# Science:Math Exam Resources/Courses/MATH100/December 2016/Question 14 (a)/Solution 1

Note that ${\displaystyle \sin \left({\frac {\pi }{2}}\right)=1,}$ ${\displaystyle \sin \left({\frac {3\pi }{2}}\right)=-1.}$

Combining this with the second condition, we have

${\displaystyle \sin x-f(x)\leq |\sin x-f(x)|=|f(x)-\sin x|\leq {\frac {1}{3}}\implies f\left({\frac {\pi }{2}}\right)\geq \sin \left({\frac {\pi }{2}}\right)-{\frac {1}{3}}={\frac {2}{3}}>0}$ and

${\displaystyle f(x)-\sin x\leq |f(x)-\sin x|\leq {\frac {1}{3}}\implies f\left({\frac {3\pi }{2}}\right)\leq \sin \left({\frac {3\pi }{2}}\right)+{\frac {1}{3}}=-{\frac {2}{3}}<0}$.

So, by the intermediate value theorem (${\displaystyle f}$ is continuous by the first condition), ${\displaystyle f}$ has at least one zero in the interval ${\displaystyle (\pi /2,3\pi /2)}$, which is contained in the interval ${\displaystyle (0,2\pi ).}$.