Question

The denominator of a fraction number is greater than $16$ of the square of the numerator, then the least value of the number is

• A. $\frac{-1}{4}$
• B. $-\frac{1}{8}$
• C. $\frac{1}{2}$
• D. $\frac{1}{16}$

Answer 1

The correct option is B

$-\frac{1}{8}$

Explanation for the correct option:

Step 1: Forming the function and its derivative

Let the numerator of the given number be $x$.

Then the given number is $\frac{x}{x^2+16}$

Let $f\left(x\right)=\frac{x}{x^2+16}$

Finding the value of $x$ by taking $f^{\text{'}}\left(x\right)=0$

Now $f^{\text{'}}\left(x\right)=\frac{\left(x^2+16\right)1-x\left(2x\right)}{{\left(x^2+16\right)}^2}\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=u\text{'}v-v\text{'}u\right]$

$=\frac{-x^2+16}{{\left(x^2+16\right)}^2}$

Step 2: Applying the condition of maxima and minima

Substituting $f^{\text{'}}\left(x\right)=0$

$\Rightarrow \frac{-x^2+16}{{\left(x^2+16\right)}^2}=0$

$\Rightarrow x^2=16$

$\Rightarrow x=\pm 4$

Step 3: Finding the least value of function using the double derivative rule

Finding $f^{\text{'}\text{'}}\left(x\right)$

$f^{\text{'}\text{'}}\left(x\right)=\frac{{\left(x^2+16\right)}^2\left(-2x\right)-\left(-x^2+16\right)2\left(x^2+16\right)2x}{{\left(x^2+16\right)}^4}\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=u\text{'}v-v\text{'}u\right]$

$=\frac{-2x{\left(x^2+16\right)}^2-4x\left({16}^2-x^4\right)}{{\left(x^2+16\right)}^4}$

Finding the least value of $f\left(x\right)$

For $x=4,f^{\text{'}\text{'}}\left(x\right)<0$

For $x=-4,f^{\text{'}\text{'}}\left(x\right)>0$

Hence, the least value of $f\left(x\right)$ occurs for $x=-4$

Substituting $x=-4$in $f\left(x\right)$

$$\begin{gathered} \frac{x}{x^2+16}=\frac{-4}{{\left(-4\right)}^2+16} \\ =\frac{-4}{16+16} \\ =\frac{-4}{32} \\ =\frac{-1}{8} \end{gathered}$$

Therefore, the least value of $f\left(x\right)$ is $-\frac{1}{8}$

Hence, option (B) is the correct answer.