WAEC SSCE - Further Mathematics - 2008
Tambaya 1 Rahoto
If \(f(x) = 2x^{2} - 3x - 1\), find the value of x for which f(x) is minimum.
Bayanin Amsa
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.
Tambaya 2 Rahoto
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?
Bayanin Amsa
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.
Tambaya 3 Rahoto
If \(\begin{vmatrix} 4 & x \\ 5 & 3 \end{vmatrix} = 32\), find the value of x.
Bayanin Amsa
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.
Tambaya 4 Rahoto
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.
Bayanin Amsa
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) \(\frac{167}{576}\).
Tambaya 5 Rahoto
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 \(f^{-1}\).
Bayanin Amsa
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 \(\boxed{\textbf{(A) }f^{-1} : x \to \frac{2x+3}{x-1}, x \neq 1}\).
Tambaya 6 Rahoto
| 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.
Bayanin Amsa
Tambaya 7 Rahoto
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 \(5 * x = 37\).
Bayanin Amsa
Tambaya 8 Rahoto
Express \(r = (12, 210°)\) in the form \(a i + b j\).
Bayanin Amsa
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: \(6(-\sqrt{3}i - j)\).
Tambaya 9 Rahoto
The sum of the first n terms of a linear sequence is \(S_{n} = n^{2} + 2n\). Determine the general term of the sequence.
Bayanin Amsa
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}.
Tambaya 10 Rahoto
Express \(\frac{1}{1 - \sin 45°}\) in surd form.
Bayanin Amsa
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: \(2 + \sqrt{2}\).
Tambaya 11 Rahoto
Find the locus of points which is equidistant from P(4, 5) and Q(-6, -1).
Bayanin Amsa
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.
Tambaya 12 Rahoto
Simplify \(\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}}\)
Bayanin Amsa
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 \(\frac{5\sqrt{2}}{2}\).
Tambaya 13 Rahoto
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"?
Bayanin Amsa
Tambaya 14 Rahoto
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.
Tambaya 15 Rahoto
The polynomial \(2x^{3} + x^{2} - 3x + p\) has a remainder of 20 when divided by (x - 2). Find the value of constant p.
Bayanin Amsa
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).
Tambaya 16 Rahoto
Find the value of the constant k for which \(a = 4 i - k j\) and \(b = 3 i + 8 j\) are perpendicular.
Bayanin Amsa
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.
Tambaya 17 Rahoto
Given that \(\overrightarrow{AB} = 5i + 3j\) and \(\overrightarrow{AC} = 2i + 5j\), find \(\overrightarrow{BC}\).
Bayanin Amsa
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.
Tambaya 18 Rahoto
Express \(\frac{7\pi}{6}\) radians in degrees.
Bayanin Amsa
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.
Tambaya 19 Rahoto
If n items are arranged two at a time, the number obtained is 20. Find the value of n.
Bayanin Amsa
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.
Tambaya 20 Rahoto
Two forces \(F_{1} = (10N, 020°)\) and \(F_{2} = (7N, 200°)\) act on a particle. Find the resultant force.
Bayanin Amsa
Tambaya 21 Rahoto
The distance between P(x, 7) and Q(6, 19) is 13 units. Find the values of x.
Bayanin Amsa
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.
Tambaya 22 Rahoto
If \(y = x^{2} - 6x + 11\) is written in the form \(y = a(x - h)^{2} + k\), find the value of \((a + h + k)\).
Bayanin Amsa
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.
Tambaya 24 Rahoto
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?
Bayanin Amsa
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.
Tambaya 25 Rahoto
Which of the following quadratic curves will not intersect with the x- axis?
Bayanin Amsa
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 \(y = 3x^{2} - 2x + 4\).
Tambaya 26 Rahoto
A rectangle has a perimeter of 24m. If its area is to be maximum, find its dimension.
Bayanin Amsa
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.
Tambaya 27 Rahoto
| 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?
Bayanin Amsa
Tambaya 28 Rahoto
| 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?
Bayanin Amsa
Tambaya 29 Rahoto
If events A and B are independent and \(P(A) = \frac{7}{12}\) and \(P(A \cap B) = \frac{1}{4}\), find P(B).
Bayanin Amsa
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.
Tambaya 30 Rahoto
If \(p = \begin{pmatrix} 2 \\ -2 \end{pmatrix} \) and \(q = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\), find \(|q - \frac{1}{2}p|\).
Bayanin Amsa
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: \(\sqrt{29}\).
Tambaya 31 Rahoto
If \(P = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}\) and \(Q = \begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}\), find PQ.
Bayanin Amsa
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: \(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\).
Tambaya 32 Rahoto
Find the range of values of x for which \(2x^{2} + 7x - 15 > 0\).
Tambaya 33 Rahoto
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.
Bayanin Amsa
Tambaya 34 Rahoto
What is the coordinate of the centre of the circle \(5x^{2} + 5y^{2} - 15x + 25y - 3 = 0\)?
Bayanin Amsa
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) \((\frac{3}{2}, -\frac{5}{2})\).
Tambaya 35 Rahoto
Find the derivative of \(3x^{2} + \frac{1}{x^{2}}\)
Bayanin Amsa
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) \(6x - \frac{2}{x^{3}}\).
Tambaya 36 Rahoto
The sum of the first n terms of a linear sequence is \(S_{n} = n^{2} + 2n\). Find the common difference of the sequence.
Tambaya 38 Rahoto
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.
Bayanin Amsa
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 \(\frac{1}{2}\).
Tambaya 39 Rahoto
If \(2\log_{4} 2 = x + 1\), find the value of x.
Tambaya 40 Rahoto
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%.
Tambaya 41 Rahoto
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 \(\cos (A + B)\).
Tambaya 42 Rahoto
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.
Tambaya 43 Rahoto
(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.
Tambaya 44 Rahoto
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.
Tambaya 45 Rahoto
(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.
Bayanin Amsa
None
Tambaya 46 Rahoto
Differentiate, with respect to x, \(x^{3} + 2x\) from the first principle.
Tambaya 47 Rahoto
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.
Tambaya 48 Rahoto
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.
Tambaya 49 Rahoto
(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.
Bayanin Amsa
None
Tambaya 50 Rahoto
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.
Tambaya 51 Rahoto
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?
Bayanin Amsa
None
Tambaya 52 Rahoto
(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.
Tambaya 53 Rahoto
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 \(f(-4)\).
Bayanin Amsa
None
Tambaya 54 Rahoto
(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.
Tambaya 55 Rahoto
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.
Tambaya 56 Rahoto
(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.
Bayanin Amsa
None
Tambaya 57 Rahoto
(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.
