Short Questions (Question 26 & 27)

Question 26 (3 marks):
Find the value of
$\begin{array}{l}\left(\text{a}\right)\underset{x\to 1}{\lim }\left(7-x^2\right),\\ \left(\text{b}\right)f\text{'}\text{'}\left(2\right)\text{ if }f\text{'}\left(x\right)=2x^3-4x+3.\end{array}$

Solution:
(a)
$\begin{array}{l}\underset{x\to 1}{\lim }\left(7-x^2\right)\\ =7-{\left(1\right)}^2\\ =6\end{array}$

(b)
$\begin{array}{l}f\text{'}\left(x\right)=2x^3-4x+3\\ f\text{'}\text{'}\left(x\right)=6x^2-4\\ f\text{'}\text{'}\left(2\right)=6{\left(2\right)}^2-4\\ =24-4\\ =20\end{array}$

Question 27 (4 marks):
It is given that L = 4t – t² and x = 3 + 6t.
(a) Express $\frac{dL}{dx}$ in terms of t.
(b) Find the small change in x, when L changes from 3 to 3.4 at the instant t = 1.

Solution:
(a)
$\begin{array}{l}\text{Given }L=4t-t^2\text{ and }x=3+6t\\ L=4t-t^2\\ \frac{dL}{dt}=4-2t\\ \\ x=3+6t\\ \frac{dx}{dt}=6\\ \\ \frac{dL}{dx}=\frac{dL}{dt}\times \frac{dt}{dx}\\ \frac{dL}{dx}=\left(4-2t\right)\times \frac{1}{6}\\ =\frac{4-2t}{6}\\ =\frac{2-t}{3}\end{array}$


(b)
$\begin{array}{l}\delta L=3.4-3=0.4\\ \frac{\delta L}{\delta x}\approx \frac{dL}{dx}\\ \delta x=\delta L÷\frac{\delta L}{\delta x}\\ \delta x=\delta L\times \frac{\delta x}{\delta L}\\ =0.4\times \frac{3}{2-t}\\ =\frac{2}{5}\times \frac{3}{2-t}\\ =\frac{6}{5\left(2-t\right)}\\ \\ \text{When }t=1,\\ \delta x=\frac{6}{5\left(2-1\right)}=\frac{6}{5}\end{array}$