JEE 2023 — Mathematics PYQ

Step 1: Identify the boundaries of the region.

The region is bounded by:

• The hyperbola: $y=\frac{8}{x}$ (from $xy=8$)

• The parabola: $y=x^2$

• The horizontal line: $y=1$

Step 2: Find the points of intersection.

• Intersection of $y=1$ and $y=x^2$:

  $x^2=1⟹x=1$ (considering the first quadrant where the curves meet).

• Intersection of $y=x^2$ and $y=\frac{8}{x}$:

  $x^2=\frac{8}{x}⟹x^3=8⟹x=2$.

• Intersection of $y=1$ and $y=\frac{8}{x}$:

  $1=\frac{8}{x}⟹x=8$.

Step 3: Set up the integral for the area.

The area can be split into two parts along the x-axis:

1. From $x=1$ to $x=2$, the upper curve is $y=x^2$ and the lower curve is $y=1$.

2. From $x=2$ to $x=8$, the upper curve is $y=\frac{8}{x}$ and the lower curve is $y=1$.

Step 4: Calculate the integrals.

• First Part:

  $${\left[\frac{x^3}{3}-x\right]}_1^2=\left(\frac{8}{3}-2\right)-\left(\frac{1}{3}-1\right)=\frac{2}{3}-\left(-\frac{2}{3}\right)=\frac{4}{3}$$

• Second Part:

  $${\left[8{\log }_{e}x-x\right]}_2^8=(8{\log }_{e}8-8)-(8{\log }_{e}2-2)$$

  Using ${\log }_{e}8=3{\log }_{e}2$:

  $$(24{\log }_{e}2-8)-(8{\log }_{e}2-2)=16{\log }_{e}2-6$$

Step 5: Total Area.

Correct Option: (2)