WAEC SSCE - Further Mathematics - 2010…

Ajụjụ 1 Ripọtì

What is the angle between $a = (3i - 4j)$ and $b = (6i + 4j)$?

13°

87° Azịza ziri ezi

100°

110°

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Trigonometry (WAEC)

Trigonometry

Lee Azịza

Ajụjụ 2 Ripọtì

Calculate, correct to one decimal place, the length of the line joining points X(3, 5) and Y(5, 1).

4.0

4.2

4.5 Azịza ziri ezi

5.0

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Lee Azịza

Ajụjụ 3 Ripọtì

Two particles are fired together along a smooth horizontal surface with velocities 4 m/s and 5 m/s. If they move at 60° to each other, find the distance between them in 2 seconds.

$2\sqrt{61}$

$\sqrt{42}$

$2\sqrt{21}$ Azịza ziri ezi

$2\sqrt{10}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Trigonometry (WAEC)

Trigonometry

Lee Azịza

Ajụjụ 4 Ripọtì

Two forces (2i - 5j)N and (-3i + 4j)N act on a body of mass 5kg. Find in $ms^{-2}$, the magnitude of the acceleration of the body.

$\frac{\sqrt{2}}{5}$ Azịza ziri ezi

$5\sqrt{2}$

$2\sqrt{5}$

$\frac{5\sqrt{2}}{2}$

Akọwa Nkọwa

To find the magnitude of acceleration, we need to use Newton's Second Law, which states that the force acting on an object is equal to its mass multiplied by its acceleration. In this case, we have two forces: (2i - 5j)N and (-3i + 4j)N. To find the net force, we add these two vectors by adding their corresponding components: (2i - 5j)N + (-3i + 4j)N = -i - j N So the net force is (-i - j)N. Now, using Newton's Second Law, we can find the acceleration: F = ma where F is the net force, m is the mass, and a is the acceleration. In our case, the net force is (-i - j)N and the mass is 5kg, so we have: (-i - j)N = 5kg * a Solving for a, we get: a = (-i - j)N / 5kg To find the magnitude of a, we take the square root of the sum of the squares of its components: |a| = sqrt((-1)^2 + (-1)^2) N/kg = sqrt(2) N/kg Converting to $ms^{-2}$ by dividing by 5, we get: |a| = sqrt(2)/5 $ms^{-2}$ Therefore, the answer is $\frac{\sqrt{2}}{5}$.

Gụọ akwụkwọ nkuzi na Vectors (WAEC)

Vectors

Lee Azịza

Ajụjụ 5 Ripọtì

The coefficient of the 7th term in the binomial expansion of $(2 - \frac{x}{3})^{10}$ in ascending powers of x is

$\frac{560}{243}$

$\frac{841}{243}$

$\frac{1120}{243}$ Azịza ziri ezi

$\frac{4481}{243}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Binomial Theorem (WAEC)

Binomial Theorem

Lee Azịza

Ajụjụ 6 Ripọtì

If $\sin x = -\sin 70°, 0° < x < 360°$, determine the two possible values of x.

110°, 250°

110°, 290°

200°, 250°

250°, 290° Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Trigonometry (WAEC)

Trigonometry

Lee Azịza

Ajụjụ 7 Ripọtì

Given that $P = {x : \text{x is a factor of 6}}$ is the domain of $g(x) = x^{2} + 3x - 5$, find the range of x.

{-1, 5, 13}

{5, 13, 49}

{1, 2, 3, 6}

{-1, 5, 13, 49} Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Polynomial Functions (WAEC)

Gụọ akwụkwọ nkuzi na Functions (WAEC)

Polynomial Functions

Functions

Lee Azịza

Ajụjụ 8 Ripọtì

Find the value of $\cos(60° + 45°)$ leaving your answer in surd form.

$\frac{6 + \sqrt{2}}{4}$

$\frac{3 + \sqrt{6}}{4}$

$\frac{\sqrt{2} - \sqrt{6}}{4}$ Azịza ziri ezi

$\frac{3 - \sqrt{6}}{4}$

Akọwa Nkọwa

Using the sum to product formula, we have: \begin{align*} \cos(60° + 45°) &= \cos 60° \cos 45° - \sin 60° \sin 45° \\ &= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2} - \sqrt{6}}{4} \end{align*} Therefore, the correct answer is option C: $\frac{\sqrt{2} - \sqrt{6}}{4}$.

Gụọ akwụkwọ nkuzi na Surds (WAEC)

Surds

Lee Azịza

Ajụjụ 9 Ripọtì

Find the domain of $f(x) = \frac{x}{3 - x}, x \in R$, the set of real numbers.

${x : x \in R, x \neq 3}$ Azịza ziri ezi

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

${x : x \in R, x \neq 0}$

${x : x \in R, x\neq -3}$

Akọwa Nkọwa

The function $f(x) = \frac{x}{3-x}$ is defined for all real numbers except when the denominator is zero, i.e., when $3-x=0$. Therefore, the domain of the function is all real numbers except $x=3$, or in interval notation: ${x : x \in R, x \neq 3}$. So the correct option is: ${x : x \in R, x \neq 3}$.

Gụọ akwụkwọ nkuzi na Rational Functions (WAEC)

Rational Functions

Lee Azịza

Ajụjụ 10 Ripọtì

The initial velocity of an object is $u = \begin{pmatrix} -5 \\ 3 \end{pmatrix} ms^{-1}$. If the acceleration of the object is $a = \begin{pmatrix} 3 \\ -4 \end{pmatrix} ms^{-2}$ and it moved for 3 seconds, find the final velocity.

$\begin{pmatrix} -14 \\ 15 \end{pmatrix} ms^{-1}$

$\begin{pmatrix} -2 \\ 1 \end{pmatrix} ms^{-1}$

$\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}$ Azịza ziri ezi

$\begin{pmatrix} 14 \\ -9 \end{pmatrix} ms^{-1}$

Akọwa Nkọwa

The final velocity of an object can be found using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we have: v = (-5, 3) + 3(3, -4) v = (-5, 3) + (9, -12) v = (4, -9) Therefore, the final velocity of the object is $\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}$. The correct option is (C).

Gụọ akwụkwọ nkuzi na Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Lee Azịza

Ajụjụ 11 Ripọtì

Simplify $\frac{\log_{5} 8}{\log_{5} \sqrt{8}}$.

-2

$\frac{-1}{2}$

$\frac{1}{2}$

2 Azịza ziri ezi

Akọwa Nkọwa

We can use the properties of logarithms to simplify this expression. Recall that $\log_{a} b^{c} = c\log_{a} b$ and $\log_{a} \sqrt{b} = \frac{1}{2}\log_{a} b$. So, \[\frac{\log_{5} 8}{\log_{5} \sqrt{8}} = \frac{\log_{5} 8}{\frac{1}{2}\log_{5} 8} = \frac{2\log_{5} 8}{\log_{5} 8} = \boxed{2}.\] Therefore, the answer is 2.

Gụọ akwụkwọ nkuzi na Indices And Logarithmic Functions (WAEC)

Indices And Logarithmic Functions

Lee Azịza

Ajụjụ 12 Ripọtì

The mean age of 15 pupils in a class is 14.2 years. One new pupil joined the class and the mean changed to 14.1 years. Calculate the age of the new pupil.

12.4 years

12.6 years Azịza ziri ezi

13.2 years

14.1 years

Akọwa Nkọwa

Let's use the formula for the mean (average) of a set of numbers: \[\text{mean} = \frac{\text{sum of all numbers}}{\text{number of numbers}}\] Let the age of the new pupil be x. Before the new pupil joined the class, the sum of ages of the 15 pupils would be: \[15 \times 14.2 = 213\] After the new pupil joined the class, the sum of ages of the 16 pupils would be: \[15 \times 14.2 + x = 213 + x\] We know that the new mean is 14.1, so we can set up an equation: \[\frac{15 \times 14.2 + x}{16} = 14.1\] Multiplying both sides by 16, we get: \[15 \times 14.2 + x = 16 \times 14.1\] Simplifying the left side, we get: \[213 + x = 225.6\] Subtracting 213 from both sides, we get: \[x = 12.6\] Therefore, the age of the new pupil is 12.6 years. Answer: (b) 12.6 years.

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Statistics

Lee Azịza

Ajụjụ 13 Ripọtì

The probabilities that a husband and wife will be alive in 15 years time are m and n respectively. Find the probability that only one of them will be alive at that time.

m + n

m + n - 2mn Azịza ziri ezi

1 - mn

Akọwa Nkọwa

The probability that only one of the husband and wife will be alive in 15 years can happen in two ways: either the husband is alive and the wife is not, or the wife is alive and the husband is not. The probability of the first case is (m)(1-n) = m - mn, and the probability of the second case is (n)(1-m) = n - mn. Therefore, the total probability that only one of them will be alive is (m - mn) + (n - mn) = m + n - 2mn. Hence, the correct answer is: m + n - 2mn.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 14 Ripọtì

Find the maximum value of $2 + \sin (\theta + 25)$.

3 Azịza ziri ezi

Akọwa Nkọwa

The value of $\sin(\theta + 25)$ varies between -1 and 1. Therefore, the maximum value of $2 + \sin(\theta + 25)$ occurs when $\sin(\theta + 25) = 1$. This occurs when $\theta + 25 = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer. Thus, $\theta = \frac{\pi}{2} - 25 + 2n\pi$. Substituting this value of $\theta$ into the expression for $2 + \sin(\theta + 25)$, we get: \[2 + \sin\left(\frac{\pi}{2} + 2n\pi\right) = 2 + 1 = 3\] Therefore, the maximum value of $2 + \sin(\theta + 25)$ is 3.

Gụọ akwụkwọ nkuzi na Trigonometry (WAEC)

Trigonometry

Lee Azịza

Ajụjụ 15 Ripọtì

A stone is thrown vertically upwards and its height at any time t seconds is $h = 45t - 9t^{2}$. Find the maximum height reached.

45. 25 m

45.50 m

56.00 m

56.25 m Azịza ziri ezi

Akọwa Nkọwa

The height of the stone at any time t seconds is given by $h = 45t - 9t^{2}$. To find the maximum height reached, we need to find the vertex of the parabolic function $h$, since the vertex represents the highest point of the parabola. The vertex of the parabola is given by the formula $\frac{-b}{2a}$, where $a$ and $b$ are the coefficients of the quadratic function. In this case, $a = -9$ and $b = 45$. Substituting these values into the formula for the vertex, we get: \[\frac{-b}{2a} = \frac{-45}{2(-9)} = \frac{45}{18} = 2.5.\] Therefore, the stone reaches its maximum height after 2.5 seconds. To find the maximum height, we substitute $t = 2.5$ into the equation for $h$: \[h = 45(2.5) - 9(2.5)^2 = 56.25\text{ m}.\] Therefore, the maximum height reached by the stone is 56.25 meters. The answer is (D) 56.25 m.

Gụọ akwụkwọ nkuzi na Differentiation (WAEC)

Differentiation

Lee Azịza

Ajụjụ 16 Ripọtì

Calculate, correct to one decimal place, the acute angle between the lines 3x - 4y + 5 = 0 and 2x + 3y - 1 = 0.

70.6° Azịza ziri ezi

50.2°

39.8°

19.4°

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Matrices And Linear Transformation (WAEC)

Gụọ akwụkwọ nkuzi na Co-ordinate Geometry (WAEC)

Matrices And Linear Transformation

Co-ordinate Geometry

Lee Azịza

Ajụjụ 17 Ripọtì

Find the least value of n for which $^{3n}C_{2} > 0, n \in R$.

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{2}{3}$ Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Permutation And Combinations (WAEC)

Permutation And Combinations

Lee Azịza

Ajụjụ 18 Ripọtì

If $\begin{vmatrix} 3 & x \\ 2 & x - 2 \end{vmatrix} = -2$, find the value of x.

-8 Azịza ziri ezi

-4

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Lee Azịza

Ajụjụ 19 Ripọtì

If $h(x) = x^{3} - \frac{1}{x^{3}}$, evaluate $h(a) - h(\frac{1}{a})$.

-1

$2a^{3} - \frac{2}{a^{3}}$ Azịza ziri ezi

$\frac{2}{a^{3}} - 2a^{3}$

Akọwa Nkọwa

We can start by computing the values of $h(a)$ and $h(\frac{1}{a})$, and then find their difference. First, we have $h(a) = a^{3} - \frac{1}{a^{3}}$. Next, we have $h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}$. Therefore, \begin{align*} h(a) - h(\frac{1}{a}) &= \left(a^{3} - \frac{1}{a^{3}}\right) - \left(\frac{1}{a^{3}} - a^{3}\right) \\ &= a^{3} - \frac{1}{a^{3}} - \frac{1}{a^{3}} + a^{3} \\ &= 2a^{3} - \frac{2}{a^{3}}. \end{align*} So the answer is $2a^{3} - \frac{2}{a^{3}}$. Therefore, is the correct answer.

Gụọ akwụkwọ nkuzi na Polynomial Functions (WAEC)

Polynomial Functions

Lee Azịza

Ajụjụ 20 Ripọtì

Find the direction cosines of the vector $4i - 3j$.

$\frac{9}{10}, \frac{27}{10}$

$\frac{17}{27}, -\frac{17}{27}$

$\frac{4}{5}, -\frac{3}{5}$ Azịza ziri ezi

$\frac{4}{7}, \frac{-3}{7}$

Akọwa Nkọwa

Direction cosines are the cosines of the angles that a vector makes with the positive x, y and z axes. Let's denote the vector by $\mathbf{v} = 4\mathbf{i} - 3\mathbf{j} + 0\mathbf{k}$. Then the direction cosines of $\mathbf{v}$ are: \begin{align*} \cos\alpha &= \frac{4}{5} \\ \cos\beta &= \frac{-3}{5} \\ \cos\gamma &= 0 \end{align*} Therefore, the correct option is: - $\frac{4}{5}, -\frac{3}{5}$

Gụọ akwụkwọ nkuzi na Vectors (WAEC)

Vectors

Lee Azịza

Ajụjụ 21 Ripọtì

If $y = 2(2x + \sqrt{x})^{2}$, find $\frac{\mathrm d y}{\mathrm d x}$.

$2\sqrt{x}(2x + \sqrt{2})$

$4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})$ Azịza ziri ezi

$4(2x + \sqrt{x})(2 + \sqrt{x})$

$8(2x + \sqrt{x})(2 + \sqrt{x})$

Akọwa Nkọwa

To differentiate the given function with respect to x, we will use the chain rule and the power rule of differentiation. Let's start by using the power rule of differentiation to find the derivative of $(2x + \sqrt{x})^{2}$ with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})\] Next, we use the chain rule to differentiate $\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})$ with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x}) = 2 + \frac{1}{2\sqrt{x}}\] Substituting this into the previous equation, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \left(2 + \frac{1}{2\sqrt{x}}\right)\] Expanding this expression, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 4(2x + \sqrt{x}) + 2\sqrt{x}(2x + \sqrt{x})\] Simplifying further, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 8x + 6\sqrt{x}\] Finally, using the constant multiple rule of differentiation, we can find the derivative of the given function: \[\frac{\mathrm d y}{\mathrm d x} = 2 \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(8x + 6\sqrt{x}) = 16x + 12\sqrt{x}\] Therefore, the correct answer is: $4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})$.

Lee Azịza

Ajụjụ 22 Ripọtì

If (x - 3) is a factor of $2x^{2} - 2x + p$, find the value of constant p.

-12 Azịza ziri ezi

-6

Akọwa Nkọwa

If (x-3) is a factor of the quadratic expression, then the expression can be factored into the form: $$(x - 3)(ax + b)$$ where a and b are constants. Multiplying out the brackets gives: $$2x^{2} - 2x + p = (x - 3)(ax + b) = ax^{2} + (b - 3a)x - 3b$$ Since the coefficients of the quadratic expression are equal to those of the factored expression, we can equate the corresponding coefficients to get a system of equations: $$a = 2$$ $$b - 3a = -2$$ $$-3b = p$$ Solving these equations simultaneously, we get: $$a = 2, b = 3a - 2 = 4, p = -3b = -12$$ Therefore, the constant p is -12.

Gụọ akwụkwọ nkuzi na Polynomial Functions (WAEC)

Polynomial Functions

Lee Azịza

Ajụjụ 23 Ripọtì

For what values of x is $\frac{x^{2} - 9x + 18}{x^{2} + 2x - 35}$ undefined?

6 or 3

-18 or -9

-7 or 5 Azịza ziri ezi

-5 or 7

Akọwa Nkọwa

The expression is undefined when the denominator is equal to zero because division by zero is undefined. Therefore, we can find the values of x that make the denominator zero by setting it equal to zero and solving for x: $$x^{2} + 2x - 35 = 0$$ We can factor this quadratic equation as: $$(x + 7)(x - 5) = 0$$ So, the denominator is equal to zero when x = -7 or x = 5. Therefore, the expression is undefined for these values of x. So the correct answer is: - -7 or 5

Gụọ akwụkwọ nkuzi na Rational Functions (WAEC)

Rational Functions

Lee Azịza

Ajụjụ 24 Ripọtì

If $\overrightarrow{OA} = 3i + 4j$ and $\overrightarrow{OB} = 5i - 6j $ where O is the origin and M is the midpoint of AB, find OM.

-2i - 10j

-2i + 2j

4i - j Azịza ziri ezi

4i + j

Akọwa Nkọwa

We can find the midpoint M of AB by using the midpoint formula: \[\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}\] Substituting the values we have: \[\overrightarrow{OM} = \frac{(3i + 4j) + (5i - 6j)}{2} = \frac{8i - 2j}{2} = 4i - j\] Therefore, the answer is 4i - j.

Gụọ akwụkwọ nkuzi na Vectors (WAEC)

Vectors

Lee Azịza

Ajụjụ 25 Ripọtì

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

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

$-\frac{15}{16}$

$\frac{15}{16}$

$1\frac{1}{2}$ Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Integration (WAEC)

Integration

Lee Azịza

Ajụjụ 26 Ripọtì

The third of geometric progression (G.P) is 10 and the sixth term is 80. Find the common ratio.

2 Azịza ziri ezi

Akọwa Nkọwa

We can use the formula for the nth term of a geometric progression to solve this problem. Let a be the first term and r be the common ratio, then the third term is ar^2 and the sixth term is ar^5. We are given that ar^2 = 10 and ar^5 = 80. Dividing the second equation by the first, we get: (ar^5)/(ar^2) = 80/10 Simplifying and canceling a, we get: r^3 = 8 Taking the cube root of both sides, we get: r = 2 Therefore, the common ratio is 2, and the answer is (a) 2.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Sets

Lee Azịza

Ajụjụ 27 Ripọtì

P, Q, R, S are points in a plane such that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j and PS = xi + yj. Find (x, y).

(-6, -15)

(-6, 5)

(20, 5) Azịza ziri ezi

(20, 15)

Akọwa Nkọwa

Given that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j, and PS = xi + yj. Since PQ + QR + RS = PS, we have: (8i - 5j) + (5i + 7j) + (7i + 3j) = xi + yj Simplifying the left-hand side, we get: 20i + 5j = xi + yj Equating the corresponding components, we have: 20 = x and 5 = y Therefore, the coordinates of point S are (20, 5). Hence, the correct answer is (20, 5).

Gụọ akwụkwọ nkuzi na Vectors (WAEC)

Vectors

Lee Azịza

Ajụjụ 28 Ripọtì

Simplify $(1 + 2\sqrt{3})^{2} - (1 - 2\sqrt{3})^{2}$

$8\sqrt{3}$ Azịza ziri ezi

$2 - 4\sqrt{3}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Surds (WAEC)

Surds

Lee Azịza

Ajụjụ 29 Ripọtì

If $\frac{5}{\sqrt{2}} - \frac{\sqrt{8}}{8} = m\sqrt{2}$, where m is a constant. Find m.

$1\frac{1}{2}$

$1\frac{1}{4}$

$2\frac{1}{4}$ Azịza ziri ezi

$2\frac{1}{2}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Surds (WAEC)

Surds

Lee Azịza

Ajụjụ 30 Ripọtì

In a class of 50 pupils, 35 like Science and 30 like History. What is the probability of selecting a pupil who likes both Science and History?

0.10

0.30 Azịza ziri ezi

0.60

0.70

Akọwa Nkọwa

We can use the formula: P(A and B) = P(A) + P(B) - P(A or B) where A is the event of liking Science, B is the event of liking History, P(A and B) is the probability of selecting a pupil who likes both Science and History, and P(A or B) is the probability of selecting a pupil who likes either Science or History or both. We know that there are 50 pupils in the class, 35 like Science and 30 like History. This means that the maximum number of pupils who could like both Science and History is 30 (since 30 already like History). Therefore, we have: P(A and B) <= 30/50 = 0.6 Also, we know that P(A or B) = P(A) + P(B) - P(A and B) (as shown above). Therefore: P(A and B) = P(A) + P(B) - P(A or B) = 35/50 + 30/50 - P(A or B) = 65/50 - P(A or B) We also know that P(A or B) <= 1, since it is the probability of selecting a pupil who likes either Science or History or both. Therefore: P(A and B) >= 65/50 - 1 = 15/50 = 0.3 Thus, the probability of selecting a pupil who likes both Science and History is between 0.3 and 0.6. The only option that falls within this range is 0.30, so the answer is: - 0.30

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 31 Ripọtì

Find the axis of symmetry of the curve $y = x^{2} - 4x - 12$.

x = -2

y = -2

x = 2 Azịza ziri ezi

y = 2

Akọwa Nkọwa

To find the axis of symmetry of a quadratic equation in the form of $y = ax^2 + bx + c$, we need to use the formula: $$x = \frac{-b}{2a}$$ Comparing the given equation $y = x^2 - 4x - 12$ with the standard form of the quadratic equation, we have: $$a = 1, b = -4, c = -12$$ Substituting the values in the formula, we get: $$x = \frac{-(-4)}{2(1)} = 2$$ Therefore, the axis of symmetry of the curve is the vertical line passing through the point $(2,0)$ and the equation of this line is: $$x = 2$$ Hence, the correct answer is "x = 2".

Lee Azịza

Ajụjụ 32 Ripọtì

Find the equation of the tangent to the curve $y = 4x^{2} - 12x + 7$ at point (2, -1).

y + 4x - 9 = 0

y - 4x - 9 = 0

y - 4x + 9 = 0 Azịza ziri ezi

y + 4x + 9 = 0

Akọwa Nkọwa

Lee Azịza

Ajụjụ 33 Ripọtì

A company took delivery of 12 vehicles made up of 7 buses and 5 saloon cars for two of its departments; Personnel and General Administration. If the Personnel department is to have at least 3 saloon cars, in how many ways can these vehicles be distributed equally between the departments?

350

455

462 Azịza ziri ezi

571

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Permutation And Combinations (WAEC)

Permutation And Combinations

Lee Azịza

Ajụjụ 34 Ripọtì

Yomi was asked to label four seats S, R, P, Q. What is the probability he labelled them in alphabetical order?

$\frac{1}{24}$ Azịza ziri ezi

$\frac{1}{6}$

$\frac{2}{13}$

$\frac{1}{4}$

Akọwa Nkọwa

There are 4 seats to be labeled, so Yomi can label the first seat with any of the 4 letters. After labeling the first seat, there are 3 letters left to choose from for the second seat, 2 letters for the third seat, and only 1 letter left for the last seat. Thus, there are a total of 4 x 3 x 2 x 1 = 24 possible ways to label the seats. Out of these 24 possible ways, only one arrangement is in alphabetical order (i.e., S, P, Q, R). Therefore, the probability that Yomi labeled the seats in alphabetical order is 1/24. Therefore, the correct option is (a) $\frac{1}{24}$.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Co-ordinate Geometry (WAEC)

Probability

Co-ordinate Geometry

Lee Azịza

Ajụjụ 35 Ripọtì

The distance s metres of a particle from a fixed point at time t seconds is given by $s = 7 + pt^{3} + t^{2}$, where p is a constant. If the acceleration at t = 3 secs is $8 ms^{-2}$, find the value of p.

$\frac{1}{3}$ Azịza ziri ezi

$\frac{4}{9}$

$\frac{5}{9}$

$1$

Akọwa Nkọwa

Given, distance of a particle from a fixed point at time t seconds is given by, $$s = 7 + pt^{3} + t^{2}$$ We can find the acceleration of the particle by differentiating the distance equation twice with respect to time. $$\frac{d}{dt} s = \frac{d}{dt} (7 + pt^{3} + t^{2})$$ $$\Rightarrow \frac{ds}{dt} = 3pt^2 + 2t$$ $$\frac{d^{2}}{dt^{2}} s = \frac{d}{dt} (3pt^2 + 2t)$$ $$\Rightarrow \frac{d^{2}s}{dt^{2}} = 6pt + 2$$ We are given that the acceleration at t = 3 sec is $8 ms^{-2}$, so we can substitute the values in the second derivative of the distance equation to get, $$6p(3) + 2 = 8$$ $$\Rightarrow 18p = 6$$ $$\Rightarrow p = \frac{6}{18} = \frac{1}{3}$$ Hence, the value of p is $\frac{1}{3}$. Therefore, is correct.

Gụọ akwụkwọ nkuzi na Polynomial Functions (WAEC)

Polynomial Functions

Lee Azịza

Ajụjụ 36 Ripọtì

The roots of a quadratic equation are $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$. Find its equation.

$x^{2} - 6x - 9 = 0$

$x^{2} - 6x + 6 = 0$ Azịza ziri ezi

$x^{2} + 6x - 9 = 0$

$x^{2} + 6x + 6 = 0$

Akọwa Nkọwa

The roots of a quadratic equation of the form $ax^{2} + bx + c = 0$ are given by the formula: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ If the roots are $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$, then we have: $$x = 3 - \sqrt{3} \quad \text{or} \quad x = 3 + \sqrt{3}$$ Therefore, we can write the equation as: $$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0$$ Expanding the brackets, we get: $$x^{2} - (3 - \sqrt{3} + 3 + \sqrt{3})x + (3 - \sqrt{3})(3 + \sqrt{3}) = 0$$ Simplifying, we get: $$x^{2} - 6x + 6 = 0$$ Hence, the equation of the quadratic with roots $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$ is $x^{2} - 6x + 6 = 0$. Therefore, the answer is.

Gụọ akwụkwọ nkuzi na Surds (WAEC)

Surds

Lee Azịza

Ajụjụ 37 Ripọtì

Given that $\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4$ and y = 6 when x = 3, find the equation for y.

$x^{3} - 4x - 9$ Azịza ziri ezi

$x^{3} - 4x + 9$

$x^{3} + 4x - 9$

$x^{3} + 4x + 9$

Akọwa Nkọwa

Lee Azịza

Ajụjụ 38 Ripọtì

A bicycle wheel of diameter 70 cm covered a distance of 350 cm in 2 seconds. How many radians per second did it turn?

5 Azịza ziri ezi

Akọwa Nkọwa

To find the radians per second that the bicycle wheel turns, we need to find the angle in radians that the wheel turns in 1 second. The distance covered by the bicycle wheel in 1 second can be found by dividing the distance covered in 2 seconds by 2: \[\frac{350 \, \text{cm}}{2 \, \text{s}} = 175 \, \text{cm/s}\] To convert this to radians per second, we need to know the circumference of the wheel in centimeters. The circumference of the wheel is given by: \[\pi \times \text{diameter} = \pi \times 70 \, \text{cm} = 220 \pi \, \text{cm}\] So, in one revolution, the wheel covers a distance of 220π cm. The angle in radians that the wheel turns in 1 second can be found by dividing the distance covered in 1 second by the circumference of the wheel: \[\frac{175 \, \text{cm/s}}{220 \pi \, \text{cm/rev}} = \frac{5}{2\pi} \, \text{rev/s}\] To convert revolutions per second to radians per second, we multiply by 2π: \[\frac{5}{2\pi} \times 2\pi = 5 \, \text{rad/s}\] Therefore, the bicycle wheel turns at a rate of 5 radians per second. So the answer is (A) 5.

Gụọ akwụkwọ nkuzi na Vectors (WAEC)

Vectors

Lee Azịza

Ajụjụ 39 Ripọtì

If $16^{3x} = \frac{1}{4}(32^{x - 1})$, find the value of x.

$-1$ Azịza ziri ezi

$\frac{-1}{3}$

$\frac{-3}{7}$

$\frac{-5}{19}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Surds (WAEC)

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Surds

Sets

Lee Azịza

Ajụjụ 40 Ripọtì

Two forces $F_{1} = (7i + 8j)N$ and $F_{2} = (3i + 4j)N$ act on a particle. Find the magnitude and direction of $F_{1} - F_{2}$.

$(4\sqrt{2} N, 000°)$

$(4\sqrt{2} N, 045°)$ Azịza ziri ezi

$(4\sqrt{2} N, 090°)$

$(4\sqrt{2} N, 180°)$

Akọwa Nkọwa

To find the magnitude and direction of $F_{1} - F_{2}$, we need to subtract the components of $F_{2}$ from $F_{1}$. \begin{align*} F_{1} - F_{2} &= (7i + 8j) - (3i + 4j)\\ &= 4i + 4j\\ \end{align*} The magnitude of $F_{1} - F_{2}$ is given by: \begin{align*} |F_{1} - F_{2}| &= \sqrt{(4)^2 + (4)^2}\\ &= 4\sqrt{2} N\\ \end{align*} The direction of $F_{1} - F_{2}$ is given by: \begin{align*} \theta &= \tan^{-1}\left(\frac{4j}{4i}\right)\\ &= \tan^{-1}(1)\\ &= 45°\\ \end{align*} Therefore, the magnitude and direction of $F_{1} - F_{2}$ are $(4\sqrt{2} N, 045°)$. Answer is correct.

Gụọ akwụkwọ nkuzi na Statics (WAEC)

Statics

Lee Azịza






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WAEC SSCE - Further Mathematics - 2010 (Objective)