# Find this unique real number!

Calculus Level pending

$$f(x)$$ is an odd function ( that is $$f(-x)=-f(x)$$ ) and $$f$$ is twice continuously differentiable in $$\mathbb{R}$$. $$f(1)=1$$. We can prove that there is a unique real number $$a$$ such that for any function $$f(x)$$ that satisfies those conditions there exists $$0<b<1$$ such that $$f''(b)+f'(b)=a$$ ($$b$$ can depend on $$f$$) what is the value of $$a$$?

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