1
JEE Main 2023 (Online) 29th January Evening Shift
+4
-1

The area of the region $$A = \left\{ {(x,y):\left| {\cos x - \sin x} \right| \le y \le \sin x,0 \le x \le {\pi \over 2}} \right\}$$ is

A
$$\sqrt 5 + 2\sqrt 2 - 4.5$$
B
$$1 - {3 \over {\sqrt 2 }} + {4 \over {\sqrt 5 }}$$
C
$$\sqrt 5 - 2\sqrt 2 + 1$$
D
$${3 \over {\sqrt 5 }} - {3 \over {\sqrt 2 }} + 1$$
2
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1

Let $$\Delta$$ be the area of the region $$\left\{ {(x,y) \in {R^2}:{x^2} + {y^2} \le 21,{y^2} \le 4x,x \ge 1} \right\}$$. Then $${1 \over 2}\left( {\Delta - 21{{\sin }^{ - 1}}{2 \over {\sqrt 7 }}} \right)$$ is equal to

A
$$2\sqrt 3 - {1 \over 3}$$
B
$$2\sqrt 3 - {2 \over 3}$$
C
$$\sqrt 3 - {4 \over 3}$$
D
$$\sqrt 3 - {2 \over 3}$$
3
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1

Let $$[x]$$ denote the greatest integer $$\le x$$. Consider the function $$f(x) = \max \left\{ {{x^2},1 + [x]} \right\}$$. Then the value of the integral $$\int\limits_0^2 {f(x)dx}$$ is

A
$${{5 + 4\sqrt 2 } \over 3}$$
B
$${{4 + 5\sqrt 2 } \over 3}$$
C
$${{8 + 4\sqrt 2 } \over 3}$$
D
$${{1 + 5\sqrt 2 } \over 3}$$
4
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1

Let $$A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\}$$ and

$$B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text {. }$$.

Then the ratio of the area of A to the area of B is

A
$$\frac{\pi}{\pi+1}$$
B
$$\frac{\pi-1}{\pi+1}$$
C
$$\frac{\pi}{\pi-1}$$
D
$$\frac{\pi+1}{\pi-1}$$
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