Further Mathematics/Mathematics (Elective) Objective Tests

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These are the solutions to the Further Mathematics (Elective Mathematics) multiple-choice questions on the Objective Tests.
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(1.) A binary operation * is defined on the set R of real numbers by $p * q = (p + q)^2 + 2p$, where p, q ∈ R.
Find the value of $(-3 * 7) + 5$

$ A.\;\; 27 \\[3ex] B.\;\; 15 \\[3ex] C.\;\; -15 \\[3ex] D.\;\; -27 \\[3ex] $

$ p * q = (p + q)^2 + 2p \\[3ex] (-3 * 7) + 5 \\[3ex] = (-3 + 7)^2 + 2(-3) + 5 \\[3ex] = 4^2 - 6 + 5 \\[3ex] = 16 - 6 + 5 \\[3ex] = 15 $

(2.) Given that $(3x - 1), (x + 1)$ and $(x - 1)$ are the first 3 consecutive terms of an exponential sequence (G.P) where x > 0.
Find the value of x.

$ A.\;\; 6 \\[3ex] B.\;\; 3 \\[3ex] C.\;\; 2 \\[3ex] D.\;\; 1 \\[3ex] $

$ (3x - 1), (x + 1), (x - 1) \\[3ex] \text{common ratio, } r = \dfrac{x + 1}{3x - 1} = \dfrac{x - 1}{x + 1} \\[5ex] (3x - 1)(x - 1) = (x + 1)(x + 1) \\[3ex] 3x^2 - 3x - x + 1 = x^2 + x + x + 1 \\[3ex] 3x^2 - 4x + 1 = x^2 + 2x + 1 \\[3ex] 3x^2 - x^2 - 4x - 2x + 1 - 1 = 0 \\[3ex] 2x^2 - 6x = 0 \\[3ex] 2x(x - 3) = 0 \\[3ex] 2x = 0 \hspace{1em}OR\hspace{1em} x - 3 = 0 \\[3ex] x = \dfrac{0}{2} \hspace{1em}OR\hspace{1em} x = 3 \\[5ex] x = 0, 3 \\[3ex] But\;\; x \gt 0 \\[3ex] Hence\;\; x = 3 $

(3.) Solve $\left(3^{x^2}\right) = \dfrac{1}{81}\left(3^{5x}\right)$

$ A.\;\; 1 \hspace{1em}or\hspace{1em} 4 \\[3ex] B.\;\; 1 \hspace{1em}or\hspace{1em} -4 \\[3ex] C.\;\; -1 \hspace{1em}or\hspace{1em} 4 \\[3ex] D.\;\; -1 \hspace{1em}or\hspace{1em} -4 \\[3ex] $

$ \left(3^{x^2}\right) = \dfrac{1}{81}\left(3^{5x}\right) \\[6ex] 3^{x^2} = \dfrac{1}{3^4}\left(3^{5x}\right) \\[7ex] 3^{x^2} = 3^{5x - 4}...Law\;2...Exp \\[5ex] \text{same base; equate exponents} \\[3ex] x^2 = 5x - 4 \\[3ex] x^2 - 5x + 4 = 0 \\[3ex] (x - 1)(x - 4) = 0 \\[3ex] x - 1 = 0 \hspace{3em}OR\hspace{3em} x - 4 = 0 \\[3ex] x = 1 \hspace{3em}OR\hspace{3em} x = 4 \\[3ex] $ Check

$x = 1, 4$
LHS	RHS
$ 3^{x^2} \\[4ex] x = 1 \\[3ex] 3^{1^2} \\[4ex] 3^1 $ $ 3^{x^2} \\[4ex] x = 4 \\[3ex] 3^{4^2} \\[4ex] 3^{16} $	$ \dfrac{1}{81}\left(3^{5x}\right) \\[6ex] x = 1 \\[3ex] \dfrac{1}{3^4}\left(3^{5(1)}\right)...Law\;5...Exp \\[7ex] \dfrac{3^5}{3^4} \\[7ex] 3^{5 - 4} ...Law\;2...Exp \\[5ex] 3^1 $ $ \dfrac{1}{81}\left(3^{5x}\right) \\[6ex] x = 4 \\[3ex] \dfrac{1}{3^4}\left(3^{5(4)}\right)...Law\;5...Exp \\[7ex] \dfrac{3^{20}}{3^4} \\[7ex] 3^{20 - 4} ...Law\;2...Exp \\[5ex] 3^{16} $

(4.) Find the equation of the axis of symmetry of $y = -4x^2 + 9x - 13$.

$ A.\;\; x = \dfrac{9}{8} \\[5ex] B.\;\; x = \dfrac{8}{9} \\[5ex] C.\;\; x = -\dfrac{8}{9} \\[5ex] D.\;\; x = -\dfrac{9}{8} \\[5ex] $

The equation of the axis of symmetry of a quadratic function is the x-coordinate of the vertex

$ y = -4x^2 + 9x - 13 \\[3ex] Compare\;\;to: y = ax^2 + bx + c \\[3ex] a = -4 \\[3ex] b = 9 \\[3ex] x-coordinate \;\;of\;\;the\;\;vertex \\[3ex] = -\dfrac{b}{2a} \\[5ex] = -\dfrac{9}{2(-4)} \\[5ex] = \dfrac{9}{8} \\[3ex] $ This is the axis of symmetry of the quadratic function.

(5.) If $x = \dfrac{p\sqrt{q}}{r}$, which of the following represents $\log_{10}x$?

$ A.\;\; \dfrac{\log_{10}p \times \log_{10}\sqrt{q}}{\log_{10}r} \\[6ex] B.\;\; \log_{10} p + \dfrac{1}{2}\log_{10}q + \log_{10}r \\[5ex] C.\;\; \dfrac{\log_{10} p + \dfrac{1}{2}\log_{10} q}{\log_{10} r} \\[7ex] D.\;\; \log_{10} p + \dfrac{1}{2}\log_{10} q - \log_{10} r \\[5ex] $

$ \log_{10} x = \log x \\[3ex] x = \dfrac{p\sqrt{q}}{r} \\[3ex] \text{Introduce log to both sides} \\[3ex] \log x = \log\left(\dfrac{p\sqrt{q}}{r}\right) \\[5ex] = \log p + \log \sqrt{q} - \log r ...Laws\;1\;\;and\;\;2...Log \\[4ex] = \log p + \log q^{\dfrac{1}{2}} - \log r ...Law\;7...Exp \\[5ex] = \log p + \dfrac{1}{2}\log q - \log r ...Law\;5...Log \\[5ex] = \log_{10} p + \dfrac{1}{2}\log_{10}q + \log_{10}r $

(6.) Given that $f(x) = \dfrac{5x + 8}{7 - 3x},\;\;x \ne \dfrac{7}{3}$, find $f^{-1}(2)$

$ A.\;\; \dfrac{11}{6} \\[5ex] B.\;\; \dfrac{6}{11} \\[5ex] C.\;\; -\dfrac{6}{11} \\[5ex] D.\;\; -\dfrac{11}{6} \\[5ex] $

$ f(x) = \dfrac{5x + 8}{7 - 3x},\;\;x \ne \dfrac{7}{3} \\[5ex] y = \dfrac{5x + 8}{7 - 3x} \\[5ex] \text{Interchange x and y} \\[3ex] x = \dfrac{5y + 8}{7 - 3y} \\[5ex] \text{Solve for y} \\[3ex] 5y + 8 = x(7 - 3y) \\[3ex] 5y + 8 = 7x - 3xy \\[3ex] 5y + 3xy = 7x - 8 \\[3ex] y(5 + 3x) = 7x - 8 \\[3ex] y = \dfrac{7x - 8}{5 + 3x} \\[5ex] \text{This value of y is the inverse function} \\[3ex] y^{-1} = \dfrac{7x - 8}{5 + 3x} \\[5ex] f^(-1)(x) = \dfrac{7x - 8}{3x + 5} \\[5ex] f^(-1)(2) = \dfrac{7(2) - 8}{3(2) + 5} \\[5ex] = \dfrac{14 - 8}{6 + 5} \\[5ex] = \dfrac{6}{11} $

(7.) If $\displaystyle\int_4^9 \left(\dfrac{1}{\sqrt{x}}\right) dx = y$, find the value of y

$ A.\;\; \dfrac{1}{6} \\[5ex] B.\;\; \dfrac{2}{3} \\[5ex] C.\;\; 2 \\[3ex] D.\;\; 3 \\[3ex] $

$ \dfrac{1}{\sqrt{x}} \\[3ex] = \dfrac{1}{x^{\dfrac{1}{2}}}...Law\;7...Exp \\[7ex] = x^{-\dfrac{1}{2}} ...Law\;6...Exp \\[5ex] \implies \\[3ex] \displaystyle\int \dfrac{1}{\sqrt{x}} dx \\[5ex] = \displaystyle\int x^{-\dfrac{1}{2}} dx \\[5ex] = \dfrac{x^{-\dfrac{1}{2} + 1}}{-\dfrac{1}{2} + 1}...Power\;\;Rule \\[7ex] = \dfrac{x^{\dfrac{1}{2}}}{\dfrac{1}{2}} \\[7ex] = x^{\dfrac{1}{2}} \div \dfrac{1}{2} \\[5ex] = \sqrt{x} * \dfrac{2}{1} \\[5ex] = 2\sqrt{x} \\[5ex] \implies \\[3ex] \displaystyle\int_4^9 \left(\dfrac{1}{\sqrt{x}}\right) dx \\[7ex] = \left[2\sqrt{x}\right]_4^9 \\[5ex] = 2\sqrt{9} - 2\sqrt{4} \\[3ex] = 2(\sqrt{9} - \sqrt{4}) \\[3ex] = 2(3 - 2) \\[3ex] = 2(1) \\[3ex] = 2 \\[3ex] y = 2 $

(8.) Find the sum to infinity of $5 + \dfrac{10}{3} + \dfrac{20}{9} + $

$ A.\;\; \dfrac{5}{3} \\[5ex] B.\;\; 3 \\[3ex] C.\;\; 5 \\[3ex] D.\;\; 15 \\[3ex] $

$ 5 + \dfrac{10}{3} + \dfrac{20}{9} + \\[5ex] \text{first term, } a = 5 \\[3ex] \text{common ratio, } r = \dfrac{10}{3} \div 5 \\[5ex] r = \dfrac{10}{3} * \dfrac{1}{5} \\[5ex] r = \dfrac{2}{3} \\[5ex] \text{Sum to infinity, } S_\infty = \dfrac{a}{1 - r} \\[5ex] = a \div (1 - r) \\[3ex] = 5 \div \left(1 - \dfrac{2}{3}\right) \\[5ex] = 5 \div \dfrac{1}{3} \\[5ex] = 5 * 3 \\[3ex] = 15 $

(9.) Given that $^{(n + 1)}P_2 = 12$, find the value of n.

$ A.\;\; 12 \\[3ex] B.\;\; 6 \\[3ex] C.\;\; 4 \\[3ex] D.\;\; 3 \\[3ex] $

$ ^nP_r = \dfrac{n!}{(n - r)!} \\[5ex] ^{(n + 1)}P_2 = 12 \\[5ex] \dfrac{(n + 1)!}{(n + 1 - 2)!} = 12 \\[5ex] \dfrac{(n + 1) \cdot n \cdot (n - 1)!}{(n - 1)!} = 2 \\[5ex] n^2 + n = 12 \\[3ex] n^2 + n - 12 = 0 \\[3ex] (n + 4)(n - 3) = 0 \\[3ex] n + 4 = 0 \hspace{1em}OR\hspace{1em} n - 3 = 0 \\[3ex] n = -4 \hspace{1em}OR\hspace{1em} n = 3 \\[3ex] n \;\; \text{cannot be negative because it has to be countable} \\[3ex] \therefore n = 3 $

(11.) If $\begin{bmatrix} 4 & -2 \\[2ex] x & 6 \end{bmatrix}\begin{bmatrix} 2 \\[2ex] -5 \end{bmatrix} = 4\begin{bmatrix} 4.5 \\[2ex] -11 \end{bmatrix}$, find the value of x.

$ A.\;\; -9 \\[3ex] B.\;\; -7 \\[3ex] C.\;\; 7 \\[3ex] D.\;\; 9 \\[3ex] $

$ \begin{bmatrix} 4 & -2 \\[2ex] x & 6 \end{bmatrix}\begin{bmatrix} 2 \\[2ex] -5 \end{bmatrix} = 4\begin{bmatrix} 4.5 \\[2ex] -11 \end{bmatrix} $

$ \begin{bmatrix} 4(2) + (-2)(-5) \\[2ex] x(2) + 6(-5) \end{bmatrix} = \begin{bmatrix} 4(4.5) \\[2ex] 4(-11) \end{bmatrix} $

$ \begin{bmatrix} 8 + 10 \\[2ex] 2x - 30 \end{bmatrix} = \begin{bmatrix} 18 \\[2ex] -44 \end{bmatrix} $

$ 2x - 30 = -44 \\[3ex] 2x = -44 + 30 \\[3ex] 2x = -14 \\[3ex] x = -\dfrac{14}{2} \\[5ex] x = -7 $

(12.) If the curve $y = 3x^2 + 10x + 2$ passes through (x, −6), find the least value of x.

$ A.\;\; -5 \\[3ex] B.\;\; -4 \\[3ex] C.\;\; -3 \\[3ex] D.\;\; -2 \\[3ex] $

The curve $y = 3x^2 + 10x + 2$ passes through (x, −6)
We do not know the value of x (we need to find it)
But we do know the value of y = −6

$ y = 3x^2 + 10x + 2 \\[3ex] -6 = 3x^2 + 10x + 2 \\[3ex] 3x^2 + 10x + 2 = -6 \\[3ex] 3x^2 + 10x + 2 + 6 = 0 \\[3ex] 3x^2 + 10x + 8 = 0 \\[3ex] 3x^2 + 6x + 4x + 8 = 0 \\[3ex] 3x(x + 2) + 4(x + 2) = 0 \\[3ex] (x + 2)(3x + 4) = 0 \\[3ex] x + 2 = 0 \hspace{1em}OR\hspace{1em} 3x + 4 = 0 \\[3ex] x = -2 \hspace{1em}OR\hspace{1em} 3x = -4 \\[3ex] x = -2 \hspace{1em}OR\hspace{1em} x = -\dfrac{4}{3} \\[5ex] -2 \lt -\dfrac{4}{3} \\[5ex] $ ∴ the least value of x = −2

(13.) For what values of x is $11x + 12 \gt x^2$

$ A.\;\; -1 \lt x \lt 12 \\[3ex] B.\;\; -1 \lt x \lt \dfrac{1}{12} \\[5ex] C.\;\; x \le -1 \;\;or\;\; x \ge \dfrac{1}{12} \\[5ex] D.\;\; x \le -1 \;\;or\;\; x \ge 12 \\[3ex] $

$ 11x + 12 \gt x^2 \\[3ex] x^2 \lt 11x + 12 \\[3ex] x^2 - 11x - 12 \lt 0 \\[3ex] (x + 1)(x - 12) \lt 0 \\[3ex] Assume:\;\;(x + 1)(x - 12) = 0 \\[3ex] x + 1 = 0 \;\;\;OR\;\;\; x - 12 = 0 \\[3ex] x = -1 \;\;\;OR\;\;\; x = 12 \\[3ex] Test\;\;Intervals\;\;are: \\[3ex] x \lt -1 \\[3ex] -1 \lt x \lt 12 \\[3ex] x \gt 12 \\[3ex] $

Test Intervals: Let:	x < −1 x = −2	−1 < x < 12 x = 0	x > 12 x = 13
$x + 1$	−	+	+
$x - 12$	−	−	+
$(x + 1)(x - 12)$	+	−	+

The second test interval gives the solution of the inequality because a negative number is less than 0
Set Notation: {x | −1 < x < 12}
Interval Notation: (−1, 12)

Check

(−1, 12)
Let x = 0
LHS	RHS
$ 11x + 12 \\[3ex] 11(0) + 12 \\[3ex] 0 + 12 \\[3ex] 12 $	$ x^2 \\[3ex] 0^2 \\[3ex] 0 $
12 > 0

(14.) The table shows the distribution of marks obtained by students in a Mathematics test.

Marks	1 — 5	6 — 10	11 — 15	16 — 20
Number of Students	5	5	x	8
Cumulative Frequency	5	10	12	y

Find the value of $(2x + y)$

$ A.\;\; 38 \\[3ex] B.\;\; 34 \\[3ex] C.\;\; 28 \\[3ex] D.\;\; 24 \\[3ex] $

Based on the Cumulative Frequency:

$ 5 \\[3ex] 5 + 5 = 10 \\[3ex] 10 + x = 12 \implies x = 12 - 10 = 2 \\[3ex] 12 + 8 = y \implies y = 20 \\[5ex] 2x + y \\[3ex] = 2(2) + 20 \\[3ex] = 4 + 20 \\[3ex] = 24 $

(15.) If $\cos(y) = -\dfrac{1}{3}$, where $90^\circ \lt y \lt 180^\circ$, find cosec(y)

$ A.\;\; \dfrac{3}{4}\sqrt{2} \\[5ex] B.\;\; \dfrac{4}{3}\sqrt{2} \\[5ex] C.\;\; -\dfrac{3}{4}\sqrt{2} \\[5ex] D.\;\; -\dfrac{4}{3}\sqrt{2} \\[5ex] $

$ \sin^2 y + \cos^2 y = 1 ...\text{Pythagorean Identity} \\[3ex] \sin^2 y = 1 - \cos^2 y \\[3ex] = 1 - \left(-\dfrac{1}{3}\right)^2 \\[5ex] = 1 - \dfrac{1}{9} \\[5ex] = \dfrac{8}{9} \\[5ex] \sin y = \pm \sqrt{\dfrac{8}{9}} \\[5ex] \text{But } 90^\circ \lt y \lt 180^\circ \;\;\text{is in the 2nd Quadrant} \\[3ex] \text{In the 2nd Quadrant, sine is positive} \\[3ex] \implies \\[3ex] \sin y = \sqrt{\dfrac{8}{9}} \\[5ex] = \dfrac{\sqrt{4 * 2}}{3} \\[5ex] = \dfrac{2\sqrt{2}}{3} \\[5ex] \csc y = \dfrac{1}{\sin y} ...\text{Reciprocal Identity} \\[5ex] \csc y = 1 \div \sin y \\[3ex] = 1 \div \dfrac{2\sqrt{2}}{3} \\[5ex] = 1 * \dfrac{3}{2\sqrt{2}} \\[5ex] = \dfrac{3}{2\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \\[5ex] = \dfrac{3\sqrt{2}}{2 * 2} \\[5ex] = \dfrac{3\sqrt{2}}{4} $

(16.) A ball is thrown vertically upwards with a velocity of 12$ms^{-1}$.
Calculate the maximum height reached.
[Take g = 10$ms^{-2}$]

$ A.\;\; 10.0 m \\[3ex] B.\;\; 9.6 m \\[3ex] C.\;\; 7.2 m \\[3ex] D.\;\; 6.4 m \\[3ex] $

We shall use one of the equations of motion:

$ v^2 = u^2 + 2as \\[3ex] $ When the ball is thrown vertically upwards:
(1.) The final velocity, v is zero.
But we have an initial velocity, u = 12$ms^{-1}$

(2.) The acceleration, a is negative because it is working against gravity
Hence, it is the negative value of the acceleration due to gravity
a = −g

(3.) We shall take the distance, s to be the maximum height, H
So, we have

$ v^2 = u^2 + 2as \\[3ex] becomes \\[3ex] v^2 = u^2 - 2gH \\[3ex] 0^2 = 12^2 - 2(10) * H \\[3ex] 0 = 144 - 20H \\[3ex] 20H = 144 \\[3ex] H = \dfrac{144}{20} \\[5ex] H = 7.2\;m $

(17.) Find the equation of a circle with centre (6, −11) and radius 8 units.

$ A.\;\; x^2 + y^2 - 12x + 22y + 93 = 0 \\[3ex] B.\;\; x^2 + y^2 - 12x - 22y + 93 = 0 \\[3ex] C.\;\; x^2 + y^2 - 12x + 22y - 93 = 0 \\[3ex] D.\;\; x^2 + y^2 - 12x - 22y - 93 = 0 \\[3ex] $

The equation of a circle in the (x, y) coordinate plane with center (h, k) and radius, r is:

$ (x - h)^2 + (y - k)^2 = r^2 \\[3ex] Center:\;\; (h, k) = (6, -11) \\[3ex] h = 6 \\[3ex] k = -11 \\[3ex] r = 8 \\[3ex] \implies \\[3ex] (x - 6)^2 + (y - -11)^2 = 8^2 \\[3ex] (x - 6)^2 + (y + 11)^2 = 64 \\[3ex] [(x - 6)(x - 6)] + [(y + 11)(y + 11)] - 64 = 0 \\[3ex] (x^2 - 6x - 6x + 36) + (y^2 + 11y + 11y + 121) - 64 = 0 \\[3ex] x^2 - 12x + 36 + y^2 + 22y + 121 - 64 = 0 \\[3ex] x^2 + y^2 - 12x + 22y + 93 = 0 $

(19.) A committee of two is to be selected from 5 boys and 3 girls.
What is the probability that it will consist of a boy and a girl?

$ A.\;\; \dfrac{15}{28} \\[5ex] B.\;\; \dfrac{5}{28} \\[5ex] C.\;\; \dfrac{3}{28} \\[5ex] D.\;\; \dfrac{2}{7} \\[5ex] $

total number = 5 boys and 3 girls = 8 people
selected number = 1 boy and 1 girl = 2 people

Let:
S = sample space
$n(S) = C(8, 2)$

A = event of selecting 1 boy from 5 boys
$n(A) = C(5, 1)$

B = event of selecting 1 girl from 3 girls
$n(B) = C(3, 1)$

$ C(n, r) = \dfrac{n!}{(n - r) r!} \\[5ex] n(S) = C(8, 2) = \dfrac{8!}{(8 - 2)! * 2!} \\[5ex] = \dfrac{8 * 7 * 6!}{6! * 2 * 1} \\[5ex] = 4 * 7 \\[3ex] = 28 \\[5ex] n(A) = C(5, 1) = \dfrac{5!}{(5 - 1)! * 1!} \\[5ex] = \dfrac{5 * 4!}{4! * 1} \\[5ex] = 5 \\[5ex] n(B) = C(3, 1) = \dfrac{3!}{(3 - 1)! * 1!} \\[5ex] = \dfrac{3 * 2!}{2! * 1} \\[5ex] = 3 \\[5ex] $ Probability of selecting 1 boy from 5 boys AND 1 girl from 3 girls is:

$ = \dfrac{n(A) * n(B)}{n(S)} \\[5ex] = \dfrac{5 * 3}{28} \\[5ex] = \dfrac{15}{28} $

(20.) Two bodies having masses of 6 kg and 4 kg moving in the same direction with velocities 2 $ms^{-1}$ and 4 $ms^{-1}$ respectively, collide with each other.
If they stick together after collision, find their common velocity.

$ A.\;\; 1.6ms^{-1} \\[3ex] B.\;\; 2.8ms^{-1} \\[3ex] C.\;\; 3.2ms^{-1} \\[3ex] D.\;\; 3.6ms^{-1} \\[3ex] $

Let the:
First body be P: mass of 6 kg and velocity of 2 m/s
Second body be Q: mass of 4 kg and velocity of 4 m/s
Momentum = mass × velocity

$ \underline{\text{Before Collision}} \\[3ex] \text{Momentum of Body P} = 6 \times 2 = 12\;kgm/s \\[3ex] \text{Momentum of Body Q} = 4 \times 4 = 16\;kgm/s \\[3ex] \text{Total Momentum} = 12 + 16 = 28\;kgm/s \\[5ex] \underline{\text{After Collision}} \\[3ex] \text{Mass of P and Q after they stick together} = 6 + 4 = 10\;kg \\[3ex] \text{Let the common velocity} = v\;m/s \\[3ex] \text{Total Momentum} = 10 \times v = 10v\;kgm/s \\[5ex] \text{Based on the Principle of Conservation of Momentum:} \\[3ex] 28 = 10v \\[3ex] 10v = 28 \\[3ex] v = \dfrac{28}{10} \\[5ex] v = 2.8\;kgms^{-1} $

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