"7. If \$ f ( x + y ) = f ( x y ) \\vee x , y \\in R \$ then prove that \$ f \$ is aconsant function."

Open in App

Solution

Verified by Toppr

Was this answer helpful?

Similar Questions

f (x + y) = f (x y) \forall x, y ϵ R

then prove that

f

is a constant function.

View Solution

Let f and g be real valued functions such that f(x+y)+f(x-y)=2f(x).g(y)

\forall x, y ϵ R

. Prove that, if f(x) is not identically zero and

| f (x) | \leq 1 \forall x ϵ R

, then

| g (y) | \leq 1 \forall y ϵ R

View Solution

Let $$ f(x+y)=f(x)+f(y)+2 x y-1 $$ for all real $$ x $$ and $$ y $$ and $$ f(x) $$ be differentiable function.
If $$ f^{\prime}(0)=\cos \alpha, $$ then prove that $$ f(x)>0 \forall x \in R $$

View Solution

If $$f(x,y)=\dfrac{1}{\sqrt {x^{2}+y^{2}}}$$ then, prove that $$x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=-f$$

View Solution

F (x) = ⎡ ⎢ ⎣ \begin{matrix} cos x & - sin x & 0 sin x & cos x & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

then show that

F (x) . F (y) = F (x + y)

. Hence prove that

[F (x)]^{- 1} = F (- x)

View Solution

"7. If \\( f ( x + y ) = f ( x y ) \\vee x , y \\in R \\) then prove that \\( f \\) is aconsant function."