Question

# Let $$f$$ and $$g$$ be real-valued functions such that $$f(x+y)+f(x-y)=2f(x)\cdot g(y)\forall x,y\epsilon R$$if $$f$$ is not identically zero and $$f\left | (x) \right |\leq 1,\forall x\epsilon R,$$ then $$\left | g(y) \right |\leq 1,\forall y\epsilon R$$.If true enter 1 else enter 0

A
True
B
False

Solution

## The correct option is A TrueLet $$\displaystyle max\left | f(x) \right |=M$$ where $$0< M\leq 1$$.(Since, f is not identically zero and $$\left | f(x) \right |\leq 1\:\forall x\in R$$)Now, $$\displaystyle f(x+y)+f(x-y)=2f(x).g(y)$$$$\Rightarrow \displaystyle \left | 2f(x).g(y) \right |=\left | f(x+y)+f(x-y) \right |$$$$\Rightarrow \displaystyle \left | 2f(x) \right |.\left | g(y) \right |\leq \left | f(x+y) \right |+\left | f(x-y) \right |\leq M+M$$$$\Rightarrow \displaystyle 2\left | f(x) \right |\left | g(y) \right |\leq 2M$$$$\Rightarrow \left | g(y) \right |\leq 1$$ for $$y\in R$$Mathematics

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