# 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 \(a\)?