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

To find a zero of $f$, we use the intermediate value theorem. Note that by the first condition, $f$ is continuous.

To this end, it is enough to find two values of $x$ at which $f$ has a negative and positive function value, respectively.

Expanding the inequality in the second condition, we get

$$-{\frac {1}{3}}\leq f(x)-\sin x\leq {\frac {1}{3}}\implies \sin x-{\frac {1}{3}}\leq f(x)\leq \sin x+{\frac {1}{3}}.$$

Since $\sin \left({\frac {\pi }{2}}\right)=1,$ and $\sin \left({\frac {3\pi }{2}}\right)=-1,$ plugging these numbers into the inequalities, we get

$$f\left({\frac {\pi }{2}}\right)\geq \sin \left({\frac {\pi }{2}}\right)-{\frac {1}{3}}={\frac {2}{3}}>0$$

and

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

.

Therefore, by the intermediate value theorem, $f$ has at least one zero in the interval $(\pi /2,3\pi /2)$, which is contained in the interval $(0,2\pi ).$.