Question

The value of ${}^{\prime }{c}^{\prime }$ in Lagrange's mean value theorem for $f\left(x\right)=l{x}^2+mx+n,(l\ne 0)$ on $[a,b]$ is

• A. $\frac{a}{2}$
• B. $\frac{b}{2}$
• C. $\frac{(a-b)}{2}$
• D. $\frac{(a+b)}{2}$

Answer 1

The correct option is D $\frac{(a+b)}{2}$

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 $f\left(x\right)=l{x}^2+mx+n$ and $[a,b]=[a,b]$

$⟹f^{{}^{\prime }}\left(x\right)=2lx+m$

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

$⟹2lc+m=\frac{l(b^2-a^2)+m(b-a)}{b-a}$

$⟹2lc+m=\frac{\left(l\right(b+a)+m)(b-a)}{b-a}$

$⟹2lc+m=l(b+a)+m$

$⟹2lc=l(b+a)$

$⟹2c=(b+a)$

$⟹c=\frac{(b+a)}{2}$