Question

Suppose a, b, c are three distinct real numbers. Let
$P\left(x\right)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$
When simplified, $P\left(x\right)$ becomes

• A. $1$
• B. $x$
• C. $\frac{x^2+(a+b+c)(ab+bc+ca)}{(a-b)(b-c)(c-a)}$
• D. $0$

Answer 1

Answer: (A)

$1$

$P\left(x\right)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$
Let $f\left(x\right)=P\left(x\right)-1$
$f\left(a\right)=1+0+0-1=0$
$f\left(b\right)=0+1+0-1=0$
$f\left(c\right)=0+0+1-1=0$
$f\left(x\right)$ is a polynomial of degree atmost 2, and also attains same value (i.e. , 0) for 3 distinct values of $x$ (i.e. a, b, c)
$\therefore$ $f\left(x\right)$ is an identity with only value equal to zero.
$\therefore$ $f\left(x\right)=0\forall \times ϵR\Rightarrow P\left(x\right)=1,\forall \times ϵR$