Question

Value of $c$ of Lagrange's mean value theorem for $f\left(x\right)=2x-x^2$ in $[0,1]$

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. $-1$

Answer 1

The correct option is C $\frac{1}{2}$

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

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

Therefore, $f^{{}^{\prime }}\left(c\right)=\frac{(2b-b^2)-(2a-a^2)}{b-a}$

$⟹2-2c=\frac{(2-1)-0}{1-0}$

$⟹2-2c=1$

$⟹2c=1$

$⟹c=\frac{1}{2}$