If f is a polynomial function satisfying $2 + f (x) \cdot f (y) = f (x) + f (y) + f (x y) \forall x, y \in R$ and if $f (2) = 5$ then the value of $f (f (1))$ is
$\Rightarrow [f (1)]^{2} - 3. f (1) + 2 = 0 \Rightarrow [f (1) - 2] [f (1) - 1] = 0$
As $f (2) = 5 \Rightarrow f (1) \neq = 1 \Rightarrow f (1) = 2$

Question

If f is a polynomial function satisfying $2 + f (x) \cdot f (y) = f (x) + f (y) + f (x y) \forall x, y \in R$ and if $f (2) = 5$ then the value of $f (f (1))$ is

$\Rightarrow [f (1)]^{2} - 3. f (1) + 2 = 0 \Rightarrow [f (1) - 2] [f (1) - 1] = 0$

As $f (2) = 5 \Rightarrow f (1) \neq = 1 \Rightarrow f (1) = 2$

TestHub · Accepted Answer

2 + f ( x ) ⋅ f ( y ) = f ( x ) + f ( y ) + f ( x y ) ∀ x , y ∈ R …… . ( i ) f f ( 2 ) = 5 Put y = x 1 ​ ⇒ 2 + f ( x ) ⋅ f ( x 1 ​ ) = f ( x ) + f ( x 1 ​ ) + f ( 1 ) … ( ii ) Put x = 1 ⇒ 2 + f ( 1 ) ⋅ f ( 1 ) = f ( 1 ) + f ( 1 ) + f ( 1 )

Mathematics - Functions Question with Solution | TestHub

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