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

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MATH100 December 2018

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Question 11 (a)
Let $g(x)$ be a twice differentialble function with a continuous second derivative and also, satisfying the property that ${\frac {x}{2}}\leq g(x)\leq {\frac {x}{2}}+1$ for each positive real number $x$ . We let $f(x)=g(x)+\sin(x)$ . Prove that there exist infinitely many real numbers $c$ such that $f(c)={\frac {c+1}{2}}$ .

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Start by letting $h(x)=f(x)-{\frac {x+1}{2}}=g(x)+\sin(x)-{\frac {x+1}{2}}$ . Find two numbers $c_{1}$ and $c_{2}$ such that $h(c_{1})>0$ and $h(c_{2})<0$ . Which theorem can we use now?

Hint 2
Remember to check the necessary conditions for the theorem you want to apply.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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Solution
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 $h(c)=0$ . $\color {blue}{\text{So there are infinitely many real numbers }}c{\text{ such that }}f(c)={\frac {c+1}{2}}$ .