Let be the area of the region { ( x , y ) 2 : x 2 + y 2 21 , y 2 4 x , x 1 } . Then 1 2 ( - 21 sin - 1 2 7 ) is equal to [2023]

Question

$Let Δ be the area of the region {(x, y) \in ℝ^{2} : x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1} .$ $Then \frac{1}{2} (Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}}) is equal to$ [2023]

Smart Achievers · Accepted Answer

(4)

If $Δ$ be the area of region ${(x, y) \in ℝ^{2} : x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1} .$ Then we have,

$Area = 2 \int_{1}^{3} 2 \sqrt{x} d x + 2 \int_{3}^{\sqrt{21}} \sqrt{(21 - x^{2})} d x$

$Δ = \frac{8}{3} \cdot (3 \sqrt{3} - 1) + 21 \sin^{- 1} (\frac{2}{\sqrt{7}}) - 6 \sqrt{3}$

$\Rightarrow [Δ - 21 \sin^{- 1} (\frac{2}{\sqrt{7}})] = \frac{2 \sqrt{3} - \frac{8}{3}}{2}$

$\Rightarrow \frac{1}{2} [Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}}] = \sqrt{3} - \frac{4}{3}$

Solution

Q.
$Let Δ be the area of the region {(x, y) \in ℝ^{2} : x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1} .$ $Then \frac{1}{2} (Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}}) is equal to$ [2023]

1 $\sqrt{3} - \frac{2}{3}$

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

3 $2 \sqrt{3} - \frac{1}{3}$

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

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let Δ be the area of the region {(x,y)∈ℝ2:x2+y2≤21, y2≤4x, x≥1}. Then 12(Δ-21sin-127) is equal to [2023]

1 3-23

2 23-23

3 23-13

4 3-43