If f (x) + f (y) + f (xy) = 2 + f (x) . f (y) , for all real values of x and y and f (x) is a polynomial function with f (4) = 17 and f(1) ≠ 1, then find the value of f (5) .
Text Solution
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Sol. Let x = y = 1
f(x) + f(y) + f(xy) = 2 + f(x) . f(y)
3f (1) = 2 + (f(1)) 2 ⇒ f(1) = 1, 2. But given that
f(1) ≠ 1 so f(1) = 2
Now put y = 
f(x) + f 
+ f(1) = 2 + f(x) . f 
⇒ f(x) + f 
= f(x) . f 
so f(x) = ± x n + 1
Now f(4) = 17 ⇒ ± (4) n + 1 = 17 ⇒ n = 2
f(x) = +(x) 2 + 1.
∴ f(5) = 5 2 + 1 = 26
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