Question

For the function $f\left(x\right)=(x-1)(x-2)(x-3)$ in $[0,4]$ value of ‘c’ in Lagrange's mean value theorem is

• A. $2\pm \frac{2}{\sqrt{3}}$
• B. $1-\frac{\sqrt{21}}{6}$
• C. $1+\frac{\sqrt{21}}{6}$
• D. $4-2\sqrt{3}$

Answer 1

The correct option is A $2\pm \frac{2}{\sqrt{3}}$

Lagrange's mean value theorem states that if $f\left(x\right)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists some $c$ between $a$ and $b$ such that $f^{{}^{\prime }}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Given $f\left(x\right)=(x-1)(x-2)(x-3)$ and $[a,b]=[0,4]$

$⟹f^{{}^{\prime }}\left(x\right)=(x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)$

$⟹f^{{}^{\prime }}\left(x\right)=x^2-5x+6+x^2-4x+3+x^2-3x+2$

$⟹f^{{}^{\prime }}\left(x\right)=3x^2-12x+11$

Therefore, $f^{{}^{\prime }}\left(c\right)=\frac{(4-1)(4-2)(4-3)-(0-1)(0-2)(0-3)}{4-0}$

$⟹3c^2-12c+11=\frac{6+6}{4}$

$⟹3c^2-12c+11=3$

$⟹3c^2-12c+8=0$

$⟹c=\frac{12\pm \sqrt{{12}^2-4\left(3\right)\left(8\right)}}{2\left(3\right)}$

$⟹c=\frac{12\pm \sqrt{144-96}}{6}$

$⟹c=\frac{12\pm \sqrt{48}}{6}$

$⟹c=\frac{12\pm 4\sqrt{3}}{6}$

$⟹c=2\pm \frac{2\sqrt{3}}{3}$

$⟹c=2\pm \frac{2}{\sqrt{3}}$