Applications Of Derivatives Question 266
Question: If the function $ f(x)=x^{3}-6ax^{2}+5x $ satisfies the conditions of Lagrange’s mean value theorem for the interval [1, 2] and the tangent to the curve $ y=f(x) $ at $ x=\frac{7}{4} $ is parallel to the chord that joins the points of intersection of the curve with the ordinates $ x=1 $ and $ x=2 $ . Then the value of $ a $ is
[MP PET 1998]
Options:
A) $ \frac{35}{16} $
B) $ \frac{35}{48} $
C) $ \frac{7}{16} $
D) $ \frac{5}{16} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(b)=f(2)=8-24a+10=18-24a $
$ f(a)=f(1)=1-6a+5=6-6a $
$ f’(x)=3x^{2}-12ax+5 $
From Lagrange’s mean value theorem,
$ f’(x)=\frac{f(b)-f(a)}{b-a} $
$ =\frac{18-24a-6+6a}{2-1} $
$ \therefore f’(x)=12-18a $
At $ x=\frac{7}{4},\ 3\times \frac{49}{16}-12a\times \frac{7}{4}+5=12-18a $
therefore $ 3a=\frac{147}{16}-7 $
therefore $ 3a=\frac{35}{16} $
therefore $ a=\frac{35}{48} $ .
BETA
