2.5 SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Question 5:

(a) The following table shows the corresponding values of x and y for the equation

y = \frac{24}{x}

x	–4	–3	–2	–1	1	1.5	2	3	4
y	–6	k	–12	–24	24	n	12	8	6

Calculate the value of k and n.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.

By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of

y = \frac{24}{x}

for –4 ≤ x ≤ 4.

(c) From your graph, find

(i) The value of y when x = 3.5,

(ii) The value of x when y = –17.

(d) Draw a suitable straight line on your graph to find the value of x which satisfy the equation 2x² + 5x = 24 for –4 ≤ x ≤ 4.

Solution:

(a)

\begin{array}{l} y = \frac{24}{x} \\ when x = - 3, \\ k = \frac{24}{- 3} = - 8 \\ when x = 1.5, \\ n = \frac{24}{1.5} = 16 \end{array}

(b)

(c)

(i) From the graph, when x = 3.5, y = 7

(ii) From the graph, when y = –17, x = –1.4

(d)

\begin{array}{l} y = \frac{24}{x} -------------- (1) \\ 2 x^{2} + 5 x = 24 ------ (2) \\ (2) \div x, 2 x + 5 = \frac{24}{x} -------- (3) \\ (1) - (3), y - (2 x + 5) = 0 \\ y = 2 x + 5 \end{array}

The suitable straight line is y = 2x + 5.

Determine the x-coordinate of the point of intersection of the curve and the straight line y = 2x + 5.

x	0	3
y = 2x + 5	5	11

From the graph, x = 2.5