Short Question 9 & 10

Posted on May 18, 2020 by Myhometuition

Question 9 (2 marks):

\begin{array}{l} (a) Given {}^{6}C_{n} > 1, list out all the \\ possible values of n . \\ (b) Given {}^{y}C_{m} = {}^{y}C_{n}, express y in \\ terms of m and n . \end{array}

Solution:
(a)
n = 1, 2, 3, 4, 5

(b)
y = m + n

Question 10 (4 marks):
Danya has a home decorations shop. One day, Danya received 14 sets of cups from a supplier. Each set contained 6 pieces of cups of different colours.

(a) Danya chooses 3 sets of cups at random to be checked.
Find the number of different ways that Danya uses to choose those sets of cups.

(b) Danya takes a set of cups to display by arranging it in a row.
Find the number of different ways the cups can be arranged such that the blue cup is not displaced next to the red cup.

Solution:
(a)
Number of different ways 3 sets of cups at random to be checked
= ¹⁴C₃=364

(b)

Number of ways (Blue cup and red cup are next to each other)
= 5! × 2!
= 240

Number of different ways the cups can be arranged such that the blue cup is not displaced next to the red cup
= 6! – 240
= 720 – 240
= 480

Short Question 19

Posted on May 18, 2020 by Myhometuition

Question 19 (4 marks):
It is given that cos α = t where t is a constant and 0^o≤ α ≤ 90^o.
Express in terms of t
(a) sin (180^o + α),
(b) sec 2α.

Solution:
(a)

\begin{array}{l} sin (180^{o} + α) \\ = \sin 180 \cos α + \cos 180 \sin α \\ = 0 - \sin α \\ = - \sin α \\ = - \sqrt{1 - t^{2}} \end{array}

(b)

\begin{array}{l} \sec 2 α = \frac{1}{\cos 2 α} \\ = \frac{1}{2 \cos^{2} α - 1} \\ = \frac{1}{2 t^{2} - 1} \end{array}

Short Questions (Question 5 & 6)

Posted on May 18, 2020 by Myhometuition

Question 5 (4 marks):
Diagram shows a circle with centre O.

Diagram

PR and QR are tangents to the circle at points P and Q respectively. It is given that the length of minor arc PQ is 4 cm and

O R = \frac{5}{α} cm .

Express in terms of α,
(a) the radius, r, of the circle,
(b) the area, A, of the shaded region.

Solution:
(a)

\begin{array}{l} Given s_{P Q} = 4 \\ r α = 4 \\ r = \frac{4}{α} cm \end{array}

(b)

\begin{array}{l} P R = \sqrt{{(\frac{5}{α})}^{2} - {(\frac{4}{α})}^{2}} \\ P R = \sqrt{\frac{9}{α^{2}}} \\ P R = \frac{3}{α} \\ A = Area of shaded region \\ A = Area of quadrilateral O P R Q - \\ Area of sector O P Q \\ = 2 (Area of △ O P R) - \frac{1}{2} r^{2} θ \\ = 2 [\frac{1}{2} \times \frac{3}{α} \times \frac{4}{α}] - [\frac{1}{2} \times {(\frac{4}{α})}^{2} \times α] \\ = \frac{12}{α^{2}} - \frac{8}{α} \\ = \frac{12 - 8 α}{α^{2}} {cm}^{2} \end{array}

Question 6 (3 marks):
Diagram shows two sectors AOD and BOC of two concentric circles with centre O.

Diagram

The angle subtended at the centre O by the major arc AD is 7α radians and the perimeter of the whole diagram is 50 cm.
Given OB = r cm, OA = 2OB and ∠BOC = 2α, express r in terms of α.

Solution:

\begin{array}{l} Length of major arc A O D \\ = 2 r \times 7 α \\ = 14 r α \\ Length of minor arc B O C \\ = r \times 2 α \\ = 2 r α \\ Perimeter of the whole diagram \\ = 50 cm \\ 14 r α + 2 r α + r + r = 50 \\ 16 r α + 2 r = 50 \\ 8 r α + r = 25 \\ r (8 α + 1) = 25 \\ r = \frac{25}{8 α + 1} \end{array}

SPM Practice Question 15 & 16

Posted on May 18, 2020 by Myhometuition

Question 15 (2 marks):
It is given that the n^th term of a geometric progression is

T_{n} = \frac{3 r^{n - 1}}{2}, r \neq k .

State
(a) the value of k,
(b) the first term of progression.

Solution:
(a)
k = 0, k = 1 or k = -1 (Any one of these answer).

(b)

\begin{array}{l} T_{n} = \frac{3}{2} r^{n - 1} \\ T_{1} = \frac{3}{2} r^{1 - 1} \\ = \frac{3}{2} r^{0} \\ = \frac{3}{2} (1) \\ = \frac{3}{2} \end{array}

Question 16 (4 marks):
It is given that p, 2 and q are the first three terms of a geometric progression.
Express in terms of q
(a) the first term and the common ratio of the progression.
(b) the sum to infinity of the progression.

Solution:
(a)

\begin{array}{l} T_{1} = p, T_{2} = 2, T_{3} = q \\ \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}} \\ \frac{2}{p} = \frac{q}{2} \\ p = \frac{4}{q} \\ First term, T_{1} = p = \frac{4}{q} \\ Common ratio = \frac{q}{2} \end{array}

(b)

\begin{array}{l} a = \frac{4}{q}, r = \frac{q}{2} \\ S_{\infty} = \frac{a}{1 - r} \\ = \frac{\frac{4}{q}}{1 - \frac{q}{2}} \\ = \frac{4}{q} \div [1 - \frac{q}{2}] \\ = \frac{4}{q} \div [\frac{2 - q}{2}] \\ = \frac{4}{q} \times \frac{2}{2 - q} \\ = \frac{8}{2 q - q^{2}} \end{array}

Quadratic Functions, SPM Practice (Short Questions)

Posted on May 18, 2020 by Myhometuition

Question 10 (3 marks):
Diagram shows the graph y = a (x – p)² + q, where a, p and q are constants. The straight line y = –8 is the tangent to the curve at point H.

Diagram

(a) State the coordinates of H.
(b) Find the value of a.

Solution:
(a)

\begin{array}{l} Coordinate x of H \\ = \frac{- 1 + 7}{2} \\ = \frac{6}{2} \\ = 3 \\ Thus, coordinates of H = (3, - 8) . \end{array}

(b)

\begin{array}{l} y = a {(x - p)}^{2} + q \\ y = a {(x - 3)}^{2} + (- 8) \\ y = a {(x - 3)}^{2} - 8 ......... (1) \\ Substitute (7, 0) into (1) : \\ 0 = a {(7 - 3)}^{2} - 8 \\ 0 = 16 a - 8 \\ 16 a = 8 \\ a = \frac{1}{2} \end{array}

Question 11 (3 marks):
Faizal has a rectangular plywood with a dimension 3x metre in length and 2x metre in width. He cuts part of the plywood into a square shape with sides of x metre to make a table surface.
Find the range of values of x if the remaining area of the plywood is at least (x² + 4) metre².

Solution:

Area of plywood – area of square ≥ (x² + 4)
3x(2x) – x² ≥ x² + 4
6x² – x² – x² ≥ 4
4x² ≥ 4
x² – 1 ≥ 0
(x + 1)(x – 1) ≥ 0
x ≤ –1 or x ≥ 1
Thus, x ≥ 1 (length is > 0)

1.6.4 Function, SPM Practice (Short Question)

Posted on May 18, 2020 by Myhometuition

Question 10 (4 marks):
Given the function g : x → 2x – 8, find

\begin{array}{l} (a) g^{- 1} (x), \\ (b) the value of p such that g^{2} (\frac{3 p}{2}) = 30. \end{array}

Solution:
(a)

\begin{array}{l} Let y = g (x) \\ = 2 x - 8 \\ 2 x - 8 = y \\ 2 x = y + 8 \\ x = \frac{y + 8}{2} \\ Thus, g^{- 1} (x) = \frac{x + 8}{2} \end{array}

(b)

\begin{array}{l} g (x) = 2 x - 8 \\ g^{2} (x) = g [g (x)] \\ = g (2 x - 8) \\ = 2 (2 x - 8) - 8 \\ = 4 x - 16 - 8 \\ = 4 x - 24 \\ g^{2} (\frac{3 p}{2}) = 30 \\ 4 (\frac{3 p}{2}) - 24 = 30 \\ 6 p = 54 \\ p = 9 \end{array}

Question 11 (4 marks):
Diagram 9 shows the relation between set A, set B and set C.

Diagram 9

It is given that set A maps to set B by the function

\frac{x + 1}{2}

and maps to set C by fg : x → x² + 2x + 4.
(a) Write the function which maps set A to set B by using the function notation.
(b) Find the function which maps set B to set C.

Solution:

(a)

g : x \to \frac{x + 1}{2}

(b)

\begin{array}{l} g (x) = \frac{x + 1}{2} \\ f g (x) = x^{2} + 2 x + 4 \\ f [g (x)] = x^{2} + 2 x + 4 \\ f (\frac{x + 1}{2}) = x^{2} + 2 x + 4 \\ Let \frac{x + 1}{2} = y \\ x + 1 = 2 y \\ x = 2 y - 1 \\ ∴ f (y) = {(2 y - 1)}^{2} + 2 (2 y - 1) + 4 \\ f (y) = 4 y^{2} - 4 y + 1 + 4 y - 2 + 4 \\ f (y) = 4 y^{2} + 3 \\ f (x) = 4 x^{2} + 3 \\ Thus, function which maps set B to set C \\ is f (x) = 4 x^{2} + 3 \end{array}

Short Questions (Question 15 – 17)

Posted on May 18, 2020 by Myhometuition

Question 15 (4 marks):

\begin{array}{l} (a) Given P = \log_{a} Q, state the \\ conditions of a . \\ (b) Given \log_{3} y = \frac{2}{\log_{x y} 3}, express \\ y in terms of x . \end{array}

Solution:
(a)
a > 0, a ≠ 1

(b)

\begin{array}{l} \log_{3} y = \frac{2}{\log_{x y} 3} \\ \frac{\log_{x y} y}{\log_{x y} 3} = \frac{2}{\log_{x y} 3} \\ \log_{x y} y = 2 \\ y = {(x y)}^{2} \\ y = x^{2} y^{2} \\ \frac{1}{x^{2}} = \frac{y^{2}}{y} \\ y = \frac{1}{x^{2}} \end{array}

Question 16 (3 marks):

Given \frac{25^{h + 3}}{125^{p - 1}} =1, express p in terms of h .

Solution:

\begin{array}{l} \frac{25^{h + 3}}{125^{p - 1}} = 1 \\ 25^{h + 3} = 125^{p - 1} \\ {(5^{2})}^{h + 3} = {(5^{3})}^{p - 1} \\ 5^{2 h + 6} = 5^{3 p - 3} \\ 2 h + 6 = 3 p - 3 \\ 3 p = 2 h + 9 \\ p = \frac{2 h + 9}{3} \end{array}

Question 17 (3 marks):

\begin{array}{l} Solve the equation: \\ \log_{m} 324 - \log_{\sqrt{m}} 2 m = 2 \end{array}

Solution:

\begin{array}{l} \log_{m} 324 - \log_{\sqrt{m}} 2 m = 2 \\ \log_{m} 324 - \frac{\log_{m} 2 m}{\log_{m} m^{\frac{1}{2}}} = 2 \\ \log_{m} 324 - 2 (\frac{\log_{m} 2 m}{\log_{m} m}) = 2 \\ \log_{m} 324 - 2 \log_{m} 2 m = 2 \\ \log_{m} 324 - \log_{m} {(2 m)}^{2} = l o g_{m} m^{2} \\ \log_{m} (\frac{324}{4 m^{2}}) = l o g_{m} m^{2} \\ \frac{324}{4 m^{2}} = m^{2} \\ 4 m^{4} = 324 \\ m^{4} = 81 \\ m = \pm 3 (- 3 is rejected) \end{array}

Short Question 14 & 15

Posted on May 18, 2020 by Myhometuition

Question 14 (3 marks):
It is given that

\int \frac{5}{{(2 x + 3)}^{n}} d x = \frac{p}{{(2 x + 3)}^{5}} + c

, where c, n and p are constants.
Find the value of n and of p.

Solution:

\begin{array}{l} \int \frac{5}{{(2 x + 3)}^{n}} d x = \int 5 {(2 x + 3)}^{- n} d x \\ = \frac{5 {(2 x + 3)}^{- n + 1}}{(- n + 1) \times 2} + c \\ = \frac{5}{2 (1 - n)} \times \frac{1}{{(2 x + 3)}^{n - 1}} + c \\ = \frac{5}{2 (1 - n) {(2 x + 3)}^{n - 1}} + c \\ Compare \frac{5}{2 (1 - n) {(2 x + 3)}^{n - 1}} \\ with \frac{p}{{(2 x + 3)}^{5}} \\ n - 1 = 5 \\ n = 6 \\ \frac{5}{2 (1 - n)} = p \\ \frac{5}{2 (1 - 6)} = p \\ \frac{5}{2 (- 5)} = p \\ p = - \frac{1}{2} \end{array}

Question 15 (4 marks):
Diagram shows the curve y = g(x). The straight line is a tangent to the curve.

Diagram

Given g’(x) = –4x + 8, find the equation of the curve.

Solution:

\begin{array}{l} Given g' (x) = - 4 x + 8 \\ Maximum point when g' (x) = 0 \\ - 4 x + 8 = 0 \\ 4 x = 8 \\ x = 2 \\ Thus, maximum point is (2, 11) . \\ g' (x) = - 4 x + 8 \\ \int g' (x) = \int (- 4 x + 8) d x \\ g (x) = \frac{- 4 x^{2}}{2} + 8 x + c \\ g (x) = - 2 x^{2} + 8 x + c \\ Substitute (2, 11) into g (x) : \\ 11 = - 2 {(2)}^{2} + 8 (2) + c \\ c = 3 \\ Thus, the equation of curve is \\ g (x) = - 2 x^{2} + 8 x + 3 \end{array}

Short Questions (Question 26 & 27)

Posted on May 18, 2020 by Myhometuition

Question 26 (3 marks):
Find the value of

\begin{array}{l} (a) \lim_{x \to 1} (7 - x^{2}), \\ (b) f'' (2) if f' (x) = 2 x^{3} - 4 x + 3. \end{array}

Solution:
(a)

\begin{array}{l} \lim_{x \to 1} (7 - x^{2}) \\ = 7 - {(1)}^{2} \\ = 6 \end{array}

(b)

\begin{array}{l} f' (x) = 2 x^{3} - 4 x + 3 \\ f'' (x) = 6 x^{2} - 4 \\ f'' (2) = 6 {(2)}^{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{d L}{d x}

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} Given L = 4 t - t^{2} and x = 3 + 6 t \\ L = 4 t - t^{2} \\ \frac{d L}{d t} = 4 - 2 t \\ x = 3 + 6 t \\ \frac{d x}{d t} = 6 \\ \frac{d L}{d x} = \frac{d L}{d t} \times \frac{d t}{d x} \\ \frac{d L}{d x} = (4 - 2 t) \times \frac{1}{6} \\ = \frac{4 - 2 t}{6} \\ = \frac{2 - t}{3} \end{array}

(b)

\begin{array}{l} δ L = 3.4 - 3 = 0.4 \\ \frac{δ L}{δ x} \approx \frac{d L}{d x} \\ δ x = δ L \div \frac{δ L}{δ x} \\ δ x = δ L \times \frac{δ x}{δ L} \\ = 0.4 \times \frac{3}{2 - t} \\ = \frac{2}{5} \times \frac{3}{2 - t} \\ = \frac{6}{5 (2 - t)} \\ When t = 1, \\ δ x = \frac{6}{5 (2 - 1)} = \frac{6}{5} \end{array}

Short Questions (Question 8 & 9)

Posted on May 18, 2020 by Myhometuition

Question 8 (3 marks):
The following information refers to the equation of two straight lines, AB and CD.

\begin{array}{l} A B : y - 2 k x - 3 = 0 \\ C D : \frac{x}{3 h} + \frac{y}{4} = 1 \\ where h and k are constants . \end{array}

Given the straight lines AB and CD are perpendicular to each other, express h in terms of k.

Solution:

\begin{array}{l} A B : y - 2 k x - 3 = 0 \\ y = 2 k x + 3 \\ m_{A B} = 2 k \\ C D : \frac{x}{3 h} + \frac{y}{4} = 1 \\ m_{C D} = - \frac{4}{3 h} \\ m_{A B} \times m_{C D} = - 1 \\ 2 k \times (- \frac{4}{3 h}) = - 1 \\ - 8 k = - 3 h \\ h = \frac{8}{3} k \end{array}

Question 9 (3 marks):
A straight line passes through P(3, 1) and Q(12, 7). The point R divides the line segment PQ such that 2PQ = 3RQ.
Find the coordinates of R.

Solution:

\begin{array}{l} 2 P Q = 3 R Q \\ \frac{P Q}{R Q} = \frac{3}{2} \\ Point R \\ = (\frac{1 (12) + 2 (3)}{1 + 2}, \frac{1 (7) + 2 (1)}{1 + 2}) \\ = (\frac{18}{3}, \frac{9}{3}) \\ = (6, 3) \end{array}