Long Question (Question 9)

Posted on May 17, 2020 by Myhometuition

Question 9 (7 marks):
Diagram shows the plan of a rectangular garden ABCD. The garden consists of a semicircular pond ATD and grassy area ABCDT.

It is given that DC = 6y metre and BC = 7x metre, x ≠ y. The area of the rectangular garden ABCD is 168 metre² and the perimeter of the grassy area is 60 metre. The pond with uniform depth contains 15.4 metre³ of water.
By using

$π = \frac{22}{7}$ , find the depth, in metre, of water in the pond.

Solution:

$\begin{array}{l} Area of A B C D = 168 \\ (6 y) (7 x) = 168 \\ 42 x y = 168 \\ x y = 4................... (1) \\ Perimeter of grassy area = 60 \\ 6 y + 6 y + 7 x + (\frac{1}{2} \times 2 \times \frac{22^{11}}{7} \times \frac{7 x}{2}) = 60 \\ 12 y + 18 x = 60 \\ 2 y + 3 x = 10................... (2) \\ From (1) : x y = 4 \\ x = \frac{4}{y} ................... (3) \\ Substitute (3) into (2): \\ 2 y + 3 (\frac{4}{y}) = 10 \\ 2 y^{2} + 12 = 10 y \\ y^{2} - 5 y + 6 = 0 \\ (y - 2) (y - 3) = 0 \\ y - 2 = 0 \\ y = 2 \\ or \\ y - 3 = 0 \\ y = 3 \\ Substitute the values of y into (3): \\ When y = 2 \\ x = \frac{4}{2} = 2 (ignored, x \neq y) \\ When y = 3 \\ x = \frac{4}{3} \\ Given volume of water = 15.4 \\ \frac{1}{2} (\frac{22}{7}) {(\frac{7 x}{2})}^{2} d = 15.4 \\ \frac{1}{2} (\frac{22}{7}) {[\frac{7 (\frac{4}{3})}{2}]}^{2} d = 15.4 \\ \frac{11}{7} {(\frac{14}{3})}^{2} d = 15.4 \\ \frac{308}{9} d = 15.4 \\ d = 0.45 m \\ Thus, the depth of water in the pond is 0 .45 m . \end{array}$

Long Questions (Question 5)

Posted on May 17, 2020 by Myhometuition

Question 5 (7 marks):
It is given that the equation of a curve is

$y = \frac{5}{x^{2}} .$
(a) Find the value of

$\frac{d y}{d x}$ when x = 3.
(b) Hence, estimate the value of

$\frac{5}{{(2.98)}^{2}} .$

Solution:
(a)

$\begin{array}{l} y = \frac{5}{x^{2}} = 5 x^{- 2} \\ \frac{d y}{d x} = - 10 x^{- 3} \\ = - \frac{10}{x^{3}} \\ When x = 3 \\ \frac{d y}{d x} = - \frac{10}{3^{3}} = - \frac{10}{27} \end{array}$

(b)

$\begin{array}{l} δ x = 2.98 - 3 = - 0.02 \\ δ y = \frac{d y}{d x} . δ x \\ = - \frac{10}{27} \times (- 0.02) \\ = 0.007407 \\ Values of \frac{5}{{(2.98)}^{2}} \\ = y + δ y \\ = \frac{5}{x^{2}} + (0.007407) \\ = \frac{5}{3^{2}} + (0.007407) \\ = 0.56296 \end{array}$

Long Questions (Question 4)

Posted on May 17, 2020 by Myhometuition

Question 4 (6 marks):
Diagram shows the front view of a part of a roller coaster track in a miniature park.

The curve part of the track of the roller coaster is represented by an equation

$y = \frac{1}{64} x^{3} - \frac{3}{16} x^{2}$ , with point A as the region.
Find the shortest vertical distance, in m, from the track to ground level.

Solution:

$\begin{array}{l} y = \frac{1}{64} x^{3} - \frac{3}{16} x^{2} ............... (1) \\ \frac{d y}{d x} = 3 (\frac{1}{64}) x^{2} - 2 (\frac{3}{16}) x^{1} \\ = \frac{3}{64} x^{2} - \frac{3}{8} x \\ At turning point, \frac{d y}{d x} = 0 \\ \frac{3}{64} x^{2} - \frac{3}{8} x = 0 \\ x (\frac{3}{64} x - \frac{3}{8}) = 0 \\ x = 0 or \\ \frac{3}{64} x - \frac{3}{8} = 0 \\ \frac{3}{64} x = \frac{3}{8} \\ x = \frac{3}{8} \times \frac{64}{3} \\ x = 8 \\ Substitute values of x into equation (1): \\ When x = 0, \\ y = \frac{1}{64} {(0)}^{3} - \frac{3}{16} {(0)}^{2} \\ y = 0 \\ When x = 8, \\ y = \frac{1}{64} {(8)}^{3} - \frac{3}{16} {(8)}^{2} \\ y = - 4 \\ Thus, turning points : (0, 0) and (8, - 4) \end{array}$

$\begin{array}{l} \frac{d y}{d x} = \frac{3}{64} x^{2} - \frac{3}{8} x \\ \frac{d^{2} y}{d x^{2}} = 2 (\frac{3}{64}) x - \frac{3}{8} \\ = \frac{3}{32} x - \frac{3}{8} \\ When x = 0, \\ \frac{d^{2} y}{d x^{2}} = \frac{3}{32} (0) - \frac{3}{8} \\ = - \frac{3}{8} (< 0) \\ (0, 0) is maximum point . \\ When x = 8, \\ \frac{d^{2} y}{d x^{2}} = \frac{3}{32} (8) - \frac{3}{8} \\ = \frac{3}{8} (> 0) \\ (8, - 4) is minimum point . \\ Shortest vertical distance between track \\ and ground level is at the minimum point . \\ Shortest vertical distance \\ = 5 - 4 \\ = 1 m \end{array}$

Long Questions (Question 10)

Posted on May 17, 2020 by Myhometuition

Question 10 (8 marks):
Diagram shows a circle and a sector of a circle with a common centre O. The radius of the circle is r cm.

It is given that the length of arc PQ and arc RS are 2 cm and 7 cm respectively. QR = 10 cm.
[Use θ = 3.142]
Find
(a) the value of r and of θ,
(b) the area, in cm², of the shaded region.

Solution:
(a)

$\begin{array}{l} Length of arc P Q = 2 cm \\ r θ = 2 ................. (1) \\ Length of arc R S = 7 cm \\ (r + 10) θ = 7 \\ r θ + 10 θ = 7 ................. (2) \\ Substitute (1) into (2) : \\ 2 + 10 θ = 7 \\ 10 θ = 5 \\ θ = \frac{5}{10} \\ θ = 0.5 rad \\ From (1) : \\ When θ = 0.5 rad, \\ r \times 0.5 = 2 \\ r = 4 \end{array}$

(b)

$\begin{array}{l} O S = O R = 4 + 10 = 14 cm \\ Area of shaded region \\ = area of Δ O R S - area of sector O P Q \\ = (\frac{1}{2} \times 14^{2} \times \sin 0.5 rad) - (\frac{1}{2} \times 4^{2} \times 0.5) \\ = 42.981 {cm}^{2} \end{array}$

Long Questions (Question 9)

Posted on May 17, 2020 by Myhometuition

Question 9 (7 marks):
Mathematics Society of SMK Mulia organized a competition to design a logo for the society.

Diagram shows the circular logo designed by Adrian. The three blue coloured regions are congruent. It is given that the perimeter of the blue coloured region is 20π cm.
[Use π = 3.142]
Find
(a) the radius, in cm, of the logo to the nearest integer,
(b) the area, in cm², of the yellow coloured region.

Solution:
(a)

$\begin{array}{l} 6 arcs = 20 π \\ 6 r θ = 20 π \\ 6 r [60^{o} \times \frac{π}{{180^{o}}^{3}}] = 20 π \\ 2 π r = 20 π \\ r = 10 cm \end{array}$

(b)

$\begin{array}{l} Area of yellow coloured region \\ = 3 [area of triangle O A B] - 6 [area of segment] \\ = 3 [\frac{1}{2} a b \sin C] - 6 [\frac{1}{2} r^{2} (θ - \sin θ)] \\ = 3 [\frac{1}{2} (10) (10) \sin 120^{o}] - 6 [\frac{1}{2} {(10)}^{2} (θ - \sin θ)] \\ = 3 (43.3013) - 6 [50 (1.0473 - \sin 1.0473)] \leftarrow \begin{array}{l} change to rad mode \\ θ = 60^{o} \times \frac{3.142}{180^{o}} = 1.0473 \end{array} \\ = 129.9039 - 6 (9.0612) \\ = 129.9039 - 54.3672 \\ = 75.54 {cm}^{2} \end{array}$

Long Questions (Question 10)

Posted on May 17, 2020 by Myhometuition

Question 10 (7 marks):
Solution by scale drawing is not accepted.
Diagram shows the locations of town M and town N drawn on a Cartesian plane.

PQ is a straight road such that the distance from town M and town N to any point on the road is always equal.
(a) Find the equation of PQ.

(b) Another straight road, ST with an equation y = 2x + 7 is to be built.
(i) A traffic light is to be installed at the crossroads of the two roads.
Find the coordinates of the traffic light.
(ii) Which of the two roads passes through town L

$(- \frac{4}{3}, 1) ?$

Solution:
(a)

$\begin{array}{l} T (x, y) is a point on P Q . \\ T M = T N \\ \sqrt{[x - {(- 4)}^{2}] + {[y - (- 1)]}^{2}} = \sqrt{{(x - 2)}^{2} + {(y - 1)}^{2}} \\ \sqrt{{(x + 4)}^{2} + {(y + 1)}^{2}} = \sqrt{{(x - 2)}^{2} + {(y - 1)}^{2}} \\ {(x + 4)}^{2} + {(y + 1)}^{2} = {(x - 2)}^{2} + {(y - 1)}^{2} \\ x^{2} + 8 x + 16 + y^{2} + 2 y + 1 \\ = x^{2} - 4 x + 4 + y^{2} - 2 y + 1 \\ 8 x + 2 y + 17 + 4 x + 2 y - 5 = 0 \\ 12 x + 4 y + 12 = 0 \\ 3 x + y + 3 = 0 \\ Equation of P Q : 3 x + y + 3 = 0 \end{array}$

(b)(i)

$\begin{array}{l} y = 2 x + 7 ............ (1) \\ 3 x + y + 3 = 0 ............ (2) \\ Substitute (1) into (2) : \\ 3 x + 2 x + 7 + 3 = 0 \\ 5 x = - 10 \\ x = - 2 \\ When x = - 2, \\ F r o m (1), \\ y = 2 (- 2) + 7 = 3 \\ Coordinates of traffic light = (- 2, 3) . \end{array}$

(b)(ii)

$\begin{array}{l} L (- \frac{4}{3}, 1) : x = - \frac{4}{3}, y = 1 \\ The equation of S T : y = 2 x + 7 \\ Left hand side: y = 1 \\ Right hand side: 2 (- \frac{4}{3}) + 7 = 4 \frac{1}{3} \\ Thus, the road y = 2 x + 7 does not \\ pass through L . \\ The equation of P Q : 3 x + y + 3 = 0 \\ Left hand side: \\ 3 x + y + 3 = 3 (- \frac{4}{3}) + 1 + 3 \\ = - 4 + 4 = 0 \\ Right hand side = 0 \\ Left hand side = Right hand side \\ Thus, the road 3 x + y + 3 = 0 \\ passes through L . \end{array}$

Long Questions (Question 9)

Posted on May 17, 2020 by Myhometuition

Question 9 (6 marks):
Solution by scale drawing is not accepted.
Diagram shows a triangle OCD.

Diagram

(a) Given the area of triangle OCD is 30 units², find the value of h.

(b) Point Q (2, 4) lies on the straight line CD.
(i) Find CQ : QD.
(ii) Point P moves such that PD = 2 PQ.
Find the equation of the locus P.

Solution:
(a)

$\begin{array}{l} Given Area of △ O C D = 30 \\ \frac{1}{2} | \begin{array}{l} 0 h - 6 \\ 0 2 8 \end{array} \begin{array}{l} 0 \\ 0 \end{array} | = 30 \\ | (0) (2) + (h) (8) + (- 6) (0) - (0) (h) - (2) (- 6) - (8) (0) | = 60 \\ | 0 + 8 h + 0 - 0 + 12 - 0 | = 60 \\ | 8 h + 12 | = 60 \\ 8 h + 12 = 60 \\ 8 h = 48 \\ h = 6 \\ or \\ 8 h + 12 = - 60 \\ 8 h = - 72 \\ h = - 9 (ignore) \end{array}$

(b)(i)

$\begin{array}{l} [\frac{6 (m) + (- 6) (n)}{m + n}, \frac{2 (m) + (8) (n)}{m + n}] = (2, 4) \\ \frac{6 m - 6 n}{m + n} = 2 \\ 6 m - 6 n = 2 m + 2 n \\ 4 m = 8 n \\ \frac{m}{n} = \frac{8}{4} \\ \frac{m}{n} = \frac{2}{1} \\ \frac{2 m + 8 n}{m + n} = 4 \\ 2 m + 8 n = 4 m + 4 n \\ 2 m = 4 n \\ \frac{m}{n} = \frac{4}{2} \\ \frac{m}{n} = \frac{2}{1} \\ Thus, C Q = Q D = 2 : 1 \end{array}$

(b)(ii)

$\begin{array}{l} P D = 2 P Q \\ \sqrt{{(x - 6)}^{2} + {(y - 2)}^{2}} = 2 \sqrt{{(x - 2)}^{2} + {(y - 4)}^{2}} \\ {(x - 6)}^{2} + {(y - 2)}^{2} = 4 [{(x - 2)}^{2} + {(y - 4)}^{2}] \\ x^{2} - 12 x + 36 + y^{2} - 4 y + 4 = 4 [x^{2} - 4 x + 4 + y^{2} - 8 y + 16] \\ x^{2} - 12 x + 36 + y^{2} - 4 y + 4 = 4 x^{2} - 16 x + 16 + 4 y^{2} - 32 y + 64 \\ The equation of locus P : \\ 3 x^{2} + 3 y^{2} - 4 x - 28 y + 40 = 0 \end{array}$

SPM Practice (Long Question)

Posted on May 17, 2020 by Myhometuition

Question 4:

$\begin{array}{l} The function f is denoted by f : x \to \frac{1 + x}{1 - x}, x \neq 1. \\ Find f^{2}, f^{3}, f^{4} and hence write down the functions f^{51} and f^{52} . \end{array}$

Solution:

$\begin{array}{l} f (x) = \frac{1 + x}{1 - x}, x \neq 1 \\ f^{2} (x) = f [f (x)] = f (\frac{1 + x}{1 - x}) \\ = \frac{1 + (\frac{1 + x}{1 - x})}{1 - (\frac{1 + x}{1 - x})} = \frac{\frac{1 - x + 1 + x}{1 - x}}{\frac{1 - x - 1 - x}{1 - x}} \\ = \frac{2}{- 2 x} = - \frac{1}{x} \\ f^{3} (x) = f [f^{2} (x)] = f (- \frac{1}{x}) \\ = \frac{1 + (- \frac{1}{x})}{1 - (- \frac{1}{x})} = \frac{\frac{x - 1}{x}}{\frac{x + 1}{x}} \\ = \frac{x - 1}{x + 1} \\ f^{4} (x) = f [f^{3} (x)] = f (\frac{x - 1}{x + 1}) \\ = \frac{1 + (\frac{x - 1}{x + 1})}{1 - (\frac{x - 1}{x + 1})} = \frac{\frac{x + 1 + x - 1}{x + 1}}{\frac{x + 1 - x + 1}{x + 1}} \\ = \frac{2 x}{2} = x \\ f^{5} (x) = f [f^{4} (x)] = f (x) = \frac{1 + x}{1 - x} (recurring) \\ ∴ f^{51} (x) = f^{3} [f^{48} (x)] = f^{3} (x) \\ = \frac{x - 1}{x + 1} \\ f^{52} (x) = f^{4} [f^{48} (x)] = f^{4} (x) = x \end{array}$

SPM Practice (Long Question)

Posted on May 17, 2020 by Myhometuition

Question 2:
The function f and g is defined by

$\begin{array}{l} f (x) = 3 x - 2 \\ g (x) = \frac{3}{x}, x \neq 0 \\ Find \\ (a) f^{- 1} (2), \\ (b) g f (- 3), \\ (c) function h if h f (x) = 3 x + 2, \\ (d) function k if f k (x) = 4 x - 7. \end{array}$

Solution:
(a)

$\begin{array}{l} Let f^{- 1} (2) = x, \\ thus f (x) = 2 \\ 3 x - 2 = 2 \\ 3 x = 4 \\ x = \frac{4}{3} \\ f^{- 1} (2) = \frac{4}{3} \end{array}$

(b)

$\begin{array}{l} g f (- 3) = g [3 (- 3) - 2] \\ = g (- 11) \\ = - \frac{3}{11} \end{array}$

(c)

$\begin{array}{l} h [f (x)] = 3 x + 2 \\ h (3 x - 2) = 3 x + 2 \\ Let y = 3 x - 2 \\ thus x = \frac{y + 2}{3} \\ h (y) = 3 (\frac{y + 2}{3}) + 2 \\ = y + 2 + 2 \\ = y + 4 \\ ∴ h (x) = x + 4 \end{array}$

(d)

$\begin{array}{l} f [k (x)] = 4 x - 7 \\ 3 k (x) - 2 = 4 x - 7 \\ 3 k (x) = 4 x - 5 \\ k (x) = \frac{4 x - 5}{3} \end{array}$

SPM Practice (Long Question)

Posted on May 17, 2020 by Myhometuition

Question 6 (8 marks):
It is given that g : x → 2x – 3 and h : x → 1 – 3x.
(a) Find
(i) h (5)
(ii) the value of k if

$g (k + 2) = \frac{1}{7} h (5),$
(iii) hg(x).

(b) Hence, sketch the graph of y = | hg(x) | for –1 ≤ x ≤ 3.
State the range of y.

Solution:
(a)(i)

$\begin{array}{l} h (x) = 1 - 3 x \\ h (5) = 1 - 3 (5) \\ = - 14 \end{array}$

(a)(ii)

$\begin{array}{l} g (x) = 2 x - 3 \\ g (k + 2) = \frac{1}{7} h (5) \\ 2 (k + 2) - 3 = \frac{1}{7} (- 14) \\ 2 k + 4 - 3 = - 2 \\ 2 k = - 3 \\ k = - \frac{3}{2} \end{array}$

(a)(iii)

$\begin{array}{l} g (x) = 2 x - 3, h (x) = 1 - 3 x \\ h g (x) = h (2 x - 3) \\ = 1 - 3 (2 x - 3) \\ = 1 - 6 x + 9 \\ = 10 - 6 x \end{array}$

(b)
y = |hg(x)|,
y = |10 – 6x|
Range of y : 0 ≤ y ≤ 16