Let phi(x)=(b(x-a))/(b-a)+(a(x-b))/(a-b)...

Let `phi(x)=(b(x-a))/(b-a)+(a(x-b))/(a-b)`, where `x inR` and a and b are fixed real numbers with `aneb`. Then, `phi(a+b)` is equal to

A

`phi(ab)`

B

`phi(-ab)`

C

`phi(a)+phi(b)`

D

`phi(a-b)`

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The correct Answer is:

C

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