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Suppose, a, b, c are three distinct real...

Suppose, a, b, c are three distinct real numbers. Let P (x) =`((x-b)(x-c))/((a-b)(a-c))+((x-c)(x-a))/((b-c)(b-a))+((x-a)(x-b))/((c-a)(c-b))`.
When simplified, P (x) becomes

A

1

B

x

C

`(x^(2)+(a+b+c)(ab+bc+ca))/((a-b)(b-c)(c-a))`

D

0

Text Solution

Verified by Experts

The correct Answer is:
A
`P(x)=((x-b)(x-c))/((a-b)(a-c))+((x-c)(x-a))/((b-c)(b-a))+((x-a)(x-b))/((c-a)(c-b))`
`"Let "P(a)=1+0+0=1`
`P(b) =0+1+0=1`
`P (c) =0+0+1=1`
`:. P(x)=1 " for all "x in R`.
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