Applications Of Derivatives Question 316
Question: If x is real, then greatest and least values of $ \frac{x^{2}-x+1}{x^{2}+x+1} $ are
[RPET 1999; AMU 1999; UPSEAT 2002]
Options:
A) $ 3,,-\frac{1}{2} $
B) $ 3,\frac{1}{3} $
C) $ -3,,-\frac{1}{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ y=\frac{x^{2}-x+1}{x^{2}+x+1} $
therefore $ \frac{dy}{dx}=\frac{(x^{2}+x+1),(2x-1)-(x^{2}-x+1),(2x+1)}{{{(x^{2}+x+1)}^{2}}} $
therefore $ \frac{dy}{dx}=\frac{2x^{2}-2}{{{(x^{2}+x+1)}^{2}}}=0 $
therefore $ 2x^{2}-2=0 $
therefore $ x=-1,,+1 $
$ \frac{d^{2}y}{dx^{2}}=\frac{4,(-x^{3}+3x+1)}{x^{2}+x+1} $
At $ x=-1 $ , $ \frac{d^{2}y}{dx^{2}}<0, $ the function will occupy maximum value, $ f(-1)=3 $ and at $ x=1 $
$ \frac{d^{2}y}{dx^{2}}>0, $ the function will occupy minimum value $ f(1)=\frac{1}{3} $ .
BETA
