If `f(x)=1/(4x^2+2x+1)` , then its maximum value is `4/3` (b) `2/3` (c) 1 (d) `3/4`

Options

A. `4/3`
B. 1
C. `2/3`
D. `3/4`

Answer

A

Explanation

`f(x)` is maximum when `4x^(2)+2x+1` is minimum
Let `g(x)=4x^(2)+2x+1`
`g'(x)=8x+2` and `g''(x)=8`
But `g'(x)=0`, therefore `8x+2=0impliesx=-1/4`
`:.g''(-1/4)=8gt0,g` is minimum at `x=-1/4`
Hence `f(x)` is maximum at `x=-1/4`
`:.f(x)=1/(4(1/16)-2/4+1)=1/(3/4)=4/3`