Question

Let $f\left(x\right)=(1+b^2)x^2+2bx+1$ and m (b) the minimum value of f (x) for a given b. As b varies. The range of m (b) is

• A. $[0,1]$
• B. $[0,\frac{1}{2}]$
• C. $[\frac{1}{2},1]$
• D. $(0,1]$

Answer 1

The correct option is D

$(0,1]$

$f\left(x\right)=(1+b^2)[x^2+\frac{2b}{1+b^2}x+\frac{b^2}{(1+b^2)b^2}]+1=(1+b^2){(x+\frac{b}{1+b^2})}^2+\frac{1}{1+b^2}⩾\frac{1}{1+b^2}\therefore m\left(b\right)=\frac{1}{1+b^2}.$ So, range of m(b) = (0,1].
Hence (d) is the correct answer.