If the area enclosed between the curves

Question:

If the area enclosed between the curves $y=\mathrm{k} x^{2}$ and $x=\mathrm{k} y^{2},(\mathrm{k}>0)$, is 1 square unit. Then $\mathrm{k}$ is:

  1. (1) $\frac{\sqrt{3}}{2}$

  2. (2) $\frac{1}{\sqrt{3}}$

  3. (3) $\sqrt{3}$

  4. (4) $\frac{2}{\sqrt{3}}$


Correct Option: , 2

Solution:

Two curves will intersect in the Ist quadrant at

$A\left(\frac{1}{k}, \frac{1}{k}\right)$

$\because$ area of shaded region $=1 .$

$\therefore \int_{0}^{\frac{1}{k}}\left(\frac{\sqrt{x}}{\sqrt{k}}-k x^{2}\right) d x=1$

$\Rightarrow\left(\frac{1}{\sqrt{k}} \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)_{0}^{\frac{1}{k}}-\left(k \cdot \frac{x^{3}}{3}\right)_{0}^{\frac{1}{k}}=1$

$\Rightarrow \frac{2}{3 \sqrt{k}} \cdot \frac{1}{k^{\frac{3}{2}}}-\frac{k}{3 k^{3}}=1$

$\Rightarrow \frac{2}{3 k^{2}}-\frac{1}{3 k^{2}}=1$

$\Rightarrow 3 k^{2}=1$

$\Rightarrow k=\pm \frac{1}{\sqrt{3}}$

$\therefore \quad k=\frac{1}{\sqrt{3}}$ $(\because k>0)$

 

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