WAEC SSCE - Further Mathematics - 2007…

Tambaya 1 Rahoto

A particle is projected vertically upwards with a speed of 40 m/s. At what times will it be 35m above its point of projection? $\text{Take g} = 10 ms^{-2}$

1 sec and 7 sec Amsar da ta dace

1 sec and 8 sec

2 sec and 5 sec

2 sec and 7 sec

Bayanin Amsa

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 2 Rahoto

Two functions f and g are defined on the set R of real numbers by $f : x \to 2x - 1$ and $g : x \to x^{2} + 1$. Find the value of $f^{-1} \circ g(3)$.

$\frac{11}{2}$ Amsar da ta dace

$\frac{9}{2}$

Bayanin Amsa

We first need to find the composition $f^{-1} \circ g(x)$, which means we need to find the inverse function of f, denoted by $f^{-1}(x)$. To find $f^{-1}(x)$, we solve the equation $y = 2x - 1$ for x: $y + 1 = 2x$ $x = \frac{y + 1}{2}$ Thus, $f^{-1}(x) = \frac{x + 1}{2}$. Now, we can find $f^{-1} \circ g(3)$ by first computing g(3): $g(3) = 3^{2} + 1 = 10$ Then, we can plug g(3) into the composition $f^{-1} \circ g(x)$: $(f^{-1} \circ g)(3) = f^{-1}(g(3)) = f^{-1}(10) = \frac{10 + 1}{2} = \frac{11}{2}$ Therefore, the value of $f^{-1} \circ g(3)$ is $\frac{11}{2}$, which is the correct answer.

Karanta bayanin darasi akan Functions (WAEC)

Functions

Duba Amsa

Tambaya 3 Rahoto

Given that $P = \begin{pmatrix} 4 & 9 \end{pmatrix}$ and $Q = \begin{pmatrix} -1 & -2 \\ 3 & 2 \end{pmatrix}$. Which of the following operations is possible?

$QP$

$P^{2}$

$Q^{2}P$

$PQ$ Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa

Tambaya 4 Rahoto

Calculate in surd form, the value of $\tan 15°$.

$2 + \sqrt{3}$

$1 + \sqrt{3}$

$\sqrt{3} - 1$

$2 - \sqrt{3}$ Amsar da ta dace

Bayanin Amsa

To calculate the value of $\tan 15°$, we can use the half-angle formula for tangent: \[\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}\] Letting $\theta = 30°$, we get: \[\tan 15° = \tan \frac{30°}{2} = \frac{\sin 30°}{1 + \cos 30°}\] We know that $\sin 30° = \frac{1}{2}$ and $\cos 30° = \frac{\sqrt{3}}{2}$. Substituting these values into the equation above, we get: \[\tan 15° = \frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}} = \frac{1}{2 + \sqrt{3}}\] Rationalizing the denominator, we get: \[\tan 15° = \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{1} = \boxed{2 - \sqrt{3}}\] Therefore, the answer is (d) $2 - \sqrt{3}$.

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 5 Rahoto

Given that $\log_{2} y^{\frac{1}{2}} = \log_{5} 125$, find the value of y.

64 Amsar da ta dace

Bayanin Amsa

We can simplify the left side of the equation as follows: $$\log_{2} y^{\frac{1}{2}} = \frac{1}{2}\log_{2} y$$ Similarly, we can simplify the right side of the equation as follows: $$\log_{5} 125 = \log_{5} 5^3 = 3\log_{5} 5 = 3$$ So we have: $$\frac{1}{2}\log_{2} y = 3$$ Multiplying both sides by 2, we get: $$\log_{2} y = 6$$ Using the definition of logarithms, we can write this as: $$2^6 = y$$ Simplifying, we get: $$y = 64$$ Therefore, the value of y is 64.

Karanta bayanin darasi akan Indices And Logarithmic Functions (WAEC)

Indices And Logarithmic Functions

Duba Amsa

Tambaya 6 Rahoto

Given that $p = 4i + 3j$, find the unit vector in the direction of p.

$\frac{1}{3}(4i + 3j)$

$\frac{1}{3}(3i + 4j)$

$\frac{1}{5}(3i + 4j)$

$\frac{1}{5}(4i + 3j)$ Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 7 Rahoto

A body is acted upon by two forces $F_{1} = (5 N, 060°)$ and $F_{2} = (10 N, 180°)$. Find the magnitude of the resultant force.

18.75 N

15.75 N

9.50 N

8.66 N Amsar da ta dace

Bayanin Amsa

We can use vector addition to find the resultant force. Let's first convert the given forces into their x and y components. For $F_1 = (5 N, 060°)$, the x component is $F_{1x} = F_1 \cos(60°) = 5 \cos(60°) = 2.5 N$ and the y component is $F_{1y} = F_1 \sin(60°) = 5 \sin(60°) = 4.33 N$. For $F_2 = (10 N, 180°)$, the x component is $F_{2x} = F_2 \cos(180°) = -10 N$ and the y component is $F_{2y} = F_2 \sin(180°) = 0 N$. Now we can add the x and y components separately to get the resultant force: $$F_{Rx} = F_{1x} + F_{2x} = 2.5 N - 10 N = -7.5 N$$ $$F_{Ry} = F_{1y} + F_{2y} = 4.33 N + 0 N = 4.33 N$$ The magnitude of the resultant force is then given by: $$F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2} = \sqrt{(-7.5)^2 + 4.33^2} \approx 8.66 N$$ Therefore, the answer is \boxed{8.66 N}.

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 8 Rahoto

The gradient of the line passing through the points P(4, 5) and Q(x, 9) is $\frac{1}{2}$. Find the value of x.

-4

12 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Duba Amsa

Tambaya 9 Rahoto

The sum of the first three terms of an Arithmetic Progression (A.P) is 18. If the first term is 4, find their product.

130

192 Amsar da ta dace

210

260

Bayanin Amsa

Let the common difference of the A.P be d. Then, the first three terms of the A.P can be written as 4, 4 + d, and 4 + 2d. Since the sum of the first three terms of the A.P is 18, we have: 4 + (4 + d) + (4 + 2d) = 18 Simplifying this equation, we get: 3d + 12 = 18 3d = 6 d = 2 Therefore, the first three terms of the A.P are 4, 6, and 8. The product of the first three terms of the A.P is: 4 × 6 × 8 = 192 Therefore, the answer is 192.

Karanta bayanin darasi akan Sequences And Series (WAEC)

Sequences And Series

Duba Amsa

Tambaya 10 Rahoto

Three students are working independently on a Mathematics problem. Their respective probabilities of solving the problem are 0.6, 0.7 and 0.8. What is the probability that at least one of them solves the problem?

0.024

0.336

0.664

0.976 Amsar da ta dace

Bayanin Amsa

To find the probability that at least one of the three students solves the problem, we need to calculate the probability that none of them solve the problem and then subtract that from 1 (which represents the total probability). The probability that the first student does not solve the problem is 0.4 (since the probability of solving the problem is 0.6). Similarly, the probability that the second student does not solve the problem is 0.3 and the probability that the third student does not solve the problem is 0.2. To find the probability that none of them solve the problem, we multiply these probabilities together: 0.4 x 0.3 x 0.2 = 0.024. Therefore, the probability that at least one of the three students solves the problem is 1 - 0.024 = 0.976. So, the correct answer is: 0.976.

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 11 Rahoto

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

$\frac{1}{2}$

$\frac{1}{2}\sqrt{3}$

$3$ Amsar da ta dace

$2\sqrt{3}$

Bayanin Amsa

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 12 Rahoto

The tangent to the curve $y = 4x^{3} + kx^{2} - 6x + 4$ at the point P(1, m) is parallel to the x- axis, where k and m are constants. Find the value of k.

-3 Amsar da ta dace

-2

Bayanin Amsa

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 13 Rahoto

The equation of a curve is given by $y = 2x^{2} - 5x + k$. If the curve has two intercepts on the x- axis, find the value(s) of constant k.

$k = \frac{8}{25}$

$k = \frac{25}{8}$

$k < \frac{25}{8}$ Amsar da ta dace

$k > \frac{25}{8}$

Bayanin Amsa

To find the x-intercepts of the curve, we need to set y = 0 and solve for x. Therefore, we have: $$ \begin{aligned} y &= 2x^{2} - 5x + k \\ 0 &= 2x^{2} - 5x + k \\ \end{aligned} $$ For the curve to have two x-intercepts, the discriminant of the quadratic equation must be greater than zero: $$ \begin{aligned} b^{2} - 4ac &> 0 \\ (-5)^{2} - 4(2)(k) &> 0 \\ 25 - 8k &> 0 \\ -8k &> -25 \\ k &< \frac{25}{8} \\ \end{aligned} $$ Therefore, the value of constant k must be less than $\frac{25}{8}$. Thus, the answer is $k < \frac{25}{8}$.

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 14 Rahoto

Which of the following is the same as $\sin (270 + x)°$?

$\sin x$

$\tan x$

$- \sin x $

$- \cos x$ Amsar da ta dace

Bayanin Amsa

We can use the identity $\sin (a+b) = \sin a \cos b + \cos a \sin b$ to simplify the expression $\sin (270 + x)°$: \begin{align*} \sin (270 + x)° &= \sin 270° \cos x + \cos 270° \sin x \\ &= (-1)(\cos x) + (0)(\sin x) \\ &= -\cos x \end{align*} Therefore, the answer is $-\cos x$.

Karanta bayanin darasi akan Trigonometry (WAEC)

Trigonometry

Duba Amsa

Tambaya 15 Rahoto

Marks	5 - 7	8 - 10	11 - 13	14 - 16	17 - 19	20 - 22
Frequency	4	7	26	41	14	8

The table above shows the marks obtained by 100 pupils in a test. Find the probability that a student picked at random scored at least 14 marks.

0.22

0.41

0.49

0.63 Amsar da ta dace

Bayanin Amsa

To find the probability that a student picked at random scored at least 14 marks, we need to add the frequencies of the classes with marks 14-16, 17-19, and 20-22, since all of these classes correspond to scores of at least 14. Adding the frequencies for these classes, we get: 41 + 14 + 8 = 63. Therefore, out of the 100 students, 63 scored at least 14 marks. To find the probability, we divide the number of students who scored at least 14 by the total number of students: 63/100 = 0.63 Therefore, the probability that a student picked at random scored at least 14 marks is 0.63. Answer: 0.63

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 16 Rahoto

The tangent to the curve $y = 4x^{3} + kx^{2} - 6x + 4$ at the point P(1, m) is parallel to the x- axis, where k and m are constants. Determine the coordinates of P.

(1, 2)

(1, 1)

(1, -1) Amsar da ta dace

(1, -2)

Bayanin Amsa

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 17 Rahoto

Given that $\alpha$ and $\beta$ are the roots of an equation such that $\alpha + \beta = 3$ and $\alpha \beta = 2$, find the equation.

$x^{2} - 3x + 2 = 0$ Amsar da ta dace

$x^{2} - 2x + 3 = 0$

$x^{2} - 3x - 2 = 0$

$x^{2} - 2x - 3 = 0$

Bayanin Amsa

We are given that $\alpha$ and $\beta$ are the roots of the equation, and we are also given two conditions based on these roots: 1. $\alpha + \beta = 3$ 2. $\alpha \beta = 2$ To find the equation, we can use the fact that if $\alpha$ and $\beta$ are roots of the equation, then the equation can be written in factored form as: $$(x - \alpha)(x - \beta) = 0$$ Expanding this equation, we get: $$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$ Substituting the values of $\alpha + \beta$ and $\alpha \beta$ from the given conditions, we get: $$x^2 - 3x + 2 = 0$$ Therefore, the equation with roots $\alpha$ and $\beta$ is: $$x^2 - 3x + 2 = 0$$ Hence, the correct option is (a) $x^{2} - 3x + 2 = 0$.

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 18 Rahoto

X and Y are two independent event. If $P(X) = \frac{1}{5}$ and $P(X \cap Y) = \frac{2}{15}$, find $P(Y)$.

$\frac{2}{3}$ Amsar da ta dace

$\frac{2}{5}$

$\frac{1}{3}$

$\frac{1}{5}$

Bayanin Amsa

The formula to find the probability of the intersection of two independent events is given by: $$P(X \cap Y) = P(X) \times P(Y)$$ In this case, we know that: $$P(X) = \frac{1}{5}$$ $$P(X \cap Y) = \frac{2}{15}$$ Substituting the values into the formula, we have: $$\frac{2}{15} = \frac{1}{5} \times P(Y)$$ Solving for $P(Y)$, we get: $$P(Y) = \frac{\frac{2}{15}}{\frac{1}{5}} = \frac{2}{15} \times \frac{5}{1} = \frac{2}{3}$$ Therefore, the probability that event Y occurs is $\frac{2}{3}$. Answer: (a) $\frac{2}{3}$.

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 19 Rahoto

Find the value of p for which $x^{2} - x + p$ becomes a perfect square.

$-\frac{1}{2}$

$\frac{1}{4}$ Amsar da ta dace

$\frac{1}{2}$

$1$

Bayanin Amsa

To make the expression $x^2-x+p$ a perfect square, we need to add and subtract a suitable constant term. To do this, we can start by finding the square of half the coefficient of the linear term. That is, we find the square of $\frac{1}{2}$ which is $\frac{1}{4}$. So, we have: $$x^2-x+\frac{1}{4}-\frac{1}{4}+p$$ Now, we can write this expression as a perfect square by factoring the first three terms: $$\left(x-\frac{1}{2}\right)^2+p-\frac{1}{4}$$ For this expression to be a perfect square, the last term must be zero, so we have: $$p-\frac{1}{4}=0$$ Solving for $p$, we get: $$p=\frac{1}{4}$$ Therefore, the value of $p$ that makes $x^2-x+p$ a perfect square is $\frac{1}{4}$.

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 20 Rahoto

Given that $P = \begin{pmatrix} 4 & 9 \end{pmatrix}$ and $Q = \begin{pmatrix} -1 & -2 \\ 3 & 2 \end{pmatrix}$. Evaluate $|Q|P$.

$\begin{pmatrix} 16 & 36 \end{pmatrix}$ Amsar da ta dace

$\begin{pmatrix} 8 & 18 \end{pmatrix}$

$\begin{pmatrix} 8 & 0 \end{pmatrix}$

$\begin{pmatrix} 1 & 2 \end{pmatrix}$

Bayanin Amsa

To evaluate $|Q|P$, we first need to calculate the determinant of matrix $Q$, denoted as $|Q|$. $$ \begin{aligned} |Q| &= \begin{vmatrix} -1 & -2 \\ 3 & 2 \end{vmatrix} \\ &= (-1\times2) - (-2\times3) \\ &= -2 + 6 \\ &= 4 \end{aligned} $$ Next, we multiply the matrix $P$ by the determinant of matrix $Q$. $$ \begin{aligned} |Q|P &= 4\begin{pmatrix} 4 & 9 \end{pmatrix} \\ &= \begin{pmatrix} 16 & 36 \end{pmatrix} \end{aligned} $$ Therefore, the answer is $\begin{pmatrix} 16 & 36 \end{pmatrix}$.

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa

Tambaya 21 Rahoto

Marks	5 - 7	8 - 10	11 - 13	14 - 16	17 - 19	20 - 22
Frequency	4	7	26	41	14	8

The table above shows the marks obtained by 100 pupils in a test. Find the upper class boundary of the class containing the third quartile.

13.0

16.0

16.5 Amsar da ta dace

22.5

Bayanin Amsa

Karanta bayanin darasi akan Statistics (WAEC)

Karanta bayanin darasi akan Probability (WAEC)

Statistics

Probability

Duba Amsa

Tambaya 22 Rahoto

A bag contains 2 red and 4 green sweets of the same size and shape. Two boys pick a sweet each from the box, one after the other, without replacement. What is the probability that at least a sweet with green wrapper is picked?

$\frac{1}{5}$

$\frac{2}{5}$

$\frac{8}{15}$

$\frac{14}{15}$ Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 23 Rahoto

Calculate, correct to one decimal place, the standard deviation of the numbers: -1, 5, 0, 2 and 9.

7.2

6.6

3.6 Amsar da ta dace

3.2

Bayanin Amsa

To find the standard deviation of a set of numbers, we need to first calculate the mean of the numbers, then subtract the mean from each number to get the deviations, square each deviation, find the mean of the squared deviations (variance), and finally take the square root of the variance to get the standard deviation. So first, let's find the mean: mean = (-1 + 5 + 0 + 2 + 9) / 5 = 3 Next, we'll find the deviations by subtracting the mean from each number: -1 - 3 = -4 5 - 3 = 2 0 - 3 = -3 2 - 3 = -1 9 - 3 = 6 Now we'll square each deviation: (-4)^2 = 16 2^2 = 4 (-3)^2 = 9 (-1)^2 = 1 6^2 = 36 Next, we'll find the mean of the squared deviations: (16 + 4 + 9 + 1 + 36) / 5 = 13.2 Finally, we'll take the square root of the variance to get the standard deviation: sqrt(13.2) ≈ 3.6 So the answer is (c) 3.6, correct to one decimal place.

Karanta bayanin darasi akan Statistics (WAEC)

Statistics

Duba Amsa

Tambaya 24 Rahoto

Given that $R = (4, 180°)$ and $S = (3, 300°)$, find the dot product.

$-6\sqrt{3}$

-6 Amsar da ta dace

$6\sqrt{3}$

Bayanin Amsa

Karanta bayanin darasi akan Trigonometry (WAEC)

Trigonometry

Duba Amsa

Tambaya 25 Rahoto

A group of 5 boys and 4 girls is to be chosen from a class of 8 boys and 6 girls. In how many ways can this be done?

840 Amsar da ta dace

480

408

380

Bayanin Amsa

To solve this problem, we need to use the concept of combinations. The number of ways to choose a group of r objects from a set of n distinct objects is given by the formula: C(n,r) = n! / (r! * (n-r)!) where n! denotes the factorial of n (i.e., the product of all positive integers up to and including n), and r! denotes the factorial of r. In this problem, we want to choose a group of 5 boys and 4 girls from a class of 8 boys and 6 girls. We can do this in two steps: Step 1: Choose the 5 boys from the 8 boys. This can be done in C(8,5) ways. Step 2: Choose the 4 girls from the 6 girls. This can be done in C(6,4) ways. The total number of ways to choose the group of 5 boys and 4 girls is the product of the number of ways in Step 1 and the number of ways in Step 2: C(8,5) * C(6,4) = (8! / (5! * 3!)) * (6! / (4! * 2!)) = (8*7*6 / (3*2*1)) * (6*5 / (2*1)) = 56 * 15 = 840 Therefore, the number of ways to choose a group of 5 boys and 4 girls from a class of 8 boys and 6 girls is 840.

Karanta bayanin darasi akan Permutation And Combinations (WAEC)

Permutation And Combinations

Duba Amsa

Tambaya 26 Rahoto

Find the area of the circle whose equation is given as $x^{2} + y^{2} - 4x + 8y + 11 = 0$.

$3\pi$

$6\pi$

$9\pi$ Amsar da ta dace

$12\pi$

Bayanin Amsa

To find the area of the circle whose equation is given as $x^{2} + y^{2} - 4x + 8y + 11 = 0$, we need to first complete the square for both $x$ and $y$. Rearranging the equation, we get: \begin{aligned} x^2 - 4x + y^2 + 8y &= -11 \\ x^2 - 4x + 4 + y^2 + 8y + 16 &= 9 \\ (x-2)^2 + (y+4)^2 &= 3^2 \end{aligned} Now we can see that the equation represents a circle with center $(2, -4)$ and radius $3$. The formula for the area of a circle is $A = \pi r^2$, where $r$ is the radius. In this case, the radius is $3$, so the area is: \begin{aligned} A &= \pi r^2 \\ &= \pi (3)^2 \\ &= \boxed{9\pi} \end{aligned} Therefore, the answer is $\boxed{9\pi}$.

Karanta bayanin darasi akan Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Duba Amsa

Tambaya 27 Rahoto

Two vectors m and n are defined by $m = 3i + 4j$ and $n = 2i - j$. Find the angle between m and n.

97.9°

79.7° Amsar da ta dace

63.4°

36.4°

Bayanin Amsa

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 28 Rahoto

The binary operation * is defined on the set of R, of real numbers by $x * y = 3x + 3y - xy, \forall x, y \in R$. Determine, in terms of x, the identity element of the operation.

$\frac{2x}{x - 3}, x \neq 3$ Amsar da ta dace

$\frac{2x}{x + 3}, x \neq -3$

$\frac{3x}{x - 3}, x \neq 3$

$\frac{3x}{x + 3}, x \neq -3$

Bayanin Amsa

To find the identity element of an operation, we need to find an element that when combined with any other element using the operation, results in that same element. Let's assume that the identity element is "a", so we need to find the value of "a" that satisfies the following equation for any real number "x": x * a = a * x = x Substituting the definition of the operation, we get: x * a = 3x + 3a - ax = x Simplifying this equation by subtracting 3a from both sides, we get: x - 3a = ax Dividing both sides by x, we get: 1 - 3a/x = a Now, we have an expression for "a" in terms of "x". Let's check each option to see which one satisfies the equation for any value of "x". : a = 2x/(x-3), x ≠ 3 Substituting this into our equation, we get: 1 - 3(2x/(x-3))/x = 2x/(x-3) Simplifying this expression, we get: -3x/(x-3) = 0 This equation is not true for all values of "x", since it is only true when x = 0. Therefore, is not the identity element. : a = 2x/(x+3), x ≠ -3 Substituting this into our equation, we get: 1 - 3(2x/(x+3))/x = 2x/(x+3) Simplifying this expression, we get: 3x/(x+3) - 1 = 0 This equation is true for all values of "x", except for x = -3. Therefore, is the identity element. : a = 3x/(x-3), x ≠ 3 Substituting this into our equation, we get: 1 - 3(3x/(x-3))/x = 3x/(x-3) Simplifying this expression, we get: -6x/(x-3) = 0 This equation is not true for all values of "x", since it is only true when x = 0. Therefore, is not the identity element. : a = 3x/(x+3), x ≠ -3 Substituting this into our equation, we get: 1 - 3(3x/(x+3))/x = 3x/(x+3) Simplifying this expression, we get: 6x/(x+3) - 1 = 0 This equation is true for all values of "x", except for x = -3. Therefore, is the identity element. Therefore, the identity element of the operation is: a = 2x/(x+3), x ≠ -3.

Karanta bayanin darasi akan Binary Operations (WAEC)

Binary Operations

Duba Amsa

Tambaya 29 Rahoto

Three forces $F_{1} = (8 N, 300°), F_{2} = (6 N, 090°)$ and $F_{3} = (4 N, 180°)$ act on a particle. Find the vertical component of the resultant force.

2 N

10 N

$(4 + 4\sqrt{3})N$

$(6 - 4\sqrt{3})N$ Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 30 Rahoto

Evaluate $\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3}$

4 Amsar da ta dace

Bayanin Amsa

We can evaluate the given limit by plugging in the value 3 directly into the expression, but this results in a division by zero, which is undefined. Instead, we can use algebraic techniques to simplify the expression and evaluate the limit. We can factor the numerator of the expression as follows: $x^{2} - 2x - 3 = (x - 3)(x + 1)$ Substituting this factorization into the original expression, we get: $\frac{x^{2} - 2x - 3}{x - 3} = \frac{(x - 3)(x + 1)}{x - 3} = x + 1$ Now, we can take the limit of the simplified expression as x approaches 3: $\lim \limits_{x \to 3} x + 1 = 3 + 1 = 4$ Therefore, the value of the given limit is 4, which is the correct answer.

Karanta bayanin darasi akan Differentiation (WAEC)

Differentiation

Duba Amsa

Tambaya 31 Rahoto

The equation of a circle is given by $x^{2} + y^{2} - 4x - 2y - 3$. Find the radius and the coordinates of its centre.

$3, (-1, 2)$

$2\sqrt{2}, (2, -1)$

$2\sqrt{2}, (2, 1)$ Amsar da ta dace

$9, (2, 1)$

Bayanin Amsa

The equation of a circle in standard form is $(x-a)^2 + (y-b)^2 = r^2$, where $(a,b)$ are the coordinates of the center and $r$ is the radius. We need to rewrite the given equation in this standard form by completing the square for both the $x$ and $y$ terms. \begin{align*} x^{2} + y^{2} - 4x - 2y - 3 &= 0\\ (x^{2} - 4x) + (y^{2} - 2y) &= 3\\ (x^{2} - 4x + 4) + (y^{2} - 2y + 1) &= 3 + 4 + 1\\ (x-2)^{2} + (y-1)^{2} &= 8 \end{align*} Comparing with the standard form, we see that the center of the circle is $(2,1)$ and the radius is $\sqrt{8} = 2\sqrt{2}$. Therefore, the answer is: the radius of the circle is $2\sqrt{2}$ and the coordinates of its center are $(2,1)$.

Karanta bayanin darasi akan Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Duba Amsa

Tambaya 32 Rahoto

A force of 30 N acts at an angle of 60° on a body of mass 6 kg initially at rest on a smooth horizontal plane. Find the distance covered in 10 seconds.

$125\sqrt{3} m$

$125 m$ Amsar da ta dace

$62.5 m$

$25\sqrt{3} m$

Bayanin Amsa

To solve this problem, we need to use the equations of motion. Since the force is acting on the body, it will cause an acceleration in the direction of the force. The acceleration can be found using Newton's second law of motion: $$F = ma$$ where F is the force, m is the mass of the body, and a is the acceleration. In this case, the force is 30 N and the mass is 6 kg. Therefore, the acceleration can be found as: $$a = \frac{F}{m} = \frac{30}{6} = 5\text{ m/s}^2$$ Next, we can use the equations of motion to find the distance covered by the body in 10 seconds. Since the body starts from rest, we can use the equation: $$s = \frac{1}{2}at^2$$ where s is the distance covered, a is the acceleration, and t is the time. Substituting the values, we get: $$s = \frac{1}{2} \times 5 \times 10^2 = 125\text{ m}$$ However, the force is acting at an angle of 60 degrees to the horizontal. Therefore, only the horizontal component of the force will cause motion in the horizontal direction. The horizontal component of the force is: $$F_h = F \cos \theta = 30 \cos 60^\circ = 15\text{ N}$$ We can now find the acceleration in the horizontal direction as: $$a_h = \frac{F_h}{m} = \frac{15}{6} = 2.5\text{ m/s}^2$$ Using the equation of motion, the distance covered in 10 seconds in the horizontal direction is: $$s_h = \frac{1}{2} \times 2.5 \times 10^2 = 12.5\text{ m}$$ However, we also need to find the distance covered in the vertical direction. The vertical component of the force is: $$F_v = F \sin \theta = 30 \sin 60^\circ = 15\sqrt{3}\text{ N}$$ Since there is no force acting in the vertical direction, the body will move under the influence of gravity only. The acceleration due to gravity is 9.8 m/s^2. Therefore, the distance covered in the vertical direction in 10 seconds is: $$s_v = \frac{1}{2} \times 9.8 \times 10^2 = 490\text{ m}$$ The total distance covered by the body is the hypotenuse of the right-angled triangle formed by the horizontal and vertical distances: $$s_{total} = \sqrt{s_h^2 + s_v^2} = \sqrt{(12.5)^2 + (490)^2} \approx 490\text{ m}$$ Therefore, the answer is 125 m.

Karanta bayanin darasi akan Statics (WAEC)

Statics

Duba Amsa

Tambaya 33 Rahoto

If $f(x) = mx^{2} - 6x - 3$ and $f'(1) = 12$, find the value of the constant m.

9 Amsar da ta dace

-3

-4

Bayanin Amsa

The derivative of a quadratic function is obtained by differentiating the function term by term. Thus, differentiating the given function, we have: $$f'(x) = 2mx - 6$$ We are also given that $f'(1) = 12$. Substituting $x=1$ and equating to 12, we have: $$f'(1) = 2m(1) - 6 = 12$$ Simplifying this equation, we get: $$2m = 18$$ Therefore, $m=9$. Thus, the value of the constant $m$ is 9.

Duba Amsa

Tambaya 34 Rahoto

The polynomial $2x^{3} + 3x^{2} + qx - 1$ has the same reminder when divided by $(x + 2)$ and $(x - 1)$. Find the value of the constant q.

-11

-9

-3 Amsar da ta dace

Bayanin Amsa

Duba Amsa

Tambaya 35 Rahoto

Simplify $\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0}$

n - 5

n - 3

2n - 1

2n - 3 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Permutation And Combinations (WAEC)

Permutation And Combinations

Duba Amsa

Tambaya 36 Rahoto

Simplify $\frac{\tan 80° - \tan 20°}{1 + \tan 80° \tan 20°}$

$3\sqrt{2}$

$2\sqrt{3}$

$\sqrt{3}$ Amsar da ta dace

$\sqrt{2}$

Bayanin Amsa

We can simplify this expression using the trigonometric identity: \[\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B}\] In this case, we have A = 80° and B = 20°. Therefore: \[\tan(A-B) = \tan(80°-20°) = \tan 60° = \sqrt{3}\] Using the identity, we can rewrite the expression as: \[\frac{\tan 80° - \tan 20°}{1 + \tan 80° \tan 20°} = \tan(80°-20°) = \sqrt{3}\] Therefore, the simplified expression is $\sqrt{3}$. So the correct option is (3) $\sqrt{3}$.

Karanta bayanin darasi akan Trigonometry (WAEC)

Trigonometry

Duba Amsa

Tambaya 37 Rahoto

Given that the straight lines $kx - 5y + 6 = 0$ and $mx + ny - 1 = 0$ are parallel, find a relationship connecting the constants m, n and k.

5n - km = 0

kn + 5m = 0 Amsar da ta dace

5n + km = 0

kn - 5m = 0

Bayanin Amsa

When two lines are parallel, their slopes are equal. In other words, the coefficient of x in one equation divided by the coefficient of y should be equal to the coefficient of x in the other equation divided by the coefficient of y. Therefore, we can write: k/(-5) = m/n Multiplying both sides by -5n, we get: kn = -5m This is the relationship connecting the constants m, n, and k. Therefore, the correct option is (2) kn + 5m = 0.

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa

Tambaya 38 Rahoto

Find the fourth term of the binomial expansion of $(x - k)^{5}$ in descending powers of x.

$10x^{3}k^{2}$

$5x^{3}k^{2}$

$-5x^{2}k^{3}$

$-10x^{2}k^{3}$ Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Binomial Theorem (WAEC)

Binomial Theorem

Duba Amsa

Tambaya 39 Rahoto

Two bodies of masses 8 kg and 5 kg travelling in the same direction with speeds x m/s and 2 m/s respectively collide. If after collision, they move together with a speed of 3.85 m/s, find, correct to the nearest whole number, the value of x.

5 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa

Tambaya 40 Rahoto

How many ways can 12 people be divided into three groups of 2, 7 and 3 in that order?

7920 Amsar da ta dace

792

187

Bayanin Amsa

In this problem, we need to find the number of ways of dividing 12 people into three groups of 2, 7 and 3 in that order. Firstly, we can choose the first group of 2 people in $\binom{12}{2}$ ways, which gives us 66 ways. Then, we can choose the second group of 7 people from the remaining 10 people in $\binom{10}{7}$ ways, which gives us 120 ways. Finally, the last group of 3 people is determined by the previous two groups, so there is only one way to form this group. Therefore, the total number of ways of dividing 12 people into three groups of 2, 7 and 3 in that order is the product of the number of ways of forming each group, which is equal to $66 \times 120 \times 1 = 7,920$. Thus, the answer is 7920.

Karanta bayanin darasi akan Permutation And Combinations (WAEC)

Permutation And Combinations

Duba Amsa






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WAEC SSCE - Further Mathematics - 2007 (Objective)