WAEC SSCE - Further Mathematics - 2009

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The formula for permutations is: \(^{n}P_{r} = \frac{n!}{(n-r)!}\) The formula for combinations is: \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\) We are given that: \(\frac{^{8}P_{x}}{^{8}C_{x}} = 6\) Substituting the formulas for permutations and combinations, we get: \(\frac{\frac{8!}{(8-x)!}}{\frac{8!}{x!(8-x)!}} = 6\) Simplifying, we get: \(\frac{x!}{(8-x)!} = \frac{1}{6}\) Multiplying both sides by \(\frac{(8-x)!}{x!}\), we get: \(\frac{8!}{x!(8-x)!} = 6\) Dividing both sides by 8!, we get: \(\frac{1}{^{8}C_{x}} = \frac{1}{720}\) Multiplying both sides by 720, we get: \(^{8}C_{x} = 720\) We can use the formula for combinations to find the value of x: \(^{8}C_{x} = \frac{8!}{x!(8-x)!} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{x!(8-x)!}\) Since \(^{8}C_{x} = 720\), we have: \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{x!(8-x)!} = 720\) Dividing both sides by 720, we get: \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{720} = x!(8-x)!\) Simplifying, we get: \(x!(8-x)! = 40320\) The only value of x that satisfies this equation is x = 3. Therefore, the correct answer is.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To find the coefficient of \(x^{4}\) in the binomial expansion of \((2 + x)^{6}\), we can use the formula for the binomial expansion: \[\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}\] where in this case, \(n = 6\), \(a = 2\), and \(b = x\). So, we have \[(2+x)^6 = \sum_{k=0}^{6}\binom{6}{k}(2)^{6-k}x^{k}.\] To find the coefficient of \(x^{4}\), we need to look at the term where \(k = 4\), which is \[\binom{6}{4}(2)^{6-4}x^{4} = 15\cdot 4x^{4} = 60x^{4}.\] Therefore, the coefficient of \(x^{4}\) in the binomial expansion of \((2 + x)^{6}\) is 60. So the answer is option C, 60.

Akọwa Nkọwa

When two dice are tossed, there are 6 × 6 = 36 possible outcomes, since each die has 6 possible outcomes. Out of these 36 possible outcomes, there are 3 even numbers on each die (2, 4, 6), so the probability of rolling an even number on a single die is 3/6 = 1/2. To find the probability that both dice show even numbers, we need to multiply the probability of rolling an even number on the first die (1/2) by the probability of rolling an even number on the second die (also 1/2), since the two events are independent. So, the probability that both dice show even numbers is: 1/2 × 1/2 = 1/4 Therefore, the answer is option (B) \(\frac{1}{4}\).

Akọwa Nkọwa

To find \(\alpha^{2} + \beta^{2}\), we need to use the following identities: \[\alpha + \beta = \frac{-b}{a}\quad\text{and}\quad \alpha\beta = \frac{c}{a}\] where a, b, and c are the coefficients of the quadratic equation \(ax^{2} + bx + c = 0\). In this case, the quadratic equation is \(2x^{2} - 5x + m = 0\), so a = 2, b = -5, and c = m. Using the identity \(\alpha + \beta = \frac{-b}{a}\), we have: \[\alpha + \beta = \frac{-(-5)}{2} = \frac{5}{2}\] Using the identity \(\alpha\beta = \frac{c}{a}\), we have: \[\alpha\beta = \frac{m}{2}\] Now, we can use the identity: \[\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta\] Substituting the values we found earlier, we get: \[\alpha^{2} + \beta^{2} = \left(\frac{5}{2}\right)^{2} - 2\left(\frac{m}{2}\right)\] Simplifying, we get: \[\alpha^{2} + \beta^{2} = \frac{25}{4} - m\] Therefore, the answer is \(\frac{25}{4} - m\), which is.

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To find the angle between two vectors, we can use the dot product formula: \(\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos \theta\), where \(\mathbf{a}\) and \(\mathbf{b}\) are the vectors, \(\|\mathbf{a}\|\) and \(\|\mathbf{b}\|\) are their magnitudes, and \(\theta\) is the angle between them. So, for the given vectors, we have: \(\begin{pmatrix} 13 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 4 \end{pmatrix} = \sqrt{13^2 + 1^2} \sqrt{1^2 + 4^2} \cos \theta\) Simplifying, we get: \(53 = \sqrt{170} \sqrt{17} \cos \theta\) \(\cos \theta = \frac{53}{\sqrt{170} \sqrt{17}}\) Taking the inverse cosine of both sides, we get: \(\theta \approx 73^\circ\) So, the answer is closest to 72°.

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We can solve this problem using the concept of probability. Let's begin by finding the total number of ways to select two bulbs out of ten. We can do this using the combination formula: $${10 \choose 2} = \frac{10!}{2!(10-2)!} = 45$$ So there are 45 ways to select two bulbs out of ten. Now let's find the number of ways to select two good bulbs out of the seven good ones. We can use the combination formula again: $${7 \choose 2} = \frac{7!}{2!(7-2)!} = 21$$ So there are 21 ways to select two good bulbs out of seven. Finally, we can find the probability of selecting two good bulbs by dividing the number of ways to select two good bulbs by the total number of ways to select two bulbs: $$P(\text{both bulbs are good}) = \frac{21}{45} = \frac{7}{15}$$ Therefore, the answer is, the probability that both bulbs are good is $\frac{7}{15}$.

Akọwa Nkọwa

We can use the properties of logarithms to simplify the given expressions and then solve for x. Firstly, from the first equation, we have: \[\log_{3}(x - y) = 1\] \[x - y = 3\] Similarly, from the second equation, we have: \[\log_{3}(2x + y) = 2\] \[2x + y = 9\] Now, we have a system of two equations with two variables, x and y. We can solve for y by subtracting the first equation from the second: \[(2x + y) - (x - y) = 9 - 3\] \[x + 2y = 6\] \[y = 3 - \frac{1}{2}x\] Substituting this value of y into the first equation: \[x - (3 - \frac{1}{2}x) = 3\] \[\frac{3}{2}x = 6\] \[x = 4\] Therefore, the value of x is 4, which is.

Akọwa Nkọwa

If (x + 3) is a factor of the polynomial, then we can write the polynomial as: \[(x+3)(ax^2+bx+c)\] Expanding the above equation, we get: \[ax^3 + (b+3a)x^2 + (c+3b)x + 3c = x^3 + 3x^2 + nx - 12\] Equating the coefficients of corresponding powers of x, we get: \[a = 1, b + 3a = 3, c + 3b = n, 3c = -12\] From the last equation, we have: \[c = -4\] Using this value of c in the third equation, we have: \[n = c + 3b = -4 + 3b\] From the second equation, we have: \[b + 3a = 3\] \[b + 3(1) = 3\] \[b = 0\] Hence, we have: \[n = -4 + 3b = -4\] Therefore, the value of n is -4.

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We can begin by multiplying both the numerator and denominator by the conjugate of the denominator, which is \(3 + \sqrt{7}\). This is done to eliminate the square root in the denominator. \[\frac{2}{3 - \sqrt{7}} \times \frac{3 + \sqrt{7}}{3 + \sqrt{7}} = \frac{2(3 + \sqrt{7})}{9 - 7} = \frac{2(3 + \sqrt{7})}{2} = 3 + \sqrt{7}\] Therefore, \(\frac{2}{3 - \sqrt{7}} = 3 + \sqrt{7}\), and the answer is (B) \(3 + \sqrt{7}\).

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To evaluate the given expression, we need to perform matrix multiplication. We multiply the first row of the matrix on the left with the column matrix on the right as shown below: \[\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} (2 \times 2) + (3 \times 3) \\ (4 \times 2) + (1 \times 3) \end{pmatrix} = \begin{pmatrix} 13 \\ 11 \end{pmatrix}\] Therefore, the answer is \(\begin{pmatrix} 13 \\ 11 \end{pmatrix}\).

Akọwa Nkọwa

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In a circle, the length of an arc is given by:

l = rθ

where r is the radius of the circle, and θ is the angle subtended by the arc at the centre of the circle in radians.

In this question, we are given the length of the arc (l = 10.8 cm) and the angle subtended by the arc (θ = 1.2 radians). We can rearrange the formula above to solve for the radius r:

r = l/θ

Substituting the values given, we get:

r = 10.8/1.2 = 9

Therefore, the radius of the circle is 9 cm.

Akọwa Nkọwa

We can write 0.125 as a fraction: 0.125 = 1/8 So, we have: 2^x = 1/8 To find the value of x, we need to determine what power of 2 gives us 1/8. We can rewrite 1/8 as a power of 2 by using the fact that: 1/8 = 2^(-3) Therefore, we have: 2^x = 2^(-3) For the above equation to be true, x must be equal to -3. Therefore, the answer is option (D) -3.

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To find the acceleration of the particle, we need to take the derivative of the velocity function with respect to time (t). The derivative of velocity with respect to time gives us the acceleration. Taking the derivative of v with respect to t, we get: \[\frac{dv}{dt} = 6t - 2\] Now, we need to find the acceleration of the particle after 3 seconds. We can do this by substituting t = 3 into the expression we just found for the derivative of v: \[\frac{dv}{dt} = 6t - 2\] \[\frac{dv}{dt}\bigg|_{t=3} = 6(3) - 2\] \[\frac{dv}{dt}\bigg|_{t=3} = 16\] Therefore, the acceleration of the particle after 3 seconds is 16 \(ms^{-2}\). To summarize, we found the derivative of the velocity function with respect to time to get the acceleration function. Then, we substituted t = 3 into the acceleration function to find the acceleration of the particle after 3 seconds.

Akọwa Nkọwa

To find \((P^{2} + P)\), we first need to find the value of \(P^{2}\). \(P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix}\) Now, we can find \((P^{2} + P)\) by adding \(P^{2}\) and \(P\). \((P^{2} + P) = \begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\) Therefore, the answer is \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\).

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We can simplify this expression using the rules of exponents. First, let's simplify \((216)^{-\frac{2}{3}}\). We know that \((216)^{\frac{2}{3}}\) is the cube root of 216 squared. So, \((216)^{-\frac{2}{3}}\) is simply the reciprocal of that value. \[(216)^{-\frac{2}{3}} = \frac{1}{(216)^{\frac{2}{3}}} = \frac{1}{(6^3)^{\frac{2}{3}}} = \frac{1}{6^2}\] Next, let's simplify \((0.16)^{-\frac{3}{2}}\). We can rewrite \(0.16\) as \(\frac{16}{100}\), and then apply the exponent to both the numerator and denominator: \[(0.16)^{-\frac{3}{2}} = \left(\frac{16}{100}\right)^{-\frac{3}{2}} = \frac{(100)^{\frac{3}{2}}}{(16)^{\frac{3}{2}}}\] Now, we can multiply these two simplified expressions: \[\frac{1}{6^2} \times \frac{(100)^{\frac{3}{2}}}{(16)^{\frac{3}{2}}} = \frac{1}{36} \times \frac{1000}{64} = \frac{125}{288}\] Therefore, the answer is \(\frac{125}{288}\).

Akọwa Nkọwa

We can use the formula: $$F = ma$$ where F is the force applied, m is the mass of the body, and a is the acceleration produced. To find the acceleration, we can use the formula: $$a = \frac{v_f - v_i}{t}$$ where v_f is the final velocity, v_i is the initial velocity, and t is the time taken. Substituting the given values, we get: $$a = \frac{35 - 5}{5} = 6 \text{ m/s}^2$$ Substituting this value of acceleration and the given mass into the formula for force, we get: $$F = ma = 12 \times 6 = 72 \text{ N}$$ Therefore, the value of F is 72 N. So, the answer is 72 N.

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To calculate the standard deviation, we need to follow these steps: 1. Find the mean (average) of the ages. 2. Find the difference between each age and the mean. 3. Square each of these differences. 4. Find the average of these squared differences. 5. Take the square root of this average. Let's first find the mean of the ages: mean = (5+6+6+8+10)/5 = 7 years Now we can find the difference between each age and the mean: |5 - 7| = 2 |6 - 7| = 1 |6 - 7| = 1 |8 - 7| = 1 |10 - 7| = 3 Next, we square each of these differences: 2^2 = 4 1^2 = 1 1^2 = 1 1^2 = 1 3^2 = 9 Now we find the average of these squared differences: average = (4+1+1+1+9)/5 = 2.4 Finally, we take the square root of this average: standard deviation = sqrt(2.4) ≈ 1.5 Therefore, the correct answer is not among the given options. The closest option is (d) 1.8 years, but this is not within one decimal place of the calculated standard deviation.

Akọwa Nkọwa

To find the equation of a quadratic equation, we can use the factored form, which is given by: $$(x-r_1)(x-r_2)=0$$ where $r_1$ and $r_2$ are the roots of the equation. In this case, the given roots are -3 and 1, so we have: $$(x-(-3))(x-1)=0$$ $$(x+3)(x-1)=0$$ Expanding this expression gives us: $$x^2+3x-x-3=0$$ Simplifying, we get: $$x^2+2x-3=0$$ So the equation of the quadratic equation is $x^2+2x-3=0$. Therefore, option (C) is the correct answer.

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To find the coordinates of point P, we need to differentiate the given equation with respect to x to get the derivative of the function. The derivative of y with respect to x gives us the gradient or slope of the tangent at any point on the curve. Taking the derivative of y with respect to x, we get: \[\frac{dy}{dx} = 6x - 1\] We are given that the gradient at point P is 5. So, we can equate the derivative of y with 5 and solve for x. \[6x - 1 = 5\] \[6x = 6\] \[x = 1\] Now that we know x = 1, we can find the corresponding y-coordinate by substituting x=1 into the original equation for y. \[y = 3x^{2} - x + 3\] \[y = 3(1)^{2} - 1 + 3\] \[y = 5\] Therefore, the coordinates of point P are (1, 5). To summarize, we found the x-coordinate of point P by equating the derivative of the equation with the given gradient of 5 and solved for x. Then, we found the corresponding y-coordinate by substituting the value of x into the original equation.

Akọwa Nkọwa

We know that the equation of a circle with center coordinates \((a,b)\) and radius \(r\) is given by \((x-a)^2 + (y-b)^2 = r^2\). From the problem statement, we are given that the center of the circle is at \((-2,3)\), so we can substitute \(a=-2\) and \(b=3\) in the equation of the circle as \((x+2)^2 + (y-3)^2 = r^2\). We are also given that the area of the circle is \(25\pi cm^2\). We know that the area of a circle is given by the formula \(A = \pi r^2\), where \(r\) is the radius of the circle. We can solve for the radius by substituting the area value, so we have \begin{align*} A &= \pi r^2\\ 25\pi &= \pi r^2\\ 25 &= r^2\\ r &= 5 \end{align*} Now we have the coordinates of the center and the radius, so we can substitute them in the equation of the circle to get $$(x+2)^2 + (y-3)^2 = 25$$ Expanding the square terms, we have $$x^2+4x+4+y^2-6y+9=25$$ Simplifying, we get $$x^2+y^2+4x-6y-12=0$$ Therefore, the equation of the circle is \boxed{x^2+y^2+4x-6y-12=0}.

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The semi-interquartile range of a distribution is the measure of dispersion or spread of data around the median. It is half of the difference between the upper quartile and the lower quartile. Therefore, the correct answer is \(\frac{1}{2}(\text{Upper quartile - Lower quartile})\). To find the interquartile range (IQR), we subtract the lower quartile (Q1) from the upper quartile (Q3), i.e., IQR = Q3 - Q1. Then, the semi-interquartile range (SIQR) is half of the IQR, i.e., SIQR = (Q3 - Q1) / 2. The other options listed in the question are not measures of dispersion or spread around the median, so they are not correct answers to the question.

Akọwa Nkọwa

We can solve this problem using the equations of motion for a particle moving vertically under gravity. The key idea is that at the maximum height, the vertical velocity of the stone will be zero. We can use the following equation to find the time taken by the stone to reach the maximum height: \[-u/g = -10/10 = -1\] where u is the initial velocity and g is the acceleration due to gravity. The time taken to reach the maximum height is 1 second. We can now use the following equation to find the maximum height reached: \[h = u t - \frac{1}{2} g t^{2} = 10\times 1 - \frac{1}{2}\times 10\times 1^{2} = 5\] where h is the maximum height reached. Therefore, the maximum height reached by the stone is 5 metres above the initial height of 8 metres, which gives us a total height of 13 metres. Hence, the answer is option (A) 13 metres.

Akọwa Nkọwa

To find the magnitude of the resultant force, we first need to find the horizontal and vertical components of the two given forces. For the first force, we have: - Horizontal component: $$F_{1x} = F_{1} \cos 90° = 0$$ - Vertical component: $$F_{1y} = F_{1} \sin 90° = 10 N$$ For the second force, we have: - Horizontal component: $$F_{2x} = F_{2} \cos 180° = -6 N$$ - Vertical component: $$F_{2y} = F_{2} \sin 180° = 0$$ The horizontal and vertical components of the resultant force can be found by adding the corresponding components of the two given forces: - Horizontal component: $$F_{x} = F_{1x} + F_{2x} = 0 - 6 N = -6 N$$ - Vertical component: $$F_{y} = F_{1y} + F_{2y} = 10 N + 0 = 10 N$$ The magnitude of the resultant force can be found using the Pythagorean theorem: $$|F| = \sqrt{F_{x}^{2} + F_{y}^{2}} = \sqrt{(-6)^{2} + (10)^{2}} \approx 11.66 N$$ Therefore, the magnitude of the resultant force is approximately 11.7 N. The answer is (B) 11.7 N.

Akọwa Nkọwa

To find the function f(x) from its derivative, we need to integrate the derivative with respect to x. \[f(x) = \int f'(x) dx = \int (3x^{2} - \frac{4}{x^{5}}) dx = x^{3} + \frac{1}{x^{4}} + C\] where C is the constant of integration. We can use the given information that f(1) = 4 to find the value of C: \[f(1) = 1 + 1 + C = 4\] \[C = 2\] Substituting this value of C, we get: \[f(x) = x^{3} + \frac{1}{x^{4}} + 2\] Therefore, the answer is option (B) \(f(x) = x^{3} + \frac{1}{x^{4}} + 2\).

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We know that:

\(\sin \theta = \frac{{\text{{opposite}}}}{{\text{{hypotenuse}}}}\)

We can draw a right triangle with an angle of \(\theta\), opposite side of \(\sqrt{3}\), and hypotenuse of 2. Using the Pythagorean theorem, we can find the adjacent side to be 1. Therefore, the triangle looks like:

         /|
        / |
   1  /  |sqrt(3)
     /   |
    /____|
       2

Using the double-angle formula for tangent, we have:

\(\tan 2\theta = \frac{{2 \tan \theta}}{{1 - \tan^{2} \theta}}\)

Using the formula for tangent:

\(\tan \theta = \frac{{\text{{opposite}}}}{{\text{{adjacent}}}} = \sqrt{3}\)

Substituting in the formula for double-angle tangent, we have:

\(\tan 2\theta = \frac{{2 \sqrt{3}}}{{1 - (\sqrt{3})^{2}}} = \frac{{2 \sqrt{3}}}{{1 - 3}}\)

Simplifying, we get:

\(\tan 2\theta = \frac{{2 \sqrt{3}}}{{-2}} = -\sqrt{3}\)

Therefore, the correct answer is.

Akọwa Nkọwa

To find the probability that exactly 2 of them hit the target, we need to consider all possible combinations of two men hitting the target while the other misses. There are three such combinations: PQ, QR, and PR. For the combination PQ, the probability that P hits the target and Q misses is \(\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\). The probability that P misses and Q hits is \(\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\). Therefore, the probability that exactly PQ hit the target is \(\frac{1}{3} + \frac{1}{6} = \frac{1}{2}\). Similarly, for QR, the probability that Q hits and R misses is \(\frac{1}{3} \times \frac{1}{4} = \frac{1}{12}\). The probability that Q misses and R hits is \(\frac{2}{3} \times \frac{3}{4} = \frac{1}{2}\). Therefore, the probability that exactly QR hit the target is \(\frac{1}{12} + \frac{1}{2} = \frac{7}{12}\). Finally, for PR, the probability that P hits and R misses is \(\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\). The probability that P misses and R hits is \(\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}\). Therefore, the probability that exactly PR hit the target is \(\frac{1}{8} + \frac{3}{8} = \frac{1}{2}\). The total probability that exactly 2 of them hit the target is the sum of the probabilities for each combination, which is \(\frac{1}{2} + \frac{7}{12} + \frac{1}{2} = \frac{5}{6}\). Therefore, the answer is (C) \(\frac{5}{12}\).

Akọwa Nkọwa

Let's consider the right triangle PST. We know that angle PST = 90°, angle PSR = x°, and angle PTR = 30°. Therefore, angle PTS = 60° - x°. We also know that the ladder has length L and that it slides down the wall to a point Q such that TQ = x. Now, consider the right triangle QRT. We know that angle QRT = 90° and angle QTR = 30°. Therefore, angle QRT = 60°. We also know that TQ = x and QR = L - x. Finally, consider the right triangle NST. We know that angle NST = 90°, angle NTS = y°, and angle SNT = y° - (60° - x°) = x° + y° - 60°. We also know that NS = QR = L - x. Now, we can use the tangent function to find an expression for tan y: tan y = NS/NT = (L - x)/(TQ + QR) = (L - x)/(x + (L - x)) = (L - x)/(L) We can simplify this expression by multiplying the numerator and denominator by \(\frac{1}{\sqrt{3}}\): tan y = \(\frac{\frac{1}{\sqrt{3}} (L - x)}{\frac{1}{\sqrt{3}} L}\) = \(\frac{\sqrt{3} \tan x - 1}{\sqrt{3} + \tan x}\) Therefore, the correct answer is.

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There are different ways to approach this problem, but one possible method is to use combinations. First, we need to find the total number of ways to draw two balls from the bag, without replacement. This can be calculated using the formula for combinations: \(\text{Total number of ways} = \binom{12}{2} = \frac{12!}{2!10!} = 66\) Next, we need to count the number of ways to draw two balls of the same color. There are three possible cases: both red, both white, or both black. For two red balls, we can choose 2 balls from the 3 red balls in the bag, giving us: \(\text{Number of ways for two red balls} = \binom{3}{2} = 3\) Similarly, we can count the number of ways for two white balls and two black balls: \(\text{Number of ways for two white balls} = \binom{4}{2} = 6\) \(\text{Number of ways for two black balls} = \binom{5}{2} = 10\) Therefore, the total number of ways to draw two balls of the same color is: \(\text{Number of ways for two same color balls} = 3 + 6 + 10 = 19\) Finally, we can calculate the probability of drawing two balls of the same color by dividing the number of ways for two same color balls by the total number of ways: \(\text{Probability of two same color balls} = \frac{19}{66} \approx 0.288\) Therefore, the answer is, which is \(\frac{19}{66}\).

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Let \(A = 15^{o}\) so that \(2A = 30^{o}\) and \(\tan30^{o} = \dfrac{1}{\sqrt3}\). Put \(t = \tan15^{o}\).

\[\tan30^{o} = \frac{2\tan15^{o}}{1 - \tan^2 15^{o}} \;\Rightarrow\; \frac{1}{\sqrt3} = \frac{2t}{1 - t^2}\] \[1 - t^2 = 2\sqrt3\,t \;\Rightarrow\; t^2 + 2\sqrt3\,t - 1 = 0\] \[t = \frac{-2\sqrt3 \pm \sqrt{12 + 4}}{2} = \frac{-2\sqrt3 \pm 4}{2} = -\sqrt3 \pm 2\]

Since \(15^{o}\) is acute, \(\tan15^{o} > 0\), so we take the positive value:

\[\tan15^{o} = 2 - \sqrt3\]

Akọwa Nkọwa

Total selections of \(5\) from \(14\) people: \(\binom{14}{5} = 2002\).

(a) 3 girls and 2 boys:

\[\binom{8}{3}\binom{6}{2} = 56\times15 = 840\] \[P = \frac{840}{2002} = \frac{60}{143} \approx 0.420\]

(b) All girls or all boys:

\[\binom{8}{5} + \binom{6}{5} = 56 + 6 = 62\] \[P = \frac{62}{2002} = \frac{31}{1001} \approx 0.031\]

Akọwa Nkọwa

\(A(-2,1),\ B(4,-2),\ C(1,8)\). \(D(x,y)\) is the foot of the perpendicular from \(A\) to \(BC\).

(a) Direction of \(BC\): \(\vec{BC} = C - B = (1 - 4,\ 8 - (-2)) = (-3,\ 10)\). Vector \(\vec{AD} = (x + 2,\ y - 1)\). Since \(AD \perp BC\), their dot product is zero:

\[(x + 2)(-3) + (y - 1)(10) = 0\] \[-3x - 6 + 10y - 10 = 0 \;\Rightarrow\; 3x - 10y + 16 = 0\]

The equation connecting \(x\) and \(y\) is \(3x - 10y + 16 = 0\).

(b) \(|\vec{BC}| = \sqrt{(-3)^2 + 10^2} = \sqrt{109}\). Unit vector in the direction of \(BC\):

\[\hat{BC} = \frac{1}{\sqrt{109}}\begin{pmatrix} -3 \\ 10 \end{pmatrix} = \begin{pmatrix} -\tfrac{3}{\sqrt{109}} \\[4pt] \tfrac{10}{\sqrt{109}} \end{pmatrix}\]

Akọwa Nkọwa

(a) Let \(u = 5 - x^2\), so \(du = -2x\,dx\) and \(x\,dx = -\tfrac12\,du\).

\[\int \frac{x}{\sqrt{5 - x^2}}\,dx = -\frac12\int u^{-1/2}\,du = -\sqrt{u} = -\sqrt{5 - x^2}\] \[\int_{1}^{2}\frac{x}{\sqrt{5 - x^2}}\,dx = \Big[-\sqrt{5 - x^2}\Big]_1^2 = -\sqrt{1} + \sqrt{4} = -1 + 2 = 1\]

(b)(i) Expanding along the first row:

\[\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix} = 2(1\cdot(-3) - (-2)\cdot2) + 3(0\cdot(-3) - (-2)\cdot1) + 1(0\cdot2 - 1\cdot1)\] \[= 2(1) + 3(2) + 1(-1) = 2 + 6 - 1 = 7\]

(ii) The coefficient matrix of the system is exactly this determinant, so \(\Delta = 7\). By Cramer's rule (replacing each column with \((10, -7, -9)^T\)):

\[\Delta_x = \begin{vmatrix} 10 & -3 & 1 \\ -7 & 1 & -2 \\ -9 & 2 & -3 \end{vmatrix} = 14,\quad \Delta_y = \begin{vmatrix} 2 & 10 & 1 \\ 0 & -7 & -2 \\ 1 & -9 & -3 \end{vmatrix} = -7,\quad \Delta_z = \begin{vmatrix} 2 & -3 & 10 \\ 0 & 1 & -7 \\ 1 & 2 & -9 \end{vmatrix} = 21\] \[x = \frac{14}{7} = 2,\quad y = \frac{-7}{7} = -1,\quad z = \frac{21}{7} = 3\]

\(x = 2,\ y = -1,\ z = 3\). Check: \(2(2) - 3(-1) + 3 = 10\) and \((-1) - 2(3) = -7\).

Akọwa Nkọwa

(a)(i) With \( A(1,5),\ B(3,9),\ C(-1,1) \):

\( \vec{AB} = B-A = (2,\ 4) \) and \( \vec{BC} = C-B = (-4,\ -8) \).

Since \( \vec{BC} = -2\,\vec{AB} \), the two vectors are parallel and share the common point B. Therefore A, B and C are collinear.

(ii) \( |AB| = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt5 \) and \( |BC| = \sqrt{(-4)^2+(-8)^2} = \sqrt{80} = 4\sqrt5 \).

\[ |AB| : |BC| = 2\sqrt5 : 4\sqrt5 = 1 : 2. \]

(b) Let the beam XY, length 24 m, lie horizontally. String tensions: \(T_X\) at X and \(T_M\) at M, where M is 4 m from Y, i.e. 20 m from X. The beam weight \(10g\) acts at the centre, 12 m from X; the 50 kg load acts at N, 8 m from X (take \(g = 10\text{ ms}^{-2}\)).

Vertical equilibrium: \( T_X + T_M = (10+50)g = 60g \).

Moments about X: \[ T_M(20) = 10g(12) + 50g(8) = 120g + 400g = 520g \Rightarrow T_M = 26g = 260\text{ N}. \]

Then \( T_X = 60g - 26g = 34g = 340\text{ N} \).

The tension at X is \(340\text{ N}\) and the tension at M is \(260\text{ N}\).

Akọwa Nkọwa

(a) Scatter diagram. The ten pairs \((33,4),(31,6),(28,4),(25,10),(23,12),(22,10),(19,14),(17,15),(16,18),(14,22)\) are plotted with \(X\) on the horizontal axis and \(Y\) on the vertical axis. The points trend from the upper-left down to the lower-right, showing that \(Y\) falls as \(X\) rises.

(b) Means.

\[\bar{x}=\frac{33+31+28+25+23+22+19+17+16+14}{10}=\frac{228}{10}=22.8\] \[\bar{y}=\frac{4+6+4+10+12+10+14+15+18+22}{10}=\frac{115}{10}=11.5\]

(c) Line of best fit. A single straight line is drawn through the mean point \((\bar{x},\bar{y})=(22.8,\,11.5)\) (marked ★ on the graph) with roughly equal numbers of points on either side. Its least-squares gradient is

\(n\)	\(\sum X\)	\(\sum Y\)	\(\sum XY\)	\(\sum X^2\)
10	228	115	2293	5574

\[b=\frac{n\sum XY-\sum X\sum Y}{n\sum X^{2}-\left(\sum X\right)^{2}}=\frac{10(2293)-228(115)}{10(5574)-228^{2}}=\frac{22930-26220}{55740-51984}=\frac{-3290}{3756}=-0.88\]

The intercept follows from \(\bar{y}=a+b\bar{x}\):

\[a=\bar{y}-b\bar{x}=11.5-(-0.88)(22.8)=11.5+20.06=31.5\]

so the line of best fit is

\[Y=31.5-0.88X.\]

(d)(i) Relationship. The line has a negative gradient, so there is a negative (inverse) correlation between \(X\) and \(Y\): as \(X\) increases, \(Y\) decreases.

(d)(ii) Value of \(Y\) when \(X=24\). Reading up from \(X=24\) to the line of best fit and across to the \(Y\)-axis, or substituting into the equation:

\[Y=31.5-0.88(24)=31.5-21.1=10.4\]

so \(Y\approx 10.5\) (about 10 to 11 from a hand-drawn line).

Akọwa Nkọwa

(a) K divides LM with LK:KM = 2:3, so the position vector of K is the weighted mean \[ \vec{K} = \frac{3\vec{L} + 2\vec{M}}{5}. \] Thus \[ \vec{K} = \frac{3(5\mathbf{i}+6\mathbf{j}) + 2(13\mathbf{i}+4\mathbf{j})}{5} = \frac{(15+26)\mathbf{i} + (18+8)\mathbf{j}}{5} = \frac{41\mathbf{i}+26\mathbf{j}}{5}. \] Hence \( \vec{K} = 8.2\mathbf{i} + 5.2\mathbf{j} \).

(b) Resolve each leg into east (x) and north (y) components using bearing \( \theta \): east \(= d\sin\theta\), north \(= d\cos\theta\).

\( \vec{AB} = (8\sin 60^\circ,\ 8\cos 60^\circ) = (6.928,\ 4.000) \).
\( \vec{BC} = (12\sin 130^\circ,\ 12\cos 130^\circ) = (9.193,\ -7.713) \).

\( \vec{AC} = \vec{AB}+\vec{BC} = (16.121,\ -3.713) \).

(i) \( |AC| = \sqrt{16.121^2 + (-3.713)^2} = \sqrt{259.9+13.79} = \sqrt{273.7} = 16.5\text{ km} \) (3 s.f.).

(ii) The vector points east and south, so the bearing lies between \(90^\circ\) and \(180^\circ\): \[ \text{bearing} = 90^\circ + \tan^{-1}\!\left(\frac{3.713}{16.121}\right) = 90^\circ + 12.97^\circ \approx 103^\circ. \] The bearing of C from A is \(103^\circ\) (nearest degree).

Akọwa Nkọwa

From \(\log_2 x - \log_2 y = 2\):

\[\log_2\frac{x}{y} = 2 \;\Rightarrow\; \frac{x}{y} = 2^2 = 4 \;\Rightarrow\; x = 4y\]

From \(\log_2(x - 2y) = 3\):

\[x - 2y = 2^3 = 8\]

Substitute \(x = 4y\):

\[4y - 2y = 8 \;\Rightarrow\; 2y = 8 \;\Rightarrow\; y = 4,\quad x = 16\]

\(x = 16,\ y = 4\). Check: \(\log_2 16 - \log_2 4 = 4 - 2 = 2\) and \(\log_2(16 - 8) = \log_2 8 = 3\).

Akọwa Nkọwa

Let the probabilities of topping be \(M = 0.9,\ P = 0.75,\ B = 0.8\), with complements \(0.1,\ 0.25,\ 0.2\). "At least two" means exactly two or all three.

All three: \(0.9\times0.75\times0.8 = 0.54\).

Exactly two:

• \(M,P\) not \(B\): \(0.9\times0.75\times0.2 = 0.135\)
• \(M,B\) not \(P\): \(0.9\times0.25\times0.8 = 0.180\)
• \(P,B\) not \(M\): \(0.1\times0.75\times0.8 = 0.060\)

Sum of exactly two \(= 0.135 + 0.180 + 0.060 = 0.375\).

\[P(\text{at least two}) = 0.375 + 0.54 = \mathbf{0.915}\]

Akọwa Nkọwa

(a) For a G.P., \(T_3 = ar^2 = 2\) and \(T_6 = ar^5 = 54\).

(i) Dividing: \(\dfrac{ar^5}{ar^2} = \dfrac{54}{2} \Rightarrow r^3 = 27 \Rightarrow r = 3\).

(ii) \(ar^2 = 2 \Rightarrow 9a = 2 \Rightarrow a = \dfrac{2}{9}\).

(iii) Sum of first 10 terms:

\[S_{10} = \frac{a(r^{10} - 1)}{r - 1} = \frac{\tfrac{2}{9}(3^{10} - 1)}{3 - 1} = \frac{\tfrac{2}{9}(59049 - 1)}{2} = \frac{59048}{9} \approx 6561\]

(b) In \((1 + 2x)^n\): coefficient of \(x^3 = \binom{n}{3}2^3\); coefficient of \(x^4 = \binom{n}{4}2^4\).

\[\frac{\text{coeff }x^4}{\text{coeff }x^3} = \frac{16\binom{n}{4}}{8\binom{n}{3}} = 2\cdot\frac{n-3}{4} = \frac{n-3}{2}\]

Setting this equal to \(\dfrac{3}{1}\):

\[\frac{n - 3}{2} = 3 \;\Rightarrow\; n - 3 = 6 \;\Rightarrow\; n = 9\]

Akọwa Nkọwa

(a) Sketching the two curves on the same axes.

Curve 1: \(y = 6 - x - x^{2}\)

This is an inverted (\(\cap\)) parabola since the coefficient of \(x^{2}\) is negative.

• y-intercept: when \(x = 0,\ y = 6\), so it passes through \((0,\,6)\).
• x-intercepts: when \(y = 0,\ 6 - x - x^{2} = 0 \Rightarrow x^{2} + x - 6 = 0 \Rightarrow (x+3)(x-2) = 0\), giving \(x = -3\) and \(x = 2\), i.e. the points \((-3,\,0)\) and \((2,\,0)\).
• Turning point: \(\dfrac{dy}{dx} = -1 - 2x = 0 \Rightarrow x = -\tfrac12\). Then \(y = 6 - (-\tfrac12) - (-\tfrac12)^{2} = 6\tfrac14\). Maximum at \(\left(-\tfrac12,\ 6\tfrac14\right)\).

Curve 2: \(y = 3x^{2} - 2x + 3\)

This is an upright (\(\cup\)) parabola since the coefficient of \(x^{2}\) is positive.

• y-intercept: when \(x = 0,\ y = 3\), so it passes through \((0,\,3)\).
• x-intercepts: \(b^{2} - 4ac = (-2)^{2} - 4(3)(3) = 4 - 36 = -32 < 0\), so the curve does not cut the x-axis.
• Turning point: \(\dfrac{dy}{dx} = 6x - 2 = 0 \Rightarrow x = \tfrac13\). Then \(y = 3(\tfrac13)^{2} - 2(\tfrac13) + 3 = 2\tfrac23\). Minimum at \(\left(\tfrac13,\ 2\tfrac23\right)\).

Plotting both on the same axes, the \(\cap\) parabola sits above the \(\cup\) parabola between their two crossing points, enclosing the finite shaded region:

(b) x-coordinates of the points of intersection.

At the points of intersection the two y-values are equal:

\[6 - x - x^{2} = 3x^{2} - 2x + 3\]\[0 = 3x^{2} - 2x + 3 - 6 + x + x^{2}\]\[0 = 4x^{2} - x - 3\]

Factorising: \(4x^{2} - 4x + 3x - 3 = 4x(x-1) + 3(x-1) = (4x + 3)(x - 1) = 0\).

\[x = -\frac{3}{4} \quad\text{or}\quad x = 1\]

So the curves intersect where \(x = -\tfrac34\) and \(x = 1\).

(c) Area of the finite region bounded by the two curves.

Between \(x = -\tfrac34\) and \(x = 1\), the curve \(y = 6 - x - x^{2}\) lies above \(y = 3x^{2} - 2x + 3\), so the enclosed area is

\[A = \int_{-3/4}^{1}\big[(6 - x - x^{2}) - (3x^{2} - 2x + 3)\big]\,dx = \int_{-3/4}^{1}\big(3 + x - 4x^{2}\big)\,dx\]\[A = \left[\,3x + \frac{x^{2}}{2} - \frac{4x^{3}}{3}\,\right]_{-3/4}^{1}\]

At \(x = 1:\quad 3(1) + \dfrac{1}{2} - \dfrac{4}{3} = \dfrac{18 + 3 - 8}{6} = \dfrac{13}{6}\).

At \(x = -\tfrac34:\quad 3\left(-\tfrac34\right) + \dfrac{(-3/4)^{2}}{2} - \dfrac{4}{3}\left(-\tfrac34\right)^{3} = -\dfrac{9}{4} + \dfrac{9}{32} + \dfrac{4}{3}\cdot\dfrac{27}{64} = -\dfrac{9}{4} + \dfrac{9}{32} + \dfrac{9}{16} = -\dfrac{45}{32}\).

\[A = \frac{13}{6} - \left(-\frac{45}{32}\right) = \frac{208}{96} + \frac{135}{96} = \frac{343}{96} \approx 3.57\ \text{square units}\]

The area of the finite region bounded by the two curves is \(\dfrac{343}{96} \approx 3.57\) square units.

Akọwa Nkọwa

(a) Factorise \(2x^2 + x - 6 = (2x - 3)(x + 2)\). Roots at \(x = \dfrac{3}{2}\) and \(x = -2\). Since the parabola opens upward, it is negative between the roots:

\[-2 < x < \frac{3}{2}\]

(b) Multiply numerator and denominator by the conjugate \((3\sqrt5 - \sqrt2)\):

\[\text{Denominator} = (3\sqrt5)^2 - (\sqrt2)^2 = 45 - 2 = 43\] \[\text{Numerator} = (5 - 2\sqrt{10})(3\sqrt5 - \sqrt2) = 15\sqrt5 - 5\sqrt2 - 6\sqrt{50} + 2\sqrt{20}\]

Since \(\sqrt{50} = 5\sqrt2\) and \(\sqrt{20} = 2\sqrt5\):

\[= 15\sqrt5 - 5\sqrt2 - 30\sqrt2 + 4\sqrt5 = 19\sqrt5 - 35\sqrt2\]

Therefore

\[\frac{5 - 2\sqrt{10}}{3\sqrt5 + \sqrt2} = \frac{19\sqrt5 - 35\sqrt2}{43} = -\frac{35}{43}\sqrt2 + \frac{19}{43}\sqrt5\]

So \(m = -\dfrac{35}{43}\) and \(n = \dfrac{19}{43}\).

Akọwa Nkọwa

First close the gaps between the stated class limits to obtain the true class boundaries, and set out the mid-values, class boundaries and the working columns for the assumed-mean method (assumed mean \(A = 43\)).

Battery life (days)	Class boundaries	Mid-value \(x\)	Frequency \(f\)	\(d = x - 43\)	\(fd\)
26 - 30	25.5 - 30.5	28	4	-15	-60
31 - 35	30.5 - 35.5	33	7	-10	-70
36 - 40	35.5 - 40.5	38	13	-5	-65
41 - 45	40.5 - 45.5	43	8	0	0
46 - 50	45.5 - 50.5	48	6	5	30
51 - 55	50.5 - 55.5	53	2	10	20
Total			40		-145

(a) Histogram

The bars are drawn contiguously on the class boundaries, with frequency on the vertical axis. Because the classes are of equal width (5 days) the bar heights equal the frequencies \(4, 7, 13, 8, 6, 2\).

(b) Mode from the histogram

The tallest bar is the modal class \(36 - 40\) (boundaries \(35.5 - 40.5\)). On the histogram two lines are drawn inside this bar: one from its top-left corner to the top-left corner of the next bar, and one from its top-right corner to the top-right corner of the previous bar. A vertical line dropped from where they cross meets the axis at the mode.

This agrees with the formula, where \(L = 35.5\), \(d_1 = 13 - 7 = 6\), \(d_2 = 13 - 8 = 5\) and \(c = 5\):

\[\text{Mode} = L + \frac{d_1}{d_1 + d_2}\,c = 35.5 + \frac{6}{6 + 5}\times 5 = 35.5 + 2.73 = 38.2\text{ days}\]

(c) Mean (assumed mean \(A = 43\))

\[\bar{x} = A + \frac{\sum fd}{\sum f} = 43 + \frac{-145}{40} = 43 - 3.625 = 39.375 \approx 39.4\text{ days}\]

(d) Percentage living less than 31 days or more than 45 days

Less than 31 days is the class \(26 - 30\): \(4\) batteries. More than 45 days covers the classes \(46 - 50\) and \(51 - 55\): \(6 + 2 = 8\) batteries.

\[\text{Number} = 4 + 6 + 2 = 12\] \[\text{Percentage} = \frac{12}{40}\times 100\% = 30\%\]

Summary: mode \(\approx 38.2\) days, mean \(\approx 39.4\) days, and \(30\%\) of the batteries last under 31 days or over 45 days.

Akọwa Nkọwa

(a) Write each combination out:

\[{}^{n+1}C_3 = \frac{(n+1)n(n-1)}{6},\qquad {}^{n-1}C_3 = \frac{(n-1)(n-2)(n-3)}{6}\] \[{}^{n+1}C_3 - {}^{n-1}C_3 = \frac{n-1}{6}\Big[(n+1)n - (n-2)(n-3)\Big]\] \[= \frac{n-1}{6}\Big[(n^2 + n) - (n^2 - 5n + 6)\Big] = \frac{n-1}{6}(6n - 6) = (n - 1)^2\]

So the expression simplifies to \((n - 1)^2\).

(b) A six has \(p = \dfrac16,\ q = \dfrac56\), with \(n = 5\) throws (binomial).

(i) At most two sixes \(= P(0) + P(1) + P(2)\):

\[P(0) = \left(\tfrac56\right)^5 = \tfrac{3125}{7776},\quad P(1) = \binom{5}{1}\tfrac16\left(\tfrac56\right)^4 = \tfrac{3125}{7776},\quad P(2) = \binom{5}{2}\left(\tfrac16\right)^2\left(\tfrac56\right)^3 = \tfrac{1250}{7776}\] \[P(\le 2) = \frac{3125 + 3125 + 1250}{7776} = \frac{7500}{7776} \approx 0.965\]

(ii) Exactly three sixes:

\[P(3) = \binom{5}{3}\left(\tfrac16\right)^3\left(\tfrac56\right)^2 = 10\cdot\frac{1}{216}\cdot\frac{25}{36} = \frac{250}{7776} \approx 0.032\]

Akọwa Nkọwa

\(\mathbf{r} = 4t\,\mathbf{i} + (12 - 3t)\,\mathbf{j}\).

(a) Velocity is the time derivative:

\[\mathbf{v} = \frac{d\mathbf{r}}{dt} = 4\,\mathbf{i} - 3\,\mathbf{j}\ (\text{ms}^{-1})\]

(b) Positions at the two times:

\[\mathbf{r}(0) = 0\,\mathbf{i} + 12\,\mathbf{j},\qquad \mathbf{r}(5) = 20\,\mathbf{i} + (12 - 15)\,\mathbf{j} = 20\,\mathbf{i} - 3\,\mathbf{j}\]

Displacement \(= \mathbf{r}(5) - \mathbf{r}(0) = 20\,\mathbf{i} - 15\,\mathbf{j}\).

\[|\text{displacement}| = \sqrt{20^2 + (-15)^2} = \sqrt{400 + 225} = \sqrt{625} = 25\ \text{units}\]

Akọwa Nkọwa

Reading the diagram, the 6N force acts due East (direction 090°). Measuring anticlockwise from this 6N line, the other forces sit at these angles:

• 6N along East: \(0^\circ\)
• 8N: \(55^\circ\) above the 6N line, so \(55^\circ\)
• 4N (upper-left): a further \(100^\circ\) beyond the 8N, so \(55^\circ+100^\circ=155^\circ\)
• 4N (lower-right): \(25^\circ\) below East, so \(-25^\circ\) i.e. \(335^\circ\)
• 5N (lower-left): \(130^\circ\) round from the lower 4N, i.e. \(-25^\circ-130^\circ=-155^\circ\) i.e. \(205^\circ\)

(a) Resolve each force into components along the 6N direction (\(x\)) and perpendicular to it (\(y\)).

\[\sum F_x = 6\cos0^\circ+8\cos55^\circ+4\cos155^\circ+5\cos205^\circ+4\cos335^\circ\]

\[\sum F_x = 6+4.589-3.625-4.532+3.625 = 6.057\ \text{N}\]

\[\sum F_y = 6\sin0^\circ+8\sin55^\circ+4\sin155^\circ+5\sin205^\circ+4\sin335^\circ\]

\[\sum F_y = 0+6.553+1.690-2.113-1.690 = 4.440\ \text{N}\]

Magnitude of the resultant:

\[R=\sqrt{(\sum F_x)^2+(\sum F_y)^2}=\sqrt{6.057^2+4.440^2}=\sqrt{36.69+19.71}=\sqrt{56.40}\]

\[R\approx 7.51\ \text{N}\]

(b) Direction of the resultant. Both components are positive, so the resultant lies above the 6N (East) line.

\[\theta=\tan^{-1}\!\left(\frac{\sum F_y}{\sum F_x}\right)=\tan^{-1}\!\left(\frac{4.440}{6.057}\right)=\tan^{-1}(0.7332)\approx 36.2^\circ\]

The resultant is about 7.51 N acting at \(36.2^\circ\) above the 6N direction. Since the 6N points along a bearing of \(090^\circ\), this direction corresponds to a bearing of \(090^\circ-36.2^\circ\approx \mathbf{054^\circ}\).

Akọwa Nkọwa

(a) Five ordinates means four strips of width \(h = \dfrac{1 - 0}{4} = 0.25\). With \(f(x) = \dfrac{3}{1 + x^2}\):

\(x\)	0	0.25	0.5	0.75	1
\(f(x)\)	3.0000	2.8235	2.4000	1.9200	1.5000

\[\int_0^1 f\,dx \approx \frac{h}{2}\Big[f_0 + f_4 + 2(f_1 + f_2 + f_3)\Big] = \frac{0.25}{2}\Big[3 + 1.5 + 2(2.8235 + 2.4 + 1.92)\Big]\] \[= 0.125\big[4.5 + 2(7.1435)\big] = 0.125(18.787) = 2.348\ (\text{4 s.f.})\]

(b) \(A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\). First \(A^2\):

\[A^2 = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 7 & 4 \\ 12 & 7 \end{pmatrix}\] \[A^2 + A + 2I = \begin{pmatrix} 7 & 4 \\ 12 & 7 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 11 & 5 \\ 15 & 11 \end{pmatrix}\]

Image of \((1, 2)\):

\[\begin{pmatrix} 11 & 5 \\ 15 & 11 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 11 + 10 \\ 15 + 22 \end{pmatrix} = \begin{pmatrix} 21 \\ 37 \end{pmatrix}\]

The image is \((21,\ 37)\).

Akọwa Nkọwa

Expand and gather to standard form:

\[(x+1)(x+2) = k(3x+7)\] \[x^2 + 3x + 2 = 3kx + 7k\] \[x^2 + (3 - 3k)x + (2 - 7k) = 0\]

For equal roots the discriminant is zero:

\[(3 - 3k)^2 - 4(2 - 7k) = 0\] \[9 - 18k + 9k^2 - 8 + 28k = 0\] \[9k^2 + 10k + 1 = 0\] \[(9k + 1)(k + 1) = 0\]

Hence \(k = -\dfrac{1}{9}\) or \(k = -1\).