Mathematics - Statistics Question with Solution | TestHub

Here, $f(x)$ is continuous on the interval $[0,8]$. So it will also be continuous on 2 and 4. At $x=2$, ${\lim }_{x\to 2^{-}}f(x)={\lim }_{x\to 2^{+}}f(x)$ ${\lim }_{x\to 2^{-}}\left(x^2+\mathrm{ax}+\mathrm{b}\right)={\lim }_{x\to 2^{+}}(3x+2)$ $4+2a+b=3(2)+2$ $\therefore 2a+b=4$ ...(i) At $x=4$ ${\lim }_{x\to 4^{-}}f(x)={\lim }_{x\to 4^{+}}f(x)$ ${\lim }_{x\to 4^{-}}(3x+2)={\lim }_{x\to 4^{+}}2ax+5b$ $3(4)+2=2a(4)+5b$ $\therefore 8a+5b=14$ .......(ii) On solving Eqs. (i) and (ii), we get $a=3$ and $b=-2$.