Applications Of Derivatives Question 404
Question: In the Mean-Value theorem $ \frac{f(b)-f(a)}{b-a}=f’(c), $ if $ a=0,b=\frac{1}{2} $ and $ f(x)=x(x-1)(x-2), $ the value of c is
[MP PET 2003]
Options:
A) $ 1-\frac{\sqrt{15}}{6} $
B) $ 1+\sqrt{15} $
C) $ 1-\frac{\sqrt{21}}{6} $
D) $ 1+\sqrt{21} $
Show Answer
Answer:
Correct Answer: C
Solution:
From mean value theorem $ {f}’(c)=\frac{f(b)-f(a)}{b-a} $
$ a=0,,f(a)=0 $
therefore $ b=\frac{1}{2},,f(b)=\frac{3}{8} $
$ {f}’(x)=(x-1)(x-2)+x(x-2)+x(x-1) $
$ {f}’(c)=(c-1)(c-2)+c(c-2)+c(c-1) $ = $ c^{2}-3c+2+c^{2}-2c+c^{2}-c $
$ {f}’(c)=3c^{2}-6c+2 $
According to mean value theorem, $ {f}’(c)=\frac{f(b)-f(a)}{b-a} $
therefore $ 3c^{2}-6c+2=\frac{(3/8)-0}{(1/2)-0},=\frac{3}{4} $
therefore $ 3c^{2}-6c+\frac{5}{4}=0 $
$ c=\frac{6\pm \sqrt{36-15}}{2\times 3}=\frac{6\pm \sqrt{21}}{6} $
$ =1\pm \frac{\sqrt{21}}{6} $ .
BETA
