Exam

JEE Main

Year

2019

Date

8 April

Shift

Evening Shift

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2019 · 8 April · Evening Shift · Mathematics · Area Under Curves

Let S(α) = \ (x, y): y ^ 2 ≤ x, 0 ≤ x ≤ α\ and A(α) is area of the...

Let $S\left(\alpha\right) = \left\{\left(x, y\right): {y}^{2} \le x, 0 \le x \le \alpha\right\}$ and $A\left(\alpha\right)$ is area of the region $S\left(\alpha\right) .$ If for a $\lambda, 0 < \lambda < 4, A\left(\lambda\right): A\left(4\right) = 2: 5,$then $\lambda$ equals:

Step-by-step solution

$S$\left(\lambda\right) = 2\mathop{∫}_{0}^{\lambda}\sqrt{x}dx = 2{\left[\dfrac{{x}^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right]}_{0}^{\lambda}
$= \dfrac{4}{3}{\lambda}^{\dfrac{3}{2}}$
$∴ S$\left(4\right) = \dfrac{4}{3}{\left(4\right)}^{\dfrac{3}{2}}
Given \dfrac{S\left(\lambda\right)}{S\left(4\right)} = \dfrac{2}{5} ⇒ \dfrac{{\lambda}^{\dfrac{3}{2}}}{{4}^{\dfrac{3}{2}}} = \dfrac{2}{5}
$⇒ \lambda = 4{\left(\dfrac{4}{25}\right)}^{\dfrac{1}{3}}$