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
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
Updated On: Jun 14, 2022
- [0, 1]
- (0, 1/2]
- [1/2, 1]
- (0, 1]
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The Correct Option is D
Solution and Explanation
$f(x) = (1+b^2) x^2 + 2bx + 1$
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 $b^{2} \ge0 \Rightarrow 1 + b^{2} \ge1$
$ \Rightarrow \frac{1}{1+b^{2} } \le1 $
Thus $m (b) = (0, 1]$
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 $b^{2} \ge0 \Rightarrow 1 + b^{2} \ge1$
$ \Rightarrow \frac{1}{1+b^{2} } \le1 $
Thus $m (b) = (0, 1]$
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