JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Area Under Curves

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:

Answer: 6

Solution

For region

A_{1}

: The boundary curves are

x = \pm y^{2}

and

∣ x ∣ + 2 y = 8

. In the first quadrant,

x + 2 y = 8

intersects

x = y^{2}

y = 2

(since

y^{2} + 2 y = 8 \Rightarrow y = 2

).

Area of

A_{1}

(by symmetry, double the first quadrant area):

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}] = 2 [\frac{8}{3} + (32 - 16) - (16 - 4)] = 2 \cdot \frac{20}{3} = \frac{40}{3}

For region

A_{2}

∣ x ∣ + ∣ y ∣ \leq k

is a square with vertices at

(\pm k, 0)

and

(0, \pm k)

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

Given:

27 \cdot \frac{40}{3} = 5 \cdot 2 k^{2}

360 = 10 k^{2} \Rightarrow k^{2} = 36 \Rightarrow k = 6

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer