The area of the smaller region in first quadrant bounded by x^2+y^2=4 and x=3/2 is: Option 1: /2 Sq.units Option 2: -3sqrt(7)8-2^-1(3/4) Sq.units Option 3: 2(-2) Sq.units Option 4: /2+3sqrt(7)8 Sq.units

The answer is '-3sqrt(7)8-2^-1(3/4) Sq.units'. View full solution:

CUET Mathematics - The area of the smaller region in first quadrant bounded by x^2+y^2=4 and x=3/2 is: | PYQs + Solutions

CUET Mathematics 2026 18th May Shift 2

Calculus

Hard

core

$\dfrac{\pi}{2}$ Sq.units

$\pi-\dfrac{3\sqrt{7}}{8}-2\sin^{-1}\left(\dfrac{3}{4}\right)$ Sq.units

$2(\pi-2)$ Sq.units

$\dfrac{\pi}{2}+\dfrac{3\sqrt{7}}{8}$ Sq.units

✅ Correct Option: 2