Applications Of Derivatives Question 348
Question: v If $ y=a\log x+bx^{2}+x $ has its extremum value at $ x=1 $ and $ x=2, $ then $ (a,b) $ =
[UPSEAT 2002]
Options:
A) $ ( 1,\frac{1}{2} ) $
B) $ ( \frac{1}{2},,2 ) $
C) $ ( 2,\frac{-1}{2} ) $
D) $ ( \frac{-2}{3},\frac{-1}{6} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{dy}{dx}=\frac{a}{x}+2bx+1 $
therefore $ {{( \frac{dy}{dx} )} _{x=1}}=a+2b+1=0 $
therefore $ a=-2b-1 $ and $ {{( \frac{dy}{dx} )} _{x=2}}=\frac{a}{2}+4b+1=0 $
therefore $ \frac{-2b-1}{2}+4b+1=0 $
therefore $ -b+4b+\frac{1}{2}=0 $
therefore $ 3b=\frac{-1}{2} $
therefore $ b=\frac{-1}{6} $ and $ a=\frac{1}{3}-1=\frac{-2}{3} $ .
BETA
