Question

Let f(x) be a non constant polynomial satisfying the relation f(x).f(y)=f(x)+f(y)+f(xy)-2,for all real x and y and f(0) not equals to 1, suppose f(4)=65 then

Answer 1

Answer:

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Answer 2

Given:

f(x)f(y) = f(x) + f(y) + f(xy)- 2

f(0) not equal to 1

f(4) = 65

To Find:

n

Solution:

  putting y = 1/x

⇒ f(x)×f(1/x) = f(x) + f(1/x) + f(1)- 2

⇒  f(x)×f(1/x) = f(x) + f(1/x) + 2- 2

⇒ f(x)×f(1/x) = f(x) + f(1/x)

⇒ f(x) = $x^{n}$ + 1

Also. f(4) = 65 it is given

      f(x) = $x^{n}$ + 1

       65 = $4^{n}$ + 1

        $4^{n}$= 64

          n = 64/4

           n = 16

Therefore the value of n is 16.