Question

For certain values of $a,m$ and $b,$ the function
$f\left(x\right)=\{\begin{array}{cc}3,& x=0\\ -x^2+3x+a,& 0<x<1\\ mx+b,& 1\le x\le 2\end{array}$ satisfies the hypothesis of the mean value theorem for the interval $[0,2].$ Then the value of $a+b+m$ is

• A. $7$
• B. $8$
• C. $5$
• D. $9$

Answer 1

The correct option is B $8$

$f\left(x\right)=\{\begin{array}{cc}3,& x=0\\ -x^2+3x+a,& 0<x<1\\ mx+b,& 1\le x\le 2\end{array}$
Since function satisfies the hypothesis of the mean value theorem for the interval $[0,2],$ so function should be continuous in $[0,2].$
For continuity at $x=0,$ we have
$f\left(0\right)=f\left(0^{+}\right)$
$\Rightarrow 3=\underset{x\to 0^{+}}{lim}(-x^2+3x+a)$
$\Rightarrow 3=\underset{h\to 0}{lim}(-h^2+3h+a)$
$\therefore a=3$

For continuity at $x=1,$ we have
$f\left(1\right)=f\left(1^{-}\right)$
$\Rightarrow m+b=\underset{x\to 1^{-}}{lim}(-x^2+3x+a)$
$\Rightarrow m+b=-1+3+a$
$\Rightarrow m+b=5$
Hence, $a+b+m=8$