The denominator of a fraction is greater than 16 of the square of 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 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))`…………….i
Put `f'(x)=0` for, maxima or minima.
`f'(x)=0`
`implies16-x^(2)=0impliesx=4,-4`
Again differentiating we get
`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`
`f(x)=(-4)/(16+16)=-1/8`