Science:Math Exam Resources/Courses/MATH100/December 2018/Question 11 (a)/Solution 1

Let $h(x)=f(x)-{\frac {x+1}{2}}=g(x)+\sin(x)-{\frac {x+1}{2}}$. Note that since $g$ is differentiable (and since $\sin(x)$ is differentiable), then $h$ is also differentiable (and therefore continuous).

Let $c_{1}={\frac {\pi }{2}}$, then $h(c_{1})=g(c_{1})+\sin(c_{1})-{\frac {c_{1}+1}{2}}\geq {\frac {c_{1}}{2}}+\sin(c_{1})-{\frac {c_{1}+1}{2}}=\sin(c_{1})-{\frac {1}{2}}={\frac {1}{2}}>0$.

Let $c_{2}=-{\frac {\pi }{2}}$, then $h(c_{2})=g(c_{2})+\sin(c_{2})-{\frac {c_{2}+1}{2}}\leq {\frac {c_{2}}{2}}+1+\sin(c_{2})-{\frac {c_{2}+1}{2}}=\sin(c_{2})+{\frac {1}{2}}=-{\frac {1}{2}}<0$.

Thus, by the intermediate value theorem (which applies because, as we saw, $h$ is continuous), there exists $c\in (-{\frac {\pi }{2}},{\frac {\pi }{2}})$ such that $h(c)=0$.

Using the same method, by the periodicity of $\sin(x)$ we see that for any integer $n$, there exists $c\in (-{\frac {\pi }{2}}+2n\pi ,{\frac {\pi }{2}}+2n\pi )$ such that $h(c)=0$. $\color {blue}{\text{So there are infinitely many real numbers }}c{\text{ such that }}f(c)={\frac {c+1}{2}}$.