1
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1

The area enclosed by the curves $$y=\log _{e}\left(x+\mathrm{e}^{2}\right), x=\log _{e}\left(\frac{2}{y}\right)$$ and $$x=\log _{\mathrm{e}} 2$$, above the line $$y=1$$ is:

A
$$2+\mathrm{e}-\log _{\mathrm{e}} 2$$
B
$$1+e-\log _{e} 2$$
C
$$e-\log _{e} 2$$
D
$$1+\log _{e} 2$$
2
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1

The area of the region enclosed by $$y \leq 4 x^{2}, x^{2} \leq 9 y$$ and $$y \leq 4$$, is equal to :

A
$$\frac{40}{3}$$
B
$$\frac{56}{3}$$
C
$$\frac{112}{3}$$
D
$$\frac{80}{3}$$
3
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1

Consider a curve $$y=y(x)$$ in the first quadrant as shown in the figure. Let the area $$\mathrm{A}_{1}$$ is twice the area $$\mathrm{A}_{2}$$. Then the normal to the curve perpendicular to the line $$2 x-12 y=15$$ does NOT pass through the point.

A
(6, 21)
B
(8, 9)
C
(10, $$-$$4)
D
(12, $$-$$15)
4
JEE Main 2022 (Online) 27th July Morning Shift
+4
-1

The area of the smaller region enclosed by the curves $$y^{2}=8 x+4$$ and $$x^{2}+y^{2}+4 \sqrt{3} x-4=0$$ is equal to

A
$$\frac{1}{3}(2-12 \sqrt{3}+8 \pi)$$
B
$$\frac{1}{3}(2-12 \sqrt{3}+6 \pi)$$
C
$$\frac{1}{3}(4-12 \sqrt{3}+8 \pi)$$
D
$$\frac{1}{3}(4-12 \sqrt{3}+6 \pi)$$
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