Applications Of Derivatives Question 146
If the function $ f(x)=ax^{3}+bx^{2}+11x-6 $ satisfies the conditions of Rolle’s theorem on [1, 3] for $ x=2+\frac{1}{\sqrt{3}} $ , then values of a and b , respectively, are
Options:
A) $ -,3,2 $
B) $ 2,-4 $
C) 1, 6
D) none of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=ax^{2}+bx+c $ Satisfies condition of Rolle’s theorem in [1, 3].
Therefore, $ f(1)=f(3) $ or $ a+b+11-6=27a+9b+33-6 $ or $ -26a-8b=-11 $ …(1)
And $ f’(x)=3ax^{2}+2bx+11 $ or $ f’( 2+\frac{1}{\sqrt{3}} )=3a{{( 2+\frac{1}{\sqrt{3}} )}^{2}}+2b( 2+\frac{1}{\sqrt{3}} )+11=0 $ or $ 3a( 4+\frac{1}{3}+\frac{1}{\sqrt{3}} )+2b( 2+\frac{1}{\sqrt{3}} )+11=0 $ …(2)
From equations (1) and (2), We get
$ a=1,b=-,6. $
BETA
