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

Options

A. `-1/4`
B. `-1/8`
C. `1/12`
D. `1/16`

Answer

B

Explanation

Let the number be `x` then the function
`f(x)=x/(x^(2)+16)`
On differentiating both sides w.r.t `x` we get
`f'(x)=((x^(2)+16).1-x(2x))/((x^(2)+16)^(2))`
`=(x^(2)+16-2x^(2))/((x^(2)+16)^(2))=(16-x^(2))/((x^(2)+16)^(2))`
Put `f'(x)=0` for maxima or minima.
`16-x(2)=0impliesx=4, -4`
Again differentiating
`f''(x)=((x^(2)+16)^(2)(-2x)-(16-x^(2))2(x^(2)+16)2x)/((x^(2)+16)^(4))`
At `x=4, f''(x)lt0` and at `x=-4, f''(x)gt0`
`:.` Least value of `f(x)=(-4)/(16+16)=-1/8`