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

  • Option 1)

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

  • Option 2)

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

  • Option 3)

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

  • Option 4)

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

 

Answers (1)

y^{2}=x     

A(\lambda)=2\int _{0}^{\lambda}\sqrt{x}dx

A(\lambda)=\frac{4}{3}(\lambda)^{\frac{3}{2}}

A(4)=\frac{4}{3}(4)^{\frac{3}{2}}

\frac{A(\lambda)}{A(4)}=\frac{\frac{4}{3}(\lambda)^{\frac{3}{2}}}{\frac{4}{3}(4)^{\frac{3}{2}}}=\frac{2}{5}

(\lambda)^{\frac{3}{2}}=\frac{16}{5}

\lambda = 4\left ( \frac{4}{25} \right )^{\frac{1}{3}}


Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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