Let \(f(x)=(x-a)(x-b) \cdot q(x)+r(x)\)
Let \(r(x)=\alpha x+\beta \quad[\because \operatorname{deg} r(x) < \operatorname{deg}\). of divisor]
\(\begin{aligned}
\therefore \quad f(x) & =(x-a)(x-b) \cdot q(x)+\alpha x+\beta \\
f(a) & =\alpha a+\beta \quad \ldots (i) \\
f(b) & =\alpha b+\beta \quad \ldots (ii)
\end{aligned}\)
Subtract Eqs. (i) from (ii)
\(\begin{aligned}
f(a)-f(b) & =\alpha(a-b) \\
\alpha & =\frac{f(a)-f(b)}{a-b} \\
\alpha & =\frac{f(b)-f(a)}{b-a}
\end{aligned}\)
Put, \(\alpha\) in Eq. (i)
\(\begin{aligned}
f(a) & =\left(\frac{f(b)-f(a)}{b-a}\right) \times a+\beta \\
f(a)[b-a] & =a f(b)-a f(a)+\beta(b-a) \\
b f(a)-a f(a) & =a f(b)-a f(a)+\beta(b-a) \\
\beta & =\frac{b f(a)-a f(b)}{(b-a)} \\
r(x) & =\alpha x+\beta \\
& =\frac{f(b)-f(a) x}{b-a}+\frac{b f(a)-a f(b)}{(b-a)} \\
& =\frac{x f(b)-x f(a)+b f(a)-a f(b)}{b-a} \\
r(x) & =\frac{(x-a) f(b)+(b-x) f(a)}{b-a} \\
r(x) & =\frac{(x-a) f(b)-(x-b) f(a)}{b-a}
\end{aligned}\)
Hence, option (c) is correct.