question_answer

If the area enclosed between the curves \[y=k{{x}^{2}}\] and \[x=k{{y}^{2}},\text{ }\left( k>0 \right)\], is 1 square unit. Then k is- [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

A) \[\sqrt{3}\]                                            

B) \[\frac{\sqrt{3}}{2}\]

C) \[\frac{2}{\sqrt{3}}\]             

D)                  \[\frac{1}{\sqrt{3}}\]

Correct Answer: D

Solution :

\[y=k{{x}^{2}}\] \[x=k{{y}^{2}}\Rightarrow x=k{{(k{{x}^{2}})}^{2}}\Rightarrow \,\,x={{k}^{3}}{{x}^{4}}\Rightarrow x=0\] \[x=\frac{1}{k}\] Area \[=\,\,\int\limits_{0}^{1/k}{\sqrt{\frac{x}{k}}-k{{x}^{2}}dx}\] \[=\,\,\frac{1}{\sqrt{k}}\frac{{{x}^{3/2}}}{3/2}\,-\,k\frac{{{x}^{3}}}{3}_{0}^{1/k}\,=\,1\] \[\Rightarrow \,\,\,\frac{2}{3{{k}^{2}}}\,-\,\,\frac{1}{3{{k}^{2}}}\,\,=\,\,1\] \[{{k}^{2}}\,=\,\frac{1}{3}\,\,\Rightarrow \,\,k=\frac{1}{\sqrt{3}}\]