Functions Question 584
Question: 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 is
[IIT Screening 2001]
Options:
A) [0, 1]
B) $ ( 0,\ \frac{1}{2} ] $
C) $ [ \frac{1}{2},\ 1 ] $
D) $ (0,\ 1] $
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=(1+b^{2})x^{2}+2bx+\frac{b^{2}}{(1+b^{2})}-\frac{b^{2}}{1+b^{2}}+1 $
$ =(1+b^{2}),{{( x+\frac{b}{1+b^{2}} )}^{2}}+\frac{1}{1+b^{2}}\ge \frac{1}{1+b^{2}} $
\ $ m(b)=\frac{1}{1+b^{2}} $ , so range of $ m(b)=(0,,1] $ .
BETA
