WAEC SSCE - Further Mathematics - 2019 (Objective)

Question 1 Report

The second and fourth terms of an exponential sequence (G.P) are $\frac{2}{9}$ and $\frac{8}{81}$ respectively. Find the sixth term of the sequence

$\frac{81}{32}$

$\frac{9}{8}$

$\frac{1}{4}$

$\frac{32}{729}$

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Sets

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Question 2 Report

Rationalize; $\frac{1}{\sqrt{2 + 1}}$

$\sqrt{2}$ - 1

1 - $\sqrt{2}$

$\frac{\sqrt{2} - 1}{2}$

$\frac{1 - \sqrt{2}}{2}$

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Surds

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Question 3 Report

Evaluate tan 75$^o$; leaving the answer in surd form (radicals)

$\sqrt{3 + 2}$

$\sqrt{3 + 1}$

$\sqrt{3 - 1}$

$\sqrt{3 - 2}$

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Surds

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Question 4 Report

Solve: 8$^{x - 2}$ = 4$^{3x}$

-2

-1

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Sets

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Question 5 Report

An operation (*) is defined on the set T = {-1, 0, ...., 5} by x * y = x + y - xy. Which of the following operation(s) will give an image which is an element of T?

I. 2(*)5 II. 3(*)5 III. 3(*)4

I only

II only

I and III only

II and III only

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Sets

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Question 6 Report

How many numbers greater than 200 can be formed from the digits 1,2,3,4, 5 if no digit is to be repeated in any particular number?

288

300

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Permutation And Combinations

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Question 7 Report

Which of the following vectors is perpendicular to $\begin{pmatrix} -1 & 3 \end{pmatrix}$?

$\begin{pmatrix} -3 & 1 \end{pmatrix}$

$\begin{pmatrix} 1 & 3 \end{pmatrix}$

$\begin{pmatrix} 1 & -3 \end{pmatrix}$

$\begin{pmatrix} -1 & -3 \end{pmatrix}$

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Vectors

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Question 8 Report

Evaluate: $^{lim}_{x \to 1} \begin{pmatrix} \frac{1 - x}{x^2 - 3x + 2} \end {pmatrix}$

-1

- $\frac{1}{2}$

$\frac{1}{2}$

Answer Details

To evaluate the limit, we can simply substitute the value of x=1 in the expression inside the limit. However, direct substitution yields an indeterminate form of 0/0. Therefore, we need to manipulate the expression before substituting the value of x. We can factorize the denominator of the expression inside the limit as follows: x^2 - 3x + 2 = (x-2)(x-1) So, the expression inside the limit becomes: \[\frac{1-x}{(x-2)(x-1)}\] Now, we can simplify the expression by canceling out the common factor of (x-1) in the numerator and denominator: \[\frac{1-x}{(x-2)(x-1)} = \frac{1}{x-2}\] Now, we can substitute the value of x=1 in the simplified expression to get the limit value: \[\lim_{x \to 1}\frac{1}{x-2} = -\infty\] Therefore, the limit of the given expression as x approaches 1 is negative infinity (-∞), which is not one of the given options.

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Question 9 Report

Given that P and Q are non-empty subsets of the universal set, U. Find P $\cap$ (Q U Q`).

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To understand this problem, we need to break it down into smaller parts. First, let's define what each symbol means: - $P\cap Q$ means the intersection of sets P and Q, which consists of all the elements that are in both sets P and Q. - $Q'$ means the complement of set Q, which consists of all the elements in the universal set U that are not in set Q. - $U$ is the universal set, which contains all the possible elements that we are considering. Next, let's look at the expression $Q\cup Q'$. This means the union of set Q and its complement, which contains all the elements in set Q and all the elements that are not in set Q. In other words, it's just the universal set U. So, we can rewrite the original expression as $P\cap U$, which is just equal to set P. This is because the intersection of any set with the universal set is just the original set itself. Therefore, the answer to the problem is simply set P.

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Sets

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Question 10 Report

Which of these inequalities is represented by the shaded portion of the graph?

2y + x - 3 < 0

2y - x - 3 < 0

2y - x + 3 < 0

2y + x +3 < 0

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Question 11 Report

Find the constant term in the binomial expansion of (2x$^2$ + $\frac{1}{x^2}$)$^4$

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Binomial Theorem

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Question 12 Report

A particle starts from rest and moves in a straight line such that its velocity, V ms$^{-1}$, at time t second is given by V = 3t$^2$ - 6t. Calculate the acceleration in the 3rd second.

0 ms$^{-2}$

3 ms$^{-2}$

6 ms$^{-2}$

9 ms$^{-2}$

Answer Details

Acceleration is the rate of change of velocity, so to find the acceleration at any point in time, we need to find the derivative of the velocity function with respect to time. In this case, the velocity function is given by V = 3t$^2$ - 6t. So, taking the derivative of V with respect to time t, we get: dV/dt = 6t - 6 Now that we have the derivative, we can evaluate it at t = 3 to find the acceleration at the 3rd second. Plugging in t = 3, we get: dV/dt = 6 * 3 - 6 = 12 - 6 = 6 So, the acceleration at the 3rd second is 6 m/s$^2$.

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Question 13 Report

A particle starts from rest and moves in a straight line such that its velocity, V ms$^{-1}$, at time t second is given by V = 3t$^2$ - 6t. Calculate the acceleration in the 3rd second.

12m

16m

64m

96m

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Question 14 Report

Calculate the distance between points (-2, -5) and (-1, 3)

$\sqrt{5}$ units

$\sqrt{17}$ units

$\sqrt{65}$ units

$\sqrt{73}$ units

Answer Details

To calculate the distance between two points, we can use the distance formula: d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Where (x1, y1) and (x2, y2) are the coordinates of the two points. In this case, the two points are (-2, -5) and (-1, 3). So, we can substitute these values into the distance formula: d = $\sqrt{(-1 - (-2))^2 + (3 - (-5))^2}$ Simplifying this expression, we get: d = $\sqrt{(1)^2 + (8)^2}$ d = $\sqrt{1 + 64}$ d = $\sqrt{65}$ Therefore, the distance between the points (-2, -5) and (-1, 3) is $\sqrt{65}$ units. So, option (C) is the correct answer. Note that the distance formula can be used to find the distance between any two points in a two-dimensional plane.

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Vectors

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Question 15 Report

Evaluate $\int^0_0 \sqrt{x} dx$

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Integration

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Question 16 Report

The function f : x $\to$ x$^2$ + px + q has turning point when x = -3 and remainder of -6 when divided by (x + 2). Find the value of q.

-2

-8

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Polynomial Functions

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Question 17 Report

Find correct to the nearest degree,5 the angle between p = 12i - 5j and q = 4i +3j

59$^o$

60$^o$

75$^o$

76$^o$

Answer Details

To find the angle between two vectors, we can use the dot product formula: p $\cdot$ q = \|p\| \|q\| cos $\theta$ where p $\cdot$ q is the dot product of vectors p and q, \|p\| and \|q\| are the magnitudes of vectors p and q respectively, and $\theta$ is the angle between the two vectors. First, let's calculate the magnitudes of vectors p and q: \|p\| = $\sqrt{(12)^2 + (-5)^2}$ = $\sqrt{169}$ = 13 \|q\| = $\sqrt{(4)^2 + (3)^2}$ = $\sqrt{25}$ = 5 Next, let's calculate the dot product of vectors p and q: p $\cdot$ q = (12)(4) + (-5)(3) = 48 - 15 = 33 Substituting the values we obtained into the formula for the dot product, we get: 33 = (13)(5) cos $\theta$ Solving for cos $\theta$, we get: cos $\theta$ = $\frac{33}{65}$ Using a calculator, we can find that the inverse cosine of $\frac{33}{65}$ is approximately 59.08$^o$. Therefore, the angle between vectors p and q is approximately 59 degrees when rounded to the nearest degree. Answer: 59$^o$.

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Trigonometry

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Question 18 Report

Given that X and Y are independent events such that P(X) = 0.5, P(Y) = m and P(X U Y) = 0.75, find the value of m.

0.6

0.5

0.4

0.3

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Probability

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Question 19 Report

Find the coordinates of the point in the curve y = 3x$^2$ - 2x - 5 where the tangent is parallel to the line y = - 5 = 8x

$\begin{pmatrix} - \frac{5}{3} &, 0 \end {pmatrix}$

$\begin{pmatrix} 0, & - \frac{5}{3} \end {pmatrix}$

$\begin{pmatrix} 0, & \frac{5}{3} \end {pmatrix}$

$\begin{pmatrix} \frac{5}{3} &, 0 \end {pmatrix}$

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Differentiation

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Question 20 Report

Given that $\frac{1}{x^2 - 4} = \frac{p}{(x + 2)} + \frac{Q}{(x - 2})$

x $\neq \pm 2$

Find the value of (P + Q)

$\frac{3}{2}$

$\frac{1}{2}$

Answer Details

We can start by simplifying the right-hand side of the equation using partial fraction decomposition. To do this, we need to find the values of P and Q. We can use a common denominator on the right-hand side of the equation to get: \[\frac{1}{x^2 - 4} = \frac{p(x - 2) + Q(x + 2)}{(x + 2)(x - 2)}\] Next, we can multiply both sides of the equation by the denominator of the left-hand side to get: \[1 = \frac{p(x - 2) + Q(x + 2)}{(x + 2)(x - 2)}(x^2 - 4)\] Simplifying the right-hand side by multiplying out the terms, we get: \[1 = \frac{(p + Q)x^3 - 4p + 4Q}{(x + 2)(x - 2)}\] Since the left-hand side is just the number 1, the numerator on the right-hand side must also be equal to 1. Therefore, we can set up a system of equations to solve for P and Q: \[p + Q = 0\] \[-4p + 4Q = 1\] Solving for P and Q using the system of equations, we get: \[P = -\frac{1}{4}\] \[Q = \frac{1}{4}\] Therefore, P + Q = 0, which is option (D).

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Polynomial Functions

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Question 21 Report

The probabilities that John and Jane will pass an examination are 0.9 and 0.7 respectively. Find the probability that at least one of them will pass the examination.

0.28

0.67

0.72

0.97

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Probability

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Question 22 Report

Find the coordinates of the centre of the circle 3x$^2$ + 3y$^2$ - 6x + 9y - 5 = 0

(-3. $\frac{9}{2}$)

(-1. $\frac{3}{2}$)

(1, - $\frac{3}{2}$)

(3. -$\frac{9}{2}$)

Answer Details

To find the center of the circle, we need to rewrite the equation in the standard form: (x - h)$^2$ + (y - k)$^2$ = r$^2$ where (h,k) is the center of the circle and r is the radius. Starting with the given equation: 3x$^2$ + 3y$^2$ - 6x + 9y - 5 = 0 We can group the x and y terms together: (3x$^2$ - 6x) + (3y$^2$ + 9y) = 5 Next, we need to complete the square for both x and y. For the x terms, we can factor out a 3 from the first two terms: 3(x$^2$ - 2x) To complete the square, we need to add and subtract ($\frac{2}{2}$)$^2$ = 1 inside the parenthesis: 3(x$^2$ - 2x + 1 - 1) Then, we can simplify this expression: 3((x - 1)$^2$ - 1) = 3(x - 1)$^2$ - 3 For the y terms, we can follow the same process: 3(y$^2$ + 3y) = 3(y$^2$ + 3y + ($\frac{3}{2}$)$^2$ - ($\frac{3}{2}$)$^2$) = 3((y + $\frac{3}{2}$)$^2$ - $\frac{9}{4}$) = 3(y + $\frac{3}{2}$)$^2$ - $\frac{27}{4}$ Now we can substitute these expressions back into the original equation and simplify: 3(x - 1)$^2$ - 3 + 3(y + $\frac{3}{2}$)$^2$ - $\frac{27}{4}$ = 5 3(x - 1)$^2$ + 3(y + $\frac{3}{2}$)$^2$ = $\frac{49}{4}$ Dividing both sides by 3, we get: (x - 1)$^2$ + (y + $\frac{3}{2}$)$^2$ = ($\frac{7}{2}$)$^2$ Comparing this equation to the standard form: (x - h)$^2$ + (y - k)$^2$ = r$^2$ We can see that the center of the circle is at (1, -$\frac{3}{2}$) and the radius is ($\frac{7}{2}$). Therefore, the answer is (3). (1, -$\frac{3}{2}$)

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Co-ordinate Geometry

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Question 23 Report

Find the sum of the first 20 terms of the sequence -7-3, 1, ......

620

660

690

1240

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Sequences And Series

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Question 24 Report

Find the value of x for which 6$\sqrt{4x^2 + 1}$ = 13x, where x > 0

$\frac{6}{5}$

$\frac{25}{24}$

$\frac{24}{25}$

$\frac{5}{6}$

Answer Details

To solve for x in the equation 6$\sqrt{4x^2 + 1}$ = 13x, we need to isolate x on one side of the equation. First, we can simplify the left-hand side by squaring both sides of the equation: (6$\sqrt{4x^2 + 1}$)^2 = (13x)^2 Simplifying the left-hand side, we get: 6^2 * (4x^2 + 1) = 13^2 * x^2 Simplifying further: 144x^2 + 36 = 169x^2 Subtracting 144x^2 from both sides: 36 = 25x^2 Dividing both sides by 25: x^2 = $\frac{36}{25}$ Taking the square root of both sides: x = $\frac{6}{5}$ Therefore, the value of x that satisfies the equation is $\frac{6}{5}$.

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Surds

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Question 25 Report

If $^nC_2$ = 15, find the value of n

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Permutation And Combinations

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Question 26 Report

Calculate the mean deviation of 5, 8, 2, 9 and 6

Answer Details

To calculate the mean deviation, we need to first find the mean or average of the given values. Mean or average is calculated by adding up all the values and then dividing the sum by the total number of values. In this case, the sum of the given values is 5 + 8 + 2 + 9 + 6 = 30. Dividing this sum by the total number of values, which is 5 in this case, we get the mean or average as 30/5 = 6. Now, to find the mean deviation, we need to find the deviation of each value from the mean, which is the absolute difference between the value and the mean. For example, the deviation of 5 from the mean is |5 - 6| = 1. Similarly, the deviation of 8 from the mean is |8 - 6| = 2, and so on for all the values. Once we have found the deviation of each value from the mean, we add up all the deviations and divide the sum by the total number of values to get the mean deviation. In this case, the sum of all the deviations is 1 + 2 + 4 + 3 + 0 = 10. Dividing this sum by the total number of values, which is 5, we get the mean deviation as 10/5 = 2. Therefore, the answer is: 2.

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Statistics

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Question 27 Report

Solve, correct to three significant figures, (0.3)$^x$ = (0,5)$^8$

4.61

4.606

0.461

0.0130

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Surds

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Question 28 Report

Solve; $\frac{P}{2} + \frac{k}{3}$ = 5 and 2p = k = 6 simultaneously

p = -6, k = -6

p = -6, k = 6

p = 6, k = 6

p = 6, k = -6

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Polynomial Functions

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Question 29 Report

If P = $\begin {pmatrix} 2 & 3\\ -4 & 1 \end {pmatrix}$, Q = $\begin{pmatrix} 6 \\ 8 \end {pmatrix}$ and PQ = k $\begin {pmatrix} 45\\ -20 \end {pmatrix}$. Find the value of k.

-$\frac{5}{4}$

-$\frac{4}{5}$

$\frac{4}{5}$

$\frac{5}{4}$

Answer Details

Given matrices are: $$P = \begin {pmatrix} 2 & 3\\ -4 & 1 \end {pmatrix}, \quad Q = \begin{pmatrix} 6 \\ 8 \end {pmatrix}, \quad PQ = k \begin {pmatrix} 45\\-20 \end {pmatrix}$$ We know that for two matrices to be multiplied, the number of columns of the first matrix should be equal to the number of rows of the second matrix. In this case, the number of columns of $P$ is 2 and the number of rows of $Q$ is 2, so we can multiply them. $$PQ = \begin {pmatrix} 2 & 3\\ -4 & 1 \end {pmatrix} \begin{pmatrix} 6 \\ 8 \end {pmatrix} = \begin {pmatrix} 2(6)+3(8)\\-4(6)+1(8) \end {pmatrix} = \begin {pmatrix} 45\\-20 \end {pmatrix}$$ Comparing the above equation with the given equation, we get: $$k = \frac{\text{corresponding elements in } PQ}{\text{corresponding elements in } \begin {pmatrix} 45\\-20 \end {pmatrix}} = \frac{-20}{45} = -\frac{4}{9}$$ Therefore, the value of $k$ is -$\frac{4}{5}$. ()

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Matrices And Linear Transformation

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Question 30 Report

Point X and Y are on the same horizontal base as the foot of a building such that X is 96m due east of the building and Y is due west. If the angle of elevation of the top of that building from X is 30$^o$ and that of Y is 50$^o$, calculate the distance of Y from the base of the building.

30m

32m

42m

50m

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Vectors

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Question 31 Report

Consider the following statements:

p: Birds fly

q: The sky is blue

r: The grass is green
What is the symbolic representation of "If the grass is green and the sky is not blue, then the birds do not fly"?

A. (r ^ p) $\to$ q

(r ^ q) $\to$ p

(r ^~ p) $\to$ q

(r ~ ^ p) $\to$ q

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Logical Reasoning

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Question 32 Report

A linear transformation is defined by T: (x, y) $\to$ (-x + y, -4y). Find the image, Q`, of Q(-3, 2) under T

Q`(5, -8)

Q`(-8, 5)

Q`(5, -3)

Q`(-5, -8)

Answer Details

To find the image, Q`, of point Q(-3, 2) under the linear transformation T, we need to apply the transformation matrix to the coordinates of Q.

T: (x, y) → (-x + y, -4y)

So, we have:

T(Q) = (-(-3) + 2, -4(2)) = (5, -8)

Therefore, the image, Q`, of Q(-3, 2) under T is (5, -8).

Explanation: A linear transformation is a function that maps vectors to other vectors while preserving some properties such as linearity and proportionality. In this case, the linear transformation T takes a vector (x, y) and maps it to a new vector (-x + y, -4y). To find the image of a point under T, we simply plug in the coordinates of the point into the transformation matrix and apply the transformation. In this case, we plugged in the coordinates of Q(-3, 2) and found that the image is (5, -8).

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Matrices And Linear Transformation

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Question 33 Report

A uniform beam, PQ. is 100 m long and weighs 35 N. It is placed on a support at a point 40 cm from P. If weights of 54 N and FN are attached at P and Q respectively in order to keep it in a horizontal position, calculate, correct to the nearest whole number, the value of F.

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Matrices And Linear Transformation

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Question 34 Report

If g : r $\to$ 5 - 2r, r is a real number, find the image of -3

-1

-9

Answer Details

The given function is g : r $\to$ 5 - 2r, where r is a real number. To find the image of -3, we need to substitute -3 for r in the function and simplify: g(-3) = 5 - 2(-3) g(-3) = 5 + 6 g(-3) = 11 Therefore, the image of -3 under the function g is 11.

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Functions

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Question 35 Report

Given that g ; x $\to$ 3x and f ; x $\to$ cos x. Find the value of g$^o$ f(20$^o$)

0.50

0.94

2.60

2.82

Answer Details

The function g of x is defined as g(x) = 3x, and the function f of x is defined as f(x) = cos(x). To find the value of g(f(20°)), we need to first evaluate f(20°) and then plug that result into g. So, first let's evaluate f(20°): f(20°) = cos(20°) = cos(20 x (π/180)) = cos(π/9) Next, we plug the result into g: g(f(20°)) = g(cos(π/9)) = 3cos(π/9) = 2.82 Therefore, the value of g(f(20°)) is 2.82.

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Functions

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Question 36 Report

If the mean of 2, 5, (x + 1), (x + 2), 7 and 9 is 6, find the median.

6.5

6.0

5.5

5.0

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Statistics

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Question 37 Report

Find the coefficient of the term in the binomial expansion of [2x + $\frac{3y}{4}$]$^3$ in descending powers of x.

$\frac{27}{64}$y$^2$

$\frac{27}{8}$y$^2$

8y$^2$

9y$^2$

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Binomial Theorem

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Question 38 Report

A 35 N force acts on a body of mass 5 kg for 2 seconds. Calculate the change in momentum of the body.

70 kg ms$^{-1}$

55 kg ms$^{-1}$

50 kg ms$^{-1}$

35 kg ms$^{-1}$

Answer Details

To calculate the change in momentum of the body, we can use the formula: Change in momentum = Force × time Here, the force acting on the body is 35 N and the time for which it acts is 2 seconds. The mass of the body is given as 5 kg. So, substituting the values in the formula, we get: Change in momentum = 35 N × 2 s Change in momentum = 70 Ns Now, we know that momentum is defined as the product of mass and velocity. So, we can also calculate the change in momentum by finding the initial and final velocities of the body and then using the formula: Change in momentum = mass × (final velocity - initial velocity) However, the initial velocity is not given in the question. So, we cannot use this method to find the change in momentum. Therefore, the answer is 70 kg ms$^{-1}$.

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Matrices And Linear Transformation

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Question 39 Report

If y = (5 - x)$^{-3}$, and $\frac{dy}{dx}$

$\frac{-15}{(5 - x)^4}$

$\frac{-3}{(5 - x)^4}$

$\frac{3}{(5 - x)^4}$

$\frac{15}{(5 - x)^4}$

Answer Details

To find $\frac{dy}{dx}$, we need to differentiate y with respect to x using the chain rule and the power rule of differentiation. Using the chain rule, we get: $\frac{dy}{dx}$ = $\frac{d}{dx}$(5 - x)$^{-3}$ = -3(5 - x)$^{-4}$$\frac{d}{dx}$(5 - x) = -3(5 - x)$^{-4}$(-1) = $\frac{3}{(5 - x)^4}$ Therefore, the correct answer is (c) $\frac{3}{(5 - x)^4}$.

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Rational Functions

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Question 40 Report

Find the area between line y = x + 1 and the x-axis from x = -2 to x = 0.

5 square units

4 square units

2 square units

1 square unit

Answer Details

To find the area between the line y = x + 1 and the x-axis from x = -2 to x = 0, we need to integrate the equation of the line with respect to x over the interval [-2, 0] and take the absolute value of the result. The equation of the line y = x + 1 can be rewritten as x = y - 1, which gives us a different way to represent the line. Integrating this expression with respect to x over the interval [-2, 0] gives: ∫[-2,0] (y - 1) dx = [xy - x] from -2 to 0 = (0-0) - (-2*(-1)) = 2 Taking the absolute value of this result gives us an area of 2 square units. Therefore, the answer is 2 square units.

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Co-ordinate Geometry

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