Question

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

• A. a local maxima
• B. a local minima
• C. neither a local maxima noe a local minima
• D. none of these

Answer 1

The correct option is A a local maxima

Given $f\left(x\right)=x\sqrt{1-x^2}$

$f^{\prime }\left(x\right)=x.\frac{1}{2\sqrt{1-x^2}}.(-2x)+\sqrt{1-x^2}$

$=\frac{-2x^2}{2\sqrt{1-x^2}}+\sqrt{1-x^2}$

$=\frac{-x^2}{\sqrt{1-x^2}}+\sqrt{1-x^2}$

$=\frac{-x^2}{\sqrt{1-x^2}}+\sqrt{1-x^2}$

$=\frac{-x^2+1-x^2}{\sqrt{1-x^2}}$

If f'(x)=0

$\Rightarrow \frac{1-2x^2}{\sqrt{1-x^2}}=0$

$\Rightarrow 1-2x^2=0$

$\Rightarrow 1=2x^2$

$\Rightarrow x^2=\frac{1}{2}$

$\Rightarrow x=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow x=\frac{1}{\sqrt{2}}$ $(x=-\frac{1}{\sqrt{2}}$ is reglcted since x > 0)

$f^{\prime \prime }\left(x\right)=\frac{\sqrt{1-x^2}.(-4x)-(1-2x^2).\frac{1}{2\sqrt{1-x^2}}.-2x}{(1-x^2)}$

$=\frac{-4x\sqrt{1-x^2}+\frac{x(1-2x^2)}{\sqrt{1-x^2}}}{(1-x^2)}$

$=\frac{-4x(1-x^2)+x-2x^3}{(1-x^2)\sqrt{1-x^2}}$

$=\frac{-4x+4x^3+x-2x^3}{(1-x^2)\sqrt{1-x^2}}$

$=\frac{2x^3-3x}{(1-x^2)\sqrt{1-x^2}}$

At $x=\frac{1}{\sqrt{2}},f^{\prime \prime }\left(\frac{1}{\sqrt{2}}\right)=\frac{2(\frac{1}{\sqrt{2}}{)}^3-3(\frac{1}{\sqrt{2}})}{\{1-(\frac{1}{\sqrt{2}}{)}^2\}\sqrt{1-(\frac{1}{\sqrt{2}}{)}^2}}$

$=\frac{\frac{2}{2\sqrt{2}}-\frac{3}{\sqrt{2}}}{(1-\frac{1}{2})\sqrt{1-\frac{1}{2}}}$

$=\frac{\frac{2-6}{2\sqrt{2}}}{\frac{1}{2}.\frac{1}{\sqrt{2}}}$

= -4 < 0

$\therefore x=\frac{1}{\sqrt{2}}$ is a local maxima.