Let Δ be the area of the region (x, y) ∈ R 2: x2+y2 ≤ 21, y2 ≤ 4 x, x ≥ 1 . Then (1/2)(Δ-21 sin -1 (2/√7)) is equal to

Question

Let Δ be the area of the region (x, y) ∈ R 2: x2+y2 ≤ 21, y2 ≤ 4 x, x ≥ 1 . Then (1/2)(Δ-21 sin -1 (2/√7)) is equal to (A) √3-(2/3) (B) 2 √3-(2/3) (C) 2 √3-(1/3) (D) √3-(4/3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b258a90583aac148d658237ba23d929f-.png" /> Area 2 ∫ limits13 2 √x d x+2 ∫ limits3√21 √21-x2 d x Δ=(8/3)(3 √3-1)+21 sin -1((2/√7))-6 √3 (1/2)(Δ-21 sin -1((2/√7)))=(2 √3-(8/3)/2) =√3-(4/3)

Q. Let $Δ$ be the area of the region ${(x, y) \in R^{2} : x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1}$ . Then $\frac{1}{2} (Δ - 21 sin^{- 1} \frac{2}{7})$ is equal to

$3 - \frac{2}{3}$

$23 - \frac{2}{3}$

$23 - \frac{1}{3}$

$3 - \frac{4}{3}$

Area $2 1 \int 3 2 x d x + 2 3 \int 21 21 - x^{2} d x$
$Δ = \frac{8}{3} (33 - 1) + 21 sin^{- 1} (\frac{2}{7}) - 63$
$\frac{1}{2} (Δ - 21 sin^{- 1} (\frac{2}{7})) = \frac{2 3 - \frac{8}{3}}{2}$
$= 3 - \frac{4}{3}$

Q. Let Δ be the area of the region {(x,y)∈R2:x2+y2≤21,y2≤4x,x≥1}. Then 21​(Δ−21sin−17​2​) is equal to

3​−32​

23​−32​

23​−31​

3​−34​