Let A_1=x,y:x≤y^2,x+2y≤8 and A_2=x,y:x+y≤k. If 27 (Area A_1) =5(Area A_2), then k is equal to

Correct answer: 6. A_1=x,y:x≤y^2,x+2y≤8 and A_2=x,y:x+y≤k Area A_1=2∫02y^2dy+∫248-2ydy =2y^3302+8y-y^224 Area A_1=2×20/3=40/3 Area A_2=4×1/2k^2 Area A_2=2k^2 Given 27Area A_1=5Area A_2 ⇒27×40/3=5×2k^2 ⇒k=6

Q53 / 90 · JEE Main 2022 (27 Jun Shift 1) · Mathematics

Let $A_{1} = \{(x, y) : |x| \leq y^{2}, |x| + 2 y \leq 8\}$ and $A_{2} = \{(x, y) : |x| + |y| \leq k\} .$ If $27$ (Area $A_{1}$ ) $= 5$ (Area $A_{2}$ ), then $k$ is equal to

Correct answer: 6

Solution

$A_{1} = \{(x, y) : |x| \leq y^{2}, |x| + 2 y \leq 8\}$ and

$A_{2} = \{(x, y) : |x| + |y| \leq k\}$

Area $(A_{1}) = 2 [\int_{0}^{2} y^{2} d y + \int_{2}^{4} (8 - 2 y) d y]$

$= 2 [{(\frac{y^{3}}{3})}_{0}^{2} + {(8 y - y^{2})}_{2}^{4}]$

Area $(A_{1}) = 2 \times \frac{20}{3} = \frac{40}{3}$

Area $(A_{2}) = 4 \times \frac{1}{2} k^{2}$

Area $(A_{2}) = 2 k^{2}$

Given $27 (Area A_{1}) = 5 (Area A_{2})$

$\Rightarrow 27 \times \frac{40}{3} = 5 \times 2 k^{2}$

$\Rightarrow k = 6$

Hence, the correct answer is 6.

Let A1=x,y:x≤y2,x+2y≤8 and A2=x,y:x+y≤k. If 27 (Area A1) =5(Area A2), then k is equal to

Solution

Relevant Questions

Let $A_{1} = \{(x, y) : |x| \leq y^{2}, |x| + 2 y \leq 8\}$ and $A_{2} = \{(x, y) : |x| + |y| \leq k\} .$ If $27$ (Area $A_{1}$ ) $= 5$ (Area $A_{2}$ ), then $k$ is equal to