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Vraag 1 Verslag
Find the value of x for which 2(32x - 1) = 162
Antwoorddetails
2(32x - 1) = 162, 32x - 1 = 81
= 34
2x - 1 = 4 , 2x = 5
x = 52
Vraag 2 Verslag
In a basket, there are 6 grapes, 11 bananas and 13 oranges. If one fruit is chosen at random. What is the probability that the fruit is either a grape or a banana
Antwoorddetails
To find the probability that a fruit chosen at random is either a grape or a banana, we need to first determine the total number of fruits in the basket, which is the sum of grapes and bananas, since we are only interested in these two types of fruits. Total number of grapes and bananas = 6 + 11 = 17 Therefore, the probability of choosing either a grape or a banana is: P(grape or banana) = (number of grapes + number of bananas) / total number of fruits = 17 / (6 + 11 + 13) = 17 / 30 Hence, the probability that the fruit chosen is either a grape or a banana is 17/30. Therefore, the answer is not any of the options provided.
Vraag 3 Verslag
Find the locus of point equidistant from two straight lines y - 5 = 0 and y - 3 = 0
Antwoorddetails
Locus of point P equidistant from y - 5 = 0 and y - 3 = 0 is y = 4 i.e y - 4 = 0
Vraag 4 Verslag
Find the value of x for which the function f(x) = 2x3 - x2 - 4x + 4 has a maximum value
Antwoorddetails
f(x) = 2x3 - x2 - 4x – 4
f’(x) = 6x2 - 2x – 4
As f’(x) = 0
Implies 6x2 - 2x – 4 = 0
3x – x – 2 = 0 (By dividing by 2)
(3x – 2)(x + 1) = 0
3x – 2 = 0 implies x = -2/3
Or x + 1 = 0 implies x = -1
f’(x) = 6x2 - 2x – 4
f’’(x) = 12x – 2
At max point f’’(x) < 0
∴f’’(x) = 12x – 2 at x = -1
= 12(-1) – 2
= -12 – 2 = -14
∴Max at x = 1
Vraag 5 Verslag
\(W \alpha L^2\) and \(W = 6\) when \(L = 4\). If \(L = \sqrt{17}\), find \(W\).
Antwoorddetails
W α
L2, WL2
= k(constant)
642
= k
W(√17)2
W = 17×616
= 518
= 638
Vraag 6 Verslag
A binary operation ⊕ on real numbers is defined by x⊕y = xy + x + y for any two real numbers x and y. The value of (-3/4)⊕6 is
Antwoorddetails
The given binary operation ⊕ on real numbers is defined as x⊕y = xy + x + y for any two real numbers x and y. Substituting the values x = -3/4 and y = 6 in the given expression, we get: (-3/4)⊕6 = (-3/4)×(6) + (-3/4) + (6) = (-9/2) + (21/4) = (-18/4) + (21/4) = 3/4 Therefore, the value of (-3/4)⊕6 is 3/4. Hence, option (A) is the correct answer.
Vraag 7 Verslag
Find the sum to infinity of the series \(2 + \frac{3}{2} + \frac{9}{8} + \frac{27}{32} + \ldots\)
Vraag 8 Verslag
Find the value of x for which the function f(x) = 2x3 - x2 - 4x + 4 has a maximum value
Antwoorddetails
f(x) = 2x3 - x2 - 4x + 4
f(x) = 6x2 - 2x - 4 at turning point, f1(x) = 0
6x2 - 2x - 4 = 0, 3x2 - x - 2 = 0, 3x2 - 3x + 2x - 2 = 0
(3x + 2)(x - 1) = 0, x = -23
or 1
f11(x) = 12x - 2,
when x = 23
, f11(x) = 12(-23
) - 2 = -10 < 0
→
f(x) is maximum @ x = -23
when x = 1, f11(x) = 12(1)- 2 = 10 > 0
→
f(x) is maximum @ x = 1
Vraag 9 Verslag
Integrate \( \frac{x^2-\sqrt{x}}{x} \) with respect to x
Antwoorddetails
Integrate x2−√xx
with respect to x
∫x2−√xx
dx = ∫(x2x−√xx
)dx
= ∫(x - x12
)dx
= x22
- x1212
+ k
= x22
- 2√x
+ k
Vraag 10 Verslag
If the lines 2y - kx + 2 = 0 and y + x - k/2 = 0 Intersect at (1, -2), find the value of k
Antwoorddetails
The problem gives two equations of two lines and a point of intersection between them. We need to find the value of "k" in one of the equations. The point of intersection (1, -2) lies on both lines, so it must satisfy both equations. Substituting x=1 and y=-2 in the first equation 2y - kx + 2 = 0 gives: 2(-2) - k(1) + 2 = 0 Simplifying this equation: -4 - k + 2 = 0 -2 - k = 0 k = -2 Therefore, the value of k is -2. Option (C) is the correct answer.
Vraag 11 Verslag
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | \(y - 1\) | \(y\) | 9 | 4 | 1 |
The table above gives the frequency distribution of marks obtained by a group of students in a test. If the total mark scored is 200, the value of y is
Antwoorddetails
(3 x 5) + (4(y - 1)) + (5y) + (6 x 9) + (7 x 4) + 8 = 200
15 + 4y - 4 + 5y + 54 + 28 + 8 = 200
9y + 105 - 4 = 200
9y + 101 = 200
9y = 200 - 101
9y = 99
y = 11
Vraag 12 Verslag
The solution of the quadratic inequality (x3 + x - 12) ≥ 0 is
Antwoorddetails
(x3 + x - 12) ≥ 0
(x + 4)(x - 3) ≥ 0
Either x + 4 ≥ 0 implies x ≥ -4
Or x - 3 ≥ 0 implies x ≥ 3
∴ x ≥ 3 or x ≥ -4
Vraag 13 Verslag
If X = {all the perfect squares less than 40}
Y = {all the odd numbers fro, 1 to 15}. Find X ∩ Y.
Antwoorddetails
All the perfect squares < 40
X = {1, 4, 9, 16, 25, 36}
All the odd numbers from 1 to 15
Y = {1, 3, 5, 7, 9, 11, 13, 15}
X ∩ Y = {1, 9}
Vraag 14 Verslag
A particle p moves between points S and T such that angle SPT is always constant. Find the locus of P.
Antwoorddetails
The locus of P is a semi-circle with ST as diameter.
Vraag 15 Verslag
Find y, if \( \sqrt{12} - \sqrt{147} + y\sqrt{3} = 0 \)
Antwoorddetails
√12−√147
+ y√3
= 0
√4×3−√49×3
+ y√3
= 0
y - 5 = 0
y = 5
Vraag 16 Verslag
If log102 = x, express log1012.5 in terms of x
Antwoorddetails
We know that log base 10 of 2 is x, which means that 10 to the power of x is equal to 2. To express log base 10 of 12.5 in terms of x, we need to find a way to write 12.5 in terms of 2 and x. We can write 12.5 as 10 to the power of 1.09691 (approximately) using a calculator. Now, we can use the laws of logarithms to simplify the expression: log base 10 of 12.5 = log base 10 of (10^1.09691) = 1.09691 * log base 10 of 10 = 1.09691 Therefore, we want to find an expression among the given options that equals 1.09691 when x is substituted into it. We can check each option by substituting x into it and simplifying: : 2(1 + x) = 2 + 2x Substituting x = log base 10 of 2 gives 2 + 2(log base 10 of 2), which does not equal 1.09691. : 2 + 3x Substituting x = log base 10 of 2 gives 2 + 3(log base 10 of 2), which also does not equal 1.09691. : 2(1 - x) = 2 - 2x Substituting x = log base 10 of 2 gives 2 - 2(log base 10 of 2), which also does not equal 1.09691. : 2 - 3x Substituting x = log base 10 of 2 gives 2 - 3(log base 10 of 2), which equals 1.09691. Therefore, the correct answer is: 2 - 3x.
Vraag 17 Verslag
The nth term of the sequence \( \frac{3}{2} \), 3, 7, 16, 35, 74, ..., is
Antwoorddetails
In the given sequence, each term is obtained by adding 1 to twice the previous term, starting from 32/3, which is the same as (96/3) + (32/3) = 128/3. So, the second term of the sequence is 2(128/3) + 1 = 257/3. Similarly, the third term is 2(257/3) + 1 = 515/3. Continuing this pattern, we can find the nth term by performing the following operation: Tn = 2(Tn-1) + 1 Therefore, the nth term of the sequence is 2^n * (128/3) - (2^n + 1)/3. Thus, the correct option is 5.2n - 2 - n.
Vraag 18 Verslag
The volume of a hemispherical bowl is \(718\frac{2}{3}\,\text{cm}^3\). Find its radius.
Antwoorddetails
Volume of hemisphere = 23πr3
23πr3
= 71823
23
x 227r3
= 21563
r3 = 21563
x 32
x 722
r3 = 49 x 7
r = 3√343
= 7cm3
Vraag 19 Verslag
The sum of the ages of Musa and Lawal is 28 years. After sharing a certain sum of money in the ratio of their ages, Musa gets ₦600 and Lawal ₦800. How old is Lawal?
Antwoorddetails
Let the age of Musa be x, and the age of Lawal be y. According to the first statement, we know that: x + y = 28 ---- (Equation 1) Let the common ratio be k, so the ratio of Musa to Lawal's share is x:y. Therefore, we can write: x:y = k:k = 1:1 (since the ratio is not given) So, the total amount of money shared can be represented as: x + y = 2k ---- (Equation 2) From the second statement, we know that: x/y = 6/8 = 3/4 Multiplying both sides by 4y, we get: 4x = 3y Now we have two equations with two variables (Equations 1 and 2), which can be solved to find the values of x and y: x + y = 28 ---- (Equation 1) 2x = 3y ---- (Equation 3) Solving the equations 1 and 3 simultaneously, we get: x = 12 y = 16 Therefore, Lawal's age is 16 years.
Vraag 20 Verslag
A senatorial candidate had planned to visit seven cities prior to a primary election. However, he could only visit four of the cities. How many different itineraries could be considered?
Antwoorddetails
Number of itineraries = 7P4
=7!(7−4)!=7!3!=7×6×5×4×3!3!=840
Vraag 21 Verslag
Given
P = {1, 3, 5, 7, 9, 11}
And Q = {2, 4, 6, 8, 1, 12}. Determine the relationship between P and Q
Antwoorddetails
The given sets P and Q are two different sets containing some numbers. To determine the relationship between the sets, we need to compare the elements of the sets. By comparing the elements of P and Q, we can see that they have only one element in common, which is 1. The other elements in the sets are unique and do not appear in both sets. Therefore, we can say that P and Q have some elements that are different and some that are the same. Hence, the correct answer is P ∩ Q ≠ ∅.
Vraag 22 Verslag
If the lines 3y = 4x - 1 and qy = x + 3 are parallel to each other, the value of q is
Antwoorddetails
Two lines are parallel if their slopes are equal. Slope is defined as the ratio of the change in the y-coordinates to the change in the x-coordinates between two points on the line. The given lines are 3y = 4x - 1 and qy = x + 3. To find the slope of the first line, we can rearrange the equation in the slope-intercept form y = mx + b where m is the slope and b is the y-intercept. 3y = 4x - 1 y = (4/3)x - 1/3 So the slope of the first line is 4/3. To find the slope of the second line, we can rearrange the equation in the same way: qy = x + 3 y = (1/q)x + 3/q So the slope of the second line is 1/q. If the two lines are parallel, their slopes are equal. Therefore: 4/3 = 1/q We can solve for q by cross-multiplying: 4q = 3 q = 3/4 Therefore, the value of q is 3/4.
Vraag 23 Verslag
The area of a square is 144 sqcm. Find the length of the diagonal.
Antwoorddetails
To find the length of the diagonal of a square, we need to use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the two shorter sides (legs) is equal to the square of the longest side (hypotenuse). In this case, the two legs of the right triangle are the sides of the square, and the hypotenuse is the length of the diagonal. Let's call the length of one side of the square "x". Then, we know that the area of the square is given by: x^2 = 144 sqcm Taking the square root of both sides, we get: x = 12 cm Now, using the Pythagorean theorem, we can find the length of the diagonal: d^2 = x^2 + x^2 = 2x^2 d = sqrt(2x^2) = x * sqrt(2) = 12 * sqrt(2) cm Therefore, the length of the diagonal is 12√2 cm.
Vraag 24 Verslag
The table above shows the number of pupils in each age group in a class. What is the probability that a pupil chosen at random is at least 1 years old?
Antwoorddetails
P(At east 11 yrs) = P(11yrs) + P(12yrs)
= 27/40 + 7/40
= 34/40
= 17/20
Vraag 25 Verslag
The histogram above represents the weights of students who travelled out to their school for an examination. How many people made the trip.
Vraag 26 Verslag
Integrate \( \frac{x^2-\sqrt{x}}{x} \) with respect to x
Antwoorddetails
To integrate x² - √x/x with respect to x, we can start by factoring the expression as follows: x² - √x/x = x - 1/√x Then we can integrate each term separately: ∫(x - 1/√x) dx = ∫x dx - ∫(1/√x) dx The first integral is straightforward: ∫x dx = 1/2 x² + C1 For the second integral, we can use the substitution u = √x, du/dx = 1/(2√x), dx = 2√x du: ∫(1/√x) dx = ∫2 du = 2u + C2 = 2√x + C2 Substituting back u = √x, we get: ∫(1/√x) dx = 2√x + C2 Putting everything together, we have: ∫(x² - √x/x) dx = ∫x dx - ∫(1/√x) dx = (1/2 x² + C1) - (2√x + C2) = 1/2 x² - 2√x + C where C = C1 - C2 is the constant of integration. Therefore, the correct option is x²/2 - 2√x + K, where K = C is the constant of integration.
Vraag 27 Verslag
Factorize 2t2 + t - 15
Antwoorddetails
To factorize 2t² + t - 15, we need to find two binomials that multiply to give us 2t² + t - 15. To do this, we can use the factoring method called "AC method." First, we need to find two numbers whose product is 2(-15) = -30 and whose sum is 1. These numbers are 6 and -5. Next, we replace the middle term t with 6t - 5t: 2t² + 6t - 5t - 15 Then we group the terms: (2t² + 6t) - (5t + 15) We factor out the greatest common factor from each group: 2t(t + 3) - 5(t + 3) We notice that we have a common binomial factor of (t + 3), so we can factor it out: (t + 3)(2t - 5) Therefore, the factored form of 2t² + t - 15 is (t + 3)(2t - 5). So, the correct option is: (t + 3)(2t - 5).
Vraag 28 Verslag
Find the sum to infinity to the following series \(3 + 2 + \frac{4}{3} + \frac{8}{9} + \frac{16}{17} + .....\)
Antwoorddetails
To find the sum to infinity of this series, we need to determine if it is a converging or diverging series. We can do this by finding the common ratio between each term. The common ratio between the second and first term is 2/3. The common ratio between the third and second term is 4/3. The common ratio between the fourth and third term is 8/9, and so on. We can see that the common ratio is less than 1, so the series is converging. Therefore, we can use the formula for the sum of an infinite geometric series: S = a/(1 - r) where S is the sum, a is the first term, and r is the common ratio. In this case, the first term is 3 and the common ratio is 2/3. So, plugging these values into the formula, we get: S = 3/(1 - 2/3) = 3/(1/3) = 9 Therefore, the sum to infinity of this series is 9. So, the answer to the question is option (D) 9.
Vraag 29 Verslag
The graph above is represented by
Antwoorddetails
The roots of the graph are -2, -1 and 1
y = (x + 2)(x + 1)(x - 1) = (x + 2)(x2 - 1)
= x3 + 2x2 - x - 2
Vraag 30 Verslag
The nth term of the sequence 3/2, 3, 7, 16, 35, 74 ..... is
Antwoorddetails
The nth term of the sequence is 5 . 2n-2 - n. To understand why, we can look at how the sequence is generated. The first term is 3/2, the second term is 3, and each subsequent term is generated by doubling the previous term and subtracting its position in the sequence. For example, to get the third term, we double the second term (which is 3) to get 6, and then subtract the position of the term (which is 3) to get 3+3=6. Similarly, to get the fourth term, we double the third term (which is 6) to get 12, and then subtract the position of the term (which is 4) to get 12-4=8. Using this pattern, we can derive the general formula for the nth term: 5 . 2n-2 - n.
Vraag 31 Verslag
A man made a profit of 5% when he sold an article for N60,000.00. How much would he have sell the article to make a profit of 26%
Antwoorddetails
5% profit = 100 + 5 = 105%
26% profit = 100 + 26 = 126%
∴ 105% → N60,000
1% → 60000/15
126% = 1000/105 x 126/1
=N72,000
Vraag 32 Verslag
If \(y = (1 + x)^2\), find \(\frac{dy}{dx}\)
Antwoorddetails
To find dy/dx for y = (1 + x)^2, we can use the chain rule of differentiation. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x), where f'(g(x)) is the derivative of the outer function evaluated at g(x), and g'(x) is the derivative of the inner function evaluated at x. In this case, we can let f(x) = x^2 and g(x) = 1 + x, so that y = f(g(x)) = (1 + x)^2. Then, we can find the derivatives of f(x) and g(x) as follows: f'(x) = 2x g'(x) = 1 Using the chain rule, we can now find dy/dx: dy/dx = f'(g(x)) * g'(x) = 2(1 + x) * 1 = 2 + 2x Therefore, dy/dx = 2 + 2x.
Vraag 33 Verslag
Make L the subjects of the formula if \( \sqrt{\frac{42w}{5l}} \)
Antwoorddetails
√42w5l
square both side of the equation
d2=(√42W5l)2d2=42W5l5ld2=42Wl=42W5d2
Vraag 34 Verslag
Simplify \( \frac{3}{5} \div \left(\frac{2}{7} x \frac{4}{3} \div \frac{4}{9}\right) \)
Antwoorddetails
Vraag 35 Verslag
A binary operation \( \oplus \) on real numbers is defined by \( x \oplus y = xy + x + y \) for any two real numbers \( x \) and \( y \). The value of \( \left(-\frac{3}{4}\right) \oplus 6 \) is
Antwoorddetails
x ⊕
y = xy + x + y
(-34
) ⊕
6 = -34.6
- 34
+ 6
= -92
- 34
+ 6
= -34
Vraag 36 Verslag
x10 = 12145 find x.
Antwoorddetails
x10 = 1214 5, 1 x 53 + 2 x 52 + 1 x 51 + 4 x 5o
= 125 + 50 + 5 + 4 18410
x = 184
Vraag 37 Verslag
If the lines \(2y - kx + 2 = 0\) and \(y + x - \frac{k}{2} = 0\) intersect at (1, 2), find the value of k
Antwoorddetails
2y - kx + 2 = 0, @ x = 1, y = -2
2(-2) - k(1) + 2 = 0
-4 - k + 2 = 0
k = -2
Or y + x - k2
= 0, @ x= 1, y = -2
-2 + 1 - k2
= 0
-1 = k2
k = -1 x 2
= -2
Vraag 38 Verslag
If x10 = 12145 find x
Antwoorddetails
x10 = 12145
= x10 = 1 x 53 + 2 x 52 + 1 * 51 + 4 x 50
= 1 x 125 + 2 x 25 + 1 x 5 + 4 x 1
= 125 + 50 + 5 + 4
= 184
Vraag 39 Verslag
If 5, 8, 6 and 2 occur with frequencies 3, 2, 4 and 1 respectively, find the product of the modal and medial number.
Antwoorddetails
To find the modal and medial numbers from the given frequency distribution, we need to first determine the mode and median of the dataset. The mode is the number that occurs most frequently, and the median is the middle number when the data is arranged in order. In this case, the mode is 6, which occurs with a frequency of 4. The median can be found by arranging the numbers in order: 2, 5, 5, 5, 6, 6, 6, 6, 8, 8 The median is the middle number, which is also 6. Therefore, the modal number is 6, and the medial number is also 6. The product of the modal and medial numbers is: 6 × 6 = 36 Therefore, the answer is 36.
Vraag 40 Verslag
In the diagram P, Q, R, S are points on the circle RQS = 30o. PRS = 50o and PSQ = 20o. What is the value of xo + yo?
Antwoorddetails
Draw a line from P to Q
< PQS = < PRS (angle in the sam segment)
< PQS = 50o
Also, < QSR = < QPR(angles in the segment)
< QPR = xo
x + y + 5= = 180(angles in a triangle)
x + y = 180 - 50
x + y = 130o
Vraag 41 Verslag
Determine the value of \( \displaystyle \int_{0}^{\frac{\pi}{2}}(-2\cos x)\,dx \)
Antwoorddetails
Vraag 42 Verslag
The solution of the quadratic inequality \( (x^2 + x - 12) \ge 0 \) is
Antwoorddetails
(x2 + x - 12) ≥
0 , (x - 3)(x + 4) ≥
0
For the condition to hold, each of (x - 3) and (x + 4) must be of the same sign
.i.e. x - 3 ≥
0 and x + 4 ≥
0
or x - 3≤
0 and x + 4 ≤
0
when x ≥
3, the condition is satisfied
when x ≥
-4, the condition is not satisfied.
when x ≤
3, the condition is not satisfied
when x ≤
-4 , the condition is not satisfied. Thus, the solution of the inequality is x ≥
3 or x ≤
-4 ,
Vraag 43 Verslag
Find the value of x for which 2(32x-1) = 162
Antwoorddetails
Vraag 44 Verslag
The volume of a hemispherical bowl is \(718\frac{2}{3}\). Find its radius .
Antwoorddetails
Vraag 45 Verslag
The graph above is represented by
Antwoorddetails
x = -2, x = -1 and x = 1
then the factors; x+2, x+1 and x-1
Product of the factors; (x+2)(x+1)(x-1)
= y = (x + 2)(x2 - x + x - 1)
= y = (x+2)(x2-1)
x3 - x + 2x2 - 2 = y
x3 + 2x2 - x - 2 = y
Vraag 46 Verslag
If y = x cosx, find \( \frac{dy}{dx} \).
Antwoorddetails
If y = x cosx, find dydx
.
Using product rule dydx
= cosx - x sinx
Vraag 47 Verslag
If y = x cos x, find dy/dx
Antwoorddetails
To find dy/dx of y = x cos x, we can use the product rule of differentiation, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. In this case, we have: y = x cos x Using the product rule, we get: dy/dx = cos x - x sin x Therefore, the correct option is: cos x - x sin x. To explain it in simple terms, the derivative of x cos x is equal to cos x minus x times the derivative of cos x, which is -sin x. This gives us cos x - x sin x as the answer.
Vraag 48 Verslag
Simplify \( \frac{3}{5} \div \left(\frac{2}{7} \times \frac{4}{3} \div \frac{4}{9}\right) \)
Antwoorddetails
Vraag 49 Verslag
Each of the interior angles of a regular polygon is 140o. How many sides has the polygon?
Antwoorddetails
The sum of the interior angles of a polygon can be found using the formula: S = (n - 2) × 180o where S is the sum of the interior angles, and n is the number of sides in the polygon. For a regular polygon, all interior angles have the same measure. In this case, the interior angle of the polygon is given as 140o. Therefore, we can use the formula: S = n × 140o Substituting this into the formula for the sum of interior angles, we get: n × 140o = (n - 2) × 180o Simplifying this equation, we get: 140n = 180n - 360 Solving for n, we get: 40n = 360 n = 9 Therefore, the polygon has 9 sides. The answer is 9.
Vraag 50 Verslag
Antwoorddetails
To solve this problem, we need to use the formula for the mean of a frequency distribution: mean = (sum of (value × frequency))/total frequency From the table, we can see that the total frequency is 25, and the sum of (value × frequency) is: 5 × 4 + 6 × 6 + 7 × 7 + 8 × 5 + 9 × 3 = 20 + 36 + 49 + 40 + 27 = 172 So, the mean mark is: mean = 172/25 = 6.88 We also know that the total mark scored is 200, so we can set up an equation: total mark = mean × total frequency + y Substituting in the values we know, we get: 200 = 6.88 × 25 + y Solving for y, we get: y = 200 - 6.88 × 25 = 11 Therefore, the value of y is 11. Answer: 11.
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