Question

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

Answer 1

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

Now, $R.H.L.=4$
$\Rightarrow \underset{x\to 1^{+}}{lim}(b+ax)=4$
$\Rightarrow \underset{h\to 0}{lim}(b+a(1+h\left)\right)=4$
$\Rightarrow b+a=4\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow a=4,b=0$
$\Rightarrow a^2+b^2=4^2+0^2=16$