WAEC SSCE - Further Mathematics - 2008
Ajụjụ 1 Ripọtì
Which of the following quadratic curves will not intersect with the x- axis?
Akọwa Nkọwa
A quadratic curve is a polynomial of degree 2, which means that it can be written in the form of \(y = ax^{2} + bx + c\), where a, b, and c are constants. For a quadratic curve to intersect with the x-axis, it must have at least one x-intercept. An x-intercept is a point on the curve where the value of y is equal to zero. In other words, it is where the curve crosses the x-axis. To find the x-intercepts of a quadratic curve, we need to solve the equation \(y = ax^{2} + bx + c\) for x when y = 0. This gives us the quadratic formula, which is: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ If the discriminant (\(b^{2} - 4ac\)) is greater than or equal to zero, then the quadratic curve will intersect with the x-axis. If the discriminant is less than zero, then the quadratic curve will not intersect with the x-axis. Using this information, we can determine which of the given quadratic curves will not intersect with the x-axis: - \(y = 2 - 4x - x^{2}\) The discriminant for this quadratic is \(16 - 4(1)(2) = 8\), which is greater than zero. Therefore, this quadratic curve will intersect with the x-axis. - \(y = x^{2} - 5x -1\) The discriminant for this quadratic is \(25 + 4(1)(1) = 29\), which is greater than zero. Therefore, this quadratic curve will intersect with the x-axis. - \(y = 2x^{2} - x - 1\) The discriminant for this quadratic is \((-1)^{2} - 4(2)(-1) = 9\), which is greater than zero. Therefore, this quadratic curve will intersect with the x-axis. - \(y = 3x^{2} - 2x + 4\) The discriminant for this quadratic is \((-2)^{2} - 4(3)(4) = -44\), which is less than zero. Therefore, this quadratic curve will not intersect with the x-axis. Therefore, the quadratic curve that will not intersect with the x-axis is the one described by the equation \(y = 3x^{2} - 2x + 4\).
Ajụjụ 2 Ripọtì
The distance between P(x, 7) and Q(6, 19) is 13 units. Find the values of x.
Akọwa Nkọwa
To find the value of x, we can use the distance formula which states that the distance between two points P(x1, y1) and Q(x2, y2) is given by: d = sqrt((x2 - x1)^2 + (y2 - y1)^2) In this case, we are given the distance between P(x, 7) and Q(6, 19) as 13 units. Therefore, we can write: 13 = sqrt((6 - x)^2 + (19 - 7)^2) Simplifying this equation, we get: 169 = (6 - x)^2 + 144 25 = (6 - x)^2 Taking the square root of both sides, we get: 5 = 6 - x or 5 = x - 6 Solving for x in each case, we get: x = 1 or x = 11 Therefore, the possible values of x are 1 or 11. Note that we can also check that the distance between P(1, 7) and Q(6, 19) and between P(11, 7) and Q(6, 19) is indeed 13 units.
Ajụjụ 3 Ripọtì
Find the locus of points which is equidistant from P(4, 5) and Q(-6, -1).
Akọwa Nkọwa
To find the locus of points equidistant from two fixed points, we need to find the perpendicular bisector of the line segment joining the two points. This perpendicular bisector will be the locus of all points that are equidistant from the two given points. So, let's first find the midpoint of the line segment PQ joining P(4, 5) and Q(-6, -1). The midpoint M is given by: \begin{align*} M &= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\\ &= \left(\frac{4 + (-6)}{2}, \frac{5 + (-1)}{2}\right)\\ &= (-1, 2) \end{align*} Now, let's find the slope of the line segment PQ: \begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1}\\ &= \frac{-1 - 5}{-6 - 4}\\ &= \frac{-6}{-10}\\ &= \frac{3}{5} \end{align*} Since the perpendicular bisector of PQ will be perpendicular to PQ, its slope will be the negative reciprocal of the slope of PQ: \begin{align*} m_\perp &= -\frac{1}{m}\\ &= -\frac{1}{\frac{3}{5}}\\ &= -\frac{5}{3} \end{align*} Now we have the slope of the perpendicular bisector and a point on it, which is the midpoint M. We can use point-slope form to find the equation of the perpendicular bisector: \begin{align*} y - y_1 &= m_\perp(x - x_1)\\ y - 2 &= -\frac{5}{3}(x + 1) \end{align*} Simplifying this equation, we get: \begin{align*} 3y - 6 &= -5x - 5\\ 5x + 3y - 1 &= 0 \end{align*} Therefore, the locus of points equidistant from P(4, 5) and Q(-6, -1) is the line 5x + 3y - 1 = 0. Hence, the correct option is (d) 5x + 3y - 1 = 0.
Ajụjụ 4 Ripọtì
A rectangle has a perimeter of 24m. If its area is to be maximum, find its dimension.
Akọwa Nkọwa
To find the dimensions of a rectangle that has the maximum area with a given perimeter, we can start by expressing the perimeter in terms of the rectangle's dimensions. Let's call the length of the rectangle l and its width w. Then the perimeter P is given by: P = 2l + 2w We know that P = 24m, so we can substitute this into the above equation to get: 24 = 2l + 2w Simplifying this equation, we get: 12 = l + w Now we need to express the area of the rectangle in terms of l and w. The area A is given by: A = lw We want to find the values of l and w that will maximize A, subject to the constraint that 12 = l + w. One way to do this is to express l in terms of w using the equation 12 = l + w, and substitute this expression into the equation for A. We get: l = 12 - w A = lw = w(12 - w) = 12w - w^2 Now we have an expression for A in terms of just one variable, w. To find the maximum value of A, we can take the derivative of A with respect to w, set it equal to zero, and solve for w. We get: dA/dw = 12 - 2w = 0 w = 6 So the width of the rectangle that maximizes the area is 6m. To find the length, we can use the equation 12 = l + w, which gives us: l = 12 - w = 12 - 6 = 6 So the dimensions of the rectangle that has the maximum area with a perimeter of 24m are 6m by 6m. Therefore, the correct answer is 6, 6.
Ajụjụ 5 Ripọtì
| Marks | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| No of students | 5 | 7 | 9 | 6 | 3 | 6 | 4 |
The table above shows the distribution of marks by some candidates in a test. If a student is selected at random, what is the probability that she scored at least 6 marks?
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
The probability of Jide, Atu and Obu solving a given problem are \(\frac{1}{12}\), \(\frac{1}{6}\) and \(\frac{1}{8}\) respectively. Calculate the probability that only one solves the problem.
Akọwa Nkọwa
To calculate the probability that only one person solves the problem, we can consider the following cases: Case 1: Jide solves the problem, and Atu and Obu don't solve the problem. The probability of Jide solving the problem is \(\frac{1}{12}\), the probability of Atu not solving the problem is \(\frac{5}{6}\) (since the probability of Atu solving the problem is \(\frac{1}{6}\)), and the probability of Obu not solving the problem is \(\frac{7}{8}\) (since the probability of Obu solving the problem is \(\frac{1}{8}\)). Therefore, the probability of this case is: \[\frac{1}{12} \times \frac{5}{6} \times \frac{7}{8} = \frac{35}{576}\] Case 2: Atu solves the problem, and Jide and Obu don't solve the problem. The probability of Atu solving the problem is \(\frac{1}{6}\), the probability of Jide not solving the problem is \(\frac{11}{12}\) (since the probability of Jide solving the problem is \(\frac{1}{12}\)), and the probability of Obu not solving the problem is \(\frac{7}{8}\). Therefore, the probability of this case is: \[\frac{1}{6} \times \frac{11}{12} \times \frac{7}{8} = \frac{77}{576}\] Case 3: Obu solves the problem, and Jide and Atu don't solve the problem. The probability of Obu solving the problem is \(\frac{1}{8}\), the probability of Jide not solving the problem is \(\frac{11}{12}\), and the probability of Atu not solving the problem is \(\frac{5}{6}\). Therefore, the probability of this case is: \[\frac{1}{8} \times \frac{11}{12} \times \frac{5}{6} = \frac{55}{576}\] So the total probability of only one person solving the problem is the sum of the probabilities of these three cases: \[\frac{35}{576} + \frac{77}{576} + \frac{55}{576} = \frac{167}{576}\] Therefore, the correct option is (d) \(\frac{167}{576}\).
Ajụjụ 7 Ripọtì
| Marks | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| No of students | 5 | 7 | 9 | 6 | 3 | 6 | 4 |
The table above shows the distribution of marks by some candidates in a test. What is the median score?
Akọwa Nkọwa
Ajụjụ 8 Ripọtì
Find the value of the constant k for which \(a = 4 i - k j\) and \(b = 3 i + 8 j\) are perpendicular.
Akọwa Nkọwa
For two vectors to be perpendicular, their dot product must be zero. So, we can find the value of \(k\) by taking the dot product of the given vectors and setting it equal to zero. The dot product of two vectors, \(\vec{a} \cdot \vec{b}\), is given by: \(\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z\) where \(\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}\) and \(\vec{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k}\) are the components of the vectors along the x, y, and z axes. So, using the given vectors, we have: \(\vec{a} \cdot \vec{b} = (4\hat{i} - k\hat{j}) \cdot (3\hat{i} + 8\hat{j}) = 4 \times 3 + (-k) \times 8 = 12 - 8k\) For the vectors to be perpendicular, this dot product must be zero, so: \(12 - 8k = 0\) Solving for \(k\), we get: \(k = \frac{12}{8} = \frac{3}{2}\) Therefore, the answer is option (D) \(\frac{3}{2}\). So, the constant \(k\) must be equal to \(\frac{3}{2}\) for the vectors \(a\) and \(b\) to be perpendicular.
Ajụjụ 9 Ripọtì
The initial and final velocities of an object of mass 5 kg are \(u = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\) and \(v = \begin{pmatrix} 4 \\ 7 \end{pmatrix}\) respectively. Find the magnitude of its change in momentum.
Ajụjụ 10 Ripọtì
A binary operation ,*, is defined on the set R, of real numbers by \(a * b = a^{2} + b + ab\). Find the value of x for which \(5 * x = 37\).
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
The coefficient of the 5th term in the binomial expansion of \((1 + kx)^{8}\), in ascending powers of x is \(\frac{35}{8}\). Find the value of the constant k.
Akọwa Nkọwa
In the binomial expansion of \((1+kx)^8\), the coefficient of the 5th term in ascending powers of x is given by the expression: \[\binom{8}{4}(kx)^4(1)^{8-4}\] which simplifies to: \[\binom{8}{4}k^4x^4\] Using the formula for binomial coefficients, we can write: \[\binom{8}{4} = \frac{8!}{4!4!} = 70\] So, we can write the expression for the 5th term coefficient as: \[70k^4x^4\] We are given that this coefficient is \(\frac{35}{8}\), so we can write: \[70k^4x^4 = \frac{35}{8}\] Simplifying this equation, we get: \[k^4x^4 = \frac{1}{16}\] Dividing both sides by \(x^4\), we get: \[k^4 = \frac{1}{16}\] Taking the fourth root of both sides, we get: \[k = \pm \frac{1}{2}\] Since the expansion is of the form \((1+kx)^8\), the sign of k does not affect the coefficient of the 5th term, so we can take \(k = \frac{1}{2}\). Therefore, the value of the constant k is \(\frac{1}{2}\).
Ajụjụ 12 Ripọtì
Given that \(\overrightarrow{AB} = 5i + 3j\) and \(\overrightarrow{AC} = 2i + 5j\), find \(\overrightarrow{BC}\).
Akọwa Nkọwa
To find \(\overrightarrow{BC}\), we need to subtract \(\overrightarrow{AB}\) from \(\overrightarrow{AC}\), as both of these vectors share the point \(A\). So, \[\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}\] Substituting the given values, we get: \begin{align*} \overrightarrow{BC} &= \overrightarrow{AC} - \overrightarrow{AB} \\ &= (2i + 5j) - (5i + 3j) \\ &= 2i + 5j - 5i - 3j \\ &= \boxed{-3i + 2j} \end{align*} Therefore, the answer is option (B) -3i + 2j.
Ajụjụ 13 Ripọtì
The polynomial \(2x^{3} + x^{2} - 3x + p\) has a remainder of 20 when divided by (x - 2). Find the value of constant p.
Akọwa Nkọwa
The problem tells us that when we divide the polynomial \(2x^{3} + x^{2} - 3x + p\) by \((x - 2)\), the remainder is 20. We can use polynomial long division to find the quotient and remainder. \begin{array}{c|ccccc} \multicolumn{2}{r}{2x^2} & 5x & 7 \\ \cline{2-6} x-2 & 2x^3 & x^2 & -3x & +p & \\ \multicolumn{2}{r}{-2x^3} & +4x^2 & & &\\ \cline{2-3} \multicolumn{2}{r}{0} & 5x^2 & -3x & &\\ \multicolumn{2}{r}{} & -5x^2 & +10x & &\\ \cline{3-4} \multicolumn{2}{r}{} & 0 & 7x & +p & \\ \multicolumn{2}{r}{} & & -7x & +14 & \\ \cline{4-5} \multicolumn{2}{r}{} & & 0 & p+14 & \\ \end{array} The quotient is \(2x^{2} + 5x + 7\), and the remainder is \(p + 14\). We know that the remainder is equal to 20, so we can set up the equation: \[p + 14 = 20.\] Solving for \(p\), we get: \[p = 20 - 14 = 6.\] Therefore, the value of the constant \(p\) is 6, which is option (B).
Ajụjụ 14 Ripọtì
Express \(\frac{1}{1 - \sin 45°}\) in surd form.
Akọwa Nkọwa
We can start by using the identity \(\sin 45° = \frac{\sqrt{2}}{2}\): \begin{align*} \frac{1}{1 - \sin 45°} &= \frac{1}{1 - \frac{\sqrt{2}}{2}} \\ &= \frac{1}{\frac{2 - \sqrt{2}}{2}} \\ &= \frac{2}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} \quad \text{(rationalizing the denominator)} \\ &= \frac{2(2 + \sqrt{2})}{2^2 - (\sqrt{2})^2} \\ &= \frac{2(2 + \sqrt{2})}{2} \\ &= \boxed{2 + \sqrt{2}} \end{align*} Therefore, the answer is option A: \(2 + \sqrt{2}\).
Ajụjụ 15 Ripọtì
Express \(\frac{7\pi}{6}\) radians in degrees.
Akọwa Nkọwa
To convert radians to degrees, we can use the conversion formula: $$\text{degrees} = \frac{\text{radians}}{\pi} \times 180^\circ$$ Substituting the value of \(\frac{7\pi}{6}\) radians, we get: $$\text{degrees} = \frac{7\pi}{6} \div \pi \times 180^\circ$$ Simplifying the expression, we get: $$\text{degrees} = \frac{7}{6} \times 180^\circ$$ $$\text{degrees} = 210^\circ$$ Therefore, \(\frac{7\pi}{6}\) radians is equal to 210 degrees.
Ajụjụ 16 Ripọtì
If \(\begin{vmatrix} 4 & x \\ 5 & 3 \end{vmatrix} = 32\), find the value of x.
Akọwa Nkọwa
The determinant of a 2x2 matrix [a, b; c, d] is given by \(ad-bc\). Here we are given the determinant of the matrix \(\begin{vmatrix} 4 & x \\ 5 & 3 \end{vmatrix}\) to be 32. So, we have \((4 \times 3) - (x \times 5) = 32\). Simplifying this equation gives us \(12 - 5x = 32\). Solving for x, we get \(x = -4\). Therefore, the value of x is -4.
Ajụjụ 17 Ripọtì
Find the derivative of \(3x^{2} + \frac{1}{x^{2}}\)
Akọwa Nkọwa
To find the derivative of \(3x^{2} + \frac{1}{x^{2}}\), we will use the power rule and the chain rule of differentiation. First, using the power rule, we can differentiate the term \(3x^{2}\) as follows: \[\frac{d}{dx}(3x^{2}) = 6x.\] Next, using the chain rule, we can differentiate the term \(\frac{1}{x^{2}}\) as follows: \[\frac{d}{dx}\left(\frac{1}{x^{2}}\right) = -\frac{1}{x^{3}} \cdot \frac{d}{dx}(x^{-2}) = -\frac{1}{x^{3}} \cdot (-2x^{-3}) = \frac{2}{x^{5}}.\] Putting these together, we get: \[\frac{d}{dx}(3x^{2} + \frac{1}{x^{2}}) = 6x + \frac{2}{x^{5}}.\] Therefore, the answer is option (C) \(6x - \frac{2}{x^{3}}\).
Ajụjụ 18 Ripọtì
Express \(r = (12, 210°)\) in the form \(a i + b j\).
Akọwa Nkọwa
We can express a vector in terms of its rectangular or Cartesian coordinates by using the formulas: \[x = r\cos\theta\] \[y = r\sin\theta\] where r is the magnitude of the vector and θ is the angle it makes with the positive x-axis. Using these formulas, we have: \[x = r\cos\theta = 12\cos 210° = 12 \times \frac{-\sqrt{3}}{2} = -6\sqrt{3}\] \[y = r\sin\theta = 12\sin 210° = 12 \times \frac{-1}{2} = -6\] Therefore, we can express vector r as: \[r = -6\sqrt{3}i - 6j = 6(-\sqrt{3}i - j)\] Hence, the correct option is: \(6(-\sqrt{3}i - j)\).
Ajụjụ 19 Ripọtì
If \(2\log_{4} 2 = x + 1\), find the value of x.
Ajụjụ 20 Ripọtì
Two statements are represented by p and q as follows:
p : He is brilliant; q : He is regular in class
Which of the following symbols represent "He is regular in class but dull"?
Akọwa Nkọwa
Ajụjụ 21 Ripọtì
If n items are arranged two at a time, the number obtained is 20. Find the value of n.
Akọwa Nkọwa
The number of ways to arrange n items taken two at a time is given by the formula: \(_nP_2 = \frac{n!}{(n-2)!} = n(n-1)\) We are given that \(_nP_2 = 20\), so we can set up the equation: \begin{aligned} n(n-1) &= 20 \\ n^2 - n - 20 &= 0 \end{aligned} We can solve for n using the quadratic formula: \begin{aligned} n &= \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-20)}}{2(1)} \\ &= \frac{1 \pm \sqrt{81}}{2} \\ &= \frac{1 \pm 9}{2} \end{aligned} We discard the negative solution, and the positive solution is: \begin{aligned} n &= \frac{1+9}{2} \\ &= 5 \end{aligned} Therefore, the value of n is 5.
Ajụjụ 22 Ripọtì
If \(f(x) = 2x^{2} - 3x - 1\), find the value of x for which f(x) is minimum.
Akọwa Nkọwa
To find the value of x for which f(x) is minimum, we need to find the critical points of the function f(x) and determine whether they correspond to a minimum or maximum value. The critical points are the values of x where the derivative of f(x) is equal to zero or does not exist. Taking the derivative of f(x) with respect to x, we get: $$f'(x) = 4x - 3$$ Setting f'(x) equal to zero and solving for x, we get: $$4x - 3 = 0$$ $$x = \frac{3}{4}$$ Therefore, x = 3/4 is a critical point of f(x). To determine whether x = 3/4 corresponds to a minimum or maximum value of f(x), we can use the second derivative test. The second derivative of f(x) is: $$f''(x) = 4$$ Since f''(x) is positive for all values of x, we can conclude that x = 3/4 corresponds to a minimum value of f(x). Therefore, the value of x for which f(x) is minimum is x = 3/4.
Ajụjụ 24 Ripọtì
| Marks | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| No of students | 5 | 7 | 9 | 6 | 3 | 6 | 4 |
The table above shows the distribution of marks by some candidates in a test. Find, correct to one decimal place, the mean of the distribution.
Akọwa Nkọwa
Ajụjụ 25 Ripọtì
If \(y = x^{2} - 6x + 11\) is written in the form \(y = a(x - h)^{2} + k\), find the value of \((a + h + k)\).
Akọwa Nkọwa
We can start by completing the square to rewrite the quadratic equation in the form given. \begin{align*} y &= x^{2} - 6x + 11\\ &= (x^{2} - 6x + 9) + 2\\ &= (x - 3)^{2} + 2 \end{align*} Now we can see that the equation is in the desired form, with \(a = 1\), \(h = 3\), and \(k = 2\). Therefore, \begin{align*} (a + h + k) &= (1 + 3 + 2)\\ &= 6 \end{align*} So the value of \((a + h + k)\) is 6.
Ajụjụ 26 Ripọtì
If \(P = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}\) and \(Q = \begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}\), find PQ.
Akọwa Nkọwa
To find the product of matrices P and Q, we need to multiply the rows of the first matrix with the columns of the second matrix. Multiplying the first row of matrix P with the first column of matrix Q gives: \begin{pmatrix} 1 & -2 \end{pmatrix}\begin{pmatrix} -2 \\ 1 \end{pmatrix} = (-2\times 1) + (-2\times 1) = -4 Multiplying the first row of matrix P with the second column of matrix Q gives: \begin{pmatrix} 1 & -2 \end{pmatrix}\begin{pmatrix} 3 \\ 0 \end{pmatrix} = (1\times 3) + (-2\times 0) = 3 Similarly, multiplying the second row of matrix P with the first and second columns of matrix Q gives: \begin{pmatrix} 3 & 4 \end{pmatrix}\begin{pmatrix} -2 \\ 1 \end{pmatrix} = (-6) + 4 = -2 \begin{pmatrix} 3 & 4 \end{pmatrix}\begin{pmatrix} 3 \\ 0 \end{pmatrix} = 9 + 0 = 9 Hence, the product of matrices P and Q is: PQ = \(\begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}\)\(\begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}\) = \(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\) Therefore, the answer is option D: \(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\).
Ajụjụ 27 Ripọtì
The sum of the first n terms of a linear sequence is \(S_{n} = n^{2} + 2n\). Determine the general term of the sequence.
Akọwa Nkọwa
To find the general term of a linear sequence, we need to find the common difference between any two consecutive terms of the sequence. We know that the sum of the first n terms of the sequence is given by: $$S_{n} = n^{2} + 2n$$ If we subtract the sum of the first (n-1) terms from the sum of the first n terms, we get the nth term of the sequence: $$\begin{aligned} a_{n} &= S_{n} - S_{n-1} \\ &= (n^{2} + 2n) - [(n-1)^{2} + 2(n-1)] \\ &= 2n - 1 \end{aligned}$$ Therefore, the general term of the sequence is given by: $$a_{n} = 2n - 1$$ So, the correct option is \textbf{2n + 1}.
Ajụjụ 28 Ripọtì
Two forces \(F_{1} = (10N, 020°)\) and \(F_{2} = (7N, 200°)\) act on a particle. Find the resultant force.
Akọwa Nkọwa
Ajụjụ 29 Ripọtì
A body starts from rest and moves in a straight line with uniform acceleration of \(5 ms^{-2}\). How far, in metres, does it go in 10 seconds?
Akọwa Nkọwa
The problem gives us the initial velocity of the body, \(u=0\) (because it starts from rest), the acceleration, \(a=5 ms^{-2}\), and the time, \(t=10s\), and asks us to find the distance, \(s\), that the body travels in that time. We can use the following equation to solve the problem: \(s = ut + \frac{1}{2}at^2\) where \(u\) is the initial velocity, \(a\) is the acceleration, \(t\) is the time, and \(s\) is the distance traveled. Plugging in the values we have: \(s = 0 \times 10 + \frac{1}{2} \times 5 \times 10^2 = 250\text{ m}\) Therefore, the answer is option (B) 250 m. This equation gives us the distance traveled by an object with uniform acceleration starting from rest. It is derived by combining the equations of motion for constant acceleration.
Ajụjụ 30 Ripọtì
If \(p = \begin{pmatrix} 2 \\ -2 \end{pmatrix} \) and \(q = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\), find \(|q - \frac{1}{2}p|\).
Akọwa Nkọwa
To find the value of |q - 1/2p|, we first need to calculate q - 1/2p. q - 1/2p = \(\begin{pmatrix} 3 \\ 4 \end{pmatrix}\) - 1/2\(\begin{pmatrix} 2 \\ -2 \end{pmatrix}\) = \(\begin{pmatrix} 3 \\ 4 \end{pmatrix}\) - \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\) = \(\begin{pmatrix} 2 \\ 5 \end{pmatrix}\) Now, we need to calculate the magnitude or length of the vector (2, 5). |q - 1/2p| = \(\sqrt{(2)^2 + (5)^2}\) = \(\sqrt{4 + 25}\) = \(\sqrt{29}\) Therefore, the answer is option D: \(\sqrt{29}\).
Ajụjụ 31 Ripọtì
If events A and B are independent and \(P(A) = \frac{7}{12}\) and \(P(A \cap B) = \frac{1}{4}\), find P(B).
Akọwa Nkọwa
If events A and B are independent, then the occurrence of event A does not affect the occurrence of event B, and vice versa. We can use the formula for the probability of the intersection of two independent events: $$P(A \cap B) = P(A) \cdot P(B)$$ Given that \(P(A) = \frac{7}{12}\) and \(P(A \cap B) = \frac{1}{4}\), we can substitute these values into the formula and solve for \(P(B)\): $$\frac{1}{4} = \frac{7}{12} \cdot P(B)$$ Multiplying both sides by \(\frac{12}{7}\), we get: $$P(B) = \frac{3}{7}$$ Therefore, the probability of event B is \(\frac{3}{7}\). Option (a) is the correct answer.
Ajụjụ 32 Ripọtì
The sum of the first n terms of a linear sequence is \(S_{n} = n^{2} + 2n\). Find the common difference of the sequence.
Akọwa Nkọwa
Ajụjụ 33 Ripọtì
A function f is defined on R, the set of real numbers, by: \(f : x \to \frac{x + 3}{x - 2}, x \neq 2\), find \(f^{-1}\).
Akọwa Nkọwa
To find \(f^{-1}\), we need to solve the equation: \[y = \frac{x + 3}{x - 2}\] for x in terms of y. To do this, we start by swapping the roles of x and y: \[x = \frac{y + 3}{y - 2}\] Now we solve for y in terms of x. We begin by multiplying both sides by \(y-2\): \[x(y-2) = y + 3\] Expanding the left-hand side: \[xy - 2x = y + 3\] Rearranging the terms: \[xy - y = 2x + 3\] Factoring out y on the left-hand side: \[y(x-1) = 2x + 3\] Finally, dividing both sides by \(x-1\): \[y = \frac{2x+3}{x-1}, x\neq 1\] Therefore, the inverse function of \(f(x) = \frac{x+3}{x-2}, x\neq 2\) is: \[f^{-1}(x) = \frac{2x+3}{x-1}, x\neq 1\] Hence, the answer is \(\boxed{\textbf{(A) }f^{-1} : x \to \frac{2x+3}{x-1}, x \neq 1}\).
Ajụjụ 34 Ripọtì
What is the coordinate of the centre of the circle \(5x^{2} + 5y^{2} - 15x + 25y - 3 = 0\)?
Akọwa Nkọwa
The general equation of a circle in standard form is: \((x-h)^2 + (y-k)^2 = r^2\) where the center of the circle is located at (h, k) and has a radius of r. To get the equation of the circle into this form, we need to complete the square for both the x and y terms: \begin{aligned} 5x^{2} + 5y^{2} - 15x + 25y - 3 &= 0 \\ 5(x^{2} - 3x) + 5(y^{2} + 5y) &= 3 \\ 5(x^{2} - 3x + \frac{9}{4}) + 5(y^{2} + 5y + \frac{25}{4}) &= 3 + 5(\frac{9}{4} + \frac{25}{4}) \\ 5(x - \frac{3}{2})^{2} + 5(y + \frac{5}{2})^{2} &= 35 \\ (x - \frac{3}{2})^{2} + (y + \frac{5}{2})^{2} &= 7 \end{aligned} Comparing this to the standard form, we can see that the center of the circle is at \((\frac{3}{2}, -\frac{5}{2})\). Therefore, the answer is (B) \((\frac{3}{2}, -\frac{5}{2})\).
Ajụjụ 35 Ripọtì
Simplify \(\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}}\)
Akọwa Nkọwa
First, we simplify the term inside the square root in the numerator: \begin{align*} \sqrt{3} + \sqrt{48} &= \sqrt{3} + \sqrt{16\cdot3} \\ &= \sqrt{3} + \sqrt{16}\sqrt{3} \\ &= \sqrt{3} + 4\sqrt{3} \\ &= 5\sqrt{3}. \end{align*} Now we substitute this simplified value back into the original expression: \begin{align*} \frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}} &= \frac{5\sqrt{3}}{\sqrt{6}} \\ &= \frac{5\sqrt{3}}{\sqrt{2}\sqrt{3}} \\ &= \frac{5}{\sqrt{2}} \\ &= \frac{5\sqrt{2}}{2}. \end{align*} Therefore, the simplified expression is \(\frac{5\sqrt{2}}{2}\).
Ajụjụ 36 Ripọtì
In the diagram above, forces P, Q and 50N are acting on a body at M. If the system is in equilibrium, calculate, in terms of \(\theta\), the magnitude of P.
Akọwa Nkọwa
Ajụjụ 37 Ripọtì
A test consists of 12 questions out of which candidates are to answer 10. If the first 6 are compulsory, in how many ways can each candidate select her questions?
Akọwa Nkọwa
Out of the 12 questions, the first 6 are compulsory and each candidate must answer them. Therefore, the candidate has to choose 4 questions out of the remaining 6 questions to answer. The number of ways to choose 4 questions out of 6 is given by the formula for combinations: \[\binom{6}{4} = \frac{6!}{4!2!} = \frac{6\times 5}{2\times 1} = 15\] Therefore, each candidate can select her questions in 15 ways. Hence, the answer is 15.
Ajụjụ 40 Ripọtì
If \(\sin A = \frac{3}{5}\) and \(\cos B = \frac{15}{17}\), where A is an obtuse angle and B is acute, find the value of \(\cos (A + B)\).
Given \(\sin A=\dfrac{3}{5}\) with A obtuse, and \(\cos B=\dfrac{15}{17}\) with B acute.
Find cos A. A is obtuse (second quadrant), so cosine is negative. Using \(\cos^2A=1-\sin^2A:\)
\[\cos A=-\sqrt{1-\tfrac{9}{25}}=-\sqrt{\tfrac{16}{25}}=-\frac{4}{5}.\]
Find sin B. B is acute, so sine is positive:
\[\sin B=\sqrt{1-\tfrac{225}{289}}=\sqrt{\tfrac{64}{289}}=\frac{8}{17}.\]
Apply the compound-angle formula.
\[\cos(A+B)=\cos A\cos B-\sin A\sin B=\left(-\frac{4}{5}\right)\!\left(\frac{15}{17}\right)-\left(\frac{3}{5}\right)\!\left(\frac{8}{17}\right).\]
\[=-\frac{60}{85}-\frac{24}{85}=-\frac{84}{85}.\]
Akọwa Nkọwa
Given \(\sin A=\dfrac{3}{5}\) with A obtuse, and \(\cos B=\dfrac{15}{17}\) with B acute.
Find cos A. A is obtuse (second quadrant), so cosine is negative. Using \(\cos^2A=1-\sin^2A:\)
\[\cos A=-\sqrt{1-\tfrac{9}{25}}=-\sqrt{\tfrac{16}{25}}=-\frac{4}{5}.\]
Find sin B. B is acute, so sine is positive:
\[\sin B=\sqrt{1-\tfrac{225}{289}}=\sqrt{\tfrac{64}{289}}=\frac{8}{17}.\]
Apply the compound-angle formula.
\[\cos(A+B)=\cos A\cos B-\sin A\sin B=\left(-\frac{4}{5}\right)\!\left(\frac{15}{17}\right)-\left(\frac{3}{5}\right)\!\left(\frac{8}{17}\right).\]
\[=-\frac{60}{85}-\frac{24}{85}=-\frac{84}{85}.\]
Ajụjụ 41 Ripọtì
(a) Five female and seven male teachers applied for 4 vacancies in a Junior High School. The teachers are equally qualified. Find the number of ways of employing the 4 teachers, if : (i) there is no restriction ; (ii) at least 2 of them are females.
(b) The table shows the positions awarded to 7 contestants by Judges X and Y in a competition.
| Contestant | P | Q | R | S | T | U | V |
| Judge X | 2 | 7 | 1 | 3 | 6 | 5 | 4 |
| Judge Y | 4 | 6 | 2 | 3 | 1 | 1 | 5 |
(i) Calculate, correct to one decimal place, the Spearman's rank correlation coefficient.
(ii) Interpret your answer in b(i) above.
(a) Selecting 4 teachers from 5 female + 7 male = 12 teachers.
(i) No restriction. Choose any 4 of the 12:
\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \mathbf{495 \text{ ways}} \](ii) At least 2 females. The cases are exactly 2, 3 or 4 females (the rest male):
\[ \binom{5}{2}\binom{7}{2} + \binom{5}{3}\binom{7}{1} + \binom{5}{4}\binom{7}{0} \] \[ = (10)(21) + (10)(7) + (5)(1) = 210 + 70 + 5 = \mathbf{285 \text{ ways}} \](b) Spearman's rank correlation for 7 contestants (\(n = 7\)). Judge Y has a tie (T and U both awarded position 1), so they share positions 1 and 2, each taking the average rank \(1.5\); the remaining Y values are then re-ranked \(2\to3, 3\to4, 4\to5, 5\to6, 6\to7\). Judge X already has distinct positions 1 to 7.
| Contestant | \(R_X\) | \(R_Y\) | \(d\) | \(d^2\) |
|---|---|---|---|---|
| P | 2 | 5 | -3 | 9 |
| Q | 7 | 7 | 0 | 0 |
| R | 1 | 3 | -2 | 4 |
| S | 3 | 4 | -1 | 1 |
| T | 6 | 1.5 | 4.5 | 20.25 |
| U | 5 | 1.5 | 3.5 | 12.25 |
| V | 4 | 6 | -2 | 4 |
| Total \(\sum d^2\) | 50.5 | |||
(ii) Interpretation. \(r_s \approx 0.1\) is very close to zero, so there is only a very weak positive correlation. The two judges show almost no agreement in the way they ranked the contestants.
Akọwa Nkọwa
(a) Selecting 4 teachers from 5 female + 7 male = 12 teachers.
(i) No restriction. Choose any 4 of the 12:
\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \mathbf{495 \text{ ways}} \](ii) At least 2 females. The cases are exactly 2, 3 or 4 females (the rest male):
\[ \binom{5}{2}\binom{7}{2} + \binom{5}{3}\binom{7}{1} + \binom{5}{4}\binom{7}{0} \] \[ = (10)(21) + (10)(7) + (5)(1) = 210 + 70 + 5 = \mathbf{285 \text{ ways}} \](b) Spearman's rank correlation for 7 contestants (\(n = 7\)). Judge Y has a tie (T and U both awarded position 1), so they share positions 1 and 2, each taking the average rank \(1.5\); the remaining Y values are then re-ranked \(2\to3, 3\to4, 4\to5, 5\to6, 6\to7\). Judge X already has distinct positions 1 to 7.
| Contestant | \(R_X\) | \(R_Y\) | \(d\) | \(d^2\) |
|---|---|---|---|---|
| P | 2 | 5 | -3 | 9 |
| Q | 7 | 7 | 0 | 0 |
| R | 1 | 3 | -2 | 4 |
| S | 3 | 4 | -1 | 1 |
| T | 6 | 1.5 | 4.5 | 20.25 |
| U | 5 | 1.5 | 3.5 | 12.25 |
| V | 4 | 6 | -2 | 4 |
| Total \(\sum d^2\) | 50.5 | |||
(ii) Interpretation. \(r_s \approx 0.1\) is very close to zero, so there is only a very weak positive correlation. The two judges show almost no agreement in the way they ranked the contestants.
Ajụjụ 42 Ripọtì
The table shows the distribution of marks obtained by some candidates in a test.
| Marks | 10-14 | 15-24 | 25-29 | 30-39 | 40-44 | 45-49 |
| No of candidates | 14 | 30 | 22 | 18 | 12 | 4 |
Draw a histogram for the distribution.
Since the class intervals have unequal widths, the height of each rectangle is the frequency density:
\[\text{Frequency density}=\frac{\text{Frequency}}{\text{Class width}}\]
| Marks | Class boundaries | Class width | Frequency | Frequency density |
|---|---|---|---|---|
| 10-14 | 9.5-14.5 | 5 | 14 | 2.8 |
| 15-24 | 14.5-24.5 | 10 | 30 | 3.0 |
| 25-29 | 24.5-29.5 | 5 | 22 | 4.4 |
| 30-39 | 29.5-39.5 | 10 | 18 | 1.8 |
| 40-44 | 39.5-44.5 | 5 | 12 | 2.4 |
| 45-49 | 44.5-49.5 | 5 | 4 | 0.8 |
The required histogram is:
The rectangles are contiguous, with their bases at the stated class boundaries. Their areas are proportional to the corresponding frequencies.
Akọwa Nkọwa
Since the class intervals have unequal widths, the height of each rectangle is the frequency density:
\[\text{Frequency density}=\frac{\text{Frequency}}{\text{Class width}}\]
| Marks | Class boundaries | Class width | Frequency | Frequency density |
|---|---|---|---|---|
| 10-14 | 9.5-14.5 | 5 | 14 | 2.8 |
| 15-24 | 14.5-24.5 | 10 | 30 | 3.0 |
| 25-29 | 24.5-29.5 | 5 | 22 | 4.4 |
| 30-39 | 29.5-39.5 | 10 | 18 | 1.8 |
| 40-44 | 39.5-44.5 | 5 | 12 | 2.4 |
| 45-49 | 44.5-49.5 | 5 | 4 | 0.8 |
The required histogram is:
The rectangles are contiguous, with their bases at the stated class boundaries. Their areas are proportional to the corresponding frequencies.
Ajụjụ 43 Ripọtì
A straight line passes through the point P(-1, 3). Another line which passes through Q(-4, 4) intersects the first line at the point R(k, 5), where k is a constant. If \(<PRQ = 90°\), find the values of k.
The point of intersection is \(R(k,5),\) with \(P(-1,3)\) and \(Q(-4,4).\) The condition \(\angle PRQ=90^\circ\) means the segments \(RP\) and \(RQ\) are perpendicular at R, so the vectors \(\overrightarrow{RP}\) and \(\overrightarrow{RQ}\) have zero dot product.
\[\overrightarrow{RP}=P-R=(-1-k,\,3-5)=(-1-k,\,-2),\]
\[\overrightarrow{RQ}=Q-R=(-4-k,\,4-5)=(-4-k,\,-1).\]
Set the dot product to zero:
\[(-1-k)(-4-k)+(-2)(-1)=0.\]
Expand \((-1-k)(-4-k)=(1+k)(4+k)=k^2+5k+4,\) so
\[k^2+5k+4+2=0\Rightarrow k^2+5k+6=0.\]
Factorise:
\[(k+2)(k+3)=0\Rightarrow k=-2\ \text{or}\ k=-3.\]
Both values give a right angle at R, so the required values are \(k=-2\) and \(k=-3.\)
Akọwa Nkọwa
The point of intersection is \(R(k,5),\) with \(P(-1,3)\) and \(Q(-4,4).\) The condition \(\angle PRQ=90^\circ\) means the segments \(RP\) and \(RQ\) are perpendicular at R, so the vectors \(\overrightarrow{RP}\) and \(\overrightarrow{RQ}\) have zero dot product.
\[\overrightarrow{RP}=P-R=(-1-k,\,3-5)=(-1-k,\,-2),\]
\[\overrightarrow{RQ}=Q-R=(-4-k,\,4-5)=(-4-k,\,-1).\]
Set the dot product to zero:
\[(-1-k)(-4-k)+(-2)(-1)=0.\]
Expand \((-1-k)(-4-k)=(1+k)(4+k)=k^2+5k+4,\) so
\[k^2+5k+4+2=0\Rightarrow k^2+5k+6=0.\]
Factorise:
\[(k+2)(k+3)=0\Rightarrow k=-2\ \text{or}\ k=-3.\]
Both values give a right angle at R, so the required values are \(k=-2\) and \(k=-3.\)
Ajụjụ 44 Ripọtì
(a) Forces \(F_{1} = \begin{pmatrix} -5 & 4 \end{pmatrix} N; F_{2} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} N; F_{3} = \begin{pmatrix} 2 & -1 \end{pmatrix} N\) and \(F_{4} = \begin{pmatrix} 3 & -5 \end{pmatrix} N\) act on a body. Find the :
(i) resultant of these forces ; (ii) fifth force that will keep the body in equilibrium.
(b) A body moving at 20 ms\(^{-1}\) accelerates uniformly at 2.5 ms\(^{-2}\) for 4 seconds. It continues the journey at the speed for 8 seconds, before coming to rest in t seconds with a uniform retardation. If the ratio of the acceleration to the retardation is 3 : 4,
(i) sketch the velocity- time graph of the journey ; (ii) find t ; (iii) find the total distance of the journey.
(a)
(i) Resultant force
Writing each force as a column vector,
\[\begin{aligned}\mathbf R&=\begin{pmatrix}-5\\4\end{pmatrix}+\begin{pmatrix}2\\5\end{pmatrix}+\begin{pmatrix}2\\-1\end{pmatrix}+\begin{pmatrix}3\\-5\end{pmatrix}\\&=\begin{pmatrix}-5+2+2+3\\4+5-1-5\end{pmatrix}\\&=\begin{pmatrix}2\\3\end{pmatrix}\text{ N}.\end{aligned}\]
Hence, \(\boxed{\mathbf R=2\mathbf i+3\mathbf j\text{ N}}\).
(ii) Fifth force for equilibrium
For equilibrium, the fifth force \(\mathbf F_5\) must be the negative of the resultant:
\[\mathbf F_5=-\mathbf R=-\begin{pmatrix}2\\3\end{pmatrix}=\boxed{\begin{pmatrix}-2\\-3\end{pmatrix}\text{ N}}.\]
(b)
During the first 4 s,
\[v=20+(2.5)(4)=30\text{ m s}^{-1}.\]
Let the magnitude of the retardation be \(r\text{ m s}^{-2}\). Since
\[2.5:r=3:4,\qquad r=\frac{4}{3}(2.5)=\frac{10}{3}\text{ m s}^{-2}.\]
(i) Velocity-time graph
The graph consists of straight-line sections joining \((0,20)\), \((4,30)\), \((12,30)\), and \((21,0)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).
(ii) Time taken to come to rest
For the retardation stage, \(u=30\text{ m s}^{-1}\), \(v=0\), and acceleration \(=-\frac{10}{3}\text{ m s}^{-2}\).
\[0=30-\frac{10}{3}t\]
\[\frac{10}{3}t=30\]
\[\boxed{t=9\text{ s}}.\]
(iii) Total distance travelled
The total distance is the area under the velocity-time graph:
\[\begin{aligned}A_1&=\frac12(20+30)\times4=100\text{ m},\\A_2&=30\times8=240\text{ m},\\A_3&=\frac12\times30\times9=135\text{ m}.\end{aligned}\]
\[\text{Total distance}=100+240+135=\boxed{475\text{ m}}.\]
Akọwa Nkọwa
(a)
(i) Resultant force
Writing each force as a column vector,
\[\begin{aligned}\mathbf R&=\begin{pmatrix}-5\\4\end{pmatrix}+\begin{pmatrix}2\\5\end{pmatrix}+\begin{pmatrix}2\\-1\end{pmatrix}+\begin{pmatrix}3\\-5\end{pmatrix}\\&=\begin{pmatrix}-5+2+2+3\\4+5-1-5\end{pmatrix}\\&=\begin{pmatrix}2\\3\end{pmatrix}\text{ N}.\end{aligned}\]
Hence, \(\boxed{\mathbf R=2\mathbf i+3\mathbf j\text{ N}}\).
(ii) Fifth force for equilibrium
For equilibrium, the fifth force \(\mathbf F_5\) must be the negative of the resultant:
\[\mathbf F_5=-\mathbf R=-\begin{pmatrix}2\\3\end{pmatrix}=\boxed{\begin{pmatrix}-2\\-3\end{pmatrix}\text{ N}}.\]
(b)
During the first 4 s,
\[v=20+(2.5)(4)=30\text{ m s}^{-1}.\]
Let the magnitude of the retardation be \(r\text{ m s}^{-2}\). Since
\[2.5:r=3:4,\qquad r=\frac{4}{3}(2.5)=\frac{10}{3}\text{ m s}^{-2}.\]
(i) Velocity-time graph
The graph consists of straight-line sections joining \((0,20)\), \((4,30)\), \((12,30)\), and \((21,0)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).
(ii) Time taken to come to rest
For the retardation stage, \(u=30\text{ m s}^{-1}\), \(v=0\), and acceleration \(=-\frac{10}{3}\text{ m s}^{-2}\).
\[0=30-\frac{10}{3}t\]
\[\frac{10}{3}t=30\]
\[\boxed{t=9\text{ s}}.\]
(iii) Total distance travelled
The total distance is the area under the velocity-time graph:
\[\begin{aligned}A_1&=\frac12(20+30)\times4=100\text{ m},\\A_2&=30\times8=240\text{ m},\\A_3&=\frac12\times30\times9=135\text{ m}.\end{aligned}\]
\[\text{Total distance}=100+240+135=\boxed{475\text{ m}}.\]
Ajụjụ 45 Ripọtì
A car travelling at a velocity of 50kmh\(^{-1}\), covers a distance of 20km. If it was accelerating at 6kmh\(^{-1}\), calculate, correct to one decimal place, the time the car took to cover the distance.
Ajụjụ 46 Ripọtì
A survey indicated that 65% of the families in an area have cars. Find, correct to three decimal places, the probability that among 7 families selected at random in the area
(a) exactly 5 ;
(b) 3 or 4 ;
(c) at most 2 of them have cars.
Ajụjụ 47 Ripọtì
The magnitude of a force \(xi + 15j\) is 17N. Find the :
(a) possible values of x ;
(b) directions of the forces, correct to the nearest degree.
The force is \(x\mathbf{i}+15\mathbf{j}\) with magnitude 17 N.
(a) Possible values of x. The magnitude is \(\sqrt{x^2+15^2}:\)
\[\sqrt{x^2+225}=17\Rightarrow x^2+225=289\Rightarrow x^2=64\Rightarrow x=\pm8.\]
So \(x=8\) or \(x=-8.\)
(b) Directions (to the nearest degree), measured from the positive x-axis.
When \(x=8\): the force \(8\mathbf{i}+15\mathbf{j}\) lies in the first quadrant.
\[\theta=\tan^{-1}\!\left(\frac{15}{8}\right)\approx61.9^\circ\approx62^\circ.\]
When \(x=-8\): the force \(-8\mathbf{i}+15\mathbf{j}\) lies in the second quadrant.
\[\theta=180^\circ-\tan^{-1}\!\left(\frac{15}{8}\right)\approx180^\circ-61.9^\circ\approx118^\circ.\]
So the two forces act at approximately \(62^\circ\) and \(118^\circ\) to the positive x-axis.
Akọwa Nkọwa
The force is \(x\mathbf{i}+15\mathbf{j}\) with magnitude 17 N.
(a) Possible values of x. The magnitude is \(\sqrt{x^2+15^2}:\)
\[\sqrt{x^2+225}=17\Rightarrow x^2+225=289\Rightarrow x^2=64\Rightarrow x=\pm8.\]
So \(x=8\) or \(x=-8.\)
(b) Directions (to the nearest degree), measured from the positive x-axis.
When \(x=8\): the force \(8\mathbf{i}+15\mathbf{j}\) lies in the first quadrant.
\[\theta=\tan^{-1}\!\left(\frac{15}{8}\right)\approx61.9^\circ\approx62^\circ.\]
When \(x=-8\): the force \(-8\mathbf{i}+15\mathbf{j}\) lies in the second quadrant.
\[\theta=180^\circ-\tan^{-1}\!\left(\frac{15}{8}\right)\approx180^\circ-61.9^\circ\approx118^\circ.\]
So the two forces act at approximately \(62^\circ\) and \(118^\circ\) to the positive x-axis.
Ajụjụ 48 Ripọtì
(a) Express \(\frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x}\) in partial fractions.
(b) If \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24\), find the value of x.
(a) Partial fractions of \(\dfrac{2x^2-5x+1}{x^3-4x^2+3x}.\)
Factorise the denominator: \(x^3-4x^2+3x=x(x^2-4x+3)=x(x-1)(x-3).\)
Write
\[\frac{2x^2-5x+1}{x(x-1)(x-3)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-3}.\]
Multiply out: \(2x^2-5x+1=A(x-1)(x-3)+Bx(x-3)+Cx(x-1).\)
\(x=0:\ 1=A(-1)(-3)=3A\Rightarrow A=\dfrac{1}{3}.\)
\(x=1:\ 2-5+1=-2=B(1)(-2)=-2B\Rightarrow B=1.\)
\(x=3:\ 18-15+1=4=C(3)(2)=6C\Rightarrow C=\dfrac{2}{3}.\)
\[\frac{2x^2-5x+1}{x^3-4x^2+3x}=\frac{1}{3x}+\frac{1}{x-1}+\frac{2}{3(x-3)}.\]
(b) Solve \(\begin{vmatrix}x-3&-4&3\\5&2&2\\2&-4&6-x\end{vmatrix}=-24.\)
Expand along the first row:
\[(x-3)\big[2(6-x)-2(-4)\big]-(-4)\big[5(6-x)-2(2)\big]+3\big[5(-4)-2(2)\big].\]
\[=(x-3)(20-2x)+4(26-5x)+3(-24).\]
Now \((x-3)(20-2x)=-2x^2+26x-60,\) so the total is
\[-2x^2+26x-60+104-20x-72=-2x^2+6x-28.\]
Set equal to \(-24:\)
\[-2x^2+6x-28=-24\Rightarrow -2x^2+6x-4=0\Rightarrow x^2-3x+2=0.\]
\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2.\]
Akọwa Nkọwa
(a) Partial fractions of \(\dfrac{2x^2-5x+1}{x^3-4x^2+3x}.\)
Factorise the denominator: \(x^3-4x^2+3x=x(x^2-4x+3)=x(x-1)(x-3).\)
Write
\[\frac{2x^2-5x+1}{x(x-1)(x-3)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-3}.\]
Multiply out: \(2x^2-5x+1=A(x-1)(x-3)+Bx(x-3)+Cx(x-1).\)
\(x=0:\ 1=A(-1)(-3)=3A\Rightarrow A=\dfrac{1}{3}.\)
\(x=1:\ 2-5+1=-2=B(1)(-2)=-2B\Rightarrow B=1.\)
\(x=3:\ 18-15+1=4=C(3)(2)=6C\Rightarrow C=\dfrac{2}{3}.\)
\[\frac{2x^2-5x+1}{x^3-4x^2+3x}=\frac{1}{3x}+\frac{1}{x-1}+\frac{2}{3(x-3)}.\]
(b) Solve \(\begin{vmatrix}x-3&-4&3\\5&2&2\\2&-4&6-x\end{vmatrix}=-24.\)
Expand along the first row:
\[(x-3)\big[2(6-x)-2(-4)\big]-(-4)\big[5(6-x)-2(2)\big]+3\big[5(-4)-2(2)\big].\]
\[=(x-3)(20-2x)+4(26-5x)+3(-24).\]
Now \((x-3)(20-2x)=-2x^2+26x-60,\) so the total is
\[-2x^2+26x-60+104-20x-72=-2x^2+6x-28.\]
Set equal to \(-24:\)
\[-2x^2+6x-28=-24\Rightarrow -2x^2+6x-4=0\Rightarrow x^2-3x+2=0.\]
\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2.\]
Ajụjụ 49 Ripọtì
A function F is defined on the set R, of real numbers by \(f : x \to px^{2} + qx + 2\), where p and q are constants. If \(f(-2) = 0\) and \(f(1) = 3\), find \(f(-4)\).
Given \(f(x)=px^2+qx+2.\)
From \(f(-2)=0:\ p(-2)^2+q(-2)+2=0\Rightarrow 4p-2q+2=0\Rightarrow 2p-q=-1.\)
From \(f(1)=3:\ p+q+2=3\Rightarrow p+q=1.\)
Add the two results: \((2p-q)+(p+q)=-1+1\Rightarrow 3p=0\Rightarrow p=0.\) Then \(q=1.\)
So \(f(x)=x+2.\) Therefore
\[f(-4)=-4+2=-2.\]
Check: \(f(-2)=-2+2=0\) and \(f(1)=1+2=3,\) as required.
Akọwa Nkọwa
Given \(f(x)=px^2+qx+2.\)
From \(f(-2)=0:\ p(-2)^2+q(-2)+2=0\Rightarrow 4p-2q+2=0\Rightarrow 2p-q=-1.\)
From \(f(1)=3:\ p+q+2=3\Rightarrow p+q=1.\)
Add the two results: \((2p-q)+(p+q)=-1+1\Rightarrow 3p=0\Rightarrow p=0.\) Then \(q=1.\)
So \(f(x)=x+2.\) Therefore
\[f(-4)=-4+2=-2.\]
Check: \(f(-2)=-2+2=0\) and \(f(1)=1+2=3,\) as required.
Ajụjụ 50 Ripọtì
Differentiate, with respect to x, \(x^{3} + 2x\) from the first principle.
Differentiate \(f(x)=x^3+2x\) from first principles.
By definition,
\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.\]
Compute \(f(x+h):\)
\[f(x+h)=(x+h)^3+2(x+h)=x^3+3x^2h+3xh^2+h^3+2x+2h.\]
Then
\[f(x+h)-f(x)=3x^2h+3xh^2+h^3+2h.\]
Divide by \(h\) (for \(h\neq0\)):
\[\frac{f(x+h)-f(x)}{h}=3x^2+3xh+h^2+2.\]
Now let \(h\to0:\)
\[f'(x)=\lim_{h\to0}\left(3x^2+3xh+h^2+2\right)=3x^2+2.\]
Akọwa Nkọwa
Differentiate \(f(x)=x^3+2x\) from first principles.
By definition,
\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.\]
Compute \(f(x+h):\)
\[f(x+h)=(x+h)^3+2(x+h)=x^3+3x^2h+3xh^2+h^3+2x+2h.\]
Then
\[f(x+h)-f(x)=3x^2h+3xh^2+h^3+2h.\]
Divide by \(h\) (for \(h\neq0\)):
\[\frac{f(x+h)-f(x)}{h}=3x^2+3xh+h^2+2.\]
Now let \(h\to0:\)
\[f'(x)=\lim_{h\to0}\left(3x^2+3xh+h^2+2\right)=3x^2+2.\]
Ajụjụ 51 Ripọtì
(a) If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} + 5x - 6 = 0\), find the equation whose roots are \((\alpha - 2)\) and \((\beta - 2)\).
(b) Given that \(\int_{0} ^{k} (x^{2} - 2x) \mathrm {d} x = 4\), find the values of k.
Akọwa Nkọwa
None
Ajụjụ 52 Ripọtì
(a) An object is thrown up a smooth plane inclined at an angle of 30° to the horizontal. If the plane is 15m long and the object comes to rest at the top, find the :
(i) initial speed of the object ; (ii) time taken to reach the top.
(b) 
Force of magnitudes \(5 N, 5\sqrt{3} N, 10 N, 5\sqrt{3} N\) and \(5 N\) act on a body P, of mass 5 kg as shown in the diagram. Find the :
(i) magnitude of the resultant force ; (ii) acceleration of the body.
(a) Object on a smooth inclined plane (angle \(30^\circ\), length \(15\ \text{m}\), comes to rest at the top). On a smooth plane the only force along the plane is the component of gravity, giving a retardation
\[a=g\sin30^\circ=10\times0.5=5\ \text{m s}^{-2}\quad(\text{taking }g=10\ \text{m s}^{-2}).\]
(i) Initial speed. Using \(v^2=u^2-2as\) with \(v=0\), \(s=15\ \text{m}\):
\[0=u^2-2(5)(15)\;\Rightarrow\;u^2=150\;\Rightarrow\;u=\sqrt{150}\approx 12.25\ \text{m s}^{-1}.\]
(ii) Time to reach the top. Using \(v=u-at\) with \(v=0\):
\[0=12.25-5t\;\Rightarrow\;t=\frac{12.25}{5}\approx 2.45\ \text{s}.\]
(b) Five forces on body P. From the diagram the forces are symmetric about the horizontal \(x\)-axis. Measuring angles from the positive \(x\)-axis (each gap is \(30^\circ\)):
- \(5\ \text{N}\) at \(60^\circ\)
- \(5\sqrt3\ \text{N}\) at \(30^\circ\)
- \(10\ \text{N}\) at \(0^\circ\)
- \(5\sqrt3\ \text{N}\) at \(-30^\circ\)
- \(5\ \text{N}\) at \(-60^\circ\)
(i) Magnitude of the resultant. Resolve horizontally:
\[\sum F_x = 5\cos60^\circ+5\sqrt3\cos30^\circ+10+5\sqrt3\cos30^\circ+5\cos60^\circ\]
\[\sum F_x = 2.5+7.5+10+7.5+2.5 = 30\ \text{N}.\]
Vertically the pairs cancel by symmetry:
\[\sum F_y = 5\sin60^\circ+5\sqrt3\sin30^\circ-5\sqrt3\sin30^\circ-5\sin60^\circ = 0.\]
Hence the resultant is
\[R=\sqrt{30^2+0^2}=\mathbf{30\ \text{N}}\ \text{along the }10\text{N direction (positive }x\text{-axis).}\]
(ii) Acceleration of the body (mass \(5\ \text{kg}\)):
\[a=\frac{R}{m}=\frac{30}{5}=\mathbf{6\ \text{m s}^{-2}}\ \text{in the direction of the }10\text{N force.}\]
Akọwa Nkọwa
(a) Object on a smooth inclined plane (angle \(30^\circ\), length \(15\ \text{m}\), comes to rest at the top). On a smooth plane the only force along the plane is the component of gravity, giving a retardation
\[a=g\sin30^\circ=10\times0.5=5\ \text{m s}^{-2}\quad(\text{taking }g=10\ \text{m s}^{-2}).\]
(i) Initial speed. Using \(v^2=u^2-2as\) with \(v=0\), \(s=15\ \text{m}\):
\[0=u^2-2(5)(15)\;\Rightarrow\;u^2=150\;\Rightarrow\;u=\sqrt{150}\approx 12.25\ \text{m s}^{-1}.\]
(ii) Time to reach the top. Using \(v=u-at\) with \(v=0\):
\[0=12.25-5t\;\Rightarrow\;t=\frac{12.25}{5}\approx 2.45\ \text{s}.\]
(b) Five forces on body P. From the diagram the forces are symmetric about the horizontal \(x\)-axis. Measuring angles from the positive \(x\)-axis (each gap is \(30^\circ\)):
- \(5\ \text{N}\) at \(60^\circ\)
- \(5\sqrt3\ \text{N}\) at \(30^\circ\)
- \(10\ \text{N}\) at \(0^\circ\)
- \(5\sqrt3\ \text{N}\) at \(-30^\circ\)
- \(5\ \text{N}\) at \(-60^\circ\)
(i) Magnitude of the resultant. Resolve horizontally:
\[\sum F_x = 5\cos60^\circ+5\sqrt3\cos30^\circ+10+5\sqrt3\cos30^\circ+5\cos60^\circ\]
\[\sum F_x = 2.5+7.5+10+7.5+2.5 = 30\ \text{N}.\]
Vertically the pairs cancel by symmetry:
\[\sum F_y = 5\sin60^\circ+5\sqrt3\sin30^\circ-5\sqrt3\sin30^\circ-5\sin60^\circ = 0.\]
Hence the resultant is
\[R=\sqrt{30^2+0^2}=\mathbf{30\ \text{N}}\ \text{along the }10\text{N direction (positive }x\text{-axis).}\]
(ii) Acceleration of the body (mass \(5\ \text{kg}\)):
\[a=\frac{R}{m}=\frac{30}{5}=\mathbf{6\ \text{m s}^{-2}}\ \text{in the direction of the }10\text{N force.}\]
Ajụjụ 53 Ripọtì
(a) The sum of the first n terms of a sequence is given by \(S_{n} = \frac{5n^{2}}{2} + \frac{5n}{2}\). Write down the first four terms of the sequence and an expression for the nth term.
(b) The equation of a circle is given by \(x^{2} + y^{2} - 10x - 8y + 25 = 0\).
(i) Show that the circle touches the x- axis ; (ii) Find the coordinates of the point of contact.
(a) Sequence with \(S_n=\dfrac{5n^2}{2}+\dfrac{5n}{2}=\dfrac{5n(n+1)}{2}.\)
Each term is \(T_n=S_n-S_{n-1}.\) Compute partial sums:
\(S_1=\dfrac{5(1)(2)}{2}=5,\ S_2=\dfrac{5(2)(3)}{2}=15,\ S_3=\dfrac{5(3)(4)}{2}=30,\ S_4=\dfrac{5(4)(5)}{2}=50.\)
Terms: \(T_1=5,\ T_2=15-5=10,\ T_3=30-15=15,\ T_4=50-30=20.\)
First four terms: 5, 10, 15, 20.
General term:
\[T_n=S_n-S_{n-1}=\frac{5}{2}\big[n(n+1)-(n-1)n\big]=\frac{5}{2}\,n\,[(n+1)-(n-1)]=\frac{5}{2}\,n(2)=5n.\]
So \(T_n=5n.\)
(b) Circle \(x^2+y^2-10x-8y+25=0.\)
Complete the square:
\[(x^2-10x)+(y^2-8y)+25=0\Rightarrow (x-5)^2-25+(y-4)^2-16+25=0,\]
\[(x-5)^2+(y-4)^2=16.\]
Centre \((5,4),\) radius \(r=4.\)
(i) Touching the x-axis. The perpendicular distance from the centre \((5,4)\) to the x-axis \((y=0)\) is \(4,\) which equals the radius. Since distance \(=\) radius, the circle is tangent to (touches) the x-axis.
(ii) Point of contact. The contact point is the foot of the perpendicular from the centre to the x-axis, i.e. directly below the centre: \((5,\,0).\)
Akọwa Nkọwa
(a) Sequence with \(S_n=\dfrac{5n^2}{2}+\dfrac{5n}{2}=\dfrac{5n(n+1)}{2}.\)
Each term is \(T_n=S_n-S_{n-1}.\) Compute partial sums:
\(S_1=\dfrac{5(1)(2)}{2}=5,\ S_2=\dfrac{5(2)(3)}{2}=15,\ S_3=\dfrac{5(3)(4)}{2}=30,\ S_4=\dfrac{5(4)(5)}{2}=50.\)
Terms: \(T_1=5,\ T_2=15-5=10,\ T_3=30-15=15,\ T_4=50-30=20.\)
First four terms: 5, 10, 15, 20.
General term:
\[T_n=S_n-S_{n-1}=\frac{5}{2}\big[n(n+1)-(n-1)n\big]=\frac{5}{2}\,n\,[(n+1)-(n-1)]=\frac{5}{2}\,n(2)=5n.\]
So \(T_n=5n.\)
(b) Circle \(x^2+y^2-10x-8y+25=0.\)
Complete the square:
\[(x^2-10x)+(y^2-8y)+25=0\Rightarrow (x-5)^2-25+(y-4)^2-16+25=0,\]
\[(x-5)^2+(y-4)^2=16.\]
Centre \((5,4),\) radius \(r=4.\)
(i) Touching the x-axis. The perpendicular distance from the centre \((5,4)\) to the x-axis \((y=0)\) is \(4,\) which equals the radius. Since distance \(=\) radius, the circle is tangent to (touches) the x-axis.
(ii) Point of contact. The contact point is the foot of the perpendicular from the centre to the x-axis, i.e. directly below the centre: \((5,\,0).\)
Ajụjụ 54 Ripọtì
(a) The point P(3, -5) is rotated through an angle 60° anticlockwise about the origin. (i) Obtain the matrix for the rotation ; (ii) Find the image P' of the point P under the rotation.
(b) A linear transformation is given by \(N : (x, y) \to (2x + 3y, 3x - y)\).
(i) Write down the matrix N of the transformation ; (ii) If \(N^{2} + aN + bI = 0\), where a, b \(\in\) R, \(I\) is the \(2 \times 2\) matrix and \(0\) is the \(2 \times 2\) null matrix, find the values of a and b.
(a)(i) Rotation matrix through \(60^\circ\) anticlockwise about the origin.
\[R=\begin{pmatrix}\cos60^\circ&-\sin60^\circ\\\sin60^\circ&\cos60^\circ\end{pmatrix}=\begin{pmatrix}\tfrac{1}{2}&-\tfrac{\sqrt3}{2}\\[2pt]\tfrac{\sqrt3}{2}&\tfrac{1}{2}\end{pmatrix}.\]
(ii) Image of \(P(3,-5).\)
\[P'=R\begin{pmatrix}3\\-5\end{pmatrix}=\begin{pmatrix}\tfrac12(3)-\tfrac{\sqrt3}{2}(-5)\\[2pt]\tfrac{\sqrt3}{2}(3)+\tfrac12(-5)\end{pmatrix}=\begin{pmatrix}\tfrac{3+5\sqrt3}{2}\\[2pt]\tfrac{3\sqrt3-5}{2}\end{pmatrix}.\]
Numerically \(P'\approx(5.83,\ 0.10).\)
(b) Transformation \(N:(x,y)\to(2x+3y,\ 3x-y).\)
(i) Matrix \(N=\begin{pmatrix}2&3\\3&-1\end{pmatrix}.\)
(ii) Find a and b in \(N^2+aN+bI=0.\)
\[N^2=\begin{pmatrix}2&3\\3&-1\end{pmatrix}\begin{pmatrix}2&3\\3&-1\end{pmatrix}=\begin{pmatrix}13&3\\3&10\end{pmatrix}.\]
Then \(N^2+aN+bI=\begin{pmatrix}13+2a+b&3+3a\\3+3a&10-a+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.\)
Off-diagonal: \(3+3a=0\Rightarrow a=-1.\)
Top-left: \(13+2(-1)+b=0\Rightarrow b=-11.\)
Check bottom-right: \(10-(-1)+(-11)=0.\) Consistent.
\[a=-1,\qquad b=-11.\]
(This is the Cayley-Hamilton relation, since \(\text{trace}(N)=1\) and \(\det(N)=-11.\))
Akọwa Nkọwa
(a)(i) Rotation matrix through \(60^\circ\) anticlockwise about the origin.
\[R=\begin{pmatrix}\cos60^\circ&-\sin60^\circ\\\sin60^\circ&\cos60^\circ\end{pmatrix}=\begin{pmatrix}\tfrac{1}{2}&-\tfrac{\sqrt3}{2}\\[2pt]\tfrac{\sqrt3}{2}&\tfrac{1}{2}\end{pmatrix}.\]
(ii) Image of \(P(3,-5).\)
\[P'=R\begin{pmatrix}3\\-5\end{pmatrix}=\begin{pmatrix}\tfrac12(3)-\tfrac{\sqrt3}{2}(-5)\\[2pt]\tfrac{\sqrt3}{2}(3)+\tfrac12(-5)\end{pmatrix}=\begin{pmatrix}\tfrac{3+5\sqrt3}{2}\\[2pt]\tfrac{3\sqrt3-5}{2}\end{pmatrix}.\]
Numerically \(P'\approx(5.83,\ 0.10).\)
(b) Transformation \(N:(x,y)\to(2x+3y,\ 3x-y).\)
(i) Matrix \(N=\begin{pmatrix}2&3\\3&-1\end{pmatrix}.\)
(ii) Find a and b in \(N^2+aN+bI=0.\)
\[N^2=\begin{pmatrix}2&3\\3&-1\end{pmatrix}\begin{pmatrix}2&3\\3&-1\end{pmatrix}=\begin{pmatrix}13&3\\3&10\end{pmatrix}.\]
Then \(N^2+aN+bI=\begin{pmatrix}13+2a+b&3+3a\\3+3a&10-a+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.\)
Off-diagonal: \(3+3a=0\Rightarrow a=-1.\)
Top-left: \(13+2(-1)+b=0\Rightarrow b=-11.\)
Check bottom-right: \(10-(-1)+(-11)=0.\) Consistent.
\[a=-1,\qquad b=-11.\]
(This is the Cayley-Hamilton relation, since \(\text{trace}(N)=1\) and \(\det(N)=-11.\))
Ajụjụ 55 Ripọtì
In a hotel, the breakfast is a choice between yam (Y) or plantain (P) or both. The Venn diagram shows the choices made by 25 guests of the hotel.
(a) Find the value of x;
(b) What is the probability that a guest chosen at random chose only one of the two?
From the Venn diagram the three disjoint regions are:
- Plantain only (\(P\) region outside the overlap): \(2x+1\)
- Both yam and plantain (the overlap): \(x\)
- Yam only (\(Y\) region outside the overlap): \((x-2)^2\)
Every one of the 25 guests chose yam, plantain, or both, so the three regions add up to the total number of guests.
(a) Finding the value of \(x\)
\[ (2x+1) + x + (x-2)^2 = 25 \]Expand \((x-2)^2 = x^2 - 4x + 4\):
\[ 2x + 1 + x + x^2 - 4x + 4 = 25 \]\[ x^2 - x + 5 = 25 \]\[ x^2 - x - 20 = 0 \]Factorising:
\[ (x-5)(x+4) = 0 \]\[ x = 5 \quad \text{or} \quad x = -4 \]A number of guests cannot be negative, so we reject \(x = -4\).
\[ \boxed{x = 5} \](b) Probability that a guest chose only one of the two
Substitute \(x = 5\) into each region:
\[ \text{Plantain only} = 2(5)+1 = 11 \]\[ \text{Both} = 5 \]\[ \text{Yam only} = (5-2)^2 = 9 \]Check: \(11 + 5 + 9 = 25\), which matches the total number of guests.
Guests who chose only one of the two are those in "plantain only" or "yam only":
\[ 11 + 9 = 20 \]Therefore the required probability is:
\[ P(\text{only one}) = \frac{20}{25} = \frac{4}{5} = 0.8 \]Akọwa Nkọwa
From the Venn diagram the three disjoint regions are:
- Plantain only (\(P\) region outside the overlap): \(2x+1\)
- Both yam and plantain (the overlap): \(x\)
- Yam only (\(Y\) region outside the overlap): \((x-2)^2\)
Every one of the 25 guests chose yam, plantain, or both, so the three regions add up to the total number of guests.
(a) Finding the value of \(x\)
\[ (2x+1) + x + (x-2)^2 = 25 \]Expand \((x-2)^2 = x^2 - 4x + 4\):
\[ 2x + 1 + x + x^2 - 4x + 4 = 25 \]\[ x^2 - x + 5 = 25 \]\[ x^2 - x - 20 = 0 \]Factorising:
\[ (x-5)(x+4) = 0 \]\[ x = 5 \quad \text{or} \quad x = -4 \]A number of guests cannot be negative, so we reject \(x = -4\).
\[ \boxed{x = 5} \](b) Probability that a guest chose only one of the two
Substitute \(x = 5\) into each region:
\[ \text{Plantain only} = 2(5)+1 = 11 \]\[ \text{Both} = 5 \]\[ \text{Yam only} = (5-2)^2 = 9 \]Check: \(11 + 5 + 9 = 25\), which matches the total number of guests.
Guests who chose only one of the two are those in "plantain only" or "yam only":
\[ 11 + 9 = 20 \]Therefore the required probability is:
\[ P(\text{only one}) = \frac{20}{25} = \frac{4}{5} = 0.8 \]Ajụjụ 56 Ripọtì
The following table shows the distribution of marks obtained by some students in an examination.
| Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| Frequency | 50 | 50 | 40 | 60 | 100 | 100 | 50 | 25 | 15 | 10 |
(a) Construct a cumulative frequency table for the distribution
(b) Draw an ogive for the distribution
(c) Use your graph in (b) to determine : (i) semi- interquartile range ; (ii) number of students who failed, if the pass mark for the examination is 37 ; (iii) probability that a student selected at random scored between 20% and 60%.
(a) Cumulative frequency table
| Marks | Frequency, \(f\) | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 0–9 | 50 | −0.5–9.5 | 50 |
| 10–19 | 50 | 9.5–19.5 | 100 |
| 20–29 | 40 | 19.5–29.5 | 140 |
| 30–39 | 60 | 29.5–39.5 | 200 |
| 40–49 | 100 | 39.5–49.5 | 300 |
| 50–59 | 100 | 49.5–59.5 | 400 |
| 60–69 | 50 | 59.5–69.5 | 450 |
| 70–79 | 25 | 69.5–79.5 | 475 |
| 80–89 | 15 | 79.5–89.5 | 490 |
| 90–99 | 10 | 89.5–99.5 | 500 |
Total number of students, \(N=500\).
(b) Ogive
The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\).
(c)
(i) Semi-interquartile range
\(Q_1\) is the \(\frac{N}{4}=125\)th value and \(Q_3\) is the \(\frac{3N}{4}=375\)th value. From the ogive,
\[Q_1\approx 26.0,\qquad Q_3\approx 56.3.\]
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{56.3-26.0}{2}=\boxed{15.15\text{ marks}}.\]
(ii) Number who failed
At a mark of \(37\), the cumulative frequency read from the ogive is approximately \(190\). Hence, \(\boxed{190\text{ students}}\) failed.
(iii) Probability of scoring between \(20\%\) and \(60\%\)
From the cumulative-frequency curve, the number scoring below \(20\%\) is \(100\), while the number scoring below \(60\%\) is \(400\).
\[\text{Number scoring between }20\%\text{ and }60\%=400-100=300.\]
\[P(20\%<X<60\%)=\frac{300}{500}=\boxed{\frac35}.\]
Akọwa Nkọwa
(a) Cumulative frequency table
| Marks | Frequency, \(f\) | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 0–9 | 50 | −0.5–9.5 | 50 |
| 10–19 | 50 | 9.5–19.5 | 100 |
| 20–29 | 40 | 19.5–29.5 | 140 |
| 30–39 | 60 | 29.5–39.5 | 200 |
| 40–49 | 100 | 39.5–49.5 | 300 |
| 50–59 | 100 | 49.5–59.5 | 400 |
| 60–69 | 50 | 59.5–69.5 | 450 |
| 70–79 | 25 | 69.5–79.5 | 475 |
| 80–89 | 15 | 79.5–89.5 | 490 |
| 90–99 | 10 | 89.5–99.5 | 500 |
Total number of students, \(N=500\).
(b) Ogive
The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\).
(c)
(i) Semi-interquartile range
\(Q_1\) is the \(\frac{N}{4}=125\)th value and \(Q_3\) is the \(\frac{3N}{4}=375\)th value. From the ogive,
\[Q_1\approx 26.0,\qquad Q_3\approx 56.3.\]
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{56.3-26.0}{2}=\boxed{15.15\text{ marks}}.\]
(ii) Number who failed
At a mark of \(37\), the cumulative frequency read from the ogive is approximately \(190\). Hence, \(\boxed{190\text{ students}}\) failed.
(iii) Probability of scoring between \(20\%\) and \(60\%\)
From the cumulative-frequency curve, the number scoring below \(20\%\) is \(100\), while the number scoring below \(60\%\) is \(400\).
\[\text{Number scoring between }20\%\text{ and }60\%=400-100=300.\]
\[P(20\%<X<60\%)=\frac{300}{500}=\boxed{\frac35}.\]
Ajụjụ 57 Ripọtì
The position vectors of points 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 position vector of M ;
(b) \(|\overrightarrow{PM}|\) and \(|\overrightarrow{PQ}|\) ;
(c) the acute angle between \(\overrightarrow{PM}\) and \(\overrightarrow{PQ}\), correct to 1 decimal place ;
(d) the area of PQRM.
Position vectors: \(P=(4,-5),\ Q=(1,3),\ R=(-5,2).\) In parallelogram \(PQRM\) the diagonals \(PR\) and \(QM\) bisect each other, so \(P+R=Q+M.\)
(a) Position vector of M.
\[M=P+R-Q=(4-5-1,\ -5+2-3)=(-2,-6),\quad\text{i.e. }M=-2\mathbf{i}-6\mathbf{j}.\]
(b) \(|\overrightarrow{PM}|\) and \(|\overrightarrow{PQ}|.\)
\(\overrightarrow{PM}=M-P=(-6,-1),\ \ |\overrightarrow{PM}|=\sqrt{36+1}=\sqrt{37}\approx6.08.\)
\(\overrightarrow{PQ}=Q-P=(-3,8),\ \ |\overrightarrow{PQ}|=\sqrt{9+64}=\sqrt{73}\approx8.54.\)
(c) Acute angle between \(\overrightarrow{PM}\) and \(\overrightarrow{PQ}.\)
\[\cos\theta=\frac{\overrightarrow{PM}\cdot\overrightarrow{PQ}}{|\overrightarrow{PM}|\,|\overrightarrow{PQ}|}=\frac{(-6)(-3)+(-1)(8)}{\sqrt{37}\,\sqrt{73}}=\frac{18-8}{\sqrt{2701}}=\frac{10}{51.97}\approx0.1924.\]
\[\theta=\cos^{-1}(0.1924)\approx78.9^\circ.\]
(d) Area of PQRM. Area \(=|\overrightarrow{PM}\times\overrightarrow{PQ}|\) (magnitude of the 2D cross product of adjacent sides):
\[\text{Area}=|(-6)(8)-(-1)(-3)|=|-48-3|=51\text{ square units.}\]
Akọwa Nkọwa
Position vectors: \(P=(4,-5),\ Q=(1,3),\ R=(-5,2).\) In parallelogram \(PQRM\) the diagonals \(PR\) and \(QM\) bisect each other, so \(P+R=Q+M.\)
(a) Position vector of M.
\[M=P+R-Q=(4-5-1,\ -5+2-3)=(-2,-6),\quad\text{i.e. }M=-2\mathbf{i}-6\mathbf{j}.\]
(b) \(|\overrightarrow{PM}|\) and \(|\overrightarrow{PQ}|.\)
\(\overrightarrow{PM}=M-P=(-6,-1),\ \ |\overrightarrow{PM}|=\sqrt{36+1}=\sqrt{37}\approx6.08.\)
\(\overrightarrow{PQ}=Q-P=(-3,8),\ \ |\overrightarrow{PQ}|=\sqrt{9+64}=\sqrt{73}\approx8.54.\)
(c) Acute angle between \(\overrightarrow{PM}\) and \(\overrightarrow{PQ}.\)
\[\cos\theta=\frac{\overrightarrow{PM}\cdot\overrightarrow{PQ}}{|\overrightarrow{PM}|\,|\overrightarrow{PQ}|}=\frac{(-6)(-3)+(-1)(8)}{\sqrt{37}\,\sqrt{73}}=\frac{18-8}{\sqrt{2701}}=\frac{10}{51.97}\approx0.1924.\]
\[\theta=\cos^{-1}(0.1924)\approx78.9^\circ.\]
(d) Area of PQRM. Area \(=|\overrightarrow{PM}\times\overrightarrow{PQ}|\) (magnitude of the 2D cross product of adjacent sides):
\[\text{Area}=|(-6)(8)-(-1)(-3)|=|-48-3|=51\text{ square units.}\]
