WAEC SSCE - Further Mathematics - 2018 (Objective)

Question 1 Report

What is the probability of obtaining a head and a six when a fair coin and and a die are tossed together?

$\frac{1}{12}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{2}{3}$

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Probability

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Question 2 Report

A particle starts from rest and moves through a distance $S = 12t^{2} - 2t^{3}$ metres in time t seconds. Find its acceleration in 1 second.

24$ms^{-2}$

18$ms^{-2}$

12$ms^{-2}$

10$ms^{-2}$

Answer Details

The distance moved by a particle is given by the equation $S = 12t^{2} - 2t^{3}$, where S is the distance travelled in metres, and t is the time taken in seconds. To find the acceleration of the particle in 1 second, we need to differentiate the equation for distance with respect to time twice to obtain the equation for acceleration. First, we differentiate S with respect to time to get the velocity equation: $$\frac{dS}{dt} = \frac{d}{dt}(12t^{2} - 2t^{3}) = 24t - 6t^{2}$$ Next, we differentiate the velocity equation with respect to time to get the acceleration equation: $$\frac{d^2S}{dt^2} = \frac{d}{dt}(24t - 6t^{2}) = 24 - 12t$$ To find the acceleration in 1 second, we substitute t = 1 into the acceleration equation: $$\frac{d^2S}{dt^2} \bigg\rvert_{t=1} = 24 - 12(1) = 12ms^{-2}$$ Therefore, the acceleration of the particle in 1 second is 12$ms^{-2}$.

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Question 3 Report

Calculate the standard deviation of 30, 29, 25, 28, 32 and 24.

2.0

2.8

3.0

3.2

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Statistics

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Question 4 Report

If the polynomial $f(x) = 3x^{3} - 2x^{2} + 7x + 5$ is divided by (x - 1), find the remainder.

-17

-7

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Polynomial Functions

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Question 5 Report

Evaluate $\int_{-1}^{1} (x + 1)^{2}\mathrm {d} x$.

$\frac{8}{3}$

$\frac{7}{3}$

$\frac{5}{3}$

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Integration

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Question 6 Report

If $\log_{3}a - 2 = 3\log_{3}b$, express a in terms of b.

$a = b^{3} - 3$

$a = b^{3} - 9$

$a = 9b^{3}$

$a = \frac{b^{3}}{9}$

Answer Details

The given equation is $\log_{3}a - 2 = 3\log_{3}b$. We can use the logarithmic property that $\log_{a}b^{c} = c\log_{a}b$ to simplify the equation: $\log_{3}a - 2 = \log_{3}b^{3}$ $\log_{3}a = \log_{3}b^{3} + 2$ $\log_{3}a = \log_{3}(b^{3}\cdot 3^{2})$ Using the property that if $\log_{a}b = \log_{a}c$ then $b = c$, we get: $a = b^{3}\cdot 3^{2}$ $a = 9b^{3}$ Therefore, the value of $a$ in terms of $b$ is $a = 9b^{3}$.

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Indices And Logarithmic Functions

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Question 7 Report

If $\alpha$ and $\beta$ are the roots of $2x^{2} - 5x + 6 = 0$, find the equation whose roots are $(\alpha + 1)$ and $(\beta + 1)$.

$2x^{2} - 9x + 15 = 0$

$2x^{2} - 9x + 13 = 0$

$2x^{2} - 9x - 13 = 0$

$2x^{2} - 9x - 15 = 0$

Answer Details

To find the equation whose roots are $(\alpha + 1)$ and $(\beta + 1)$, we can use the relationship between the roots and coefficients of a quadratic equation. Let's start by finding the sum and product of the roots of the original equation, $2x^{2} - 5x + 6 = 0$. We can use the formulae: Sum of roots, $S = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}$ Product of roots, $P = \frac{c}{a} = \frac{6}{2} = 3$ Now, let's add 1 to both roots: $\alpha + 1$ and $\beta + 1$ The sum of these roots would be: $(\alpha + 1) + (\beta + 1) = \alpha + \beta + 2$ And the product would be: $(\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1$ Now, we want to find the equation whose roots are $(\alpha + 1)$ and $(\beta + 1)$. Let's call this equation $ax^{2} + bx + c = 0$. According to the relationship between roots and coefficients, we know that: $\frac{-b}{a} = \frac{5}{2}$ and $\frac{c}{a} = 3$ Solving for $b$ and $c$ in terms of $a$, we get: $b = -\frac{5a}{2}$ and $c = 3a$ Now, we can use the sum and product of the new roots to form two equations: $\frac{-b}{a} = \frac{5}{2} \implies -\frac{5a}{2a} = \frac{5}{2} \implies -5 = 5$ This equation is clearly not true, which means that the first option, $2x^{2} - 9x + 15 = 0$, is not the correct answer. Let's move on to the second option: $(\alpha + \beta) + 2 = \frac{5}{2} + 2 = \frac{9}{2}$ $(\alpha \beta + \alpha + \beta + 1) = 3 + \frac{5}{2} + \frac{5}{2} + 1 = 8$ Substituting the values of $b$ and $c$ in terms of $a$, we get: $a(x^{2} - \frac{9}{2}x + \frac{13}{2}) = 0$ Simplifying, we get: $2x^{2} - 9x + 13 = 0$ Therefore, the answer is the second option, $2x^{2} - 9x + 13 = 0$.

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Polynomial Functions

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Question 8 Report

Find the coefficient of $x^{3}$ in the binomial expansion of $(x - \frac{3}{x^{2}})^{9}$.

324

252

-252

-324

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Binomial Theorem

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Question 9 Report

If $\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} + \begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = 6$, find the value of the constant k, where k > 0.

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Matrices And Linear Transformation

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Question 10 Report

How many numbers greater than 150 can be formed from the digits 1, 2, 3, 4, 5 without repetition?

191

291

391

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Permutation And Combinations

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Question 11 Report

If $\overrightarrow{OX} = \begin{pmatrix} -7 \\ 6 \end{pmatrix}$ and $\overrightarrow{OY} = \begin{pmatrix} 16 \\ -11 \end{pmatrix}$, find $\overrightarrow{YX}$.

$\begin{pmatrix} 9 \\ -5 \end{pmatrix}$

$\begin{pmatrix} -23 \\ -5 \end{pmatrix}$

$\begin{pmatrix} 9 \\ 17 \end{pmatrix}$

$\begin{pmatrix} -23 \\ 17 \end{pmatrix}$

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Vectors

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Question 12 Report

If $4x^{2} + 5kx + 10$ is a perfect square, find the value of k.

$\frac{5\sqrt{10}}{4}$

$4\sqrt{10}$

$5\sqrt{10}$

$\frac{4\sqrt{10}}{5}$

Answer Details

We know that a perfect square trinomial has the form of $(ax + b)^2$, where a and b are constants. So, if we have a trinomial of the form $4x^2 + 5kx + 10$, we can write it as $(2x + c)^2$, where c is another constant. Expanding the square, we get: \begin{align*} (2x + c)^2 &= 4x^2 + 4cx + c^2 \\ &= 4x^2 + (4c)x + c^2 \\ \end{align*} Comparing the coefficients of the two trinomials, we get: \begin{align*} 4 &= 4 \\ 5k &= 4c \\ 10 &= c^2 \end{align*} Dividing the second equation by 4, we get: \begin{align*} \frac{5k}{4} &= c \end{align*} Since c = $ \sqrt{10} $, we get: \begin{align*} \frac{5k}{4} &= \sqrt{10} \\ \Rightarrow k &= \frac{4\sqrt{10}}{5} \end{align*} So the value of k is $ \frac{4\sqrt{10}}{5} $.

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Polynomial Functions

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Question 13 Report

The midpoint of M(4, -1) and N(x, y) is P(3, -4). Find the coordinates of N.

(2, -3)

(2, -7)

(-1, -3)

(-10, -7)

Answer Details

We know that the midpoint of a line segment is the point that is exactly halfway between the endpoints of the segment. So, to find the coordinates of point N, we need to use the midpoint formula. The midpoint formula is: Midpoint = [(x1 + x2)/2, (y1 + y2)/2] where (x1, y1) and (x2, y2) are the coordinates of the two endpoints of the segment. In this problem, we are given that the midpoint of segment MN is P(3, -4), and one endpoint is M(4, -1). Let's call the coordinates of the other endpoint N(x, y). Using the midpoint formula, we can write: [(4 + x)/2, (-1 + y)/2] = (3, -4) Now, we can solve for x and y: (4 + x)/2 = 3 => 4 + x = 6 => x = 2 (-1 + y)/2 = -4 => -1 + y = -8 => y = -7 Therefore, the coordinates of point N are (2, -7). So, the correct option is (2, -7).

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Co-ordinate Geometry

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Question 14 Report

$P = {1, 3, 5, 7, 9}, Q = {2, 4, 6, 8, 10, 12}, R = {2, 3, 5, 7, 11}$ are subsets of $U = {1, 2, 3, ... , 12}$. Which of the following statements is true?

$Q \cap R = \varnothing$

$R \subset P$

$(R \cap P) \subset (R \cap U)$

$n(P' \cap R) = 2$

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Sets

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Question 15 Report

Evaluate $\int_{\frac{1}{2}}^{1} \frac{x^{3} - 4}{x^{3}} \mathrm {d} x$.

-5.5

-2.0

2.0

5.5

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Integration

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Question 16 Report

Find the equation to the circle $x^{2} + y^{2} - 4x - 2y = 0$ at the point (1, 3).

2y - x -5 = 0

2y + x - 5 = 0

2y + x + 5 = 0

2y - x + 5 = 0

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Co-ordinate Geometry

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Question 17 Report

If $\sin\theta = \frac{3}{5}, 0° < \theta < 90°$, evaluate $\cos(180 - \theta)$.

$\frac{4}{5}$

$\frac{3}{5}$

$\frac{-3}{5}$

$\frac{-4}{5}$

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Trigonometry

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Question 18 Report

Given that $y = x(x + 1)^{2}$, calculate the maximum value of y.

-2

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Polynomial Functions

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Question 19 Report

If $\alpha$ and $\beta$ are the roots of the equation $2x^{2} + 5x + n = 0$, such that $\alpha\beta = 2$, find the value of n.

-4

-2

Answer Details

If $\alpha$ and $\beta$ are the roots of the equation $2x^{2} + 5x + n = 0$, then by the quadratic formula, we have: \[\alpha, \beta = \frac{-5 \pm \sqrt{5^{2} - 4(2)(n)}}{2(2)}\] Simplifying, we get: \[\alpha, \beta = \frac{-5 \pm \sqrt{25 - 8n}}{4}\] We are also given that $\alpha\beta = 2$. Therefore, we have: \[\alpha\beta = \frac{-5 + \sqrt{25 - 8n}}{4} \cdot \frac{-5 - \sqrt{25 - 8n}}{4} = 2\] Expanding the left-hand side, we get: \[\frac{25 - (25 - 8n)}{16} = 2\] Simplifying, we get: \[\frac{8n}{16} = 2\] \[n = 4\] Therefore, the answer is the fourth option, 4.

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Polynomial Functions

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Question 20 Report

Simplify $\frac{\sqrt{3}}{\sqrt{3} -1} + \frac{\sqrt{3}}{\sqrt{3} + 1}$

$\frac{1}{2}$

$2\sqrt{3}$

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Question 21 Report

Resolve $\frac{3x - 1}{(x - 2)^{2}}, x \neq 2$ into partial fractions.

$\frac{x}{2(x - 2)} - \frac{5}{(x - 2)^{2}}$

$\frac{5}{(x - 2)} + \frac{x}{2(x - 2)^{2}}$

$\frac{1}{2(x - 2)} + \frac{5x}{2(x- 2)^{2}}$

$\frac{-1}{2(x - 2)} + \frac{8x}{2(x - 2)^{2}}$

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Rational Functions

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Question 22 Report

The general term of an infinite sequence 9, 4, -1, -6,... is $u_{r} = ar + b$. Find the values of a and b.

a = 5, b = 14

a = -5, b = 14

a = 5, b = -14

a = -5, b = -14

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Sequences And Series

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Question 23 Report

Given that $a = i - 3j$ and $b = -2i + 5j$ and $c = 3i - j$, calculate $|a - b + c|$.

$\sqrt{13}$

$3\sqrt{13}$

$6\sqrt{13}$

$9\sqrt{13}$

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Vectors

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Question 24 Report

Find the angle between forces of magnitude 7N and 4N if their resultant has a magnitude of 9N.

39.45°

73.40°

75.34°

106.60°

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Trigonometry

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Question 25 Report

Given that $f(x) = 3x^{2} - 12x + 12$ and $f(x) = 3$, find the values of x.

1, 3

-1, -3

1, -3

-1, 3

Answer Details

To find the values of x where the function f(x) is equal to 3, we need to solve the equation:

3x^2 - 12x + 12 = 3

We can start by subtracting 3 from both sides:

3x^2 - 12x + 12 - 3 = 3 - 3
3x^2 - 12x + 9 = 0

Next, we can use the quadratic formula to find the solutions for x:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 3, b = -12, c = 9. Plugging in these values, we get:

x = (-(-12) ± √((-12)^2 - 4 * 3 * 9)) / 2 * 3
x = (12 ± √(144 - 108)) / 6
x = (12 ± √36) / 6
x = (12 ± 6) / 6

So, the two values of x that solve the equation are:

x = (12 + 6) / 6 = 18 / 6 = 3
x = (12 - 6) / 6 = 6 / 6 = 1

Therefore, the two values of x that make f(x) equal to 3 are 1 and 3.

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Polynomial Functions

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Question 26 Report

Find the stationary point of the curve $y = 3x^{2} - 2x^{3}$.

(1, 0)

(-1, 0)

(1, 1)

(-1, -1)

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Polynomial Functions

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Question 27 Report

Find the domain of $g(x) = \frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}$

${x : x \in R, x = \frac{1}{2}}$

$x: x \in R, x\neq \frac{1}{3}$

$x : x \in R, x = \frac{1}{3}$

$x: x \in R$

Answer Details

To find the domain of the given function, we need to identify any values of x that would make the denominator of the fraction equal to zero or negative, as these values would result in an undefined function. Additionally, we need to consider any other restrictions on x that may arise from the function's algebraic properties. In this case, the denominator of the function is the square root of $9x^{2} + 1$. Since the square root of a negative number is not defined in the real number system, we know that the expression inside the square root must be non-negative. This means that: $9x^{2} + 1 \geq 0$ Solving this inequality, we get: $9x^{2} \geq -1$ Dividing both sides by 9, we get: $x^{2} \geq -\frac{1}{9}$ Since the square of any real number is non-negative, we know that this inequality is satisfied for all real values of x. Therefore, there are no values of x that would make the denominator of the function equal to zero or negative, and the domain of the function is all real numbers. In other words, the correct answer is: $x: x \in R$

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Surds

Polynomial Functions

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Question 28 Report

The marks scored by 4 students in Mathematics and Physics are ranked as shown in the table below

Mathematics	3	4	2	1
Physics	4	3	1	2

Calculate the Spearmann's rank correlation coefficient.

0.2

0.5

0.6

0.7

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Read lesson note on Probability (WAEC)

Statistics

Probability

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Question 29 Report

If the determinant of the matrix $\begin{pmatrix} 2 & x \\ 3 & 5 \end{pmatrix} = 13$, find the value of x.

-2

-1

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Matrices And Linear Transformation

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Question 30 Report

Two functions f and g are defined on the set of real numbers by $f : x \to x^{2} + 1$ and $g : x \to x - 2$. Find f o g.

$x^{2} + 4x - 5$

$x^{2} - 4x + 5$

$x^{2} - 1$

$x - 1$

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Functions

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Question 31 Report

Find the constant term in the binomial expansion $(2x^{2} + \frac{1}{x})^{9}$

168

336

672

Answer Details

The constant term in a binomial expansion is the term that has no variable or a variable raised to the power of zero. To find the constant term in $(2x^{2} + \frac{1}{x})^{9}$, we need to look for the term that has no $x$ or a $x$ raised to the power of zero. To obtain the constant term, we need to choose the constant terms from each of the factors in the expansion, which are $2x^2$ and $\frac{1}{x}$, in such a way that their product is raised to a power that adds up to 9. In other words, we need to choose the constant terms that multiply to give a coefficient in the expansion of $(2x^2)^m (\frac{1}{x})^{9-m}$ that is independent of $x$. The constant term in the expansion of $(2x^2)^m (\frac{1}{x})^{9-m}$ is $\binom{9}{m}(2x^2)^m (\frac{1}{x})^{9-m}$, where $\binom{9}{m}$ is the binomial coefficient that represents the number of ways to choose $m$ items out of 9. For the constant term, we need to choose $m$ such that the powers of $x$ in the two factors add up to zero. That is, \begin{align*} 2m - (9-m) &= 0 \\ \Rightarrow m &= \frac{9}{3} \\ \Rightarrow m &= 3 \end{align*} Therefore, the constant term in the expansion of $(2x^2 + \frac{1}{x})^9$ is $\binom{9}{3}(2x^2)^3 (\frac{1}{x})^6$. Plugging in the values, we get: \begin{align*} \binom{9}{3}(2x^2)^3 (\frac{1}{x})^6 &= \frac{9!}{3!6!}(2^3x^6)(\frac{1}{x^6}) \\ &= 84(8) \\ &= 672 \end{align*} Therefore, the constant term in the binomial expansion of $(2x^2 + \frac{1}{x})^9$ is \boxed{672}.

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Binomial Theorem

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Question 32 Report

Solve $\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)$.

4.75

4.00

1.75

1.00

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Question 33 Report

Out of 70 schools, 42 of them can be attended by boys and 35 can be attended by girls. If a pupil is selected at random from these schools, find the probability that he/ she is from a mixed school.

$\frac{1}{11}$

$\frac{1}{10}$

$\frac{1}{6}$

$\frac{1}{5}$

Answer Details

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Probability

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Question 34 Report

In how many ways can the letters of the word 'ELECTIVE' be arranged?

336

1680

6720

20160

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Permutation And Combinations

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Question 35 Report

The first term of a Geometric Progression (GP) is $\frac{3}{4}$, If the product of the second and third terms of the sequence is 972, find its common ratio.

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Sets

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Question 36 Report

A body of mass 28g, initially at rest is acted upon by a force, F Newtons. If it attains a velocity of $5.4ms^{-1}$ in 18 seconds, find the value of F.

0.0082N

0.0084N

0.082N

0.084N

Answer Details

We can use the formula $F = ma$ to solve the problem, where $F$ is the force in Newtons, $m$ is the mass in kilograms, and $a$ is the acceleration in meters per second squared. We need to convert the mass from grams to kilograms and find the acceleration first. Given that the initial velocity, $u$ = 0 $ms^{-1}$, final velocity, $v$ = $5.4ms^{-1}$, and time, $t$ = 18 seconds. We can use the formula $v = u + at$ to find the acceleration: $v = u + at$ $5.4 = 0 + a\times 18$ $a = \frac{5.4}{18} = 0.3ms^{-2}$ Now, we can substitute the values of mass and acceleration in the formula to find the force: $F = ma$ $F = 0.028\times 0.3$ $F = 0.0084N$ Therefore, the value of force is 0.0084N.

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Statics

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Question 37 Report

Find the radius of the circle $x^{2} + y^{2} - 8x - 2y + 1 = 0$.

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Surds

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Question 38 Report

Express $\frac{13}{4}\pi$ radians in degrees.

495°

225°

585°

135°

Answer Details

To convert radians to degrees, we use the following formula:

degrees = radians × 180°/π

where π is approximately equal to 3.14.

Substituting the given value of radians, we get:

degrees = (13/4)π × 180°/π

The π cancels out, leaving us with:

degrees = (13/4) × 180°

Simplifying the expression, we get:

degrees = 585°

Therefore, the answer is 585°.

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Trigonometry

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Question 39 Report

A car is moving at 120$kmh^{-1}$. Find its speed in $ms^{-1}$.

33.3$ms^{-1}$

66.6$ms^{-1}$

99.9$ms^{-1}$

120.0$ms^{-1}$

Answer Details

To convert from kilometers per hour to meters per second, we need to use the following formula: 1 kilometer per hour = 0.277778 meters per second Therefore, to find the speed of the car in meters per second, we can multiply its speed in kilometers per hour by 0.277778. Speed in meters per second = 120 km/h x 0.277778 = 33.33336 m/s (rounded to 5 decimal places) Therefore, the correct answer is 33.3$ms^{-1}$.

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Probability

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Question 40 Report

A binary operation * is defined on the set of real numbers, by $a * b = \frac{a}{b} + \frac{b}{a}$. If $(\sqrt{x} + 1) * (\sqrt{x} - 1) = 4$, find the value of x.

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Binary Operations

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