\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$$
B
$$-\frac{\pi}{12}+\frac{1}{2}+\frac{\sqrt{3}}{4}$$
C
$$\frac{\pi}{6}+\frac{1}{2}-\frac{\sqrt{3}}{4}$$
D
$$\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$$
4
MHT CET 2023 11th May Morning Shift
+2
-0

Let $$\mathrm{f}(x)$$ be positive for all real $$x$$. If $$\mathrm{I}_1=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x$$ and $$\mathrm{I}_2=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x$$, where $$(2 h-1)>0$$, then $$\frac{I_1}{I_2}$$ is

A
2
B
$$\mathrm{h}$$
C
$$\frac{1}{2}$$
D
1
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