Question

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

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

Answer 1

Answer: (D)

$f(x)=(1+b^2)x^2+2bx+1$
It is a quadratic expression with coeff. of $x^2=1+b^2>0$.
$\therefore f(x)$ represents an upward parabola whose min value is
$\frac{-D}{4a},D$ being the discreminant.
$\therefore m(b)=-\frac{4b^2-4(1+b^2)}{4(1+b^2)}$
$\Rightarrow m(b)=\frac{1}{1+b^2}$
For range of m (b) :
$\frac{1}{1+b^2}>0$ also $b^2\ge 0\Rightarrow 1+b^2\ge 1$
$\Rightarrow \frac{1}{1+b^2}\le 1$
Thus $m(b)=(0,1]$