If the solve the problem


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

  1. $2\left(\frac{4}{25}\right)^{\frac{1}{3}}$

  2. $4\left(\frac{4}{25}\right)^{\frac{1}{3}}$

  3. $2\left(\frac{2}{5}\right)^{\frac{1}{3}}$

  4. $4\left(\frac{2}{5}\right)^{\frac{1}{3}}$

Correct Option:


$\mathrm{S}(\alpha)=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{y}^{2} \leq \mathrm{x}, 0 \leq \mathrm{x} \leq \alpha\right\}$

$\mathrm{A}(\alpha)=2 \int_{0}^{\alpha} \sqrt{\mathrm{x}} \mathrm{dx}=2 \alpha^{\frac{3}{2}}$

$A(4)=2 \times 4^{3 / 2}=16$

$A(\lambda)=2 \times \lambda^{3 / 2}$

$\frac{\mathrm{A}(\lambda)}{\mathrm{A}(4)}=\frac{2}{5} \Rightarrow \lambda=4 \cdot\left(\frac{4}{25}\right)^{1 / 3}$

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