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1.Relation and Function
hard
Let $f(x) = (1 + {b^2}){x^2} + 2bx + 1$ 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
$\left( {0,\;\frac{1}{2}} \right]$
C
$\left[ {\frac{1}{2},\;1} \right]$
D
$(0,\;1]$
(IIT-2001)
Solution
(d) $f(x) = (1 + {b^2}){x^2} + 2bx + \frac{{{b^2}}}{{(1 + {b^2})}} – \frac{{{b^2}}}{{1 + {b^2}}} + 1$
$ = (1 + {b^2})\,{\left( {x + \frac{b}{{1 + {b^2}}}} \right)^2} + \frac{1}{{1 + {b^2}}} \ge \frac{1}{{1 + {b^2}}}$
$\therefore$ $m(b) = \frac{1}{{1 + {b^2}}}$, so range of $m(b) = (0,\,1]$.
Standard 12
Mathematics
