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JEE Main 2019 · 8 Apr · Shift II

JEE Main 2019 Mathematics question on Definite Integrals and Areas

Let S(α)={(x,y):y2≤x,0≤x≤α}S(\alpha) = \{(x, y) : y^2 \leq x, 0 \leq x \leq \alpha\} and A(α)A(\alpha) is area of the region S(α)S(\alpha). If for λ\lambda, 0<λ<40 < \lambda < 4, A(λ):A(4)=2:5A(\lambda) : A(4) = 2 : 5, then λ\lambda equals
  1. A.
    4(425)134\left(\dfrac{4}{25}\right)^{\frac{1}{3}}
  2. B.
    4(25)134\left(\dfrac{2}{5}\right)^{\frac{1}{3}}
  3. C.
    4(425)134\left(\dfrac{4}{25}\right)^{\frac{1}{3}}
  4. D.
    4(25)134\left(\dfrac{2}{5}\right)^{\frac{1}{3}}
Show answer and solution

Correct option: C

Solution

A(α) = ∫₀^α √x dx = (2/3)α^(3/2).

So A(λ)/A(4) = λ^(3/2) / 4^(3/2) = λ^(3/2)/8 = 2/5.

Thus λ^(3/2) = 16/5, so λ = (16/5)^(2/3) = 4·(4/25)^(1/3).

[JEE Main 2019 · 8 Apr · Shift II]

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