Question

Let $f(x)$ be a polynomial and $a, b$ be distinct real numbers. Then the remainder in the division of $f(x)$ by $(x-a)(x-b)$ is

• A. $\frac{(x-a) f(a)-(x-b) f(b)}{a-b}$
• B. $\frac{(x-a) f(b)-(x-b) f(a)}{a-b}$
• C. $\frac{(x-a) f(b)-(x-b) f(a)}{b-a}$
• D. $\frac{(x-a) f(a)-(x-b) f(b)}{b-a}$

Answer 1

Answer: (C)

Let $f(x)=(x-a)(x-b) \cdot q(x)+r(x)$
Let $r(x)=\alpha x+\beta[\because d \text{egr}(x) < $ deg. of divisor $]$
$\therefore f(x)=(x-a)(x-b) \cdot q(x)+a x+\beta$
$f(a)=\alpha a+\beta \ldots .( i )$
$f(a)=\alpha a+\beta \ldots .( ii )$
Subtract Eqs. (i) from (ii)
$f(a)-f(b)=a(a-b)$
$\alpha=\frac{f(a)-f(b)}{a-b}$
$\alpha=\frac{f(b)-f(a)}{b-a}$
Put, $\alpha$ in Eq. (i)
$f(a)=\left(\frac{f(b)-f(a)}{b-a}\right) \times a+\beta$
$f(a)[b-a]=a f(b)-a f(a)+\beta(b-a)$
$b f(a)-a f(a)=a f(b)-a f(a)+\beta(b-a)$
$\beta=\frac{b f(a)-a f(b)}{(b-a)}$
$\therefore r(x)=a x+\beta$
$=\frac{f(b)-f(a) x}{b-a}+\frac{b f(a)-a f(b)}{(b-a)}$
$=\frac{x f(b)-x f(a)+b f(a)-a f(b)}{b-a}$
$r(x)=\frac{(x-a) f(b)+(b-x) f(a)}{b-a}$
$r(x)=\frac{(x-a) f(b)-(x-b) f(a)}{b-a}$