Question

The function f(x)= $x\sqrt{1-x^2},(x>0)$ has

• A. A local maxima
• B. A local minima
• C. Neither a local maxima nor a local minima
• D. None of these

Answer 1

The correct option is A A local maxima

Let $f\left(x\right)=x\sqrt{1-x^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1-2x^2}{\sqrt{1-x^2}}=0\Rightarrow x=\pm \frac{1}{\sqrt{2}}$
But as x > 0, we have $x=\frac{1}{\sqrt{2}}$
Now, again $f^{\prime \prime }\left(x\right)=\frac{\sqrt{1-x^2}(-4x)-(1-2x^2)\frac{-x}{\sqrt{1-x^2}}}{(1-x^2)}$
$=\frac{2x^3-3x}{(1-x^2{)}^{3/2}}$
$\Rightarrow f"\left(\frac{1}{\sqrt{2}}\right)=-ve$. Then f(x)is maximum at $x=\frac{1}{\sqrt{2}}$