WAEC SSCE - Further Mathematics - 2021…

Tambaya 1 Rahoto

g(x) = 2x + 3 and f(x) = 3x\(^2\) - 2x + 4

37 Amsar da ta dace

-3

-179

Bayanin Amsa

Karanta bayanin darasi akan Functions (WAEC)

Functions

Duba Amsa

Tambaya 2 Rahoto

If √5 cosx + √15sinx = 0, for 0° < x < 360°, find the values of x.

30° and 150°

150° and 210°

150° and 330° Amsar da ta dace

210° and 330°

Bayanin Amsa

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 3 Rahoto

Three forces, F\(_1\) (8N, 030°), F\(_\2) (10N, 150° ) and F\(_\3) ( KN, 240° )are in equilibrium. Find the value of N

5√3

6√2

6√3 Amsar da ta dace

9√3

Bayanin Amsa

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 4 Rahoto

A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).

Calculate the maximum height reached.

418.5m

56.0m

31.5m Amsar da ta dace

30.0m

Bayanin Amsa

Duba Amsa

Tambaya 5 Rahoto

Given that f: x --> x\(^2\) - x + 1 is defined on the Set Q = { x : 0 ≤ x < 20, x is a multiple of 5}. find the set of range of F.

{21, 91, 221} Amsar da ta dace

{21, 91, 221, 381}

{1,21, 91, 221}

{1,21, 91, 221,381}

Bayanin Amsa

Karanta bayanin darasi akan Sets (WAEC)

Sets

Duba Amsa

Tambaya 6 Rahoto

In how many ways can 8 persons be seated on a bench if only three seats are available?

100

125

336 Amsar da ta dace

427

Bayanin Amsa

The first person has 8 choices for a seat, the second person has 7 choices remaining, and the third person has 6 choices. However, the order in which they choose their seats does not matter, so we need to divide by the number of ways that the three people can be arranged. This is given by 3 factorial (3!), which is 3 x 2 x 1 = 6. Therefore, the total number of ways that 8 persons can be seated on a bench if only three seats are available is: 8 x 7 x 6 / 3! = 336 So the correct answer is (C) 336.

Karanta bayanin darasi akan Permutation And Combinations (WAEC)

Permutation And Combinations

Duba Amsa

Tambaya 7 Rahoto

If 2i +pj and 4i -2j are perpendicular, find the value of p.

4 Amsar da ta dace

Bayanin Amsa

To find the value of p, we need to use the concept of perpendicular vectors. Two vectors are perpendicular if and only if the dot product of the two vectors is equal to zero. The dot product of two vectors can be calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. If the angle between the two vectors is 90 degrees, the cosine of the angle is zero and the dot product is also zero. Therefore, to find the value of p, we need to calculate the dot product of the two vectors, 2i + pj and 4i - 2j, and set it equal to zero. The dot product of two vectors (a, b) and (c, d) is given by: (a, b) * (c, d) = ac + bd So, the dot product of 2i + pj and 4i - 2j is: (2i + pj) * (4i - 2j) = (2 * 4) + (p * -2) = 8 - 2p = 0 Therefore, 2p = 8 and p = 4. So, the value of p is 4.

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 8 Rahoto

Given that P = { x: 0 ≤ x ≤ 36, x is a factor of 36 divisible by 3} and Q = { x: 0 ≤ x ≤ 36, x is an even number and a perfect square}, find P n Q.

{1,4,9,36}

{3,9.36}

{9,36}

{36} Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Sets (WAEC)

Sets

Duba Amsa

Tambaya 9 Rahoto

If α and β are the roots of 3x\(^2\) - 7x + 6 = 0, find \(\frac{1}{α}\) + \(\frac{1}{β}\)

\(\frac{7}{6}\) Amsar da ta dace

\(\frac{7}{3}\)

\(\frac{14}{5}\)

\(\frac{14}{3}\)

Bayanin Amsa

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 10 Rahoto

A fair die is tossed 60 times and the results are recorded in the table

Number of die	1	2	3	4	5	6
Frequency	15	10	14	2	8	11

Find the probability of obtaining a prime number.

\(\frac{7}{30}\)

\(\frac{1}{6}\)

\(\frac{7}{15}\)

\(\frac{8}{15}\) Amsar da ta dace

Bayanin Amsa

A prime number is a number greater than 1 that is only divisible by 1 and itself. The prime numbers on a die are 2, 3, and 5. We know that the die is fair, which means that each number has an equal probability of being rolled. There are a total of 60 rolls, so we can use the frequency table to count the number of times each prime number appears on the die. The frequency of the number 2 is 10, the frequency of the number 3 is 14, and the frequency of the number 5 is 8. Therefore, the total number of times a prime number was rolled is: 10 + 14 + 8 = 32 The probability of rolling a prime number on a fair die is the total number of times a prime number was rolled divided by the total number of rolls: 32/60 = 8/15 Therefore, the answer is option D, 8/15.

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 11 Rahoto

If f(x) = 4x\(^3\) + px\(^2\) + 7x - 23 is divided by (2x -5), the remainder is 7. find the value of p

-7.0

-8.0 Amsar da ta dace

-9.6

Bayanin Amsa

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 12 Rahoto

A bag contains 8 red, 4 blue and 2 green identical balls. Two balls are drawn randomly from the bag without replacement. Find the probability that the balls drawn are red and blue.

A. 12/91 B. C. D.

12/91

16/91

30/91

32/91 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 13 Rahoto

If sin x = \(\frac{12}{13}\) and sin y = \(\frac{4}{5}\), where x and y are acute angles, find cos (x + y)

\(\frac{48}{65}\)

\(\frac{13}{15}\)

\(\frac{-33}{65}\) Amsar da ta dace

\(\frac{-48}{65}\)

Bayanin Amsa

Karanta bayanin darasi akan Trigonometry (WAEC)

Trigonometry

Duba Amsa

Tambaya 14 Rahoto

In △PQR, \(\overline{PQ}\) = 5i - 2j and \(\overline{QR}\) = 4i + 3j. Find \(\overline{RP}\).

-i - 5j Amsar da ta dace

-9 - j

i + 5j

-9i + j

Bayanin Amsa

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 15 Rahoto

Find the radius of the circle 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0.

17/4

17/2

17/√2 Amsar da ta dace

√17/2

Bayanin Amsa

To find the radius of a circle from its equation in general form, we need to rewrite the equation in the standard form of a circle, which is: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) Where (h, k) is the center of the circle, and r is its radius. To do this, we complete the square for both x and y terms in the given equation. We start by rearranging the terms as follows: 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0 2x\(^2\) - 4x + 2y\(^2\) - 6y = 2 Now, we need to add and subtract appropriate constants to complete the square for x and y terms separately. For the x terms, we take half of the coefficient of x (-4/2 = -2) and square it to get 4. So, we add and subtract 4 to the equation: 2x\(^2\) - 4x + 4 - 4 + 2y\(^2\) - 6y = 2 We can now group the first three terms and factor it as a perfect square: 2(x - 1)\(^2\) + 2y\(^2\) - 6y - 2 = 0 For the y terms, we take half of the coefficient of y (-6/2 = -3) and square it to get 9. So, we add and subtract 9 to the equation: 2(x - 1)\(^2\) + 2(y - 3)\(^2\) - 2 - 9 = 0 2(x - 1)\(^2\) + 2(y - 3)\(^2\) = 11 Now, we have the equation in the standard form of a circle, where the center is at (1, 3) and the radius is the square root of 11/2. We can simplify this expression to get: sqrt(11/2) = sqrt(11)/sqrt(2) = sqrt(2)*sqrt(11)/2 Therefore, the answer is option (C), 17/√2.

Duba Amsa

Tambaya 16 Rahoto

Using binomial expansion of ( 1 + x)\(^6\) = 1 + 6x + 15x\(^2\) + 20x\(^3\) + 6x\(^5\) + x)\(^6\), find, correct to three decimal places, the value of (1.998))\(^6\)

63.616

63.167 Amsar da ta dace

62.628

62.629

Bayanin Amsa

To use the binomial expansion of (1 + x)\(^6\), we substitute x = 1.998, which gives: (1 + 1.998)\(^6\) = 1 + 6(1.998) + 15(1.998)\(^2\) + 20(1.998)\(^3\) + 6(1.998)\(^5\) + (1.998)\(^6\) We are interested in finding the value of (1.998)\(^6\), which is the last term on the right-hand side of the equation. We can solve for it by subtracting the other terms from both sides of the equation: (1.998)\(^6\) = (1 + 1.998)\(^6\) - 1 - 6(1.998) - 15(1.998)\(^2\) - 20(1.998)\(^3\) - 6(1.998)\(^5\) Using a calculator, we can evaluate the right-hand side of the equation to get: (1.998)\(^6\) ≈ 63.167 Therefore, the correct answer is option B, 63.167, rounded to three decimal places. Explanation: The binomial expansion of (1 + x)\(^6\) is a formula that allows us to expand the expression into a sum of terms involving powers of x. By substituting x = 1.998, we can use the formula to find the value of (1.998)\(^6\). We then use a calculator to evaluate the expression to obtain the final answer.

Karanta bayanin darasi akan Binomial Theorem (WAEC)

Binomial Theorem

Duba Amsa

Tambaya 17 Rahoto

For what range of values of x is x\(^2\) - 2x - 3 ≤ 0

{x: -1 ≤ x ≤ 3} Amsar da ta dace

{ x: -3 ≤ x ≤ 1}

{ x: -3 ≤ x ≤ -1}

{ x: 1 ≤ x ≤ 3}

Bayanin Amsa

To solve the inequality x\(^2\) - 2x - 3 ≤ 0, we can use factoring or the quadratic formula. Factoring gives us (x - 3)(x + 1) ≤ 0, which means that the expression is less than or equal to zero when x is between or equal to -1 and 3, since the factors change sign at these values. Therefore, the correct answer is {x: -1 ≤ x ≤ 3}. Alternatively, we can use the quadratic formula to find the roots of the equation x\(^2\) - 2x - 3 = 0, which are x = -1 and x = 3. Since the quadratic function is a parabola that opens upward, it is negative in the interval between these two roots. Therefore, the expression x\(^2\) - 2x - 3 is less than or equal to zero when x is between or equal to -1 and 3.

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 18 Rahoto

The table shows the distribution of marks obtained by some students in a test

Marks	0-9	10-19	20-29	30-39	40-49
Frequency	4	12	16	6	2

Find the modal class mark.

4.5

14.5

24.5 Amsar da ta dace

34.5

Bayanin Amsa

To find the modal class mark, we need to first identify the class with the highest frequency, which in this case is the class with marks 20-29 (16 students). The modal class mark is the midpoint of this class, which is calculated by adding the lower and upper limits of the class and dividing by 2: Modal class mark = (20 + 29) / 2 = 24.5 Therefore, the modal class mark is 24.5, which is option (C).

Karanta bayanin darasi akan Statistics (WAEC)

Statistics

Duba Amsa

Tambaya 19 Rahoto

Given that \( M = \begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix} \) and \( N = \begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix} \), calculate \( (3M - 2N) \)

\(\begin{pmatrix} 1 & 6 \\ 1 & 18 \end{pmatrix}\)

\(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\) Amsar da ta dace

\(\begin{pmatrix} 1 & 6 \\ -1 & -18 \end{pmatrix}\)

\(\begin{pmatrix} -1 & -6 \\ -1 & -18 \end{pmatrix}\)

Bayanin Amsa

To calculate 3M - 2N, we need to multiply each matrix by its scalar factor and then subtract the results. First, we have: 3M = 3 x \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) Next, we have: 2N = 2 x \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) Now we can subtract these two matrices to get: 3M - 2N = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) = \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\) Therefore, the correct answer is (B) \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\).

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa

Tambaya 20 Rahoto

For what value of k is 4x\(^2\) - 12x + k, a perfect square?

-9

\(\frac{-9}{4}\)

\(\frac{9}{4}\)

9 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Sets (WAEC)

Sets

Duba Amsa

Tambaya 21 Rahoto

A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).

Calculate the distance travelled in the third second.

5.5m Amsar da ta dace

14.5m

26.0m

30.0m

Bayanin Amsa

Duba Amsa

Tambaya 22 Rahoto

Find the equation of the normal to the curve y= 2x\(^2\) - 5x + 10 at P(1, 7).

y+x-3 =0

y-x+6=0

y - x - 6=0 Amsar da ta dace

y -x+ 3 =0

Bayanin Amsa

Duba Amsa

Tambaya 23 Rahoto

The gradient ofy= 3x\(^2\) + 11x + 7 at P(x.y) is -1. Find the coordinates of P.

(-3, -2)

(-2,-3) Amsar da ta dace

(-2,3)

Bayanin Amsa

Duba Amsa

Tambaya 24 Rahoto

Simplify ( \(\frac{1}{2 - √3}\) + \(\frac{2}{2 + √3}\) )\(^{-1}\)

\(\frac{-1}{33}\)(6 + √3)

\(\frac{-1}{33}\)(6 - √3)

\(\frac{1}{33}\)(6 + √3) Amsar da ta dace

\(\frac{1}{33}\)(6 - √3)

Bayanin Amsa

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 25 Rahoto

Find the value of the derivative of y = 3x\(^2\) (2x +1) with respect to x at the point x = 2.

84 Amsar da ta dace

120

Bayanin Amsa

Duba Amsa

Tambaya 26 Rahoto

A binary operation * is defined on the set of real numbers, R, by

P * q = \(\frac{q^2 - p^2}{2pq}\). Find 3 * 2

\(\frac{13}{12}\)

\(\frac{5}{12}\)

\(\frac{-5}{12}\) Amsar da ta dace

\(\frac{-1}{2}\)

Bayanin Amsa

The binary operation * is defined as: P * q = (q^2 - p^2) / (2pq). To find 3 * 2, we need to substitute the values of p and q in the equation: 3 * 2 = (2^2 - 3^2) / (2 * 3 * 2) = (-5) / (12) = -5/12 So, the answer is -5/12.

Karanta bayanin darasi akan Binary Operations (WAEC)

Binary Operations

Duba Amsa

Tambaya 27 Rahoto

If ( 1- 2x)\(^4\) = 1 + px + qx\(^2\) - 32x\(^3\) + 16\(^4\), find the value of (q - p)

-32

-16

32 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Indices And Logarithmic Functions (WAEC)

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Indices And Logarithmic Functions

Polynomial Functions

Duba Amsa

Tambaya 28 Rahoto

Given that F\(^1\)(x) = x\(^3\) √x, find f(x)

\(\frac{2x^{9/2}}{9}\) + c Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 29 Rahoto

A committee consists of 6 boys and 4 girls. In how many ways can a sub-committee consisting of 3 boys and 2 girls be formed if one particular boy and one particular girl must be on the sub-committee?

120 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Permutation And Combinations (WAEC)

Permutation And Combinations

Duba Amsa

Tambaya 30 Rahoto

Consider the following statements:

X: Benita is polite

y: Benita is neat

z: Benita is intelligent

Which of the following symbolizes the statement: "Benita is neat if and only if she is neither polite nor intelligent"?

y⟺ ~x v z

y⟺ ~x v ~z

y⟺ ~x ^ ~z Amsar da ta dace

y⟺ ~x ^ z

Bayanin Amsa

Karanta bayanin darasi akan Logical Reasoning (WAEC)

Logical Reasoning

Duba Amsa

Tambaya 31 Rahoto

If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R

\(\frac{-32}{7}\)

\(\frac{-23}{7}\)

\(\frac{23}{7}\) Amsar da ta dace

\(\frac{32}{7}\)

Bayanin Amsa

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 32 Rahoto

If 2y\(^2\) + 7 = 3y - xy, find \(\frac{dy}{dx}\)

\(\frac{-y}{4x+y-3}\)

\(\frac{y}{4x+y-3}\)

\(\frac{-y}{4x+x-3}\)

\(\frac{y}{4x+x-3}\) Amsar da ta dace

Bayanin Amsa

Duba Amsa

Tambaya 33 Rahoto

The first term of an AP is 4 and the sum of the first three terms is 18. Find the product of the first three terms

292

272

192 Amsar da ta dace

172

Bayanin Amsa

Let the common difference of the AP be denoted by d. Then, the first three terms of the AP are 4, 4 + d, and 4 + 2d, respectively. The sum of the first three terms of the AP is given as 18. Therefore, we have: 4 + (4 + d) + (4 + 2d) = 18 Simplifying the above equation, we get: 3d + 12 = 18 3d = 6 d = 2 Hence, the common difference of the AP is 2. Therefore, the first three terms of the AP are: 4, 6, 8 The product of the first three terms is: 4 x 6 x 8 = 192 Therefore, the answer is 192.

Karanta bayanin darasi akan Sequences And Series (WAEC)

Sequences And Series

Duba Amsa

Tambaya 34 Rahoto

Simplify \(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6

log \(\frac{2}{7}\)

log 2

log \(\frac{2}{9}\) Amsar da ta dace

log 9

Bayanin Amsa

Karanta bayanin darasi akan Indices And Logarithmic Functions (WAEC)

Indices And Logarithmic Functions

Duba Amsa

Tambaya 35 Rahoto

Solve (\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\)

\(\frac{7}{5}\) Amsar da ta dace

\(\frac{6}{7}\)

\(\frac{-7}{6}\)

\(\frac{-6}{7}\)

Bayanin Amsa

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 36 Rahoto

A body of mass 15kg is placed on a smooth plane which is inclined at 60° to the horizontal. If the box is at rest,

calculate the normal reaction to the plane. [ Take g = 10m/s\(^2\) ]

129.9N

75N Amsar da ta dace

60.0N

7.5N

Bayanin Amsa

To find the normal reaction to the plane, we can use the following approach: Step 1: Draw a diagram of the situation. In this case, we have a body of mass 15kg on a smooth plane inclined at 60° to the horizontal. The weight of the body acts vertically downwards and is equal to mg, where m is the mass of the body and g is the acceleration due to gravity. Step 2: Resolve the weight of the body into two components, one perpendicular to the plane and the other parallel to the plane. The component of the weight perpendicular to the plane is equal to mg cos 60°, which is equal to (15 kg) x (10 m/s\(^2\)) x cos 60° = 75 N. Step 3: The normal reaction to the plane is equal in magnitude but opposite in direction to the component of the weight perpendicular to the plane. Therefore, the normal reaction to the plane is 75 N in the upward direction. Step 4: Check the options given in the question and select the one that matches the value obtained in Step 3. In this case, the correct option is 75N. Therefore, the normal reaction to the plane is 75 N.

Karanta bayanin darasi akan Statics (WAEC)

Statics

Duba Amsa

Tambaya 37 Rahoto

The table shows the distribution of marks obtained by some students in a test