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,
$f(x)=(x-1)(x-2)(x-3)$
$f(x)=x^3-6x^2+11x-6$
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, $f^{{}^{\prime }}(x)=3x^2-12x+11$
by Lagrange's mean value theorem
$f^{{}^{\prime }}(c)=\frac{f(4)-f(0)}{4-0}$
$\Rightarrow 3c^2-12c+11=\frac{6-(-6)}{4}$
$\Rightarrow 3c^2-12c+11=\frac{12}{4}$
$\Rightarrow 3c^2-12c+11-3=0$
$\Rightarrow 3c^2-12c+8=0$
$\Rightarrow c=\frac{12\pm \sqrt{(-12{)}^2-4(3)(8)}}{2\times 3}$
$=\frac{12\pm \sqrt{144-96}}{6}=\frac{12\pm 4\sqrt{3}}{6}$
$c=2\pm \frac{2\sqrt{3}}{3}\in [0,4)$