The area of the region given by $\left\{(\mathrm{x}, \mathrm{y}): \mathrm{xy} \leq 8,1 \leq \mathrm{y} \leq \mathrm{x}^{2}\right\}$ is :

(B) $16 \log _{e} 2-\frac{14}{3}$

(A) $8 \log _{\mathrm{e}} 2-\frac{13}{3}$

(C) $8 \log _{e} 2+\frac{7}{6}$

(D) $16 \log _{e} 2+\frac{7}{3}$

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Area Under Curves

The area of the region given by

{(x, y) : xy \leq 8, 1 \leq y \leq x^{2}}

is :

Answer: (B)

The region requires

y \geq 1

y \leq x^{2}

and

x y \leq 8

, forcing

x \geq 1

. The curve

y = x^{2}

meets

y = 1

x = 1

and meets

x y = 8

x = 2

, while

x y = 8

meets

y = 1

x = 8

. So for

1 \leq x \leq 2

the ceiling is

y = x^{2}

, and for

2 \leq x \leq 8

it is

y = \frac{8}{x}

, the floor being

y = 1

throughout. The area is

\int_{1}^{2} (x^{2} - 1) d x + \int_{2}^{8} (\frac{8}{x} - 1) d x = \frac{4}{3} + (16 ln 2 - 6)

. Therefore the area of the region is

16 lo g_{e} 2 - \frac{14}{3}

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option