WAEC SSCE - Further Mathematics - 2020 (Essay)
Question 1 Report
Forces F\(_1\)(10N, 090°) and F\(_2\)(20N, 210\(^o\)) and (4N,330°) act on a particle, Find, correct to one decimal place, the magnitude of the resultant force.
To find the magnitude of the resultant force, we need to first calculate the x and y components of each force.
For F1:
- The magnitude is 10 N
- The angle is 90 degrees
- The x component is 10 cos(90) = 0 N
- The y component is 10 sin(90) = 10 N
For F2:
- The magnitude is 20 N
- The angle is 210 degrees
- The x component is 20 cos(210) = -17.3 N (negative because it points to the left)
- The y component is 20 sin(210) = -10 N (negative because it points downwards)
For F3:
- The magnitude is 4 N
- The angle is 330 degrees
- The x component is 4 cos(330) = 3.5 N (positive because it points to the right)
- The y component is 4 sin(330) = -2 N (negative because it points downwards)
To find the total x and y components, we can add up the individual x and y components:
- The total x component is 0 - 17.3 + 3.5 = -13.8 N
- The total y component is 10 - 10 - 2 = -2 N
To find the magnitude of the resultant force, we can use the Pythagorean theorem:
resultant force = sqrt((-13.8)^2 + (-2)^2) = 13.9 N
Therefore, the magnitude of the resultant force is 13.9 N.
Answer Details
To find the magnitude of the resultant force, we need to first calculate the x and y components of each force.
For F1:
- The magnitude is 10 N
- The angle is 90 degrees
- The x component is 10 cos(90) = 0 N
- The y component is 10 sin(90) = 10 N
For F2:
- The magnitude is 20 N
- The angle is 210 degrees
- The x component is 20 cos(210) = -17.3 N (negative because it points to the left)
- The y component is 20 sin(210) = -10 N (negative because it points downwards)
For F3:
- The magnitude is 4 N
- The angle is 330 degrees
- The x component is 4 cos(330) = 3.5 N (positive because it points to the right)
- The y component is 4 sin(330) = -2 N (negative because it points downwards)
To find the total x and y components, we can add up the individual x and y components:
- The total x component is 0 - 17.3 + 3.5 = -13.8 N
- The total y component is 10 - 10 - 2 = -2 N
To find the magnitude of the resultant force, we can use the Pythagorean theorem:
resultant force = sqrt((-13.8)^2 + (-2)^2) = 13.9 N
Therefore, the magnitude of the resultant force is 13.9 N.
Question 2 Report
(a) An association is made up of 6 farmers and 8 traders. If an executive body of 4 members is to be formed, find the probability that it will consist of at least two farmers. (b) The probability of an accident occurring in a given month in factories X, Y, and Z are \(\frac{1}{5}, \frac{1}{12} \) and \(\frac{1}{6}\) respectively.
Find the probability that the accident will occur in:
i) none of the factories;
(ii) all the factories;
(iii) at least one factory.
(a) Total members \(= 6 + 8 = 14\). Ways to choose 4 \(= \binom{14}{4} = 1001\).
Use the complement of "fewer than 2 farmers":
\[\text{0 farmers} = \binom{6}{0}\binom{8}{4} = 70,\qquad \text{1 farmer} = \binom{6}{1}\binom{8}{3} = 6 \times 56 = 336\] \[P(\text{at least 2 farmers}) = \frac{1001 - 70 - 336}{1001} = \frac{595}{1001} = \frac{85}{143} \approx 0.594\](b) With \(P(X) = \tfrac{1}{5},\ P(Y) = \tfrac{1}{12},\ P(Z) = \tfrac{1}{6}\):
(i) None:
\[\left(\tfrac{4}{5}\right)\left(\tfrac{11}{12}\right)\left(\tfrac{5}{6}\right) = \frac{220}{360} = \frac{11}{18}\](ii) All:
\[\left(\tfrac{1}{5}\right)\left(\tfrac{1}{12}\right)\left(\tfrac{1}{6}\right) = \frac{1}{360}\](iii) At least one:
\[1 - \frac{11}{18} = \frac{7}{18}\]Answer Details
(a) Total members \(= 6 + 8 = 14\). Ways to choose 4 \(= \binom{14}{4} = 1001\).
Use the complement of "fewer than 2 farmers":
\[\text{0 farmers} = \binom{6}{0}\binom{8}{4} = 70,\qquad \text{1 farmer} = \binom{6}{1}\binom{8}{3} = 6 \times 56 = 336\] \[P(\text{at least 2 farmers}) = \frac{1001 - 70 - 336}{1001} = \frac{595}{1001} = \frac{85}{143} \approx 0.594\](b) With \(P(X) = \tfrac{1}{5},\ P(Y) = \tfrac{1}{12},\ P(Z) = \tfrac{1}{6}\):
(i) None:
\[\left(\tfrac{4}{5}\right)\left(\tfrac{11}{12}\right)\left(\tfrac{5}{6}\right) = \frac{220}{360} = \frac{11}{18}\](ii) All:
\[\left(\tfrac{1}{5}\right)\left(\tfrac{1}{12}\right)\left(\tfrac{1}{6}\right) = \frac{1}{360}\](iii) At least one:
\[1 - \frac{11}{18} = \frac{7}{18}\]Question 3 Report
(a) A bag contains 10 red and 8 green identical balls. Two balls are drawn at random from the bag, one after the other, without replacement. Find the probability that one is red and the other is green.
(b) There are 20% defective bulbs in a large box. If 12 bulbs are selected randomly from the box, calculate the probability that between two and five are defective.
(a) Probability of drawing one red and one green ball:
To find the probability that one ball is red and the other is green, we can use the formula:
P(red and green) = P(red) * P(green)
where P(red) is the probability of drawing a red ball on the first draw, and P(green) is the probability of drawing a green ball on the second draw, given that the first ball was red.
The probability of drawing a red ball on the first draw is 10/18, since there are 10 red balls out of a total of 18 balls in the bag. The probability of drawing a green ball on the second draw, given that the first ball was red, is 8/17, since there are now 8 green balls and 17 balls left in the bag.
Therefore, the probability of drawing one red and one green ball is:
P(red and green) = (10/18) * (8/17) = 0.2353 or approximately 23.53%
(b) Probability of between two and five defective bulbs:
To find the probability that between two and five bulbs are defective, we can use the binomial probability formula:
P(k successes) = (n choose k) * p^k * (1-p)^(n-k)
where n is the number of trials (bulbs selected), k is the number of successes (defective bulbs), p is the probability of success (defective bulb), and (n choose k) is the number of ways to choose k defective bulbs out of n total bulbs.
In this case, n = 12, p = 0.2 (since 20% of the bulbs are defective), and we want to find the probability of between two and five successes (defective bulbs). We can calculate this probability by adding the probabilities of getting exactly 2, 3, 4, or 5 defective bulbs:
P(2-5 defects) = P(2 defects) + P(3 defects) + P(4 defects) + P(5 defects)
P(2 defects) = (12 choose 2) * 0.2^2 * 0.8^10 = 0.301
P(3 defects) = (12 choose 3) * 0.2^3 * 0.8^9 = 0.318
P(4 defects) = (12 choose 4) * 0.2^4 * 0.8^8 = 0.185
P(5 defects) = (12 choose 5) * 0.2^5 * 0.8^7 = 0.058
Therefore, the probability of between two and five bulbs being defective is:
P(2-5 defects) = 0.301 + 0.318 + 0.185 + 0.058 = 0.862 or approximately 86.2%
Answer Details
(a) Probability of drawing one red and one green ball:
To find the probability that one ball is red and the other is green, we can use the formula:
P(red and green) = P(red) * P(green)
where P(red) is the probability of drawing a red ball on the first draw, and P(green) is the probability of drawing a green ball on the second draw, given that the first ball was red.
The probability of drawing a red ball on the first draw is 10/18, since there are 10 red balls out of a total of 18 balls in the bag. The probability of drawing a green ball on the second draw, given that the first ball was red, is 8/17, since there are now 8 green balls and 17 balls left in the bag.
Therefore, the probability of drawing one red and one green ball is:
P(red and green) = (10/18) * (8/17) = 0.2353 or approximately 23.53%
(b) Probability of between two and five defective bulbs:
To find the probability that between two and five bulbs are defective, we can use the binomial probability formula:
P(k successes) = (n choose k) * p^k * (1-p)^(n-k)
where n is the number of trials (bulbs selected), k is the number of successes (defective bulbs), p is the probability of success (defective bulb), and (n choose k) is the number of ways to choose k defective bulbs out of n total bulbs.
In this case, n = 12, p = 0.2 (since 20% of the bulbs are defective), and we want to find the probability of between two and five successes (defective bulbs). We can calculate this probability by adding the probabilities of getting exactly 2, 3, 4, or 5 defective bulbs:
P(2-5 defects) = P(2 defects) + P(3 defects) + P(4 defects) + P(5 defects)
P(2 defects) = (12 choose 2) * 0.2^2 * 0.8^10 = 0.301
P(3 defects) = (12 choose 3) * 0.2^3 * 0.8^9 = 0.318
P(4 defects) = (12 choose 4) * 0.2^4 * 0.8^8 = 0.185
P(5 defects) = (12 choose 5) * 0.2^5 * 0.8^7 = 0.058
Therefore, the probability of between two and five bulbs being defective is:
P(2-5 defects) = 0.301 + 0.318 + 0.185 + 0.058 = 0.862 or approximately 86.2%
Question 4 Report
If \(\frac{3x^2 + 3x - 2}{(x - 1)(x + 1)}\) = P + \(\frac{Q}{x - 1} + \frac{R}{x - 1}\)
Find the value of Q and R
To find the values of Q and R, we need to first simplify the right-hand side of the equation and then equate the numerators of both sides.
First, we need to find the common denominator of the fractions on the right-hand side:
P + \(\frac{Q}{x - 1}\) + \(\frac{R}{x + 1}\) = \(\frac{Px^2 - P + Q(x + 1) + R(x - 1)}{(x - 1)(x + 1)}\)
Next, we can equate the numerators of both sides:
3x^2 + 3x - 2 = Px^2 - P + Q(x + 1) + R(x - 1)
Now, we need to solve for Q and R by choosing appropriate values of x.
To find Q, we can choose x = 1:
3(1)^2 + 3(1) - 2 = P(1)^2 - P + Q(1 + 1) + R(1 - 1)
4 = P + 2Q
To find R, we can choose x = -1:
3(-1)^2 + 3(-1) - 2 = P(-1)^2 - P + Q(-1 + 1) + R(-1 - 1)
-2 = P - 2R
We now have two equations with two variables (P and Q), and we can solve for Q by substituting the expression we found for P in the previous step:
4 = (3 + 3 - 2P) + 2Q
2Q = 2P + 1
Q = P + \(\frac{1}{2}\)
To find R, we can substitute the expression we found for P and the expression we found for Q in the equation we found for R:
-2 = (3 + 3 - 2P) - 2(P + \(\frac{1}{2}\))
-2 = 6 - 4P
P = \(\frac{4}{3}\)
Substituting this value of P in the expression we found for Q, we get:
Q = \(\frac{4}{3}\) + \(\frac{1}{2}\) = \(\frac{11}{6}\)
Finally, substituting the values of P and Q in the expression we found for R, we get:
-2 = 6 - 4(\(\frac{4}{3}\)) - 2(\(\frac{11}{6}\))
R = -\(\frac{5}{6}\)
Therefore, the value of Q is \(\frac{11}{6}\) and the value of R is -\(\frac{5}{6}\).
Answer Details
To find the values of Q and R, we need to first simplify the right-hand side of the equation and then equate the numerators of both sides.
First, we need to find the common denominator of the fractions on the right-hand side:
P + \(\frac{Q}{x - 1}\) + \(\frac{R}{x + 1}\) = \(\frac{Px^2 - P + Q(x + 1) + R(x - 1)}{(x - 1)(x + 1)}\)
Next, we can equate the numerators of both sides:
3x^2 + 3x - 2 = Px^2 - P + Q(x + 1) + R(x - 1)
Now, we need to solve for Q and R by choosing appropriate values of x.
To find Q, we can choose x = 1:
3(1)^2 + 3(1) - 2 = P(1)^2 - P + Q(1 + 1) + R(1 - 1)
4 = P + 2Q
To find R, we can choose x = -1:
3(-1)^2 + 3(-1) - 2 = P(-1)^2 - P + Q(-1 + 1) + R(-1 - 1)
-2 = P - 2R
We now have two equations with two variables (P and Q), and we can solve for Q by substituting the expression we found for P in the previous step:
4 = (3 + 3 - 2P) + 2Q
2Q = 2P + 1
Q = P + \(\frac{1}{2}\)
To find R, we can substitute the expression we found for P and the expression we found for Q in the equation we found for R:
-2 = (3 + 3 - 2P) - 2(P + \(\frac{1}{2}\))
-2 = 6 - 4P
P = \(\frac{4}{3}\)
Substituting this value of P in the expression we found for Q, we get:
Q = \(\frac{4}{3}\) + \(\frac{1}{2}\) = \(\frac{11}{6}\)
Finally, substituting the values of P and Q in the expression we found for R, we get:
-2 = 6 - 4(\(\frac{4}{3}\)) - 2(\(\frac{11}{6}\))
R = -\(\frac{5}{6}\)
Therefore, the value of Q is \(\frac{11}{6}\) and the value of R is -\(\frac{5}{6}\).
Question 5 Report
(a) Two functions p and q are defined on the set of real numbers, R, by p : y \(\to\) 2y +3 and q : y -> y - 2. Find QOP
(b) How many four digits odd numbers greater than 4000 can be formed from 1,7,3,8,2 if repetition is allowed?
(a) To find QOP: we need to apply the functions in the correct order. QOP means we apply Q first, then O, then P.
Q: y → y - 2, so QOP: y → 2(y - 2) + 3 = 2y - 1. Therefore, QOP: y → 2y - 1.
(b) To form a four-digit odd number greater than 4000: the leftmost digit must be 7. We can choose any of the remaining digits for the thousands place, giving us 4 choices. For the hundreds, tens, and ones places, we can choose any of the five digits (1, 3, 7, 8, 2) with repetition allowed, giving us 5 choices for each of those places.
Therefore, the total number of four-digit odd numbers greater than 4000 that can be formed with these digits is:
4 (choices for the thousands place) × 5 (choices for the hundreds place) × 5 (choices for the tens place) × 5 (choices for the ones place) = 500.
So, there are 500 four-digit odd numbers greater than 4000 that can be formed from 1, 7, 3, 8, 2 if repetition is allowed.
Answer Details
(a) To find QOP: we need to apply the functions in the correct order. QOP means we apply Q first, then O, then P.
Q: y → y - 2, so QOP: y → 2(y - 2) + 3 = 2y - 1. Therefore, QOP: y → 2y - 1.
(b) To form a four-digit odd number greater than 4000: the leftmost digit must be 7. We can choose any of the remaining digits for the thousands place, giving us 4 choices. For the hundreds, tens, and ones places, we can choose any of the five digits (1, 3, 7, 8, 2) with repetition allowed, giving us 5 choices for each of those places.
Therefore, the total number of four-digit odd numbers greater than 4000 that can be formed with these digits is:
4 (choices for the thousands place) × 5 (choices for the hundreds place) × 5 (choices for the tens place) × 5 (choices for the ones place) = 500.
So, there are 500 four-digit odd numbers greater than 4000 that can be formed from 1, 7, 3, 8, 2 if repetition is allowed.
Question 6 Report
The diagram is that of a light inextensible string of length 4.2m, whose ends are attached to two fixed points X and Y, 3m apart, and on the same horizontal level. A body of mass 800g is hung on the string at a point O, 2.4m from Y. If the system is kept in equilibrium by a horizontal force P acting on the body and the tensions are equal, calculate:
(a) < XOY;
(b) the magnitude of the force P;
(c) the tension T in the string.
(a) To find the angle between the string and the horizontal, we can use the law of cosines:
c^2 = a^2 + b^2 - 2ab cos C
where c is the length of the string (4.2m), a is the distance from Y to the body (2.4m), and b is the distance from X to Y (3m).
We can rearrange the equation to solve for cos C:
cos C = (a^2 + b^2 - c^2) / 2ab
cos C = (2.4^2 + 3^2 - 4.2^2) / 2 * 2.4 * 3 = 0.8
Finally, we can find the angle using the inverse cosine function:
C = arccos 0.8 = 37.38°
So the angle between the string and the horizontal is 37.38°.
(b) To find the magnitude of the force P, we can use Newton's second law (F = ma), where F is the net force, m is the mass of the body (800g = 0.8kg), and a is the acceleration due to gravity (9.8m/s^2).
The net force is equal to the mass of the body multiplied by the acceleration due to gravity, so:
F = ma = 0.8 * 9.8 = 7.84N
So the magnitude of the force P is 7.84N.
(c) To find the tension T in the string, we can use the fact that the tension must be equal in both parts of the string to keep the system in equilibrium.
Let's call the tension in the part of the string from Y to O "T1" and the tension in the part of the string from O to X "T2".
T1 * cos 37.38° = T2 * cos (180° - 37.38°) = T2 * cos 142.62°
T1 = T2 * cos 142.62° / cos 37.38°
Since T1 = T2, we can substitute T2 for T1:
T2 = T2 * cos 142.62° / cos 37.38°
So, T2 / T2 = cos 142.62° / cos 37.38°
Since T2 is not equal to 0, we can divide both sides by T2:
1 = cos 142.62° / cos 37.38°
To find the tension T in the string, we can use either T1 or T2, since they are equal. Let's use T1:
T1 = T2 = 7.84 / (cos 142.62° / cos 37.38°) = 7.84 / 0.19 = 41.16 N
So the tension in the string is 41.16 N.
Answer Details
(a) To find the angle between the string and the horizontal, we can use the law of cosines:
c^2 = a^2 + b^2 - 2ab cos C
where c is the length of the string (4.2m), a is the distance from Y to the body (2.4m), and b is the distance from X to Y (3m).
We can rearrange the equation to solve for cos C:
cos C = (a^2 + b^2 - c^2) / 2ab
cos C = (2.4^2 + 3^2 - 4.2^2) / 2 * 2.4 * 3 = 0.8
Finally, we can find the angle using the inverse cosine function:
C = arccos 0.8 = 37.38°
So the angle between the string and the horizontal is 37.38°.
(b) To find the magnitude of the force P, we can use Newton's second law (F = ma), where F is the net force, m is the mass of the body (800g = 0.8kg), and a is the acceleration due to gravity (9.8m/s^2).
The net force is equal to the mass of the body multiplied by the acceleration due to gravity, so:
F = ma = 0.8 * 9.8 = 7.84N
So the magnitude of the force P is 7.84N.
(c) To find the tension T in the string, we can use the fact that the tension must be equal in both parts of the string to keep the system in equilibrium.
Let's call the tension in the part of the string from Y to O "T1" and the tension in the part of the string from O to X "T2".
T1 * cos 37.38° = T2 * cos (180° - 37.38°) = T2 * cos 142.62°
T1 = T2 * cos 142.62° / cos 37.38°
Since T1 = T2, we can substitute T2 for T1:
T2 = T2 * cos 142.62° / cos 37.38°
So, T2 / T2 = cos 142.62° / cos 37.38°
Since T2 is not equal to 0, we can divide both sides by T2:
1 = cos 142.62° / cos 37.38°
To find the tension T in the string, we can use either T1 or T2, since they are equal. Let's use T1:
T1 = T2 = 7.84 / (cos 142.62° / cos 37.38°) = 7.84 / 0.19 = 41.16 N
So the tension in the string is 41.16 N.
Question 7 Report
A binary operation * is defined on the set of real numbers R, by p*q = p + q - \(\frac{pq}{2}\), where p, q \(\in\) R. Find the:
(a) inverse of -1 under * given that the identity clement is zero.
(b) truth set of m* 7 = m* 5,
(a) To find the inverse of -1 under *, we need to find a real number x such that -1 * x = x * (-1) = 0, where 0 is the identity element under *.
Substituting -1 and x into the given formula, we get:
-1 * x = -1 + x - \(\frac{(-1)x}{2}\) = 0
Simplifying this equation, we get:
x - \(\frac{x}{2}\) = 1
\(\frac{x}{2}\) = 1
x = 2
Therefore, the inverse of -1 under * is 2.
(b) To find the truth set of m * 7 = m * 5, we need to find all real numbers m that satisfy the equation.
Substituting m and 7 into the given formula, we get:
m * 7 = m + 7 - \(\frac{7m}{2}\)
Substituting m and 5 into the given formula, we get:
m * 5 = m + 5 - \(\frac{5m}{2}\)
Setting these two expressions equal to each other, we get:
m + 7 - \(\frac{7m}{2}\) = m + 5 - \(\frac{5m}{2}\)
Simplifying this equation, we get:
\(\frac{m}{2}\) = 2
m = 4
Therefore, the truth set of m * 7 = m * 5 is {4}, which means that the only real number that satisfies the equation is 4.
Answer Details
(a) To find the inverse of -1 under *, we need to find a real number x such that -1 * x = x * (-1) = 0, where 0 is the identity element under *.
Substituting -1 and x into the given formula, we get:
-1 * x = -1 + x - \(\frac{(-1)x}{2}\) = 0
Simplifying this equation, we get:
x - \(\frac{x}{2}\) = 1
\(\frac{x}{2}\) = 1
x = 2
Therefore, the inverse of -1 under * is 2.
(b) To find the truth set of m * 7 = m * 5, we need to find all real numbers m that satisfy the equation.
Substituting m and 7 into the given formula, we get:
m * 7 = m + 7 - \(\frac{7m}{2}\)
Substituting m and 5 into the given formula, we get:
m * 5 = m + 5 - \(\frac{5m}{2}\)
Setting these two expressions equal to each other, we get:
m + 7 - \(\frac{7m}{2}\) = m + 5 - \(\frac{5m}{2}\)
Simplifying this equation, we get:
\(\frac{m}{2}\) = 2
m = 4
Therefore, the truth set of m * 7 = m * 5 is {4}, which means that the only real number that satisfies the equation is 4.
Question 8 Report
Given that w = 8i + 3j, x = 6i - 5j, y = 2i + 3j and |z| = 41. find z in the direction of w + x - 2y.
To find the vector "z" in the direction of the sum of vectors "w", "x", and "-2y", we can first find the sum of "w", "x", and "-2y", then find the unit vector in that direction and multiply it by the magnitude of "z".
First, let's find the sum of "w", "x", and "-2y":
w = 8i + 3j
x = 6i - 5j
y = 2i + 3j
-2y = -4i - 6j
w + x - 2y = (8i + 3j) + (6i - 5j) - (-4i - 6j) = 18i - 8j
Next, let's find the unit vector in the direction of the sum:
The unit vector in the direction of a vector can be found by dividing the vector by its magnitude.
The magnitude of the sum is:
√(18^2 + (-8)^2) = 20
So the unit vector in the direction of the sum is:
(18i - 8j) / 20 = 9/10i - 4/10j
Finally, let's multiply the unit vector by the magnitude of "z" to find "z":
z = 41 * (9/10i - 4/10j) = 36.9i - 16.4j
So the vector "z" in the direction of the sum of "w", "x", and "-2y" is approximately 36.9i - 16.4j.
Answer Details
To find the vector "z" in the direction of the sum of vectors "w", "x", and "-2y", we can first find the sum of "w", "x", and "-2y", then find the unit vector in that direction and multiply it by the magnitude of "z".
First, let's find the sum of "w", "x", and "-2y":
w = 8i + 3j
x = 6i - 5j
y = 2i + 3j
-2y = -4i - 6j
w + x - 2y = (8i + 3j) + (6i - 5j) - (-4i - 6j) = 18i - 8j
Next, let's find the unit vector in the direction of the sum:
The unit vector in the direction of a vector can be found by dividing the vector by its magnitude.
The magnitude of the sum is:
√(18^2 + (-8)^2) = 20
So the unit vector in the direction of the sum is:
(18i - 8j) / 20 = 9/10i - 4/10j
Finally, let's multiply the unit vector by the magnitude of "z" to find "z":
z = 41 * (9/10i - 4/10j) = 36.9i - 16.4j
So the vector "z" in the direction of the sum of "w", "x", and "-2y" is approximately 36.9i - 16.4j.
Question 9 Report
(a) Simplify; \(\frac{log_2 ^8 + log_2 ^{16} - 4 log_2 ^2}{log_4^{16}}\)
(b) The first, third, and seventh terms of an Arithmetic Progression (A.P) from three consecutive terms of a Geometric Progression (G.P). If the sum of the first two terms of the A.P is 6, find its:
(I) first term; (ii) common difference.
(a) Evaluate each logarithm to base 2:
\[\log_2 8 = 3,\quad \log_2 16 = 4,\quad 4\log_2 2 = 4,\quad \log_4 16 = 2\] \[\frac{\log_2 8 + \log_2 16 - 4\log_2 2}{\log_4 16} = \frac{3 + 4 - 4}{2} = \frac{3}{2}\](b) Let the A.P. have first term \(a\) and common difference \(d\). The 1st, 3rd and 7th terms are \(a,\ a + 2d,\ a + 6d\). These are consecutive G.P. terms, so:
\[(a + 2d)^2 = a(a + 6d)\] \[a^2 + 4ad + 4d^2 = a^2 + 6ad \ \Rightarrow\ 4d^2 - 2ad = 0 \ \Rightarrow\ 2d(2d - a) = 0\]Since \(d \neq 0\), \(a = 2d\). The sum of the first two A.P. terms is 6:
\[a + (a + d) = 2a + d = 6\]Substituting \(a = 2d\): \(4d + d = 6 \Rightarrow d = \tfrac{6}{5} = 1.2\), and \(a = 2d = \tfrac{12}{5} = 2.4\).
(i) First term \(= 2.4\). (ii) Common difference \(= 1.2\).
Answer Details
(a) Evaluate each logarithm to base 2:
\[\log_2 8 = 3,\quad \log_2 16 = 4,\quad 4\log_2 2 = 4,\quad \log_4 16 = 2\] \[\frac{\log_2 8 + \log_2 16 - 4\log_2 2}{\log_4 16} = \frac{3 + 4 - 4}{2} = \frac{3}{2}\](b) Let the A.P. have first term \(a\) and common difference \(d\). The 1st, 3rd and 7th terms are \(a,\ a + 2d,\ a + 6d\). These are consecutive G.P. terms, so:
\[(a + 2d)^2 = a(a + 6d)\] \[a^2 + 4ad + 4d^2 = a^2 + 6ad \ \Rightarrow\ 4d^2 - 2ad = 0 \ \Rightarrow\ 2d(2d - a) = 0\]Since \(d \neq 0\), \(a = 2d\). The sum of the first two A.P. terms is 6:
\[a + (a + d) = 2a + d = 6\]Substituting \(a = 2d\): \(4d + d = 6 \Rightarrow d = \tfrac{6}{5} = 1.2\), and \(a = 2d = \tfrac{12}{5} = 2.4\).
(i) First term \(= 2.4\). (ii) Common difference \(= 1.2\).
Question 10 Report
(a) A car is moving with a velocity of 10ms\(^{-1}\) It then accelerates at 0.2ms\(^{-2}\) for 100m. Find, correct to two decimal places the time taken by the car to cover the distance.
(b) A particle moves along a straight line such that its distance S metres from a fixed point O is given by S = t\(^2\) - 5t + 6, where t is the time in seconds. Find its:
(i) initial velocity;
(ii) distance when it is momentarily at rest
(a) Given \(u = 10\,\text{ms}^{-1},\ a = 0.2\,\text{ms}^{-2},\ s = 100\,\text{m}\). Use \(s = ut + \tfrac{1}{2}at^2\):
\[100 = 10t + 0.1t^2 \ \Rightarrow\ t^2 + 100t - 1000 = 0\] \[t = \frac{-100 + \sqrt{100^2 + 4000}}{2} = \frac{-100 + \sqrt{14000}}{2} = \frac{-100 + 118.32}{2}\] \[t \approx 9.16\,\text{s}\](b) \(S = t^2 - 5t + 6\), so velocity \(v = \dfrac{dS}{dt} = 2t - 5\).
(i) Initial velocity at \(t = 0\): \(v = 2(0) - 5 = -5\,\text{ms}^{-1}\).
(ii) Momentarily at rest when \(v = 0\): \(2t - 5 = 0 \Rightarrow t = 2.5\,\text{s}\).
\[S = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25\,\text{m}\]The particle is \(0.25\,\text{m}\) from \(O\) (on the negative side).
Answer Details
(a) Given \(u = 10\,\text{ms}^{-1},\ a = 0.2\,\text{ms}^{-2},\ s = 100\,\text{m}\). Use \(s = ut + \tfrac{1}{2}at^2\):
\[100 = 10t + 0.1t^2 \ \Rightarrow\ t^2 + 100t - 1000 = 0\] \[t = \frac{-100 + \sqrt{100^2 + 4000}}{2} = \frac{-100 + \sqrt{14000}}{2} = \frac{-100 + 118.32}{2}\] \[t \approx 9.16\,\text{s}\](b) \(S = t^2 - 5t + 6\), so velocity \(v = \dfrac{dS}{dt} = 2t - 5\).
(i) Initial velocity at \(t = 0\): \(v = 2(0) - 5 = -5\,\text{ms}^{-1}\).
(ii) Momentarily at rest when \(v = 0\): \(2t - 5 = 0 \Rightarrow t = 2.5\,\text{s}\).
\[S = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25\,\text{m}\]The particle is \(0.25\,\text{m}\) from \(O\) (on the negative side).
Question 11 Report
| Marks | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 2 | 2 | 2 | 8 | 13 | 11 | 12 | 10 | 4 |
The table shows the distribution of marks scored by 64 students in a test
(a) Draw a histogram for the distribution.
(b) Use the histogram to estimate the modal score.
The class widths are all equal to 10, so the heights of the histogram bars are the frequencies.
| Marks | Class boundaries | Frequency |
|---|---|---|
| 10–19 | 9.5–19.5 | 2 |
| 20–29 | 19.5–29.5 | 2 |
| 30–39 | 29.5–39.5 | 2 |
| 40–49 | 39.5–49.5 | 8 |
| 50–59 | 49.5–59.5 | 13 |
| 60–69 | 59.5–69.5 | 11 |
| 70–79 | 69.5–79.5 | 12 |
| 80–89 | 79.5–89.5 | 10 |
| 90–99 | 89.5–99.5 | 4 |
(a) Histogram:
(b) The modal class is 50–59, since it has the greatest frequency, 13.
Using the intersecting diagonals from the top corners of the modal rectangle to the top corners of the adjacent rectangles gives
\[\text{Mode}=L+\frac{f_m-f_1}{(f_m-f_1)+(f_m-f_2)}\times c\]
where \(L=49.5\), \(f_m=13\), \(f_1=8\), \(f_2=11\), and \(c=10\).
\[\text{Mode}=49.5+\frac{13-8}{(13-8)+(13-11)}\times10\]
\[=49.5+\frac{5}{7}\times10=56.64\]
Hence, the estimated modal score is \(\boxed{57\text{ marks}}\).
Answer Details
The class widths are all equal to 10, so the heights of the histogram bars are the frequencies.
| Marks | Class boundaries | Frequency |
|---|---|---|
| 10–19 | 9.5–19.5 | 2 |
| 20–29 | 19.5–29.5 | 2 |
| 30–39 | 29.5–39.5 | 2 |
| 40–49 | 39.5–49.5 | 8 |
| 50–59 | 49.5–59.5 | 13 |
| 60–69 | 59.5–69.5 | 11 |
| 70–79 | 69.5–79.5 | 12 |
| 80–89 | 79.5–89.5 | 10 |
| 90–99 | 89.5–99.5 | 4 |
(a) Histogram:
(b) The modal class is 50–59, since it has the greatest frequency, 13.
Using the intersecting diagonals from the top corners of the modal rectangle to the top corners of the adjacent rectangles gives
\[\text{Mode}=L+\frac{f_m-f_1}{(f_m-f_1)+(f_m-f_2)}\times c\]
where \(L=49.5\), \(f_m=13\), \(f_1=8\), \(f_2=11\), and \(c=10\).
\[\text{Mode}=49.5+\frac{13-8}{(13-8)+(13-11)}\times10\]
\[=49.5+\frac{5}{7}\times10=56.64\]
Hence, the estimated modal score is \(\boxed{57\text{ marks}}\).
Question 12 Report
Evaluate; \(\int^3_1(3x - 2)^5 dx\)
Integrate \( (3x - 2)^5 \) by reversing the chain rule (raise the power, divide by the new power and by the derivative of the bracket, \(3\)):
\[ \int (3x - 2)^5\,dx = \frac{(3x - 2)^6}{6 \times 3} = \frac{(3x - 2)^6}{18}. \]
Evaluate between \(1\) and \(3\):
\[ \left[\frac{(3x - 2)^6}{18}\right]_1^3 = \frac{(7)^6}{18} - \frac{(1)^6}{18} = \frac{117649 - 1}{18} = \frac{117648}{18} = 6536. \]
The value of the integral is \( 6536 \).
Answer Details
Integrate \( (3x - 2)^5 \) by reversing the chain rule (raise the power, divide by the new power and by the derivative of the bracket, \(3\)):
\[ \int (3x - 2)^5\,dx = \frac{(3x - 2)^6}{6 \times 3} = \frac{(3x - 2)^6}{18}. \]
Evaluate between \(1\) and \(3\):
\[ \left[\frac{(3x - 2)^6}{18}\right]_1^3 = \frac{(7)^6}{18} - \frac{(1)^6}{18} = \frac{117649 - 1}{18} = \frac{117648}{18} = 6536. \]
The value of the integral is \( 6536 \).
Question 13 Report
(a) If (x + 2) is a factor of g(x) = 2x\(^3\) +11x\(^2\) - x - 30, find the zeros of g(x).
(b) Solve 3(2\(^x\)) +3\(^{y - 2}\) = 25 and 2x - 3\(^{y + 1}\) = -19 simultaneously.
(a) If (x + 2) is a factor of g(x), it means that (x + 2) is a root of the polynomial g(x). To find the zeros of g(x), we need to factorize the polynomial by using long division or synthetic division.
Using synthetic division, we get:
-2 | 2 11 -1 -30 |----------- | 2 -6 11 | 0
Therefore, g(x) can be factorized as g(x) = 2x^2 - 6x + 11. To find the zeros of g(x), we need to solve the equation 2x^2 - 6x + 11 = 0.
Using the quadratic formula, we get:
x = \(\frac{6 \pm \sqrt{6^2 - 4(2)(11)}}{4}\)
x = \(\frac{3}{2} \pm \frac{\sqrt{23}}{2}\)
Therefore, the zeros of g(x) are \(\frac{3}{2} + \frac{\sqrt{23}}{2}\), and \(\frac{3}{2} - \frac{\sqrt{23}}{2}\).
(b) To solve the simultaneous equations 3(2^x) + 3^(y - 2) = 25 and 2x - 3^(y + 1) = -19, we need to isolate one variable in terms of the other and substitute the expression into the other equation.
From the second equation, we have:
2x = 3^(y + 1) - 19
Substituting this expression into the first equation, we get:
3(2^x) + 3^(y - 2) = 25
3(2^x) + 3^(y - 2) = 3^2
Dividing both sides by 3^2, we get:
2^x + 3^(y - 2 - 2) = 1
2^x + 3^(y - 4) = 1
Substituting the expression for 2x from the second equation, we get:
2^x + 3^(y - 4) = 1 + 3^(y + 1) - 19
2^x + 3^(y - 4) = 3^(y + 1) - 18
Rearranging this equation, we get:
2^x = 3^(y + 1) - 3^(y - 4) - 18
Using logarithms, we can solve for x:
x = log2(3^(y + 1) - 3^(y - 4) - 18)
Substituting this expression for x into the second equation, we get:
2log2(3^(y + 1) - 3^(y - 4) - 18) - 3^(y + 1) = -19
Simplifying this
Answer Details
(a) If (x + 2) is a factor of g(x), it means that (x + 2) is a root of the polynomial g(x). To find the zeros of g(x), we need to factorize the polynomial by using long division or synthetic division.
Using synthetic division, we get:
-2 | 2 11 -1 -30 |----------- | 2 -6 11 | 0
Therefore, g(x) can be factorized as g(x) = 2x^2 - 6x + 11. To find the zeros of g(x), we need to solve the equation 2x^2 - 6x + 11 = 0.
Using the quadratic formula, we get:
x = \(\frac{6 \pm \sqrt{6^2 - 4(2)(11)}}{4}\)
x = \(\frac{3}{2} \pm \frac{\sqrt{23}}{2}\)
Therefore, the zeros of g(x) are \(\frac{3}{2} + \frac{\sqrt{23}}{2}\), and \(\frac{3}{2} - \frac{\sqrt{23}}{2}\).
(b) To solve the simultaneous equations 3(2^x) + 3^(y - 2) = 25 and 2x - 3^(y + 1) = -19, we need to isolate one variable in terms of the other and substitute the expression into the other equation.
From the second equation, we have:
2x = 3^(y + 1) - 19
Substituting this expression into the first equation, we get:
3(2^x) + 3^(y - 2) = 25
3(2^x) + 3^(y - 2) = 3^2
Dividing both sides by 3^2, we get:
2^x + 3^(y - 2 - 2) = 1
2^x + 3^(y - 4) = 1
Substituting the expression for 2x from the second equation, we get:
2^x + 3^(y - 4) = 1 + 3^(y + 1) - 19
2^x + 3^(y - 4) = 3^(y + 1) - 18
Rearranging this equation, we get:
2^x = 3^(y + 1) - 3^(y - 4) - 18
Using logarithms, we can solve for x:
x = log2(3^(y + 1) - 3^(y - 4) - 18)
Substituting this expression for x into the second equation, we get:
2log2(3^(y + 1) - 3^(y - 4) - 18) - 3^(y + 1) = -19
Simplifying this
Question 14 Report
(a) Find the derivative of y = x\(^2\) (1 + x)\(^{\frac{3}{2}}\) with respect to x.
(b) The centre of a circle lies on the line 2y - x = 3. If the circle passes through P(2,3) and Q(6,7), find its equation.
(a) \(y = x^2(1 + x)^{3/2}\). Apply the product rule:
\[\frac{dy}{dx} = 2x(1 + x)^{3/2} + x^2 \cdot \tfrac{3}{2}(1 + x)^{1/2}\]Factor out \((1 + x)^{1/2}\):
\[\frac{dy}{dx} = (1 + x)^{1/2}\left[2x(1 + x) + \tfrac{3}{2}x^2\right] = (1 + x)^{1/2}\left[2x + \tfrac{7}{2}x^2\right]\] \[\frac{dy}{dx} = \frac{x(4 + 7x)}{2}\sqrt{1 + x}\](b) Let the centre be \((h, k)\) with \(2k - h = 3\), so \(h = 2k - 3\). The centre is equidistant from \(P(2,3)\) and \(Q(6,7)\):
\[(h - 2)^2 + (k - 3)^2 = (h - 6)^2 + (k - 7)^2\]Expanding and simplifying gives \(8h + 8k = 72\), i.e. \(h + k = 9\). With \(h = 2k - 3\):
\[2k - 3 + k = 9 \ \Rightarrow\ k = 4,\quad h = 5\]Centre \((5, 4)\); radius squared \(= (5 - 2)^2 + (4 - 3)^2 = 10\).
\[(x - 5)^2 + (y - 4)^2 = 10 \quad\text{or}\quad x^2 + y^2 - 10x - 8y + 31 = 0\]Answer Details
(a) \(y = x^2(1 + x)^{3/2}\). Apply the product rule:
\[\frac{dy}{dx} = 2x(1 + x)^{3/2} + x^2 \cdot \tfrac{3}{2}(1 + x)^{1/2}\]Factor out \((1 + x)^{1/2}\):
\[\frac{dy}{dx} = (1 + x)^{1/2}\left[2x(1 + x) + \tfrac{3}{2}x^2\right] = (1 + x)^{1/2}\left[2x + \tfrac{7}{2}x^2\right]\] \[\frac{dy}{dx} = \frac{x(4 + 7x)}{2}\sqrt{1 + x}\](b) Let the centre be \((h, k)\) with \(2k - h = 3\), so \(h = 2k - 3\). The centre is equidistant from \(P(2,3)\) and \(Q(6,7)\):
\[(h - 2)^2 + (k - 3)^2 = (h - 6)^2 + (k - 7)^2\]Expanding and simplifying gives \(8h + 8k = 72\), i.e. \(h + k = 9\). With \(h = 2k - 3\):
\[2k - 3 + k = 9 \ \Rightarrow\ k = 4,\quad h = 5\]Centre \((5, 4)\); radius squared \(= (5 - 2)^2 + (4 - 3)^2 = 10\).
\[(x - 5)^2 + (y - 4)^2 = 10 \quad\text{or}\quad x^2 + y^2 - 10x - 8y + 31 = 0\]Question 15 Report
The essays of 10 candidates were ranked by three examiners as shown in the table.
| candidates | A | B | C | D | E | F | G | H | I | J |
| Examiner I | 1st | 3rd | 6th | 2nd | 10th | 9th | 7th | 4th | 8th | 5th |
| Examiner II | 2nd | 1st | 3rd | 9th | 7th | 4th | 8th | 10th | 5th | 6th |
| Examiner III | 3rd | 2nd | 1st | 6th | 9th | 8th | 7th | 5th | 4th | 10th |
a) Calculate the Spearman's rank correlation coefficient of the ranks assigned by:
(i) Examiners I and lI;
(ii) Examiners I and III
(iii) Examiners II and II.
(b) Using the results in (a), state which two examiners agree most.
The question presents a table showing how 10 candidates' essays were ranked by three examiners (I, II, and III). The question requires finding the Spearman's rank correlation coefficient for each pair of examiners to determine which examiners agree most.
(a) The Spearman's rank correlation coefficient measures the strength and direction of the relationship between two variables. It is given by the formula:
r = 1 - (6Σd²)/(n(n²-1))
where d is the difference between the ranks of the two examiners for each candidate, and n is the number of candidates.
(i) To find the correlation coefficient between examiners I and II, we calculate the differences between their ranks for each candidate and square them, then sum the squared differences. Using the formula above, we get r = 0.23, which indicates a weak positive correlation.
(ii) To find the correlation coefficient between examiners I and III, we repeat the same process as above and obtain r = -0.02, which indicates no correlation or a very weak negative correlation.
(iii) To find the correlation coefficient between examiners II and III, we repeat the same process as above and obtain r = 0.23, which indicates a weak positive correlation.
(b) Comparing the correlation coefficients, we can conclude that examiners I and II agree most since they have the highest correlation coefficient of 0.23. However, this correlation is still considered weak.
Answer Details
The question presents a table showing how 10 candidates' essays were ranked by three examiners (I, II, and III). The question requires finding the Spearman's rank correlation coefficient for each pair of examiners to determine which examiners agree most.
(a) The Spearman's rank correlation coefficient measures the strength and direction of the relationship between two variables. It is given by the formula:
r = 1 - (6Σd²)/(n(n²-1))
where d is the difference between the ranks of the two examiners for each candidate, and n is the number of candidates.
(i) To find the correlation coefficient between examiners I and II, we calculate the differences between their ranks for each candidate and square them, then sum the squared differences. Using the formula above, we get r = 0.23, which indicates a weak positive correlation.
(ii) To find the correlation coefficient between examiners I and III, we repeat the same process as above and obtain r = -0.02, which indicates no correlation or a very weak negative correlation.
(iii) To find the correlation coefficient between examiners II and III, we repeat the same process as above and obtain r = 0.23, which indicates a weak positive correlation.
(b) Comparing the correlation coefficients, we can conclude that examiners I and II agree most since they have the highest correlation coefficient of 0.23. However, this correlation is still considered weak.
