# Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 08/Solution 1

The correct choice is (iii), because at the inflection point the function changes its concavity, so the tangent line moves from being above the graph to be below the graph or vice versa, so it must cross the graph from one side to the other.

(i) is not correct. See the Questions B 01; Although the zero ${\displaystyle z_{2}}$ is closer than another zero ${\displaystyle z_{1}}$ to the initial point ${\displaystyle B}$, but the newton method starting at ${\displaystyle B}$ finds the zero ${\displaystyle z_{1}}$.

(ii) is not correct. The tangent line of ${\displaystyle f(x)=\cos x}$ at a point ${\displaystyle a}$ is ${\displaystyle y=f'(a)(x-a)+f(a)=-\sin a(x-a)+\cos a}$. Assume that there is a tangent line which goes through ${\displaystyle (0,2)}$. This means that there exists a point ${\displaystyle a\in [-\pi ,\pi ]}$ such that ${\displaystyle 2=-\sin a(0-a)+\cos a=a\sin a+\cos a}$. However there's no such point ${\displaystyle a\in [-\pi ,\pi ]}$ because the maximum of ${\displaystyle a\sin a+\cos a}$ is ${\displaystyle {\frac {\pi }{2}}<2}$. This is a contradiction.

We find the maximum in the following way. Let ${\displaystyle g(a)=a\sin a+\cos a}$. Then, its derivative is ${\displaystyle g'(a)=a\cos a}$, which vanishes at ${\displaystyle a=0,\pm {\frac {\pi }{2}}}$. Indeed, these are critical points. so that the maximum in ${\displaystyle [-\pi ,\pi ]}$ is achieved either the critical points ${\displaystyle 0,\pm {\frac {\pi }{2}}}$ or the end points ${\displaystyle \pm \pi }$. Since ${\displaystyle g(0)=0,g\left({\frac {\pi }{2}}\right)={\frac {\pi }{2}},g(\pm \pi )=-1}$, the maximum is ${\displaystyle {\frac {\pi }{2}}}$.

(iv) is not correct because we can take ${\displaystyle f(x)=|x|}$, ${\displaystyle g(x)=-|x|}$, and ${\displaystyle a=0}$; Obviously, ${\displaystyle f}$ and ${\displaystyle g}$ are NOT differentiable at ${\displaystyle x=a=0}$. but ${\displaystyle h(x)=f(x)+g(x)=|x|-|x|=0}$ is a constant function and hence is differentiable at ${\displaystyle x=a}$.

(Proof for (iii)) We need to show ${\displaystyle g(x)=f(x)-({\text{tangent line}})=f(x)-f'(a)(x-a)-f(a)}$ changes the sign around ${\displaystyle x=a}$. Since ${\displaystyle g(a)=0}$, it is enough to show that ${\displaystyle g}$ is either increasing or decreasing around ${\displaystyle x=a}$. (If it is increasing the sign is changing from negative to positive, otherwise it is from positive to negative.)

For this purpose, we calculate the derivative ${\displaystyle g'(x)=f'(x)-f'(a)=\int _{a}^{x}f''(t)dt}$. Since ${\displaystyle f''}$ changes the sign around ${\displaystyle x=a}$, ${\displaystyle g'}$ doesn't change its sign around ${\displaystyle x=a}$. For example, in the case that ${\displaystyle f''}$ change its sign from negative to positive around ${\displaystyle x=a}$, then for ${\displaystyle x sufficiently close with ${\displaystyle a}$, ${\displaystyle g'(x)=\int _{a}^{x}f''(t)dt=-\int _{x}^{a}f''(t)dt>0}$ and for ${\displaystyle x sufficiently close with ${\displaystyle a}$, ${\displaystyle g'(x)=\int _{a}^{x}f''(t)dt>0}$. Therefore, we prove that ${\displaystyle g}$ is either increasing or decreasing around ${\displaystyle x=a}$.

Answer: ${\displaystyle \color {blue}(iii)}$