WAEC SSCE - Further Mathematics - 2013…

Ajụjụ 1 Ripọtì

Express $\frac{x^{2} + x + 4}{(1 - x)(x^{2} + 1)}$ in partial fractions.

$\frac{x^{2}}{x^{2} + 1} + \frac{x + 4}{1 - x}$

$\frac{3}{1 - x} + \frac{2x + 1}{x^{2} + 1}$ Azịza ziri ezi

$\frac{x^{2}}{1 - x} + \frac{x + 4}{x^{2} + 1}$

$\frac{3}{1 - x} + \frac{2x + 2}{x^{2} + 1}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Polynomial Functions (WAEC)

Polynomial Functions

Lee Azịza

Ajụjụ 2 Ripọtì

If $V = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$ and $U = \begin{pmatrix} -1 \\ 5 \end{pmatrix}$, find $|U + V|$.

$3\sqrt{10}$ Azịza ziri ezi

$\sqrt{82}$

$2\sqrt{5}$

Akọwa Nkọwa

We can find the sum of the vectors U and V by adding their corresponding components, i.e., $$ U + V = \begin{pmatrix} -1 \\ 5 \end{pmatrix} + \begin{pmatrix} -2 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 9 \end{pmatrix}. $$ Then, the magnitude (length) of the vector U + V is given by the formula $$ |U + V| = \sqrt{(-3)^2 + 9^2} = \sqrt{90} = 3\sqrt{10}. $$ Therefore, the answer is $3\sqrt{10}$.

Gụọ akwụkwọ nkuzi na Vectors (WAEC)

Vectors

Lee Azịza

Ajụjụ 3 Ripọtì

Evaluate $\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x$.

$-16$

$\frac{-16}{3}$

$\frac{16}{3}$ Azịza ziri ezi

$16$

Akọwa Nkọwa

To evaluate $\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x$, we first need to find the antiderivative of the integrand with respect to x. \[\int (8x - 4x^{2}) \mathrm {d} x = 4x^{2} - \frac{4}{3} x^{3} + C\] where C is the constant of integration. Using the limits of integration, we can evaluate the definite integral as follows: \[\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x = \left[4x^{2} - \frac{4}{3} x^{3}\right]_{0}^{2}\] \[= \left[4(2)^{2} - \frac{4}{3}(2)^{3}\right] - \left[4(0)^{2} - \frac{4}{3}(0)^{3}\right]\] \[= \frac{16}{3}\] Therefore, the answer is $\frac{16}{3}$. Option (c) is the correct answer.

Gụọ akwụkwọ nkuzi na Integration (WAEC)

Integration

Lee Azịza

Ajụjụ 4 Ripọtì

An object is thrown vertically upwards from the top of a cliff with a velocity of $25ms^{-1}$. Find the time, in seconds, when it is 20 metres above the cliff. $[g = 10ms^{-2}]$.

0 and 1

0 and 4

0 and 5

1 and 4 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Differentiation (WAEC)

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Differentiation

Probability

Lee Azịza

Ajụjụ 5 Ripọtì

The equation of a circle is $x^{2} + y^{2} - 8x + 9y + 15 = 0$. Find its radius.

$\frac{1}{2}\sqrt{15}$

$\frac{1}{2}\sqrt{85}$ Azịza ziri ezi

$\sqrt{85}$

Akọwa Nkọwa

To find the radius of a circle given its equation, we need to complete the square for both the x and y terms. We can do this by rearranging the equation as follows: \begin{align*} x^2 - 8x + y^2 + 9y &= -15 \\ (x^2 - 8x + 16) + (y^2 + 9y + 20.25) &= -15 + 16 + 20.25 \\ (x - 4)^2 + (y + 4.5)^2 &= 21.25 \end{align*} We can now see that the equation of the circle is in the standard form: \begin{equation*} (x - a)^2 + (y - b)^2 = r^2 \end{equation*} where the center of the circle is at point (a, b) and the radius is r. From the completed square form, we can identify that the center of the circle is at point (4, -4.5) and the radius is the square root of 21.25, which simplifies to: \begin{equation*} r = \sqrt{21.25} = \frac{1}{2} \sqrt{85} \end{equation*} Therefore, the correct answer is option (C), $\frac{1}{2}\sqrt{85}$.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Sets

Lee Azịza

Ajụjụ 6 Ripọtì

A particle is acted upon by two forces 6N and 3N inclined at an angle of 120° to each other. Find the magnitude of the resultant force.

$18\sqrt{3}$ N

$27$ N

$24$ N

$3\sqrt{3}$ N Azịza ziri ezi

Akọwa Nkọwa

To find the magnitude of the resultant force, we can use the Law of Cosines, which states that in any triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of those two sides and the cosine of the included angle. In this case, let the two forces be a and b, with an included angle of 120°. Then, the magnitude of the resultant force, which we'll call R, is: $$R^2 = a^2 + b^2 - 2ab\cos{120^\circ}$$ Since cosine of 120° is -1/2, we can simplify this to: $$R^2 = a^2 + b^2 + ab$$ Substituting the values given in the problem, we get: $$R^2 = (6\text{ N})^2 + (3\text{ N})^2 + (6\text{ N})(3\text{ N}) = 54\text{ N}^2$$ Taking the square root of both sides, we get: $$R = \sqrt{54\text{ N}^2} = 3\sqrt{6}\text{ N} \approx 7.746\text{ N}$$ Therefore, the correct answer is (D) $3\sqrt{3}$ N.

Gụọ akwụkwọ nkuzi na Dynamics (WAEC)

Gụọ akwụkwọ nkuzi na Statics (WAEC)

Dynamics

Statics

Lee Azịza

Ajụjụ 7 Ripọtì

Given that $\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}$, find y.

$2x^{\frac{3}{2}} + c$

$\frac{2}{3}x^{\frac{3}{2}} + c$ Azịza ziri ezi

$\frac{3}{2}x^{\frac{3}{2}} + c$

$\frac{2}{3}x^{2} + c$

Akọwa Nkọwa

Given that $\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}$, we can integrate both sides with respect to x to obtain the original function y(x). $$ \frac{\mathrm d y}{\mathrm d x} = \sqrt{x} \\ \int \frac{\mathrm d y}{\mathrm d x} \mathrm d x= \int \sqrt{x} \mathrm d x\\ y= \int x^{\frac{1}{2}} \mathrm d x \\ y= \frac{2}{3}x^{\frac{3}{2}} + c $$ where c is an arbitrary constant of integration. Therefore, the correct option is (b) $\frac{2}{3}x^{\frac{3}{2}} + c$.

Gụọ akwụkwọ nkuzi na Differentiation (WAEC)

Differentiation

Lee Azịza

Ajụjụ 8 Ripọtì

If $2\sin^{2}\theta = 1 + \cos \theta, 0° \leq \theta \leq 90°$, find $\theta$.

30°

45°

60° Azịza ziri ezi

90°

Akọwa Nkọwa

We can use trigonometric identities to solve the equation. First, we know that $\cos^2 \theta + \sin^2 \theta = 1$, so we can write: \begin{align*} 2\sin^2 \theta &= 1 + \cos \theta \\ 2(1 - \cos^2 \theta) &= 1 + \cos \theta \\ 2 - 2\cos^2 \theta &= 1 + \cos \theta \\ 2\cos^2 \theta + \cos \theta - 1 &= 0 \\ (2\cos \theta - 1)(\cos \theta + 1) &= 0 \\ \end{align*} Therefore, either $2\cos \theta - 1 = 0$ or $\cos \theta + 1 = 0$. Solving for $\theta$ in each case, we get: \begin{align*} 2\cos \theta - 1 &= 0 \\ \cos \theta &= \frac{1}{2} \\ \theta &= 60^\circ \\ \\ \cos \theta + 1 &= 0 \\ \cos \theta &= -1 \\ \theta &= 180^\circ \\ \end{align*} However, we are given that $0^\circ \leq \theta \leq 90^\circ$, so the only valid solution is $\theta = 60^\circ$. Therefore, the answer is (C) 60°.

Gụọ akwụkwọ nkuzi na Trigonometry (WAEC)

Trigonometry

Lee Azịza

Ajụjụ 9 Ripọtì

Calculate the mean deviation of 1, 2, 3, 4, 5, 5, 6, 7, 8, 9.

2 Azịza ziri ezi

Akọwa Nkọwa

To calculate the mean deviation, we first need to find the mean of the given data set. Mean = (1 + 2 + 3 + 4 + 5 + 5 + 6 + 7 + 8 + 9) / 10 = 50 / 10 = 5 Next, we find the deviation of each value from the mean by subtracting the mean from each value: 1 - 5 = -4 2 - 5 = -3 3 - 5 = -2 4 - 5 = -1 5 - 5 = 0 5 - 5 = 0 6 - 5 = 1 7 - 5 = 2 8 - 5 = 3 9 - 5 = 4 To find the mean deviation, we need to find the average of the absolute values of these deviations. Mean Deviation = (|-4| + |-3| + |-2| + |-1| + |0| + |0| + |1| + |2| + |3| + |4|) / 10 = 20 / 10 = 2 Therefore, the mean deviation of the given data set is 2. The answer is (a) 2.

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Statistics

Lee Azịza

Ajụjụ 10 Ripọtì

A binary operation * is defined on the set of real numbers, R, by $x * y = x + y - xy$. If the identity element under the operation * is 0, find the inverse of $x \in R$.

$\frac{-x}{1 - x}, x \neq 1$ Azịza ziri ezi

$\frac{1}{1 - x}, x \neq 1$

$\frac{-1}{1 - x}, x \neq 1$

$\frac{x}{1 - x}, x \neq 1$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Binary Operations (WAEC)

Binary Operations

Lee Azịza

Ajụjụ 11 Ripọtì

A body is kept at rest by three forces $F_{1} = (10N, 030°), F_{2} = (10N, 150°)$ and $F_{3}$. Find $F_{3}$.

(12N, 090°)

(10N, 270°) Azịza ziri ezi

(10N, 180°)

(10N, 120°)

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Vectors (WAEC)

Vectors

Lee Azịza

Ajụjụ 12 Ripọtì

If $\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}$, find the possible values of x.

1 and -1

-1 and 2

1 and 2

0 and -1 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Surds (WAEC)

Surds

Lee Azịza

Ajụjụ 13 Ripọtì

If $\frac{^{n}C_{3}}{^{n}P_{2}} = 1$, find the value of n.

8 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Permutation And Combinations (WAEC)

Permutation And Combinations

Lee Azịza

Ajụjụ 14 Ripọtì

Forces 50N and 80N act on a body as shown in the diagram. Find, correct to the nearest whole number, the horizontal component of the resultant force.

13N Azịza ziri ezi

43N

57N

95N

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Statics (WAEC)

Statics

Lee Azịza

Ajụjụ 15 Ripọtì

Find the equation of the line that is perpendicular to $2y + 5x - 6 = 0$ and bisects the line joining the points P(4, 3) and Q(-6, 1).

y + 5x + 3 = 0

2y - 5x - 9 = 0

5y + 2x - 8 = 0

5y - 2x - 12 = 0 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Surds (WAEC)

Surds

Lee Azịza

Ajụjụ 16 Ripọtì

Differentiate $x^{2} + xy - 5 = 0$.

$\frac{-(2x + y)}{x}$ Azịza ziri ezi

$\frac{(2x - y)}{x}$

$\frac{-x}{2x + y}$

$\frac{(2x + y)}{x}$

Akọwa Nkọwa

To differentiate the equation, we need to find the derivative of each term in it, using the rules of differentiation. \begin{align*} \frac{d}{dx}(x^{2} + xy - 5) &= \frac{d}{dx}(0) \\ \frac{d}{dx}(x^{2}) + \frac{d}{dx}(xy) - \frac{d}{dx}(5) &= 0 \\ 2x + y\frac{dx}{dx} + x\frac{dy}{dx} - 0 &= 0 \\ 2x + y + x\frac{dy}{dx} &= 0 \\ x\frac{dy}{dx} &= -(2x + y) \\ \frac{dy}{dx} &= -\frac{2x+y}{x} \end{align*} Therefore, the correct answer is $\frac{-(2x + y)}{x}$.

Gụọ akwụkwọ nkuzi na Differentiation (WAEC)

Differentiation

Lee Azịza

Ajụjụ 17 Ripọtì

Given that $P = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix}; Q = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}; R = \begin{pmatrix} -5 & 25 \\ -8 & 26 \end{pmatrix}$ and PQ = R, find the value of x.

-5

-2

5 Azịza ziri ezi

Akọwa Nkọwa

To find x such that PQ = R, we need to find the matrix product of P and Q, and then compare it to R. The matrix product of P and Q is given by: $$ PQ = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix} \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} (3\times 1) + (4\times -2) & (3\times 3) + (4\times 4) \\ (2\times 1) + (x\times -2) & (2\times 3) + (x\times 4) \end{pmatrix} = \begin{pmatrix} -5 & 25 \\ 2-2x & 22+4x \end{pmatrix} $$ We are given that PQ = R, which means we can equate the corresponding entries of the two matrices. In particular, the entry in the first row and first column of PQ must be equal to the entry in the first row and first column of R, i.e. -5. This gives us the equation: $$ -5 = -5 $$ Next, we can equate the corresponding entries in the second row and first column of PQ and R, i.e. 2-2x = -8. Solving for x, we get: $$ 2-2x = -8 \Rightarrow x = 5 $$ Therefore, the value of x that satisfies the equation PQ = R is 5.

Gụọ akwụkwọ nkuzi na Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Lee Azịza

Ajụjụ 18 Ripọtì

If $s = 3i - j$ and $t = 2i + 3j$, find $(t - 3s).(t + 3s)$.

-77 Azịza ziri ezi

-71

-53

-41

Akọwa Nkọwa

First, let's find the value of the expressions inside the parentheses: \begin{align*} t - 3s &= 2i + 3j - 3(3i - j)\\ &= 2i + 3j - 9i + 3j\\ &= -7i + 6j\\ \\ t + 3s &= 2i + 3j + 3(3i - j)\\ &= 2i + 3j + 9i - 3j\\ &= 11i \end{align*} Now we can substitute these expressions into the dot product: \begin{align*} (t - 3s).(t + 3s) &= (-7i + 6j).(11i)\\ &= -77i + 66j\\ \end{align*} Therefore, the answer is -77.

Gụọ akwụkwọ nkuzi na Dynamics (WAEC)

Gụọ akwụkwọ nkuzi na Statics (WAEC)

Dynamics

Statics

Lee Azịza

Ajụjụ 19 Ripọtì

Two out of ten tickets on sale for a raffle draw are winning tickets. If a guest bought two tickets, what is the probability that both tickets are winning tickets?

$\frac{1}{80}$

$\frac{1}{45}$ Azịza ziri ezi

$\frac{1}{20}$

$\frac{1}{10}$

Akọwa Nkọwa

There are ten tickets on sale for the raffle draw, and two of them are winning tickets. If a guest buys two tickets, there are a total of $_{10}C_2$ ways to choose two tickets out of the ten. This is because we can choose any two tickets out of ten in $_{10}C_2$ ways, and each combination is equally likely. The number of ways to choose two winning tickets out of the two available winning tickets is $_2C_2$, which is equal to 1. The probability of selecting two winning tickets is therefore: $$ \frac{\text{number of ways to choose two winning tickets}}{\text{number of ways to choose any two tickets}} = \frac{\binom{2}{2}}{\binom{10}{2}} = \frac{1}{\frac{10\times9}{2\times1}} = \frac{1}{45} $$ Therefore, the probability that both tickets are winning tickets is $\frac{1}{45}$.

Gụọ akwụkwọ nkuzi na Vectors (WAEC)

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Vectors

Probability

Lee Azịza

Ajụjụ 20 Ripọtì

Two bodies of masses 3kg and 5kg moving with velocities 2 m/s and V m/s respectively in opposite directions collide. If they move together after collision with velocity 3.5 m/s in the direction of the 5kg mass, find the value of V.

7.8 m/s

6.8 m/s Azịza ziri ezi

5.6 m/s

4.6 m/s

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Lee Azịza

Ajụjụ 21 Ripọtì

The function $f : F \to R$

= $f(x) = \begin{cases} 3x + 2 : x > 4 \\ 3x - 2 : x = 4 \\ 5x - 3 : x < 4 \end{cases}$. Find f(4) - f(-3).

28 Azịza ziri ezi

-26

-28

Akọwa Nkọwa

The function f is defined piecewise as follows: For x > 4, f(x) = 3x + 2 For x = 4, f(x) = 3x - 2 For x < 4, f(x) = 5x - 3 To find f(4) - f(-3), we first need to find f(4) and f(-3) separately. Since f(4) is defined as 3x - 2 when x = 4, we have f(4) = 3(4) - 2 = 10. Since f(x) is defined as 5x - 3 when x < 4 and we are looking for f(-3), we substitute -3 for x to get f(-3) = 5(-3) - 3 = -18. Now we can calculate f(4) - f(-3) as: f(4) - f(-3) = 10 - (-18) = 28 Therefore, the answer is 28.

Gụọ akwụkwọ nkuzi na Functions (WAEC)

Functions

Lee Azịza

Ajụjụ 22 Ripọtì

If $\alpha$ and $\beta$ are the roots of the equation $2x^{2} - 6x + 5 = 0$, evaluate $\frac{\beta}{\alpha} + \frac{\alpha}{\beta}$.

$\frac{24}{5}$

$\frac{8}{5}$ Azịza ziri ezi

$\frac{5}{8}$

$\frac{5}{24}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Polynomial Functions (WAEC)

Polynomial Functions

Lee Azịza

Ajụjụ 23 Ripọtì

P and Q are the points (3, 1) and (7, 4) respectively. Find the unit vector along PQ.

$\begin{pmatrix} 4 \\ 3 \end{pmatrix}$

$\begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}$

$\begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$ Azịza ziri ezi

$\begin{pmatrix} -0.8 \\ 0.6 \end{pmatrix}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Vectors (WAEC)

Vectors

Lee Azịza

Ajụjụ 24 Ripọtì

Find the coordinates of the point which divides the line joining P(-2, 3) and Q(4, 9) internally in the ratio 2 : 3.

$(5\frac{2}{3}, \frac{2}{5})$

$(\frac{2}{5}, 5\frac{2}{5})$ Azịza ziri ezi

$(\frac{2}{5}, 2\frac{2}{5})$

$(\frac{-2}{5}, 5\frac{2}{5})$

Akọwa Nkọwa

To solve this problem, we will use the section formula. Let the point which divides the line segment PQ internally in the ratio 2:3 be R(x,y). Then we have: \begin{align*} x = \frac{3\times(-2) + 2\times4}{2+3} = \frac{6}{5} \\ y = \frac{3\times3 + 2\times9}{2+3} = \frac{24}{5} \end{align*} Therefore, the coordinates of R are $(\frac{6}{5},\frac{24}{5})$, which corresponds to option (B).

Gụọ akwụkwọ nkuzi na Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Lee Azịza

Ajụjụ 25 Ripọtì

Find the range of values of x for which $x^{2} + 4x + 5$ is less than $3x^{2} - x + 2$

$x > \frac{-1}{2}, x > 3$

$x < \frac{-1}{2}, x > 3$ Azịza ziri ezi

$\frac{-1}{2} \leq x \leq 3$

$\frac{-1}{2} < x < 3$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Polynomial Functions (WAEC)

Polynomial Functions

Lee Azịza

Ajụjụ 26 Ripọtì

If $g(x) = \frac{x + 1}{x - 2}, x \neq -2$, find $g^{-1}(2)$.

$\frac{3}{4}$

-3 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Rational Functions (WAEC)

Rational Functions

Lee Azịza

Ajụjụ 27 Ripọtì

Given that $f(x) = 2x^{3} - 3x^{2} - 11x + 6$ and $f(3) = 0$, factorize f(x).

(x - 3)(x - 2)(2x + 2)

(x + 3)(x - 2)(x - 1)

(x - 3)(x + 2)(2x -1) Azịza ziri ezi

(x + 3)(x - 2)(2x - 1)

Akọwa Nkọwa

To factorize the given expression, we can use synthetic division or long division to divide $f(x)$ by $x - 3$ which is one of its factors as given. The result of this division is a quadratic expression that can be factored easily. Using synthetic division, we get: \begin{array}{c|cccc} & 2 & -3 & -11 & 6 \\ \hline 3 & & 6 & 9 & -6 \\ & & & -6 & 27 \\ \hline & 2 & 3 & -2 & 21 \end{array} Thus, $f(x) = (x - 3)(2x^{2} + 3x - 2)$. Now we need to factorize the quadratic expression $2x^{2} + 3x - 2$. We can use the quadratic formula or factorization by grouping to factorize the quadratic expression. Factorization by grouping is simpler and can be done as follows: \begin{align*} 2x^{2} + 3x - 2 &= 2x^{2} + 4x - x - 2 \\ &= 2x(x + 2) - 1(x + 2) \\ &= (2x - 1)(x + 2) \end{align*} Therefore, the complete factorization of $f(x)$ is: $$f(x) = (x - 3)(2x - 1)(x + 2)$$ Hence, the correct option is (C) (x - 3)(x + 2)(2x - 1).

Gụọ akwụkwọ nkuzi na Polynomial Functions (WAEC)

Polynomial Functions

Lee Azịza

Ajụjụ 28 Ripọtì

If $f(x) = x^{2}$ and $g(x) = \sin x$, find g o f.

$\sin^{2} x$

$\sin x^{2}$ Azịza ziri ezi

$(\sin x)x^{2}$

$x \sin x$

Akọwa Nkọwa

To find g o f, we need to substitute f(x) into g(x). Therefore, we have: g o f = g(f(x)) = g(x^2) = sin(x^2) So the answer is $\sin x^{2}$. Option (B).

Gụọ akwụkwọ nkuzi na Functions (WAEC)

Functions

Lee Azịza

Ajụjụ 29 Ripọtì

A circular ink blot on a piece of paper increases its area at the rate $4mm^{2}/s$. Find the rate of the radius of the blot when the radius is 8mm. $[\pi = \frac{22}{7}]$.

0.20 mm/s

0.08 mm/s Azịza ziri ezi

0.25 mm/s

0.05 mm/s

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Differentiation (WAEC)

Differentiation

Lee Azịza

Ajụjụ 30 Ripọtì

Find the third term in the expansion of $(a - b)^{6}$ in ascending powers of b.

$-15a^{4}b^{2}$

$15a^{4}b^{2}$ Azịza ziri ezi

$-15a^{3}b^{3}$

$15a^{3}b^{3}$

Akọwa Nkọwa

To find the third term in the expansion of $(a-b)^6$ in ascending powers of b, we can use the binomial theorem, which states that: \[(a-b)^6 = \sum_{n=0}^{6} {6 \choose n} a^{6-n}(-b)^n\] The third term is the term where n=2, so we plug in n=2 into the above formula and simplify: \[{6 \choose 2} a^{6-2}(-b)^2 = 15a^4b^2\] Therefore, the third term in the expansion of $(a-b)^6$ in ascending powers of b is $15a^4b^2$. So the correct answer is option B, "15a^4b^2".

Gụọ akwụkwọ nkuzi na Binomial Theorem (WAEC)

Binomial Theorem

Lee Azịza

Ajụjụ 31 Ripọtì

Find the equation of the straight line that passes through (2, -3) and perpendicular to the line 3x - 2y + 4 = 0.

2y - 3x = 0

3y - 2x + 5 = 0

3y + 2x + 5 = 0 Azịza ziri ezi

2y - 3x - 5 = 0

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Lee Azịza

Ajụjụ 32 Ripọtì

The angle subtended by an arc of a circle at the centre is $\frac{\pi}{3} radians$. If the radius of the circle is 12cm, calculate the perimeter of the major arc.

$4(6 + 5\pi)$ Azịza ziri ezi

$4(6 + 2\pi)$

$4(3 + 3\pi)$

$4(3 + 5\pi)$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Trigonometry (WAEC)

Trigonometry

Lee Azịza

Ajụjụ 33 Ripọtì

The sales of five salesgirls on a certain day are as follows; GH¢ 26.00, GH¢ 39.00, GH¢ 33.00, GH¢ 25.00 and GH¢ 37.00. Calculate the standard deviation if the mean sale is GH¢ 32.00.

GH¢ 5.65

GH¢ 5.66 Azịza ziri ezi

GH¢ 6.5

GH¢ 6.56

Akọwa Nkọwa

To calculate the standard deviation, we need to first find the deviations of each sale from the mean sale, square them, sum them up, divide by the total number of sales, and then take the square root of the result. The mean sale is GH¢ 32.00, so the deviations of each sale from the mean are: - GH¢ -6.00 - GH¢ 7.00 - GH¢ 1.00 - GH¢ -7.00 - GH¢ 5.00 Squaring these deviations, we get: - 36 - 49 - 1 - 49 - 25 Summing these up, we get: 36 + 49 + 1 + 49 + 25 = 160 Dividing by the total number of sales (which is 5), we get: 160/5 = 32 Taking the square root of this result, we get: sqrt(32) = 5.66 Therefore, the standard deviation is GH¢ 5.66. So the answer is.

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Statistics

Lee Azịza

Ajụjụ 34 Ripọtì

Solve: $\sin \theta = \tan \theta$

200°

90°

60°

0° Azịza ziri ezi

Akọwa Nkọwa

We can use the trigonometric identity $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ to rewrite the given equation as $\sin \theta = \dfrac{\sin \theta}{\cos \theta}$. Multiplying both sides by $\cos \theta$ and simplifying, we get $\cos \theta = 1$. Therefore, the possible solutions are those angles whose cosine is 1, which are multiples of 360 degrees. Among the given options, only 0 degrees (option D) is a multiple of 360 degrees. So, the answer is $\theta = 0^{\circ}$.

Gụọ akwụkwọ nkuzi na Trigonometry (WAEC)

Trigonometry

Lee Azịza

Ajụjụ 35 Ripọtì

The fourth term of an exponential sequence is 192 and its ninth term is 6. Find the common ratio of the sequence.

$\frac{1}{3}$

$\frac{1}{2}$ Azịza ziri ezi

$2$

$3$

Akọwa Nkọwa

Let's assume that the first term of the exponential sequence is "a" and the common ratio is "r". Then, we can write the fourth and ninth terms as: Fourth term: ar^3 = 192 Ninth term: ar^8 = 6 We can then divide the two equations to eliminate "a" and obtain a relationship between the two powers of "r": (ar^8) / (ar^3) = 6/192 r^5 = 1/32 Taking the fifth root of both sides, we get: r = (1/32)^(1/5) = 1/2 Therefore, the common ratio of the exponential sequence is $\frac{1}{2}$.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Sets

Lee Azịza

Ajụjụ 36 Ripọtì

Express $\log \frac{1}{8} + \log \frac{1}{2}$ in terms of $\log 2$.

3 log 2

4 log 2

-3 log 2

-4 log 2 Azịza ziri ezi

Akọwa Nkọwa

We know that: \[\log a + \log b = \log ab\] Using this property, we can rewrite the given expression as: \[\log \left(\frac{1}{8} \cdot \frac{1}{2}\right) = \log \frac{1}{16}\] Now, we use another property of logarithms: \[\log a^b = b\log a\] to rewrite $\log \frac{1}{16}$ in terms of $\log 2$: \[\log \frac{1}{16} = \log 2^{-4} = -4\log 2\] Therefore, $\log \frac{1}{8} + \log \frac{1}{2} = -4\log 2$ and the answer is: $-4 \log 2$.

Gụọ akwụkwọ nkuzi na Indices And Logarithmic Functions (WAEC)

Indices And Logarithmic Functions

Lee Azịza

Ajụjụ 37 Ripọtì

Given that $P = \begin{pmatrix} y - 2 & y - 1 \\ y - 4 & y + 2 \end{pmatrix}$ and |P| = -23, find the value of y.

-4

-3 Azịza ziri ezi

-1

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Lee Azịza

Ajụjụ 38 Ripọtì

Find the upper quartile of the following scores: 41, 29, 17, 2, 12, 33, 45, 18, 43 and 5.

41 Azịza ziri ezi

Akọwa Nkọwa

To find the upper quartile, we first need to arrange the numbers in order from smallest to largest: 2, 5, 12, 17, 18, 29, 33, 41, 43, 45 The upper quartile divides the data set into quarters, with 25% of the data falling above this point. To find the upper quartile, we need to find the median of the upper half of the data set. The upper half of the data set is: 29, 33, 41, 43, 45 The median of this data set is the middle number, which is 41. Therefore, the upper quartile is 41. So the correct answer is.

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Statistics

Lee Azịza

Ajụjụ 39 Ripọtì

Given that $a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1$, solve for n.

-6.00

-1.20

0.83

1.20 Azịza ziri ezi

Akọwa Nkọwa

We can simplify the left side of the equation as follows: $$a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = a^{\frac{5}{6} - \frac{1}{n}}$$ Since this expression equals 1, we have: $$a^{\frac{5}{6} - \frac{1}{n}} = 1$$ We can rewrite this as: $$\frac{5}{6} - \frac{1}{n} = 0$$ Solving for n, we get: $$\frac{1}{n} = \frac{5}{6}$$ $$n = \frac{6}{5} = 1.2$$ Therefore, the value of n is 1.20, which is.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Sets

Lee Azịza

Ajụjụ 40 Ripọtì

A committee consists of 5 boys namely: Kofi, John, Ojo, Ozo and James and 3 girls namely: Rose, Ugo and Ama. In how many ways can a sub-committee consisting of 3 boys and 2 girls be chosen, if Ozo must be on the sub-committee?

18 Azịza ziri ezi

Akọwa Nkọwa

The problem states that a sub-committee must be formed consisting of 3 boys and 2 girls. Also, Ozo must be on the sub-committee. This means that we already have one of the three boys chosen. We need to choose two more boys from the remaining four boys, and two girls from the three girls. The number of ways to choose two boys from the remaining four boys is given by the combination formula: C(4,2) = 6. (Alternatively, we can list all the possible combinations of two boys from the remaining four boys: Kofi and John, Kofi and Ojo, Kofi and James, John and Ojo, John and James, Ojo and James. This gives us a total of 6 combinations.) Similarly, the number of ways to choose two girls from the three girls is given by the combination formula: C(3,2) = 3. (Alternatively, we can list all the possible combinations of two girls from the three girls: Rose and Ugo, Rose and Ama, Ugo and Ama. This gives us a total of 3 combinations.) To find the total number of ways to choose the sub-committee, we multiply the number of ways to choose two boys and two girls: 6 x 3 = 18. Therefore, the answer is 18.

Gụọ akwụkwọ nkuzi na Permutation And Combinations (WAEC)

Permutation And Combinations

Lee Azịza






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WAEC SSCE - Further Mathematics - 2013 (Objective)