Calculus

Application of Integrals

Hard

The smaller area enclosed by the curve $y = |x|$ and the circle $(x-a)^2 + y^2 = a^2$ is:

$\frac{a^2}{4}(\pi - 1)$

$\frac{a^2}{4}(\pi - 2)$

$\frac{a^2}{2}(\pi - 2)$

$\frac{a^2}{2}(\pi - 1)$

✅ Correct Option: 2

Circle is centered at $(a, 0)$ with radius $a$ . Line $y = x$ meets circle at $(0,0)$ and $(a, a)$ , subtending $90$ degrees at center. Sector area = $\frac{\pi a^2}{4}$ , triangle area = $\frac{a^2}{2}$ . Smaller (segment) area = $\frac{\pi a^2}{4} - \frac{a^2}{2} = \frac{a^2}{4}(\pi - 2)$ .

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The area of the region bounded by the curve $y=|x|$ , x-axis and the lines $x=-5$ and $x=5$ is

The smaller area enclosed by the curve y=∣x∣y = |x|y=∣x∣ and the circle (x−a)2+y2=a2(x-a)^2 + y^2 = a^2(x−a)2+y2=a2 is:

Solution

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Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The smaller area enclosed by the curve $y = |x|$ and the circle $(x-a)^2 + y^2 = a^2$ is: