Question

The values of $a,m$ and $b$, for which Lagrange's mean value theorem is applicable to the function $f\left(x\right)$ for $x\in [0,2]$ is

$f\left(x\right)=\{\begin{array}{cc}3,& x=0\\ -x^2+a,& 0<x<1\\ mx+b,& 1\le x\le 2\end{array}$

• A. $a=3,m=-2,b=0$
• B. No such, $a,m,b$ exist
• C. $a=3,m=2,b=0$
• D. $a=3,m=-2,b=4$

Answer 1

The correct option is D $a=3,m=-2,b=4$

$f\left(x\right)=[\begin{array}{cc}3& x=0\\ a-x^2& 0<x<1\\ mx+b& 1\le x\le 2\end{array}$
For Lagranges mean value theorem, $f\left(x\right)$ must be continuous in $[0,2]$ and differentiable in $(0,2)$.
Hence, $a=3$
$m+b=2$
$f^{\prime }\left(x\right)=[\begin{array}{cc}-2x& 0<x<1\\ m& 1<x\le 2\end{array}$
$\Rightarrow m=-2$
and hence $b=4$