Question

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

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

Answer 1

The correct option is A

A local maxima

The explanation for the correct answer.

Step 1: Solve for the point of extremum of a function $x\sqrt{1-x^2},(x>0)$

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

$\frac{d}{dx}\left(u\times v\right)=u\times \frac{dv}{dx}+v\times \frac{du}{dx}$

$$\begin{gathered} f\text{'}\left(x\right)=x\times \frac{1}{2}\times \frac{(-2x)}{\sqrt{1-x^2}}+\sqrt{1-x^2} \\ f\text{'}\left(x\right)=\frac{1-2x^2}{\sqrt{1-x^2}} \end{gathered}$$

Equating $f\text{'}\left(x\right)$ with $0$ we get

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

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

But as $x>0,$we have $x=\frac{1}{\sqrt{2}}$

Step 2: Solve for the second derivative.

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{\left[\sqrt{1-x^2}\right]\left(-4x\right)-\left(1-2x^2\right)\left[{\displaystyle \frac{-x}{\sqrt{1-x^2}}}\right]}{1-x^2} \\ f\text{'}\text{'}\left(x\right)=\frac{\left(1-x^2\right)(-4x)-(1-2x^2)(-x)}{(1-x^2)\sqrt{1-x^2}} \\ f\text{'}\text{'}\left(x\right)=\frac{\left[2x^3-3x\right]}{{\left(1-x^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ f\text{'}\text{'}\left(\frac{1}{\sqrt{2}}\right)=\frac{-\sqrt{2}}{{\displaystyle \frac{1}{2\sqrt{2}}}}=-4 \\ f\text{'}\text{'}\left(\frac{1}{\sqrt{2}}\right)=-ve \end{gathered}$$ $...\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\times {\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}\right]$

Hence, conditions for maxima are satisfied

Then $f\left(x\right)$ is maximum at $x=\frac{1}{\sqrt{2}}$

Hence, option(A) is the correct answer.