SATHEE: Functions 1 Question 5

Functions 1 Question 5

5.

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

(2001, 1M)

(a) $[0, 1]$

(b) $0, \frac{1}{2}$

(c) $\frac{1}{2}, 1$

(d) $[0, 1]$

Show Answer

Answer:

Correct Answer: 5. (d)

Solution:

Given, $f (x) = (1 + b^{2}) x^{2} + 2 b x + 1$

$= (1 + b^{2}) x + {\frac{b}{1 + b^{2}}}^{2} + 1 - \frac{b^{2}}{1 + b^{2}}$

$m (b) =$ minimum value of $f (x) = \frac{1}{1 + b^{2}}$ is positive

and $m (b)$ varies from 1 to 0 , so range $= [0, 1]$

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