Let `f(x)=(1+b^(2))x^(2)+2bx+1` and let m(b) be the minimum value of f (x). As b varies, the range of m (b) is

Options

A. [0,1]
B. (0, 1/2]
C. [1 /2 , 1]
D. [0, 1]

Answer

D

Explanation

We have `f(x)=(1+b^2)x^2+2bx+1`
`f(x)=(1+b^2){x^2+(2b)/(1-b^2)x+=1/(1+b^2)}`
`rArr f(x)=(1+b^2){x+b/(1+b^2)}^2+1/((1+b^2))` Is is evident from this the minimum value of `f(x) is 1/(1+b^2)`
Which it attains at x- `(b)/(1+b^2)`
Clearly `0 lt m (b) le 1`
Hence , range of m(b) is (0, 1]