WAEC SSCE - Further Mathematics - 2006

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

We have the equation \(\begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -6 \\ P \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix}\). We can solve for P by performing matrix multiplication and solving the resulting system of linear equations: \(\begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -6 \\ P \end{pmatrix} = \begin{pmatrix} -6 - 3P \\ P + 4P \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix}\) This gives us two equations: \(-6 - 3P = 3\) \(5P = -26\) Solving for P, we get: \(-6 - 3P = 3 \Rightarrow -3P = 9 \Rightarrow P = -3\) \(5P = -26 \Rightarrow P = \frac{-26}{5}\) Therefore, the value of P is -3. Note that the system of linear equations is consistent, meaning that there is a unique solution.

Akọwa Nkọwa

The binary operation ♦ is defined as \(a ♦ b = \frac{ab}{4}\). So, to find the value of \(\sqrt{2} ♦ \sqrt{6}\), we just need to substitute \(\sqrt{2}\) for \(a\) and \(\sqrt{6}\) for \(b\), and then simplify the expression. \[\sqrt{2} ♦ \sqrt{6} = \frac{\sqrt{2}\cdot\sqrt{6}}{4} = \frac{\sqrt{12}}{4} = \frac{\sqrt{4}\cdot\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}\] Therefore, the value of \(\sqrt{2} ♦ \sqrt{6}\) is \(\frac{\sqrt{3}}{2}\).

Akọwa Nkọwa

To find the center of the given circle, we need to write the equation in standard form, which is: \[(x - h)^2 + (y - k)^2 = r^2\] Where (h, k) is the center of the circle and r is the radius. We start by completing the square for both x and y terms, by moving the constant term to the right side and grouping the x and y terms together: \[4x^2 - 5x + 4y^2 + 3y = 2\] To complete the square for the x terms, we need to add and subtract \(\frac{5}{4}\) times the square of the coefficient of x (which is 2), inside the first bracket: \[4\left(x^2 - \frac{5}{4}x + \left(\frac{5}{4}\right)^2\right) + 4y^2 + 3y = 2 + 4\left(\frac{5}{4}\right)^2\] We do the same for the y terms, by adding and subtracting \(\frac{3}{8}\) times the square of the coefficient of y (which is \(\frac{3}{2}\)), inside the second bracket: \[4\left(x^2 - \frac{5}{4}x + \left(\frac{5}{4}\right)^2\right) + 4\left(y^2 + \frac{3}{8}y + \left(\frac{3}{16}\right)^2\right) = 2 + 4\left(\frac{5}{4}\right)^2 + 4\left(\frac{3}{16}\right)^2\] Now we have the equation in standard form, and we can read off the center and radius: Center = \(\left(\frac{5}{4}, -\frac{3}{8}\right)\) Radius = \(\sqrt{\frac{129}{16}} = \frac{3\sqrt{43}}{4}\) Therefore, the correct answer is option (C) \((\frac{5}{8}, -\frac{3}{8})\).

Akọwa Nkọwa

We can simplify the expression by first rewriting each of the bases as powers of 2: $$8^{n} \times 2^{2n} \div 4^{3n} = (2^{3})^{n} \times 2^{2n} \div (2^{2})^{3n}.$$ Using the laws of exponents, we can simplify this to: $$(2^{3n}) \times 2^{2n} \div 2^{6n} = 2^{3n+2n-6n} = 2^{-n}.$$ Therefore, the simplified expression is \(2^{-n}\).

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To solve this problem, we want to find a vector c that satisfies the equation 4a + 3c = b. We can rearrange this equation to get: 3c = b - 4a Now we just need to solve for c. We can do this by dividing both sides by 3: c = (b - 4a) / 3 Substituting the values of a and b given in the problem, we get: c = (\begin{pmatrix} -3 \\ 5 \end{pmatrix} - 4\begin{pmatrix} 3 \\ 2 \end{pmatrix}) / 3 c = (\begin{pmatrix} -3 \\ 5 \end{pmatrix} - \begin{pmatrix} 12 \\ 8 \end{pmatrix}) / 3 c = \begin{pmatrix} -15/3 \\ -3/3 \end{pmatrix} c = \begin{pmatrix} -5 \\ -1 \end{pmatrix} Therefore, the answer is option B, \(\begin{pmatrix} -5 \\ -1 \end{pmatrix}\).

Akọwa Nkọwa

To form a committee of 4 with a chairman, we first need to select a man to be the chairman from the 5 men available. This can be done in 5 ways. Then, we need to select 3 people from the remaining 7 people (4 men and 3 women) to complete the committee. This can be done in 7 choose 3 ways, which is equal to 35. Therefore, the total number of ways to form a committee of 4 with a chairman from a group of 5 men and 3 women is: 5 * 35 = 175 So the correct answer is option (D) 175.

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To find the time it takes for the particle to come to rest, we can use the formula: time = (initial velocity) / (deceleration) where the deceleration is the rate at which the particle's velocity decreases due to the force acting against it. The deceleration can be calculated using Newton's Second Law, which states that force is equal to mass times acceleration: force = mass x acceleration Rearranging this formula, we get: acceleration = force / mass Substituting the given values, we get: acceleration = 10 N / 2.5 kg = 4 m/s^2 Therefore, the time it takes for the particle to come to rest is: time = 12 m/s / 4 m/s^2 = 3.0 s So, the answer is 3.0 s.

Akọwa Nkọwa

Akọwa Nkọwa

To solve this problem, we need to use the formula for the area of a sector of a circle, which is: Area of sector = (θ/2) × r² where θ is the central angle subtended by the sector, r is the radius of the circle, and the angle is measured in radians. In this problem, we are given the area of the sector and the central angle, so we can plug in these values and solve for the radius: 3 = (1.5/2) × r² Multiplying both sides by 2/1.5, we get: 4 = r² Taking the square root of both sides, we get: r = 2 Therefore, the radius of the circle is 2 cm. We can check our answer by plugging it back into the formula for the area of the sector: Area of sector = (1.5/2) × 2² = 1.5 × 2 = 3 This confirms that our answer is correct.

Akọwa Nkọwa

Akọwa Nkọwa

We can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration. The formula can be written as F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the force acting on the object is 32 N, and the acceleration produced is 8 \(ms^{-2}\). Since the object is at rest, its initial velocity is zero. So, we can use the formula F = ma to find the value of m. Rearranging the formula, we get m = F/a. Substituting the given values, we get m = 32 N / 8 \(ms^{-2}\) = 4 kg. Therefore, the value of m is 4 kg.

Akọwa Nkọwa

Akọwa Nkọwa

The sum of the roots of the quadratic equation \(ax^{2} + bx + c = 0\) is given by the formula \(-\frac{b}{a}\), and the product of the roots is given by the formula \(\frac{c}{a}\). Therefore, for the quadratic equation \(2x^{2} + kx + 5 = 0\), we have: \[\alpha + \beta = -\frac{k}{2}\] \[\alpha \beta = \frac{5}{2}\] Squaring the first equation, we get: \[(\alpha + \beta)^{2} = \alpha^{2} + 2\alpha\beta + \beta^{2} = \frac{k^{2}}{4}\] Substituting the given value of \(\alpha^{2} + \beta^{2} = -1\), we have: \[-1 + 2\alpha\beta = \frac{k^{2}}{4}\] Substituting the value of \(\alpha \beta = \frac{5}{2}\), we have: \[-1 + 2\left(\frac{5}{2}\right) = \frac{k^{2}}{4}\] Simplifying the left-hand side, we get: \[4 = \frac{k^{2}}{4}\] Multiplying both sides by 4, we get: \[16 = k^{2}\] Taking the square root of both sides, we get: \[k = \pm 4\] Therefore, the values of k are \(\pm 4\).

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To solve this inequality, we can start by finding the roots of the quadratic equation \(2x^2+5x-3=0\). We can do this by factoring the quadratic or by using the quadratic formula. Factoring the quadratic, we get: \begin{align*} 2x^2+5x-3&=0\\ (2x-1)(x+3)&=0 \end{align*} So the roots are \(x=\frac{1}{2}\) and \(x=-3\). These are the values of \(x\) that make the left-hand side of the inequality equal to zero. Now, we can use these roots to determine the sign of the expression \(2x^2+5x-3\) for different values of \(x\). We can do this by testing values of \(x\) in the intervals between the roots and outside the roots. For example, if we pick a value of \(x\) less than \(-3\), say \(x=-4\), then we have: \begin{align*} 2x^2+5x-3&=2(-4)^2+5(-4)-3\\ &=32-20-3\\ &=9 \end{align*} Since \(9\) is positive, this means that any value of \(x\) less than \(-3\) makes the expression \(2x^2+5x-3\) positive. Similarly, if we pick a value of \(x\) between \(-3\) and \(\frac{1}{2}\), say \(x=0\), then we have: \begin{align*} 2x^2+5x-3&=2(0)^2+5(0)-3\\ &=-3 \end{align*} Since \(-3\) is negative, this means that any value of \(x\) between \(-3\) and \(\frac{1}{2}\) makes the expression \(2x^2+5x-3\) negative. Finally, if we pick a value of \(x\) greater than \(\frac{1}{2}\), say \(x=1\), then we have: \begin{align*} 2x^2+5x-3&=2(1)^2+5(1)-3\\ &=4+5-3\\ &=6 \end{align*} Since \(6\) is positive, this means that any value of \(x\) greater than \(\frac{1}{2}\) makes the expression \(2x^2+5x-3\) positive. Putting all of this together, we can see that the inequality \(2x^2+5x-3\geq 0\) is true when \(x\leq -3\) or \(x\geq \frac{1}{2}\). Therefore, the correct option is: - \(x \leq -3\) or \(x \geq \frac{1}{2}\)

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We can use the binomial theorem to find the coefficient of \(x^{4}\) in the expansion of \((1 - 2x)^{6}\). The binomial theorem states that the \(n\)th power of a binomial \((a + b)^{n}\) can be expressed as the sum of the terms of the form \(a^{n-r}b^{r}\), where \(r\) ranges from 0 to \(n\). In this case, the binomial is \((1 - 2x)\), so we can expand it using the binomial theorem as follows: \[(1 - 2x)^{6} = \sum_{r=0}^{6} {6 \choose r} (1)^{6-r} (-2x)^{r}\] where \({6 \choose r}\) is the binomial coefficient. To find the coefficient of \(x^{4}\), we need to find the term where \(r\) is 2. So, we can substitute \(r = 2\) into the formula above to get: \[{6 \choose 2} (1)^{4} (-2x)^{2} = 15 \times 4x^{2} = 60x^{2}\] However, we are looking for the coefficient of \(x^{4}\), not \(x^{2}\). Since the exponent of \(x\) is 2 in the term we just calculated, we need to multiply it by another power of \(x^{2}\) to get \(x^{4}\). This means we need to find the term where \(r\) is 4. Substituting \(r = 4\) into the formula above, we get: \[{6 \choose 4} (1)^{2} (-2x)^{4} = 15 \times 16x^{4} = 240x^{4}\] Therefore, the coefficient of \(x^{4}\) in the binomial expansion of \((1 - 2x)^{6}\) is 240. Hence, the answer is 240.

Akọwa Nkọwa

Akọwa Nkọwa

Let's denote the sum of ages of the n men in the club by S. Then we know that: Mean age of n men = S/n = 50 Multiplying both sides by n, we get: S = 50n After two men aged 55 and 63 years leave the club, the sum of ages of the remaining men becomes: S' = S - 55 - 63 = 50n - 118 The new mean age is reduced by 1 year, so we have: S'/(n-2) = 50 - 1 Simplifying this equation and substituting S', we get: (50n - 118)/(n-2) = 49 Multiplying both sides by n-2, we get: 50n - 118 = 49(n-2) Simplifying this equation, we get: n = 20 Therefore, there were originally 20 men in the club.

Akọwa Nkọwa

If \((x - 3)\) is a factor of \(2x^{3} + 3x^{2} - 17x - 30\), then we can use long division or synthetic division to divide the polynomial by \((x - 3)\) and find the remaining factors.

Using long division, we have:

          2x^2 + 9x + 10
    ------------------------
x - 3 | 2x^3 + 3x^2 - 17x - 30
         - (2x^3 - 6x^2)
         --------------
                  9x^2 - 17x
                  - (9x^2 - 27x)
                  -----------
                           10x - 30
                           - (10x - 30)
                           ---------
                                  0

The result of the division is \(2x^{2} + 9x + 10\), which is a quadratic polynomial. Therefore, the remaining factors are given by factoring this quadratic polynomial. We can factor it as follows:

Therefore, the complete factorization of \(2x^{3} + 3x^{2} - 17x - 30\) is:

So, the correct option is (2x + 5)(x + 2).

Akọwa Nkọwa

We can use the change of base formula to evaluate \(\log_{0.25} 8\). The change of base formula states that for any positive numbers \(a\), \(b\), and \(c\) with \(a \neq 1\), we have: \(\log_{a} b = \frac{\log_{c} b}{\log_{c} a}\) In this case, we can choose any base we like, but it is convenient to use base 2 because 8 and 0.25 are powers of 2. So, we can rewrite the expression as: \(\log_{0.25} 8 = \frac{\log_{2} 8}{\log_{2} 0.25}\) Now, we can evaluate the logarithms on the right-hand side using the rules of logarithms. We know that \(2^{3} = 8\) and \(2^{-2} = 0.25\), so we have: \(\log_{2} 8 = 3\) and \(\log_{2} 0.25 = -2\) Substituting these values into the equation above, we get: \(\log_{0.25} 8 = \frac{3}{-2} = -\frac{3}{2}\) Therefore, the value of \(\log_{0.25} 8\) is \(-\frac{3}{2}\). Hence, the answer is \(-\frac{3}{2}\).

Akọwa Nkọwa

The equation given is in the form y = mx + b, where m is the slope and b is the y-intercept of the line of best fit. In this case, the slope is 0.42 and the y-intercept is 19.33. To estimate the value of y when x = 15, we simply substitute x = 15 into the equation and solve for y: y = 19.33 + 0.42(15) y = 19.33 + 6.3 y = 25.63 Therefore, the estimated value of y when x = 15 is 25.63.

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To evaluate the limit \(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\), we can start by substituting \(x = 1\) directly into the expression. However, this results in an expression of the form \(\frac{0}{0}\), which is undefined. To evaluate such a limit, we can use algebraic techniques to simplify the expression before taking the limit. In this case, we can factor the denominator to obtain: \[\frac{1 - x}{x^{2} - 3x + 2} = \frac{1 - x}{(x - 1)(x - 2)} = -\frac{1}{x - 2}\] Now, we can take the limit as \(x\) approaches 1 by substituting \(x = 1\) directly into the simplified expression. This gives: \[\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2} = \lim \limits_{x \to 1} -\frac{1}{x - 2} = -\frac{1}{-1} = 1\] Therefore, the limit of the expression as \(x\) approaches 1 is equal to 1. So, the correct option is 1.

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The equation of a line can be represented in the form y = mx + b, where m is the slope of the line and b is the y-intercept (the point where the line intersects the y-axis). Since the line is parallel to the y-axis, its slope is undefined (division by zero is not defined). Therefore, the equation of the line cannot be represented in the form y = mx + b. However, we know that the line passes through the point (0, -1). This means that the x-coordinate of every point on the line is 0 (since the line is parallel to the y-axis and does not move horizontally). Therefore, the equation of the line is x = 0. This equation means that every point on the line has an x-coordinate of 0, which is the y-axis. So, the line is simply the y-axis itself. Thus, the answer is x = 0.

Akọwa Nkọwa

\(P = {x : 1 \leq x \leq 6}\) means that P is the set of all real numbers between 1 and 6 (including 1 and 6). Similarly, \(Q = {x : 2 < x < 9}\) means that Q is the set of all real numbers between 2 and 9 (excluding 2 and 9). To find \(P \cap Q\), we need to find the set of all elements that are common to both P and Q. From the definitions of P and Q, we can see that the range of values that satisfy both P and Q is from 2 to 6 (including 2 and 6). Therefore, \(P \cap Q\) is the set of all real numbers between 2 and 6 (including 2 and 6). Hence, the correct option is (d) \({x : 2 < x \leq 6}\).

Akọwa Nkọwa

Akọwa Nkọwa

To find the angle between two vectors, we can use the dot product formula: cos θ = (A dot B) / (|A| |B|) where A and B are the two vectors, A dot B is their dot product, and |A| and |B| are their magnitudes. Let's apply this formula to the given vectors: A = 5i + 12j B = -2i + 3j A dot B = (5*-2) + (12*3) = -10 + 36 = 26 |A| = sqrt(5^2 + 12^2) = 13 |B| = sqrt((-2)^2 + 3^2) = sqrt(13) cos θ = (26) / (13 * sqrt(13)) Using a calculator, we can find that cos θ ≈ 0.801 To find the angle θ itself, we can take the inverse cosine (cos^-1) of this value: θ ≈ cos^-1(0.801) ≈ 37.8° However, this is only half of the actual angle between the two vectors. To get the full angle, we need to take into account the direction of the vectors. If we draw the two vectors on a coordinate plane, we can see that they are in different quadrants: A is in the second quadrant and B is in the fourth quadrant. The angle between them will be the difference between their angles relative to the positive x-axis. Using the inverse tangent (tan^-1) function, we can find the angles of each vector: Angle of A = tan^-1(12/5) ≈ 67.4° Angle of B = tan^-1(-3/2) ≈ -56.3° (Note that we use a negative angle for B because it is in the fourth quadrant, where angles are negative.) The angle between the vectors will be the difference between these two angles: θ = |67.4° - (-56.3°)| = 123.7° However, this is the full angle between the vectors, and the question only asks for the acute angle (the smaller of the two possible angles). To find the acute angle, we subtract the full angle from 180°: Acute angle = 180° - 123.7° ≈ 56.3° Therefore, the answer is option B: 56.3°.

Akọwa Nkọwa

The given series is 96 + 24 + 6 + ... , and we need to find its sum. First, we can observe that each term in the series is obtained by dividing the previous term by 4. So, the common ratio between consecutive terms is 1/4. Let S be the sum of the series. Then we have: S = 96 + 24 + 6 + ... Dividing both sides by 4, we get: S/4 = 24 + 6 + 1.5 + ... Now, if we subtract the second equation from the first, we get: S - S/4 = 96 Simplifying, we get: 3S/4 = 96 Multiplying both sides by 4/3, we get: S = 128 Therefore, the sum of the given series is 128. So, the correct option is (B) 128.

Akọwa Nkọwa

The expression \(\frac{^{n}P_{4}}{^{n}C_{4}}\) represents the number of ways to choose and arrange 4 objects out of a total of n objects, divided by the number of ways to choose 4 objects out of n objects without arranging them. The number of ways to choose and arrange 4 objects out of n objects is given by the formula for permutations, which is \(^{n}P_{4} = \frac{n!}{(n-4)!}\). The number of ways to choose 4 objects out of n objects without arranging them is given by the formula for combinations, which is \(^{n}C_{4} = \frac{n!}{4!(n-4)!}\). So, \(\frac{^{n}P_{4}}{^{n}C_{4}} = \frac{\frac{n!}{(n-4)!}}{\frac{n!}{4!(n-4)!}} = \frac{4!}{1!} = 24\). Therefore, the answer is 24.

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The probability that Kojo hits the target is \(\frac{2}{5}\), which means that the probability that he misses the target is \(1-\frac{2}{5}=\frac{3}{5}\). Similarly, the probability that Kwame misses the target is \(1-\frac{1}{3}=\frac{2}{3}\). Now, we want to find the probability that none of them hit the target. Since they are shooting independently, we can multiply their probabilities of missing the target: Probability that both miss the target = Probability that Kojo misses the target AND Probability that Kwame misses the target = \(\frac{3}{5} \times \frac{2}{3} = \frac{6}{15} = \frac{2}{5}\) Therefore, the probability that none of them hit the target is \(\frac{2}{5}\).

Akọwa Nkọwa

We can use the kinematic equation: h = ut + 1/2gt^2 where h is the initial height (45m), u is the initial velocity (40 m/s), g is the acceleration due to gravity (10 m/s^2), and t is the time taken for the particle to hit the ground. At the point where the particle hits the ground, h = 0, so we can solve for t: 0 = 40t - 1/2(10)t^2 Simplifying this equation gives: 0 = 5t(8 - t) This equation has two solutions: t = 0 and t = 8. We can ignore the t = 0 solution because it corresponds to the initial moment when the particle is projected upward. So, the particle will hit the ground after 8 seconds. Therefore, the answer is: 9s (since the particle was projected from a height of 45m, it takes 8 seconds to fall and hit the ground. However, the total time from the projection until it hits the ground would be 9 seconds).

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To find the point of intersection of the diagonals of a parallelogram, we need to find the midpoint of the segment that connects the two diagonals. This midpoint is the point of intersection of the diagonals. Let's start by finding the equation of the line containing the diagonal PR. The slope of PR is: $$m_{PR}=\frac{y_R-y_P}{x_R-x_P}=\frac{6-3}{2-(-2)}=\frac{3}{4}$$ Using point-slope form, the equation of line PR is: $$y-3=\frac{3}{4}(x+2)$$ Simplifying, we get: $$y=\frac{3}{4}x+\frac{15}{4}$$ Now let's find the equation of the line containing the diagonal QS. The slope of QS is: $$m_{QS}=\frac{y_S-y_Q}{x_S-x_Q}=\frac{5-4}{-1-1}=\frac{-1}{2}$$ Using point-slope form, the equation of line QS is: $$y-4=\frac{-1}{2}(x-1)$$ Simplifying, we get: $$y=-\frac{1}{2}x+\frac{9}{2}$$ To find the point of intersection of the diagonals, we need to find the midpoint of the segment that connects the two diagonals. Let's call this midpoint M. The midpoint M is the average of the coordinates of the endpoints of the segment, which are the intersection points of the diagonals with their respective sides. To find the intersection point of PR with side QS, we solve the system of equations formed by the equations of lines PR and QS: $$\begin{cases}y=\frac{3}{4}x+\frac{15}{4}\\y=-\frac{1}{2}x+\frac{9}{2}\end{cases}$$ Solving for x and y, we get: $$\begin{cases}x=\frac{6}{5}\\y=\frac{39}{10}\end{cases}$$ So the intersection point of PR with side QS is (\(\frac{6}{5}, \frac{39}{10}\)). To find the intersection point of QS with side PR, we solve the system of equations formed by the equations of lines QS and PR: $$\begin{cases}y=-\frac{1}{2}x+\frac{9}{2}\\y=\frac{3}{4}x+\frac{15}{4}\end{cases}$$ Solving for x and y, we get: $$\begin{cases}x=-\frac{6}{5}\\y=\frac{21}{10}\end{cases}$$ So the intersection point of QS with side PR is (-\(\frac{6}{5}\), \(\frac{21}{10}\)). The midpoint M is the average of these two points, so we have: $$M=(\frac{\frac{6}{5}-\frac{6}{5}}{2}, \frac{\frac{39}{10}+\frac{21}{10}}{2})=(0, \frac{3}{2})$$ Therefore, the coordinates of the point of intersection of the diagonals is (0, \(\frac{3}{2}\)). So the answer is option (C), \((0, 4\frac{1}{2})\).

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To find \(f^{-1}(-\frac{1}{2})\), we need to find the value of x such that \(f(x) = -\frac{1}{2}\). We know that \(f(x) = \frac{1}{2 - x}\), so we set \(\frac{1}{2 - x} = -\frac{1}{2}\) and solve for x: \[\frac{1}{2 - x} = -\frac{1}{2}\] \[2 - x = -2\] \[x = 4\] Therefore, \(f^{-1}(-\frac{1}{2}) = 4\). Note that we were able to solve for \(f^{-1}(y)\) by first setting \(y = f(x)\), then solving for x. This is because the inverse function of f, denoted by \(f^{-1}\), is defined such that \(f^{-1}(f(x)) = x\) for all x in the domain of f.

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(a) Triangles from 6 points in a plane. A triangle needs any 3 of the points (assuming no three are collinear), and order does not matter, so use combinations:

\[\binom{6}{3}=\frac{6!}{3!\,3!}=\frac{6\times5\times4}{3\times2\times1}=20.\]

So 20 triangles can be formed.

(b) Seating 6 people in a row with father and mother NOT together.

Total arrangements of 6 people in a row: \(6!=720.\)

Arrangements with father and mother together: treat the pair as a single block, giving \(5!\) arrangements, and the two of them can swap within the block in \(2!\) ways:

\[5!\times2!=120\times2=240.\]

Therefore arrangements with them NOT together:

\[720-240=480.\]

Akọwa Nkọwa

(a) Let masses be \(m_P\) and \(m_Q\). Taking the original direction of P as positive, after collision P reverses to \(-\tfrac12 u\) and Q moves \(+u\). Conservation of momentum:

Ratio \(m_P:m_Q=\textbf{2:5}\).

(b) For forces 3 N and 7 N with resultant 5 N at angle \(\theta\) between them:

(c) \(\vec{AB}=\begin{pmatrix}-4\\6\end{pmatrix},\ \vec{CB}=\begin{pmatrix}2\\-3\end{pmatrix}\). V is the midpoint of AB, so \(\vec{BV}=-\tfrac12\vec{AB}=\begin{pmatrix}2\\-3\end{pmatrix}\). Then

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None

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(a) Weight \(W=mg=15\times 10=150\ \text{N}\). String \(XP\) makes \(60^\circ\) and \(YP\) makes \(40^\circ\) with the downward vertical. The angles between the three forces at P are: between the two tensions \(60^\circ+40^\circ=100^\circ\); between \(T_{XP}\) and the weight \(180^\circ-60^\circ=120^\circ\); between \(T_{YP}\) and the weight \(180^\circ-40^\circ=140^\circ\). By Lami's theorem:

Tension in \(XP\approx\textbf{97.91 N}\), tension in \(YP\approx\textbf{131.91 N}\).

(b) \(h=2+20t+kt^2\), so \(\dfrac{dh}{dt}=20+2kt\). At the highest point the velocity is zero at \(t=2\):

(ii) Highest point: \(h(2)=2+20(2)+(-5)(2)^2=2+40-20=\textbf{22 m}\) above the point of throw.

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(a) Cumulative-frequency table

Time (hours)	Frequency	Class boundaries	Cumulative frequency
20–29	8	19.5–29.5	8
30–39	11	29.5–39.5	19
40–49	23	39.5–49.5	42
50–59	25	49.5–59.5	67
60–69	8	59.5–69.5	75
70–79	5	69.5–79.5	80

Plot cumulative frequency against the upper class boundary, beginning at .5, 0. The resulting less-than ogive is:

(b) Total number of employees, N=80.

1. The lower quartile corresponds to cumulative frequency \(\frac{N}{4}=20\). From the ogive, \(Q_1\approx 40\) hours.
2. The median corresponds to cumulative frequency \(\frac{N}{2}=40\). From the ogive, median \(\approx 49\) hours.
3. The 40th percentile corresponds to cumulative frequency \(0.40\times80=32\). From the ogive, \(P_{40}\approx45.5\) hours.
4. At 50 hours 30 minutes, that is, 50.5 hours, the cumulative frequency read from the ogive is approximately 45. Hence, the number who spent at least 50.5 hours at work is \[80-45=\boxed{35\text{ employees}}.\]

Akọwa Nkọwa

(a) Simplify \(\dfrac{\sqrt{75}-3}{\sqrt{3}+1}\).

Since \(\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}\), the expression is \(\dfrac{5\sqrt{3}-3}{\sqrt{3}+1}\). Rationalise by multiplying by \(\dfrac{\sqrt{3}-1}{\sqrt{3}-1}\):

So \(a=9,\ b=-4,\ c=3\).

(b) Let the circle be \(x^2+y^2+2gx+2fy+c=0\).

\((7,3):\ 58+14g+6f+c=0\)  (i)
\((-3,3):\ 18-6g+6f+c=0\)  (ii)
\((2,8):\ 68+4g+16f+c=0\)  (iii)

(i) - (ii): \(40+20g=0\Rightarrow g=-2\).

(ii): \(18-6(-2)+6f+c=0\Rightarrow 6f+c=-30\)  (iv).
(iii): \(68+4(-2)+16f+c=0\Rightarrow 16f+c=-60\)  (v).

(v) - (iv): \(10f=-30\Rightarrow f=-3\), then \(c=-30-6(-3)=-12\).

(i) Equation: \(x^2+y^2-4x-6y-12=0\), i.e. \((x-2)^2+(y-3)^2=25\).

(ii) Radius: centre \((-g,-f)=(2,3)\), \(r=\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=\sqrt{25}=\textbf{5}\).

Akọwa Nkọwa

Given \(2^{2x-3y}=32\) and \(\log_y x=2.\)

First equation. Since \(32=2^5,\) equate indices:

\[2x-3y=5.\quad(1)\]

Second equation. \(\log_y x=2\) means \(x=y^2.\quad(2)\)

Substitute (2) into (1):

\[2y^2-3y=5\Rightarrow 2y^2-3y-5=0.\]

Factorise: \((2y-5)(y+1)=0,\) so \(y=\dfrac{5}{2}\) or \(y=-1.\)

A logarithm base must be positive and not equal to 1, so \(y=-1\) is rejected. Hence

\[y=\frac{5}{2},\qquad x=y^2=\frac{25}{4}.\]

Check: \(2x-3y=2\!\left(\tfrac{25}{4}\right)-3\!\left(\tfrac{5}{2}\right)=\tfrac{25}{2}-\tfrac{15}{2}=5.\) Correct.

Akọwa Nkọwa

(a) For \(y=4x^3\), the gradient is \(\dfrac{dy}{dx}=12x^2\). Set it equal to 108:

At \(x=3:\ y=4(3)^3=108\); at \(x=-3:\ y=4(-3)^3=-108\).

So \(P(3,\ 108)\) and \(Q(-3,\ -108)\).

(b)(i) Use \(A=45^\circ,\ B=30^\circ\) so that \(A+B=75^\circ\). Since \(\sin 15^\circ=\cos 75^\circ\) and \(\cos 15^\circ=\sin 75^\circ\):

Hence \(\sin 15^\circ=\cos 75^\circ=\dfrac{\sqrt6-\sqrt2}{4}\) and \(\cos 15^\circ=\sin 75^\circ=\dfrac{\sqrt6+\sqrt2}{4}\), as required.

(ii) \(\tan 15^\circ=\dfrac{\sin 15^\circ}{\cos 15^\circ}=\dfrac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\). Rationalise:

So \(\tan 15^\circ=2-\sqrt3\).

Akọwa Nkọwa

Given \(f(x)=6x^4+mx^3-13x^2+nx+14,\) with \((x+2)\) and \((x-1)\) as factors, so \(f(-2)=0\) and \(f(1)=0.\)

(a) Values of m and n.

\(f(-2)=6(16)+m(-8)-13(4)+n(-2)+14=96-8m-52-2n+14=58-8m-2n=0.\)

So \(8m+2n=58\Rightarrow 4m+n=29.\quad(1)\)

\(f(1)=6+m-13+n+14=7+m+n=0\Rightarrow m+n=-7.\quad(2)\)

Subtract (2) from (1): \(3m=36\Rightarrow m=12.\) Then \(n=-7-12=-19.\)

\[m=12,\qquad n=-19.\]

(b) Remainder when f(x) is divided by \((x+1).\)

By the remainder theorem the remainder is \(f(-1).\) With \(f(x)=6x^4+12x^3-13x^2-19x+14:\)

\[f(-1)=6(1)+12(-1)-13(1)-19(-1)+14=6-12-13+19+14=14.\]

Remainder \(=14.\)

Akọwa Nkọwa

(a) Let \(u=5-x^2\Rightarrow \dfrac{du}{dx}=-2x\), so \(x\,dx=-\tfrac12\,du\). Limits: \(x=1\Rightarrow u=4\); \(x=2\Rightarrow u=1\).

The value of the integral is 1.

(b) With \(y=px^2+qx\), \(\dfrac{dy}{dx}=2px+q\) and \(\dfrac{d^2y}{dx^2}=2p\).

Comparing \(\dfrac{dy}{dx}=6x+7\): \(2p=6\Rightarrow p=3\) and \(q=7\). The second derivative \(2p=6\) confirms this.

So \(p=3,\ q=7\).

Akọwa Nkọwa

(a) Choosing 2 from 4 boys and 5 girls (9 pupils):

(b) Let \(p=P(\text{standard})=0.8\) and \(q=P(\text{below standard})=0.2\), with \(n=6\). This is binomial \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\).

(i) exactly 2 standard: \(\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.0154\).

(ii) exactly 1 standard: \(\binom{6}{1}(0.8)(0.2)^5=6(0.8)(0.00032)=0.0015\).

(iii) at least 2 standard: \(1-P(0)-P(1)\), where \(P(0)=(0.2)^6=0.000064\).

(iv) at most 2 standard: \(P(0)+P(1)+P(2)=0.000064+0.001536+0.01536=0.0170\).

Akọwa Nkọwa

The assumed-mean (working-mean) method for grouped data uses \(\bar{x}=A+\dfrac{\sum fd}{\sum f}\), where \(d=x-A\), \(x\) is the class mid-value and \(A=24.5\) is the assumed mean.

Each class has width 4, so the mid-value is the average of the class limits, e.g. \(\tfrac{15+18}{2}=16.5\).

\[\bar{x}=A+\frac{\sum fd}{\sum f}=24.5+\frac{-300}{120}=24.5-2.5=22.0\]

The mean age of the distribution is 22 years.

Akọwa Nkọwa

(a) Mass 5 kg, weight \(W=50\ \text{N}\), smooth plane at \(30^\circ\).

(i) Component of weight parallel to (down) the plane: \(W\sin 30^\circ=50\times 0.5=\textbf{25 N}\).

(ii) To keep the body in equilibrium a force equal and opposite to this must act up the plane: \(\textbf{25 N}\).

(b) Plank \(PQ=10\ \text{m}\), mass \(m\), \(|PA|=|BQ|=1\ \text{m}\), so \(|AB|=8\ \text{m}\). The plank's weight \(mg\) acts at its centre, 5 m from P i.e. 4 m from A. Load 8 kg (\(=80\ \text{N}\)) at C with \(|AC|=3.5\ \text{m}\). Given reaction at B, \(R_B=100\ \text{N}\).

Take moments about A (clockwise = anticlockwise):

(ii) Vertical equilibrium: \(R_A+R_B=mg+80=130+80=210\).

Akọwa Nkọwa

Take upward as positive, initial velocity \(u = 80\ \text{m/s}\), \(g = 10\ \text{m/s}^2\).

(a) Maximum height. At the highest point the velocity is zero. Using \(v^2 = u^2 - 2gH\) with \(v = 0\):

The maximum height reached is 320 m.

(b) Time to return to the point of projection. By symmetry the total time is twice the time to reach the top. Time up: \(v = u - gt \Rightarrow 0 = 80 - 10t \Rightarrow t = 8\ \text{s}\).

The object returns after 16 s. (Equivalently \(T = \dfrac{2u}{g} = \dfrac{160}{10} = 16\ \text{s}\).)

Akọwa Nkọwa

Let the first term be \(a\) and common difference \(d.\) The \(n\)th term is \(T_n=a+(n-1)d.\)

Set up the equations. Sum of 2nd and 5th terms is 42:

\[(a+d)+(a+4d)=42\Rightarrow 2a+5d=42.\]

Difference between 6th and 3rd terms is 12:

\[(a+5d)-(a+2d)=12\Rightarrow 3d=12.\]

(i) Common difference. \(3d=12\Rightarrow d=4.\)

(ii) First term. Substitute into \(2a+5d=42:\)

\[2a+5(4)=42\Rightarrow 2a=22\Rightarrow a=11.\]

(iii) 20th term.

\[T_{20}=a+19d=11+19(4)=11+76=87.\]

Akọwa Nkọwa

A body of mass \(m=5\text{ kg}\) on a smooth horizontal plane (so no friction) is acted on by the forces \(6\mathbf{i}+2\mathbf{j},\ 5\mathbf{i}+4\mathbf{j},\ 4\mathbf{i}-\mathbf{j}\) (in newtons). Assume it starts from rest.

Resultant force.

\[\mathbf{F}=(6+5+4)\mathbf{i}+(2+4-1)\mathbf{j}=15\mathbf{i}+5\mathbf{j}\text{ N}.\]

Acceleration from Newton's second law \(\mathbf{a}=\dfrac{\mathbf{F}}{m}:\)

\[\mathbf{a}=\frac{15\mathbf{i}+5\mathbf{j}}{5}=3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\]

(a) Velocity after 4 s. Starting from rest, \(\mathbf{v}=\mathbf{a}t:\)

\[\mathbf{v}=(3\mathbf{i}+\mathbf{j})\times4=12\mathbf{i}+4\mathbf{j}\text{ m/s}.\]

(b) Magnitude of the velocity after 4 s.

\[|\mathbf{v}|=\sqrt{12^2+4^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}\approx12.65\text{ m/s}.\]

(Here part (a) is taken as the velocity vector after 4 s, obtained via the resultant force and acceleration \(3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\))

Akọwa Nkọwa

Find the tangent to \(y=\dfrac{x-1}{2x+1}\) at \((1,0).\)

Differentiate using the quotient rule with \(u=x-1\ (u'=1)\) and \(v=2x+1\ (v'=2):\)

\[\frac{dy}{dx}=\frac{u'v-uv'}{v^2}=\frac{(1)(2x+1)-(x-1)(2)}{(2x+1)^2}=\frac{2x+1-2x+2}{(2x+1)^2}=\frac{3}{(2x+1)^2}.\]

Gradient at the point. At \(x=1:\)

\[\frac{dy}{dx}=\frac{3}{(2(1)+1)^2}=\frac{3}{9}=\frac{1}{3}.\]

Equation of the tangent through \((1,0)\) with gradient \(\tfrac13:\)

\[y-0=\frac{1}{3}(x-1)\ \Rightarrow\ 3y=x-1\ \Rightarrow\ x-3y-1=0.\]