Applications Of Derivatives Question 341
Question: If $ f(x)=\frac{1}{4x^{2}+2x+1} $ , then its maximum value is
[RPET 2002]
Options:
A) 4/3
B) 2/3
C) 1
D) ¾
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=\frac{1}{4x^{2}+2x+1} $
therefore $ {f}’(x)=\frac{-(8x+2)}{{{(4x^{2}+2x+1)}^{2}}} $ Put $ {f}’(x)=0 $
therefore $ 8x+2=0 $
therefore $ x=-1/4 $ . $ {f}’’,(x)=\frac{-[{{(4x^{2}+2x+1)}^{2}}8-(8x+2),2,(4x^{2}+2x+1)(8x+2)]}{{{(4x^{2}+2x+1)}^{4}}} $
$ {f}’’(-1/4)=-ve $ (point of maxima) $ f{{(-1/4)} _{max\text{.}}} $ = $ \frac{1}{4\times \frac{1}{16}-2\times \frac{1}{4}+1}=\frac{4}{3} $ .
BETA
