Question

The denominator of a fraction exceeds the square of the numerator by 16, then the least value of the fraction is

• A. $-\frac{1}{4}$
• B. $-\frac{1}{8}$
• C. $\frac{1}{12}$
• D. $\frac{1}{16}$

Answer 1

Answer: (B)

Let the numerator be $x,$ then $f(x)=\frac{x}{x^2+16}$
On differentiating w.r.t. $x,$ we get
$f^{\prime }(x)=\frac{\left(x^2+16\right)\cdot 1-x(2x)}{{\left(x^2+16\right)}^2}$
$=\frac{x^2+16-2x^2}{{\left(x^2+16\right)}^2}=\frac{16-x^2}{{\left(x^2+16\right)}^2}\dots$ (i)
Putting $f^{\prime }(x)=0$ for maxima or minima, we get,
$16-x^2=0\Rightarrow x=4,-4$
Again, on differentiating w.r.t. $x,$ we get
$f^{\prime \prime }(x)=\frac{{\left(x^2+16\right)}^2-(-2x)-\left(16-x^2\right)2\left(x^2+16\right)2x}{{\left(x^2+16\right)}^4}$
At $x=4,f^{\prime \prime }(x)<0$
$\therefore f(x)$ is maximum at $x=4$
And at $x=-4,f^{\prime \prime }(x)>0\Rightarrow f(x)$ is minimum
$\therefore$ Least value of $f(x)=\frac{-4}{16+16}=-\frac{1}{8}$