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4-2.Quadratic Equations and Inequations
normal
Suppose $a, b, c$ are three distinct real numbers, let $P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$ When simplified, $P(x)$ becomes
A
$1$
B
$x$
C
$\frac{x^2+(a+b+c)(a b+b c+c a)}{(a-b)(b-c)(c-a)}$
D
$0$
(KVPY-2011)
Solution
(a)
Given,
$P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$
$P(a)=1+0+0=1$
$P(b)=0+1+0=1$
$P(c)=0+0+1=1$
$P(x)$ is a polynomial of degree atmost $2$ and also attains same value i.e.$1$ for distinct values of $x$ (i.e. $a, b, c$ ).
$\therefore P(x)$ is an identity with only value equal to $1$ for all $R$.
$\therefore \frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}=1$
Standard 11
Mathematics
