Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 05/Solution 1

Note that since $T'(t)=k(E-T)$ we have $T''(t)=k(E-T)'=k(-T')=-k^{2}(E-T)$. This will help us to determine the concavity and therefore, the comparison between slope of the tangent line and the slope of the secant line.

• 1st case: A pizza is out of the oven, so it is cooling down. i.e., the temperature is decreasing. This implies that the slope of the tangent line at $t=5$ and $t=12$, and the slope of the secant line through the two endpoints are negative. Since in this case $E<T$, $T''(t)>0$, if we sketch an exponential type graph which is concave up and decreasing, we see that the absolute value of the slope of the tangent line at $(5,T(5))$ is greater than the absolute value of the slope of the secant line i.e. $|T'(5)|>|A|$, however, the slope itself is more negative than the secant line i.e. $T'(5)<A$, similarly we have $|T'(12)|<|A|$, while $T'(12)>A$

• 2nd case: A milk is out of the fridge, so it is heating up. i.e, the temperature is increasing. This implies that the slope of the tangent line at $t=5$ and $t=12$, and the slope of the secant line through the two endpoints are positive. Since in this case $E>T$, $T''(t)<0$, this means that an exponential type graph which is concave down and increasing, so the slope of the tangent line at $(5,T(5))$ is greater than the slope of the secant line i.e. $T'(5)>A$, similarly we have $T'(12)<A$.

The only choice matches with these scenarios is $\color {blue}(ii)$ .