Question

If $f(x)=(x-1)(x-2)(x-3)$ for $x \in[0,4]$, then the value of $c \in(0,4)$ satisfying Lagrange's mean value theorem, is

• A. $3 \pm \frac{\sqrt{2}}{3}$
• B. $2 \pm \frac{2 \sqrt{3}}{3}$
• C. $2 \pm \frac{\sqrt{3}}{2}$
• D. $3 \pm \frac{\sqrt{3}}{3}$

Answer 1

Answer: (B)

We have,
$f(x)=(x-1)(x-2)(x-3)$
$f(x)=x^{3}-6 x^{2}+11 x-6$
Given function is algebraic function so. it is continuous and difference in $[0,4]$
Now,
$ f(4) =(4)^{3}-6(4)^{2}+11(4)-6 $
$=64-96+44-6=6 $
$f(0) =-6$
Now, $f^{'}(x)=3 x^{2}-12 x+11$
by Lagrange's mean value theorem
$f^{'}(c)=\frac{f(4)-f(0)}{4-0}$
$\Rightarrow 3 c^{2}-12 c+11=\frac{6-(-6)}{4}$
$\Rightarrow 3 c^{2}-12 c+11=\frac{12}{4}$
$\Rightarrow 3 c^{2}-12 c+11-3=0$
$\Rightarrow 3 c^{2}-12 c+8=0$
$\Rightarrow c=\frac{12 \pm \sqrt{(-12)^{2}-4(3)(8)}}{2 \times 3}$
$=\frac{12 \pm \sqrt{144-96}}{6}=\frac{12 \pm 4 \sqrt{3}}{6} $
$ c =2 \pm \frac{2 \sqrt{3}}{3} \in[0,4) $