Applications Of Derivatives Question 402
Question: If the function $ f(x)=x^{3}-6x^{2}+ax+b $ satisfies Rolle-s theorem in the interval $ [1,,3] $ and $ f’( \frac{2\sqrt{3}+1}{\sqrt{3}} )=0 $ , then
[MP PET 2002]
Options:
A) $ a=-11 $
B) $ a=-6 $
C) $ a=6 $
D) $ a=11 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=x^{3}-6x^{2}+ax+b $
therefore $ {f}’(x)=3x^{2}-12x+a $
therefore $ {f}’(c)=0 $
therefore $ {f}’( 2+\frac{1}{\sqrt{3}} )=0 $
therefore $ 3{{( 2+\frac{1}{\sqrt{3}} )}^{2}}-12( 2+\frac{1}{\sqrt{3}} )+a=0 $
therefore $ 3( 4+\frac{1}{3}+\frac{4}{\sqrt{3}} )-12( 2+\frac{1}{\sqrt{3}} )+a=0 $
$ 12+1+4\sqrt{3}-24-4\sqrt{3}+a=0 $
therefore $ a=11 $ .
BETA
