Question 3:
In a geometric progression, the first term is 217 and the common ratio is 2/3.
Find
(a) the maximum value of n such that the sum of the first n terms of the progression is less than 650, [2 marks]
(b) the difference between the 5th term and the sum of all the terms of the progression. [3 marks]
Answer:
(a) $$ \begin{aligned} S_n & <650 \\ 217\left[\frac{1-\left(\frac{2}{3}\right)^n}{1-\frac{2}{3}}\right] & <650 \\ 1-\left(\frac{2}{3}\right)^n & <\frac{650}{651} \\ 1-\frac{650}{651} & <\left(\frac{2}{3}\right)^n \end{aligned} $$
$$ \begin{aligned} \frac{1}{651} & <\left(\frac{2}{3}\right)^n \\ \left(\frac{2}{3}\right)^n & >\frac{1}{651} \\ n & <\frac{\lg \left(\frac{1}{651}\right)}{\lg \left(\frac{2}{3}\right)} \\ n & <15.97 \\ n=15 & \end{aligned} $$
(b) $$ \begin{aligned} S_{\infty}-T_5 & =\frac{217}{1-\frac{2}{3}}-217\left(\frac{2}{3}\right)^4 \\ & =651-\frac{3472}{81} \\ & =608 \frac{11}{81} \end{aligned} $$
In a geometric progression, the first term is 217 and the common ratio is 2/3.
Find
(a) the maximum value of n such that the sum of the first n terms of the progression is less than 650, [2 marks]
(b) the difference between the 5th term and the sum of all the terms of the progression. [3 marks]
Answer:
(a) $$ \begin{aligned} S_n & <650 \\ 217\left[\frac{1-\left(\frac{2}{3}\right)^n}{1-\frac{2}{3}}\right] & <650 \\ 1-\left(\frac{2}{3}\right)^n & <\frac{650}{651} \\ 1-\frac{650}{651} & <\left(\frac{2}{3}\right)^n \end{aligned} $$
$$ \begin{aligned} \frac{1}{651} & <\left(\frac{2}{3}\right)^n \\ \left(\frac{2}{3}\right)^n & >\frac{1}{651} \\ n & <\frac{\lg \left(\frac{1}{651}\right)}{\lg \left(\frac{2}{3}\right)} \\ n & <15.97 \\ n=15 & \end{aligned} $$
(b) $$ \begin{aligned} S_{\infty}-T_5 & =\frac{217}{1-\frac{2}{3}}-217\left(\frac{2}{3}\right)^4 \\ & =651-\frac{3472}{81} \\ & =608 \frac{11}{81} \end{aligned} $$
Question 4:
It is given that the equation of the curve is y = x + 2x-2.
(a) Find the gradient function of the curve. [1 mark]
(b) A straight line touches the curve at point P. The straight line is perpendicular to the straight line y + 2x = 0.
Find the coordinates of P. [3 marks]
Answer:
(a) $$ \frac{\mathrm{d} y}{\mathrm{~d} x}=1-\frac{4}{x^3} $$
(b) $$ \begin{aligned} & y=-2 x \\ & m_1=-2 \\ & m_2=\frac{1}{2} \\ & 1-\frac{4}{x^3}=\frac{1}{2} \\ & \frac{1}{2}=\frac{4}{x^3} \\ & x=2 \end{aligned} $$
$$ \begin{aligned} &\text { Substitute } x=2 \text { into }\\ &\begin{aligned} y & =x+\frac{2}{x^2} \\ & =2+\frac{2}{x^2} \\ & =2+\frac{2}{4} \\ & =\frac{5}{2} \\ P & \left(2, \frac{5}{2}\right) \end{aligned} \end{aligned} $$
It is given that the equation of the curve is y = x + 2x-2.
(a) Find the gradient function of the curve. [1 mark]
(b) A straight line touches the curve at point P. The straight line is perpendicular to the straight line y + 2x = 0.
Find the coordinates of P. [3 marks]
Answer:
(a) $$ \frac{\mathrm{d} y}{\mathrm{~d} x}=1-\frac{4}{x^3} $$
(b) $$ \begin{aligned} & y=-2 x \\ & m_1=-2 \\ & m_2=\frac{1}{2} \\ & 1-\frac{4}{x^3}=\frac{1}{2} \\ & \frac{1}{2}=\frac{4}{x^3} \\ & x=2 \end{aligned} $$
$$ \begin{aligned} &\text { Substitute } x=2 \text { into }\\ &\begin{aligned} y & =x+\frac{2}{x^2} \\ & =2+\frac{2}{x^2} \\ & =2+\frac{2}{4} \\ & =\frac{5}{2} \\ P & \left(2, \frac{5}{2}\right) \end{aligned} \end{aligned} $$