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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
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B
False
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Solution

The correct option is A True
Let $$\displaystyle max\left | f(x) \right |=M$$ 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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