If the area enclosed between the curves y = k | vedclass

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(B) Given curves are $y = kx^2$ and $x = ky^2$ where $k > 0$.To find the intersection points,substitute $y = kx^2$ into $x = ky^2$:$x = k(kx^2)^2 = k^

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$\frac{1}{\sqrt{3}}$

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(B) Given curves are $y = kx^2$ and $x = ky^2$ where $k > 0$.To find the intersection points,substitute $y = kx^2$ into $x = ky^2$:$x = k(kx^2)^2 = k^3 x^4$$x(k^3 x^3 - 1) = 0$So,$x = 0$ or $x^3 = \frac{1}{k^3}$,which gives $x = 0$ or $x = \frac{1}{k}$.The intersection points are $(0, 0)$ and $(\frac{1}{k}, \frac{1}{k})$.The area $A$ enclosed between the curves is given by:$A = \int_{0}^{1/k} (\sqrt{\frac{x}{k}} - kx^2) dx = 1$$A = \left[ \frac{1}{\sqrt{k}} \cdot \frac{x^{3/2}}{3/2} - \frac{kx^3}{3} \right]_{0}^{1/k} = 1$$A = \left( \frac{2}{3\sqrt{k}} \cdot (\frac{1}{k})^{3/2} - \frac{k}{3} \cdot (\frac{1}{k})^3 \right) = 1$$A = \frac{2}{3k^2} - \frac{1}{3k^2} = \frac{1}{3k^2} = 1$$3k^2 = 1 \Rightarrow k^2 = \frac{1}{3}$Since $k > 0$,we have $k = \frac{1}{\sqrt{3}}$.

If the area enclosed between the curves $y = kx^2$ and $x = ky^2$ $(k > 0)$ is $1$ square unit,then $k$ is

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