If the area enclosed between the curves y=kx2 and x=ky2,(k>0), is 1 square unit. Then k is
A
√32
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B
1√3
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C
√3
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D
2√3
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Solution
The correct option is B1√3 Point of intersection of the two curves are - x=k(kx2)2⇒x=0,1k⇒y=0,1k Area =1k∫0√x√kdx−1k∫0kx2dx=1 ⇒1√k⋅⎡⎢
⎢
⎢⎣x3232⎤⎥
⎥
⎥⎦1k0−k⋅[x33]1k0=1 ⇒23√k⋅1k32−k3k3=1 ⇒23k2−13k2=1 ⇒k=±1√3 As k>0, so k=1√3