Let f(x) = (1 + {b^2}){x^2} + 2bx + 1 and m(b) minimum value of f(x) for a given b . As b varies, ra

English

Gujarati

Hindi

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

$[0, 1]$

$\left( {0,\;\frac{1}{2}} \right]$

$\left[ {\frac{1}{2},\;1} \right]$

$(0,\;1]$

(IIT-2001)

(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

hard

(IIT-2002)

hard

(JEE MAIN-2021)

hard

(IIT-1992)

normal

normal