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}

Short Question 10 & 11

Posted on May 18, 2020 by Myhometuition

Question 10 (3 marks):
Diagram 3 shows vectors

\vec{O P}, \vec{O Q} and \vec{O M}

drawn on a square grid.

Diagram

\begin{array}{l} (a) Express \vec{O M} in the form h \underset{˜}{p} + k \underset{˜}{q}, \\ where h and k are constants . \\ (b) On the diagram 3, mark and label \\ the point N such that \vec{M N} + \vec{O Q} = 2 \vec{O P} . \end{array}

Solution:
(a)

\vec{O M} = \underset{˜}{p} + 2 \underset{˜}{q}

(b)

\begin{array}{l} \vec{M N} + \vec{O Q} = 2 \vec{O P} \\ \vec{M N} = 2 \vec{O P} - \vec{O Q} \\ = 2 \underset{˜}{p} - \underset{˜}{q} \end{array}

Question 11 (4 marks):

\begin{array}{l} A (2, 3) and B (- 2, 5) lie on a \\ Cartesian plane . \\ It is given that 3 \vec{O A} = 2 \vec{O B} + \vec{O C} . \\ Find \\ (a) the coordinates of C, \\ (b) | \vec{A C} | \end{array}

Solution:

\begin{array}{l} Given A (2, 3) and B (- 2, 5) \\ Thus, \vec{O A} = 2 \underset{˜}{i} + 3 \underset{˜}{j} and \vec{O B} = - 2 \underset{˜}{i} + 5 \underset{˜}{j} \end{array}

(a)

\begin{array}{l} 3 \vec{O A} = 2 \vec{O B} + \vec{O C} \\ \vec{O C} = 3 \vec{O A} - 2 \vec{O B} \\ = 3 (2 \underset{˜}{i} + 3 \underset{˜}{j}) - 2 (- 2 \underset{˜}{i} + 5 \underset{˜}{j}) \\ = 6 \underset{˜}{i} + 9 \underset{˜}{j} + 4 \underset{˜}{i} - 10 \underset{˜}{j} \\ = 10 \underset{˜}{i} - \underset{˜}{j} \\ Thus, coordinate of C is (10, - 1) \end{array}

(b)

\begin{array}{l} \vec{A C} = \vec{A O} + \vec{O C} \\ = - \vec{O A} + \vec{O C} \\ = - (2 \underset{˜}{i} + 3 \underset{˜}{j}) + 10 \underset{˜}{i} - \underset{˜}{j} \\ = - 2 \underset{˜}{i} + 10 \underset{˜}{i} - 3 \underset{˜}{j} - \underset{˜}{j} \\ = 8 \underset{˜}{i} - 4 \underset{˜}{j} \\ | \vec{A C} | = \sqrt{8^{2} + {(- 4)}^{2}} \\ = \sqrt{80} units \\ = \sqrt{16 \times 5} units \\ = 4 \sqrt{5} units \end{array}

Short Question 8 & 9

Posted on May 18, 2020 by Myhometuition

Question 8 (3 marks):
Diagram shows vectors

\vec{A B}, \vec{A C} and \vec{A D}

drawn on a square grid with sides of 1 unit.

Diagram

\begin{array}{l} (a) Find | - \vec{B A} | . \\ (b) Given \vec{A B} = \underset{˜}{b} and \vec{A C} = \underset{˜}{c}, \\ express in terms of \underset{˜}{b} and \underset{˜}{c} \\ (i) \vec{B C}, \\ (i i) \vec{A D} \end{array}

Solution:
(a)

| - \vec{B A} | = \sqrt{3^{2} + 4^{2}} = 5 units

(b)(i)

\begin{array}{l} \vec{B C} = \vec{B A} + \vec{A C} \\ = - \underset{˜}{b} + \underset{˜}{c} \\ = \underset{˜}{c} - \underset{˜}{b} \end{array}

(b)(ii)

\begin{array}{l} \vec{A D} = \vec{A B} + \vec{B D} \\ = \underset{˜}{b} + 2 \vec{B C} \\ = \underset{˜}{b} + 2 (\underset{˜}{c} - \underset{˜}{b}) \\ = 2 \underset{˜}{c} - \underset{˜}{b} \end{array}

Question 9 (3 marks):
Diagram shows a trapezium ABCD.

Diagram

\begin{array}{l} Given \underset{˜}{p} = (\begin{array}{l} 3 \\ 4 \end{array}) and \underset{˜}{q} = (\begin{array}{l} k - 1 \\ 2 \end{array}), \\ where k is a constant, find value of k . \end{array}

Solution:

\begin{array}{l} \underset{˜}{p} = m \underset{˜}{q} \\ (\begin{array}{l} 3 \\ 4 \end{array}) = m (\begin{array}{l} k - 1 \\ 2 \end{array}) \\ (\begin{array}{l} 3 \\ 4 \end{array}) = (\begin{array}{l} m k - m \\ 2 m \end{array}) \\ m k - m = 3 .......... (1) \\ 2 m = 4 ................ (2) \\ From (2) : \\ 2 m = 4 \\ m = 2 \\ Substitute m = 2 into (1) : \\ 2 k - 2 = 3 \\ 2 k = 3 + 2 \\ 2 k = 5 \\ k = \frac{5}{2} \end{array}

Short Questions (Question 7 – 9)

Posted on May 18, 2020 by Myhometuition

Question 7 (2 marks):
Table shows the information about a set of data.

Table

State
(a) the value of p if m = 20,
(b) the value of q if p = 2.5.

Solution:
(a)
New standard deviation = original standard deviation × p
20 = 5 × p
p = 4

(b)
New median = [original median × p] + 1
q = 2p × 1
q = 2(2.5) + 1
q = 5 + 1
q = 6

Question 8 (3 marks):
Table shows the distribution of scores obtained by a group of students in a competition.

Table

(a) State the minimum value of x if the mode score is 4.
(b) Find the mean score of the distribution if x = 1.

Solution:
(a)
Minimum value of x = 8

(b)

\begin{array}{l} Mean \\ = \frac{1 (3) + 2 (6) + 3 (7) + 4 (1) + 5 (1)}{3 + 6 + 7 + 1 + 1} \\ = \frac{45}{18} \\ = 2.5 \end{array}

Question 9 (4 marks):
Table shows the distribution of marks for 40 students in an Additional Mathematics test. The number of students for the class interval 40 – 59 is not stated.

Table

(a) State the modal class.
(b) Puan Zainon, the subject teacher, intends to give a reward to the top ten students. Those students who achieve the minimum mark in the top ten placing will be considered to receive the reward. Elina obtains 74 marks.
Does Elina qualify to be considered to receive the reward? Give your reason.

Solution:
(a)
4 + 10 + x + 8 + 7 = 40
x + 29 = 40
x = 11

Modal class = 40 – 59

(b)

\begin{array}{l} The top ten placings are T_{31}, T_{32}, T_{33}, ... T_{40} \\ T_{31} = 59.5 + \frac{5}{8} (79.5 - 59.5) \\ = 59.5 + 12.5 \\ = 72 \\ A student has to achieve a minimum mark \\ of 72 . \\ Elina qualifies for the reward because her \\ marks > 72 marks . \end{array}

Short Questions (Question 3 & 4)

Posted on May 18, 2020 by Myhometuition

Question 3 (4 marks):
Diagram shows a standard normal distribution graph.

Diagram

The probability represented by the area of the shaded region is 0.2881.
(a) Find the value of h.
(b) X is a continuous random variable which is normally distributed with a mean, μ and a variance of 16.
Find the value of μ if the z-score of X = 58.8 is h.

Solution:
(a)
P(X < h) = 0.5 – 0.2881
P(X < h) = 0.2119
P(X < –0.8) = 0.2119
h = –0.8

(b)

\begin{array}{l} X = 58.8 \\ \frac{X - μ}{σ} = \frac{58.8 - μ}{σ} \\ Z = \frac{58.8 - μ}{4} \\ h = \frac{58.8 - μ}{4} \\ - 0.8 = \frac{58.8 - μ}{4} \\ - 3.2 = 58.8 - μ \\ μ = 58.8 + 3.2 \\ μ = 62 \end{array}

Question 4 (4 marks):
A voluntary body organizes a first aid course 4 times per month, every Saturday from March until September.
[Assume there are four Saturdays in every month]

Salmah intends to join the course but she might need to spare a Saturday per month to accompany her mother to the hospital. The probability that Salmah will attend the course each Saturday is 0.8. Salmah will be given a certificate of monthly attendance if she can attend the course at least 3 times a month.

(a) Find the probability that Salmah will be given the certificate of monthly attendance.

(b) Salmah will qualify to sit for the first aid test if she obtains more than 5 certificates of monthly attendance.
Find the probability that Salmah qualifies to take the first aid test.

Solution:
(a)

\begin{array}{l} P (X = r) = {}^{n}C_{r} p^{r} q^{n - r} \\ p = 0.8, q = 0.2, n = 4, r = 3, 4 \\ P (X \geq 3) \\ = P (X = 3) + P (X = 4) \\ = {}^{4}C_{3} {(0.8)}^{3} {(0.2)}^{1} + {}^{4}C_{4} {(0.8)}^{4} {(0.2)}^{0} \\ = 0.4096 + 0.4096 \\ = 0.8192 \end{array}

(b)

\begin{array}{l} P (X = r) = {}^{n}C_{r} p^{r} q^{n - r} \\ p = 0.8192, q = 0.1808, n = 7, r = 6, 7 \\ P (X > 5) \\ = P (X = 6) + P (X = 7) \\ = {}^{7}C_{6} {(0.8192)}^{6} {(0.1808)}^{1} + \\ {}^{7}C_{7} {(0.8192)}^{7} {(0.1808)}^{0} \\ = 0.3825 + 0.2476 \\ = 0.6301 \end{array}