We have f (x) = $\begin{cases} a+bx & x<1 \\ 4 & x=1 \\ b-ax & x>1 \end{cases}$ and $\lim\limits_{x\to 1}$ f (x) = f (1) ... (i) So first we have to find $\lim\limits_{x\to 1}$ f (x) Now, $\lim\limits_{x\to 1^{-}}$ f (x) = $\lim\limits_{x\to 1^{-}}$ (a + bx) = a + b and $\lim\limits_{x\to 1^{+}}$ f (x) = $\lim\limits_{x\to 1^{+}}$ (b - ax) = b - a Also, f (1) = 4 By (i) , $\lim\limits_{x\to 1^{-}}$ f (x) = $\lim\limits_{x\to 1^{+}}$ f (x) ⇒ a + b = 4 ... (ii) and b - a = 4 ... (iii) Adding (ii) and (iii), we get 2b = 8 ⇒ b = 4 Substituting the value of b in (iii), we get 4 – a = 4 ⇒ a = 0 Thus a = 0 and b = 4.