Question

Let $f\left(x\right)=\{\begin{array}{cc}4a-bx,& x<1\\ 3,& x=1\\ 4x-b{x}^2,& x>1\end{array}$. If $f\left(x\right)$ is continuous at $x=1$, then the absolute value of $a-b$ is

Answer 1

Since $f\left(x\right)$ is continuous at $x=1$
$\Rightarrow L.H.L.=R.H.L.=f\left(1\right)=3$
Now, $L.H.L.=3$
$\Rightarrow \underset{x\to 1^{-}}{lim}(4a-bx)=3$
$\Rightarrow \underset{h\to 0}{lim}(4a-b(1-h\left)\right)=3$
$\Rightarrow 4a-b=3\cdots \left(1\right)$

Now, $R.H.L.=3$
$\Rightarrow \underset{x\to 1^{+}}{lim}(4x-b{x}^2)=3$
$\Rightarrow \underset{h\to 0}{lim}\left(4\right(1+h)-b(1+h{)}^2)=3$
$\Rightarrow 4-b=3\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow a=1,b=1$
$\therefore a-b=1-1=0$