Suppose a, b, c are three distinct real numbe | vedclass

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Question

(a)

Given,

$P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$

$P(a)=1+0+0=1$

$P(b)=0+1+0=1$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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. $a, b, c$ ).

$	herefore P(x)$ is an identity with only value equal to $1$ for all $R$.

$	herefore \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$

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

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