WAEC SSCE - Further Mathematics - 2021

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

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

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

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

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

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

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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.

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

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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.

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

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

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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.

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

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

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

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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?

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

What is the upper class boundary of the upper quartile class?

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Find the inverse of  \(\begin{pmatrix} 4 & 2 \\ -3 & -2 \end{pmatrix}\)

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

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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.

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

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

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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.

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

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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.

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.

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

Find the modal class mark.

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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).

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))\(^6\)

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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.

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

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

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

Find the probability of obtaining a prime number.

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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.

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

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

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

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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.

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

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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"?

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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\(^2\) ]

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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.

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

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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.

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

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

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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.

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

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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.

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

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(a) A girl threw a stone horizontally with a velocity of 30m/s from the top of a cliff 50m high. How far from the foot of the cliff does the stone strike the ground? [Take g= 10m/s\(^2\) 

(b) A body A, of mass 2kg is held in equilibrium by means of two strings AP and AR. AP is inclined at 56° to the upward vertical and AR is horizontal.

Find the tensions T\(_1\), and T\(_2\), in the strings [Take g= 10ms\(^2\)]

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(a)

Time taken to reach the ground

Using S = ut + 1/2gt2 ${}^2$

 (u= 0 for a body falling from rest)

50 = 0 x t + 1/2 x 10t2 ${}^2$

5t2 ${}^2$ = 50;

t2 ${}^2$ =50/5 = 10

 t = √10 =3.16

Distance when it strikes the ground This journey is independent of gravity

d =vt =30 x 3.16 = 94.8m

(b)

Using Lami's rule,

20sin34 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ = T1sin90 $\frac{{\mathrm{<br>\; <br>}}_{}}{\mathrm{<br>}}$ = T2sin56 $\frac{{\mathrm{<br>\; <br>}}_{}}{}$

= T1 ${}_1$ 20.sin90sin34 $\frac{\mathrm{<br>}\mathrm{<br>}}{\mathrm{<br>}}$ = T2 ${}_2$ 20.sin56sin34 $\frac{\mathrm{<br>}\mathrm{<br>}}{\mathrm{<br>}}$ 

T1 ${}_1$ = 35.79N and T2 ${}_2$ = 29.65N

The table shows the frequency distribution of heights (in cm) of pupils in a certain school.

 (a) (i) Construct a cumulative frequency table. (ii) Use the table to draw a cumulative frequency curve.

(b) Using the curve, estimate the: (i)median height; (ii) inter quartile range (iii) percentage of students whose heights are most 130cm.

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(a) (i) To construct a cumulative frequency table, we need to add up the frequencies for each height interval and all the intervals before it. The cumulative frequency for each interval is shown in the table below:

(a) (ii) To draw a cumulative frequency curve, we plot the cumulative frequency for each interval against the upper limit of the interval. We then join the points with straight lines to form the curve, as shown below:

(b) (i) To estimate the median height, we locate the value on the cumulative frequency curve that corresponds to the 50th percentile. This is the point where the curve intersects the line at y = 0.5. We draw a line down to the x-axis to find the corresponding height. From the graph, we can see that the median height is around 124-125cm.

(b) (ii) To estimate the interquartile range (IQR), we need to find the heights corresponding to the 25th and 75th percentiles on the cumulative frequency curve. We draw lines down from these points to the x-axis to find the corresponding heights. The IQR is then the difference between these heights. From the graph, we can estimate the 25th percentile at around 116-117cm and the 75th percentile at around 134-135cm. Therefore, the IQR is around 18-19cm.

(b) (iii) To estimate the percentage of students whose heights are most 130cm, we locate the point on the cumulative frequency curve that corresponds to 130cm. We then draw a line across to the y-axis to find the corresponding percentile. From the graph, we can see that the percentage of students whose heights are most 130cm is around 60-65%.

Given that x =  \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\) calculate, correct to the nearest degree, the angle between the vectors

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x =  (−43) $\left(\begin{array}{c}-4\\ 3\end{array}\right)$ and y= (−915) $\left(\begin{array}{c}-9\\ 15\end{array}\right)$

Changing x and y to the form xi+yj

we have:

x= -4i + 3j and y = -9i - 15j

using:

cosØ = xy|x|ly| $\frac{𝑥𝑦}{|𝑥|𝑙𝑦|}$

where Ø is the angle between x and y

xy = (-4i+3j) (-9i - 15j)

-36 + 60 x 0 - 27 x 0- 45 = -81

|x| = √(42 ${}^2$ +32 ${}^2$) =√(16 +9) = √25 = 5

|y| = √(-9)2 ${}^2$ + (-15)2 ${}^2$ = √(81 +225) = √306

cosØ =−815√306 $\frac{-81}{5\surd 306}$

= −9√306170 $\frac{-9\surd 306}{170}$

17.49∗9170 $\frac{17.49\ast 9}{170}$

Ø = cos−1 ${}^{-1}$(-0.1029);

Ø = 95.91°

Ø = 96°

The position vectors of P, Q and R with respect to the origin are (4i-5j), (i+3j) and (-5i+2j) respectively. If PQRM is a parallelogram, find:

(a) the coordinates of M;

(b) the acute angle between \(\overline{PM}\) and \(\overline{PQ}\), correct to the nearest degree.

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To solve this problem, we first need to understand what a parallelogram is. A parallelogram is a four-sided shape with opposite sides that are parallel to each other. This means that if we draw lines from opposite corners of the parallelogram, these lines will intersect at the midpoint of the two sides they are drawn from.

Now let's look at the given position vectors of P, Q, and R. We can use these to find the position vector of M, which is opposite to P in the parallelogram. To do this, we need to use the fact that the position vector of the midpoint of a line segment is the average of the position vectors of the two endpoints. So, the position vector of M is:

¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ =  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ - (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ = (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

MR¯¯¯¯¯¯¯¯¯ $\overline{𝑀𝑅}$ =  (−52) $\left(\begin{array}{c}-5\\ 2\end{array}\right)$ - (xy) $\left(\begin{array}{c}𝑥\\ 𝑦\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

 (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

⇒ -5 - x = -3; x = -2

⇒ 2 - y = 8; y = -6
The coordinate of M (-2,-6)

(b) PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$ = OM¯¯¯¯¯¯¯¯¯ $\overline{𝑂𝑀}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=   (−2−6) $\left(\begin{array}{c}-2\\ -6\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−6−1) $\left(\begin{array}{c}-6\\ -1\end{array}\right)$

    PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ = OQ¯¯¯¯¯¯¯¯ $\overline{𝑂𝑄}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

|PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$| = √(-62 ${}^2$ + [-12 ${}^2$]) = √37

|PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$| = √(-32 ${}^2$ + 82 ${}^2$) = √73

cosØ = PM.PQ|PM¯¯¯¯¯¯¯¯¯¯||PQ¯¯¯¯¯¯¯¯| $\frac{𝑃𝑀.𝑃𝑄}{|\overline{𝑃𝑀}||\overline{𝑃𝑄}|}$

cosØ = [−6i−j].[−3i+8j]√37.√73 $\frac{[-6𝑖-𝑗].[-3𝑖+8𝑗]}{\surd 37.\surd 73}$ = 10√2710 $\frac{10}{\surd 2710}$

⇒ Ø = cos−1 ${}^{-1}$ 10√2710 $\frac{10}{\surd 2710}$

: Ø = 78.91° ≈ 79°

A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement.

Calculate the probability of picking:

(a) same color of bulbs; (6) different color of bulbs (c) at least one red bulb.

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Total bulbs:

n(Red)=5, n(B) = 7, n(G) = 4

(5+7+4) = 16

p(R)= 5/16, p(B) = 7/16, p(G) = 4/16

(a) p(All same colour of bulbs) 

= p(RR) Or p(BB) or p(GG)

516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 716 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 615 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 416 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 315