WAEC SSCE - Further Mathematics - 2007

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To make the expression \(x^2-x+p\) a perfect square, we need to add and subtract a suitable constant term. To do this, we can start by finding the square of half the coefficient of the linear term. That is, we find the square of \(\frac{1}{2}\) which is \(\frac{1}{4}\). So, we have: $$x^2-x+\frac{1}{4}-\frac{1}{4}+p$$ Now, we can write this expression as a perfect square by factoring the first three terms: $$\left(x-\frac{1}{2}\right)^2+p-\frac{1}{4}$$ For this expression to be a perfect square, the last term must be zero, so we have: $$p-\frac{1}{4}=0$$ Solving for \(p\), we get: $$p=\frac{1}{4}$$ Therefore, the value of \(p\) that makes \(x^2-x+p\) a perfect square is \(\frac{1}{4}\).

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We can evaluate the given limit by plugging in the value 3 directly into the expression, but this results in a division by zero, which is undefined. Instead, we can use algebraic techniques to simplify the expression and evaluate the limit. We can factor the numerator of the expression as follows: \(x^{2} - 2x - 3 = (x - 3)(x + 1)\) Substituting this factorization into the original expression, we get: \(\frac{x^{2} - 2x - 3}{x - 3} = \frac{(x - 3)(x + 1)}{x - 3} = x + 1\) Now, we can take the limit of the simplified expression as x approaches 3: \(\lim \limits_{x \to 3} x + 1 = 3 + 1 = 4\) Therefore, the value of the given limit is 4, which is the correct answer.

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We can simplify this expression using the trigonometric identity: \[\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B}\] In this case, we have A = 80° and B = 20°. Therefore: \[\tan(A-B) = \tan(80°-20°) = \tan 60° = \sqrt{3}\] Using the identity, we can rewrite the expression as: \[\frac{\tan 80° - \tan 20°}{1 + \tan 80° \tan 20°} = \tan(80°-20°) = \sqrt{3}\] Therefore, the simplified expression is \(\sqrt{3}\). So the correct option is (3) \(\sqrt{3}\).

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To find the probability that a student picked at random scored at least 14 marks, we need to add the frequencies of the classes with marks 14-16, 17-19, and 20-22, since all of these classes correspond to scores of at least 14. Adding the frequencies for these classes, we get: 41 + 14 + 8 = 63. Therefore, out of the 100 students, 63 scored at least 14 marks. To find the probability, we divide the number of students who scored at least 14 by the total number of students: 63/100 = 0.63 Therefore, the probability that a student picked at random scored at least 14 marks is 0.63. Answer: 0.63

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To find the identity element of an operation, we need to find an element that when combined with any other element using the operation, results in that same element. Let's assume that the identity element is "a", so we need to find the value of "a" that satisfies the following equation for any real number "x": x * a = a * x = x Substituting the definition of the operation, we get: x * a = 3x + 3a - ax = x Simplifying this equation by subtracting 3a from both sides, we get: x - 3a = ax Dividing both sides by x, we get: 1 - 3a/x = a Now, we have an expression for "a" in terms of "x". Let's check each option to see which one satisfies the equation for any value of "x". : a = 2x/(x-3), x ≠ 3 Substituting this into our equation, we get: 1 - 3(2x/(x-3))/x = 2x/(x-3) Simplifying this expression, we get: -3x/(x-3) = 0 This equation is not true for all values of "x", since it is only true when x = 0. Therefore, is not the identity element. : a = 2x/(x+3), x ≠ -3 Substituting this into our equation, we get: 1 - 3(2x/(x+3))/x = 2x/(x+3) Simplifying this expression, we get: 3x/(x+3) - 1 = 0 This equation is true for all values of "x", except for x = -3. Therefore, is the identity element. : a = 3x/(x-3), x ≠ 3 Substituting this into our equation, we get: 1 - 3(3x/(x-3))/x = 3x/(x-3) Simplifying this expression, we get: -6x/(x-3) = 0 This equation is not true for all values of "x", since it is only true when x = 0. Therefore, is not the identity element. : a = 3x/(x+3), x ≠ -3 Substituting this into our equation, we get: 1 - 3(3x/(x+3))/x = 3x/(x+3) Simplifying this expression, we get: 6x/(x+3) - 1 = 0 This equation is true for all values of "x", except for x = -3. Therefore, is the identity element. Therefore, the identity element of the operation is: a = 2x/(x+3), x ≠ -3.

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We can use vector addition to find the resultant force. Let's first convert the given forces into their x and y components. For \(F_1 = (5 N, 060°)\), the x component is \(F_{1x} = F_1 \cos(60°) = 5 \cos(60°) = 2.5 N\) and the y component is \(F_{1y} = F_1 \sin(60°) = 5 \sin(60°) = 4.33 N\). For \(F_2 = (10 N, 180°)\), the x component is \(F_{2x} = F_2 \cos(180°) = -10 N\) and the y component is \(F_{2y} = F_2 \sin(180°) = 0 N\). Now we can add the x and y components separately to get the resultant force: $$F_{Rx} = F_{1x} + F_{2x} = 2.5 N - 10 N = -7.5 N$$ $$F_{Ry} = F_{1y} + F_{2y} = 4.33 N + 0 N = 4.33 N$$ The magnitude of the resultant force is then given by: $$F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2} = \sqrt{(-7.5)^2 + 4.33^2} \approx 8.66 N$$ Therefore, the answer is \boxed{8.66 N}.

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When two lines are parallel, their slopes are equal. In other words, the coefficient of x in one equation divided by the coefficient of y should be equal to the coefficient of x in the other equation divided by the coefficient of y. Therefore, we can write: k/(-5) = m/n Multiplying both sides by -5n, we get: kn = -5m This is the relationship connecting the constants m, n, and k. Therefore, the correct option is (2) kn + 5m = 0.

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To calculate the value of \(\tan 15°\), we can use the half-angle formula for tangent: \[\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}\] Letting \(\theta = 30°\), we get: \[\tan 15° = \tan \frac{30°}{2} = \frac{\sin 30°}{1 + \cos 30°}\] We know that \(\sin 30° = \frac{1}{2}\) and \(\cos 30° = \frac{\sqrt{3}}{2}\). Substituting these values into the equation above, we get: \[\tan 15° = \frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}} = \frac{1}{2 + \sqrt{3}}\] Rationalizing the denominator, we get: \[\tan 15° = \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{1} = \boxed{2 - \sqrt{3}}\] Therefore, the answer is (d) \(2 - \sqrt{3}\).

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To evaluate \(|Q|P\), we first need to calculate the determinant of matrix \(Q\), denoted as \(|Q|\). $$ \begin{aligned} |Q| &= \begin{vmatrix} -1 & -2 \\ 3 & 2 \end{vmatrix} \\ &= (-1\times2) - (-2\times3) \\ &= -2 + 6 \\ &= 4 \end{aligned} $$ Next, we multiply the matrix \(P\) by the determinant of matrix \(Q\). $$ \begin{aligned} |Q|P &= 4\begin{pmatrix} 4 & 9 \end{pmatrix} \\ &= \begin{pmatrix} 16 & 36 \end{pmatrix} \end{aligned} $$ Therefore, the answer is \(\begin{pmatrix} 16 & 36 \end{pmatrix}\).

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Let the common difference of the A.P be d. Then, the first three terms of the A.P can be written as 4, 4 + d, and 4 + 2d. Since the sum of the first three terms of the A.P is 18, we have: 4 + (4 + d) + (4 + 2d) = 18 Simplifying this equation, we get: 3d + 12 = 18 3d = 6 d = 2 Therefore, the first three terms of the A.P are 4, 6, and 8. The product of the first three terms of the A.P is: 4 × 6 × 8 = 192 Therefore, the answer is 192.

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To solve this problem, we need to use the concept of combinations. The number of ways to choose a group of r objects from a set of n distinct objects is given by the formula: C(n,r) = n! / (r! * (n-r)!) where n! denotes the factorial of n (i.e., the product of all positive integers up to and including n), and r! denotes the factorial of r. In this problem, we want to choose a group of 5 boys and 4 girls from a class of 8 boys and 6 girls. We can do this in two steps: Step 1: Choose the 5 boys from the 8 boys. This can be done in C(8,5) ways. Step 2: Choose the 4 girls from the 6 girls. This can be done in C(6,4) ways. The total number of ways to choose the group of 5 boys and 4 girls is the product of the number of ways in Step 1 and the number of ways in Step 2: C(8,5) * C(6,4) = (8! / (5! * 3!)) * (6! / (4! * 2!)) = (8*7*6 / (3*2*1)) * (6*5 / (2*1)) = 56 * 15 = 840 Therefore, the number of ways to choose a group of 5 boys and 4 girls from a class of 8 boys and 6 girls is 840.

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The formula to find the probability of the intersection of two independent events is given by: $$P(X \cap Y) = P(X) \times P(Y)$$ In this case, we know that: $$P(X) = \frac{1}{5}$$ $$P(X \cap Y) = \frac{2}{15}$$ Substituting the values into the formula, we have: $$\frac{2}{15} = \frac{1}{5} \times P(Y)$$ Solving for $P(Y)$, we get: $$P(Y) = \frac{\frac{2}{15}}{\frac{1}{5}} = \frac{2}{15} \times \frac{5}{1} = \frac{2}{3}$$ Therefore, the probability that event Y occurs is $\frac{2}{3}$. Answer: (a) \(\frac{2}{3}\).

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To find the standard deviation of a set of numbers, we need to first calculate the mean of the numbers, then subtract the mean from each number to get the deviations, square each deviation, find the mean of the squared deviations (variance), and finally take the square root of the variance to get the standard deviation. So first, let's find the mean: mean = (-1 + 5 + 0 + 2 + 9) / 5 = 3 Next, we'll find the deviations by subtracting the mean from each number: -1 - 3 = -4 5 - 3 = 2 0 - 3 = -3 2 - 3 = -1 9 - 3 = 6 Now we'll square each deviation: (-4)^2 = 16 2^2 = 4 (-3)^2 = 9 (-1)^2 = 1 6^2 = 36 Next, we'll find the mean of the squared deviations: (16 + 4 + 9 + 1 + 36) / 5 = 13.2 Finally, we'll take the square root of the variance to get the standard deviation: sqrt(13.2) ≈ 3.6 So the answer is (c) 3.6, correct to one decimal place.

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We can use the identity \(\sin (a+b) = \sin a \cos b + \cos a \sin b\) to simplify the expression \(\sin (270 + x)°\): \begin{align*} \sin (270 + x)° &= \sin 270° \cos x + \cos 270° \sin x \\ &= (-1)(\cos x) + (0)(\sin x) \\ &= -\cos x \end{align*} Therefore, the answer is \(-\cos x\).

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In this problem, we need to find the number of ways of dividing 12 people into three groups of 2, 7 and 3 in that order. Firstly, we can choose the first group of 2 people in \(\binom{12}{2}\) ways, which gives us 66 ways. Then, we can choose the second group of 7 people from the remaining 10 people in \(\binom{10}{7}\) ways, which gives us 120 ways. Finally, the last group of 3 people is determined by the previous two groups, so there is only one way to form this group. Therefore, the total number of ways of dividing 12 people into three groups of 2, 7 and 3 in that order is the product of the number of ways of forming each group, which is equal to \(66 \times 120 \times 1 = 7,920\). Thus, the answer is 7920.

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To find the probability that at least one of the three students solves the problem, we need to calculate the probability that none of them solve the problem and then subtract that from 1 (which represents the total probability). The probability that the first student does not solve the problem is 0.4 (since the probability of solving the problem is 0.6). Similarly, the probability that the second student does not solve the problem is 0.3 and the probability that the third student does not solve the problem is 0.2. To find the probability that none of them solve the problem, we multiply these probabilities together: 0.4 x 0.3 x 0.2 = 0.024. Therefore, the probability that at least one of the three students solves the problem is 1 - 0.024 = 0.976. So, the correct answer is: 0.976.

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To solve this problem, we need to use the equations of motion. Since the force is acting on the body, it will cause an acceleration in the direction of the force. The acceleration can be found using Newton's second law of motion: $$F = ma$$ where F is the force, m is the mass of the body, and a is the acceleration. In this case, the force is 30 N and the mass is 6 kg. Therefore, the acceleration can be found as: $$a = \frac{F}{m} = \frac{30}{6} = 5\text{ m/s}^2$$ Next, we can use the equations of motion to find the distance covered by the body in 10 seconds. Since the body starts from rest, we can use the equation: $$s = \frac{1}{2}at^2$$ where s is the distance covered, a is the acceleration, and t is the time. Substituting the values, we get: $$s = \frac{1}{2} \times 5 \times 10^2 = 125\text{ m}$$ However, the force is acting at an angle of 60 degrees to the horizontal. Therefore, only the horizontal component of the force will cause motion in the horizontal direction. The horizontal component of the force is: $$F_h = F \cos \theta = 30 \cos 60^\circ = 15\text{ N}$$ We can now find the acceleration in the horizontal direction as: $$a_h = \frac{F_h}{m} = \frac{15}{6} = 2.5\text{ m/s}^2$$ Using the equation of motion, the distance covered in 10 seconds in the horizontal direction is: $$s_h = \frac{1}{2} \times 2.5 \times 10^2 = 12.5\text{ m}$$ However, we also need to find the distance covered in the vertical direction. The vertical component of the force is: $$F_v = F \sin \theta = 30 \sin 60^\circ = 15\sqrt{3}\text{ N}$$ Since there is no force acting in the vertical direction, the body will move under the influence of gravity only. The acceleration due to gravity is 9.8 m/s^2. Therefore, the distance covered in the vertical direction in 10 seconds is: $$s_v = \frac{1}{2} \times 9.8 \times 10^2 = 490\text{ m}$$ The total distance covered by the body is the hypotenuse of the right-angled triangle formed by the horizontal and vertical distances: $$s_{total} = \sqrt{s_h^2 + s_v^2} = \sqrt{(12.5)^2 + (490)^2} \approx 490\text{ m}$$ Therefore, the answer is 125 m.

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To find the area of the circle whose equation is given as \(x^{2} + y^{2} - 4x + 8y + 11 = 0\), we need to first complete the square for both \(x\) and \(y\). Rearranging the equation, we get: \begin{aligned} x^2 - 4x + y^2 + 8y &= -11 \\ x^2 - 4x + 4 + y^2 + 8y + 16 &= 9 \\ (x-2)^2 + (y+4)^2 &= 3^2 \end{aligned} Now we can see that the equation represents a circle with center \((2, -4)\) and radius \(3\). The formula for the area of a circle is \(A = \pi r^2\), where \(r\) is the radius. In this case, the radius is \(3\), so the area is: \begin{aligned} A &= \pi r^2 \\ &= \pi (3)^2 \\ &= \boxed{9\pi} \end{aligned} Therefore, the answer is \(\boxed{9\pi}\).

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We can simplify the left side of the equation as follows: $$\log_{2} y^{\frac{1}{2}} = \frac{1}{2}\log_{2} y$$ Similarly, we can simplify the right side of the equation as follows: $$\log_{5} 125 = \log_{5} 5^3 = 3\log_{5} 5 = 3$$ So we have: $$\frac{1}{2}\log_{2} y = 3$$ Multiplying both sides by 2, we get: $$\log_{2} y = 6$$ Using the definition of logarithms, we can write this as: $$2^6 = y$$ Simplifying, we get: $$y = 64$$ Therefore, the value of y is 64.

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To find the x-intercepts of the curve, we need to set y = 0 and solve for x. Therefore, we have: $$ \begin{aligned} y &= 2x^{2} - 5x + k \\ 0 &= 2x^{2} - 5x + k \\ \end{aligned} $$ For the curve to have two x-intercepts, the discriminant of the quadratic equation must be greater than zero: $$ \begin{aligned} b^{2} - 4ac &> 0 \\ (-5)^{2} - 4(2)(k) &> 0 \\ 25 - 8k &> 0 \\ -8k &> -25 \\ k &< \frac{25}{8} \\ \end{aligned} $$ Therefore, the value of constant k must be less than \(\frac{25}{8}\). Thus, the answer is \(k < \frac{25}{8}\).

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We first need to find the composition \(f^{-1} \circ g(x)\), which means we need to find the inverse function of f, denoted by \(f^{-1}(x)\). To find \(f^{-1}(x)\), we solve the equation \(y = 2x - 1\) for x: \(y + 1 = 2x\) \(x = \frac{y + 1}{2}\) Thus, \(f^{-1}(x) = \frac{x + 1}{2}\). Now, we can find \(f^{-1} \circ g(3)\) by first computing g(3): \(g(3) = 3^{2} + 1 = 10\) Then, we can plug g(3) into the composition \(f^{-1} \circ g(x)\): \((f^{-1} \circ g)(3) = f^{-1}(g(3)) = f^{-1}(10) = \frac{10 + 1}{2} = \frac{11}{2}\) Therefore, the value of \(f^{-1} \circ g(3)\) is \(\frac{11}{2}\), which is the correct answer.

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The equation of a circle in standard form is \((x-a)^2 + (y-b)^2 = r^2\), where \((a,b)\) are the coordinates of the center and \(r\) is the radius. We need to rewrite the given equation in this standard form by completing the square for both the \(x\) and \(y\) terms. \begin{align*} x^{2} + y^{2} - 4x - 2y - 3 &= 0\\ (x^{2} - 4x) + (y^{2} - 2y) &= 3\\ (x^{2} - 4x + 4) + (y^{2} - 2y + 1) &= 3 + 4 + 1\\ (x-2)^{2} + (y-1)^{2} &= 8 \end{align*} Comparing with the standard form, we see that the center of the circle is \((2,1)\) and the radius is \(\sqrt{8} = 2\sqrt{2}\). Therefore, the answer is: the radius of the circle is \(2\sqrt{2}\) and the coordinates of its center are \((2,1)\).

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We are given that \(\alpha\) and \(\beta\) are the roots of the equation, and we are also given two conditions based on these roots: 1. \(\alpha + \beta = 3\) 2. \(\alpha \beta = 2\) To find the equation, we can use the fact that if \(\alpha\) and \(\beta\) are roots of the equation, then the equation can be written in factored form as: $$(x - \alpha)(x - \beta) = 0$$ Expanding this equation, we get: $$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$ Substituting the values of \(\alpha + \beta\) and \(\alpha \beta\) from the given conditions, we get: $$x^2 - 3x + 2 = 0$$ Therefore, the equation with roots \(\alpha\) and \(\beta\) is: $$x^2 - 3x + 2 = 0$$ Hence, the correct option is (a) \(x^{2} - 3x + 2 = 0\).

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The derivative of a quadratic function is obtained by differentiating the function term by term. Thus, differentiating the given function, we have: $$f'(x) = 2mx - 6$$ We are also given that $f'(1) = 12$. Substituting $x=1$ and equating to 12, we have: $$f'(1) = 2m(1) - 6 = 12$$ Simplifying this equation, we get: $$2m = 18$$ Therefore, $m=9$. Thus, the value of the constant $m$ is 9.

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(a) Angle between \(p=\binom{3}{-1}\) and \(q=\binom{3}{4}\).

(b) Forces on square A(0,0), B(2,0), C(2,2), D(0,2).

Write each force as its magnitude times the unit vector of its line of action:

• \(10\,\text{N}\) along \(\overrightarrow{BA}=(-1,0):\ (-10,0)\)
• \(15\,\text{N}\) along \(\overrightarrow{BC}=(0,1):\ (0,15)\)
• \(20\,\text{N}\) along \(\overrightarrow{DC}=(1,0):\ (20,0)\)
• \(5\,\text{N}\) along \(\overrightarrow{AD}=(0,1):\ (0,5)\)

Resultant components:

(i) Magnitude:

(ii) Direction (measured from the positive x-axis, i.e. along AB):

The resultant is about \(22.4\,\text{N}\) at \(63.4^{\circ}\) to AB.

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Resolve each force into components with bearings measured clockwise from north, using \(x = F\sin\theta\) (east) and \(y = F\cos\theta\) (north).

• \(F_1 = 10\,\text{N at }090^\circ:\ x = 10\sin 90^\circ = 10,\ y = 10\cos 90^\circ = 0\)
• \(F_2 = 20\,\text{N at }210^\circ:\ x = 20\sin 210^\circ = -10,\ y = 20\cos 210^\circ = -17.32\)
• \(F_3 = 4\,\text{N at }330^\circ:\ x = 4\sin 330^\circ = -2,\ y = 4\cos 330^\circ = 3.46\)

Sum the components:

Magnitude of the resultant:

The magnitude of the resultant force is \(14.0\,\text{N}\) (to 1 decimal place).

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Class width \(=5\); mid-values \(19,24,\dots,54\). Code with \(u=\dfrac{x-39}{5}\), \(A=39\).

(a)(i) Mean.\[\bar{x}=39+\frac{\sum fu}{\sum f}\times 5=39+\frac{-151}{190}\times 5=39-3.97=35.0\text{ years}\]

(a)(ii) Standard deviation.\[\sigma=c\sqrt{\frac{\sum fu^2}{\sum f}-\left(\frac{\sum fu}{\sum f}\right)^2}=5\sqrt{\frac{693}{190}-\left(\frac{-151}{190}\right)^2}\]\[=5\sqrt{3.647-0.632}=5\sqrt{3.015}=5(1.737)=8.68\text{ years}\]

(b) Probability at most 36 years old. Ages \(\le 36\) cover the classes \(17-21,\,22-26,\,27-31,\,32-36\), giving \(12+24+30+37=103\) workers out of 190.\[P(\text{at most }36)=\frac{103}{190}=0.542\]

Mean \(\approx\) 35.0 years, standard deviation \(\approx\) 8.68 years, and \(P=0.542\).

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Let the winning probabilities be \(p_1 = \tfrac{1}{5},\ p_2 = \tfrac{1}{4},\ p_3 = \tfrac{1}{3},\ p_4 = \tfrac{1}{2}\), with losing probabilities \(q_1 = \tfrac{4}{5},\ q_2 = \tfrac{3}{4},\ q_3 = \tfrac{2}{3},\ q_4 = \tfrac{1}{2}\).

"At most two win" is easier via the complement: \(1 - P(3\text{ win}) - P(4\text{ win})\).

All four win:

Exactly three win (one loses at a time, summed):

So \(P(3\text{ or }4) = \dfrac{10}{120} + \dfrac{1}{120} = \dfrac{11}{120}\), and:

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None

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(a) Find k given the determinant equals \(-207\).

Expand along the first row:

Set equal to \(-207\):

So \(k=-\dfrac{1}{5}\) or \(k=-4\).

(b) Curve \(x(y^{2}+1)-y(x^{2}+1)+4=0\).

Expand: \(xy^{2}+x-x^{2}y-y+4=0\). Differentiate implicitly:

(i) Gradient at any point:

(ii) At \((-1,-3)\):

Tangent through \((-1,-3)\):

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Differentiate the curve \(y = 2x^2 + x - 3\):

At \((2, 7)\) the gradient of the tangent is \(4(2) + 1 = 9\). The normal is perpendicular, so its gradient is:

Equation of the normal through \((2, 7)\):

The normal meets the \(x\)-axis where \(y = 0\):

Therefore \(P = (65,\ 0)\).

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(a) Solve \(\sqrt{4x-3}-\sqrt{2x-5}=2\).

Isolate one surd and square:

Square again:

So \(x=3\) or \(x=7\). Check both in the original equation:

• \(x=3:\ \sqrt{9}-\sqrt{1}=3-1=2\) ✓
• \(x=7:\ \sqrt{25}-\sqrt{9}=5-3=2\) ✓

Both are valid, so \(x=3\) or \(x=7\).

(b) Area enclosed by \(y^{2}=4x\) and \(y+x=0\).

The line is \(x=-y\). Substitute into the parabola:

So \(y=0\) (giving \((0,0)\)) and \(y=-4\) (giving \((4,-4)\)). Integrating with respect to \(y\), the line \(x=-y\) lies to the right of the parabola \(x=\tfrac{y^{2}}{4}\) on \(-4\le y\le 0\):

The finite area is \(\dfrac{8}{3}\approx 2.67\) square units.

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Let \(u = x^{1/3}\), so \(x^{2/3} = u^2\). The equation becomes a quadratic:

Return to \(x\):

So \(x = 8\) or \(x = 27\).

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None

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Five different digits are available: \(4, 5, 6, 7, 8\).

(a)(i) Repetition allowed: each of the 5 positions can be any of the 5 digits:

(ii) Repetition not allowed: arrange all 5 distinct digits:

(b) A number is odd if its last digit is odd. Among the digits, the odd ones are \(5\) and \(7\), giving 2 choices for the units position. With no repetition, the remaining 4 positions are filled by the remaining 4 digits in \(4!\) ways:

So 48 of the numbers are odd.

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Let the first term be \(a\) and common ratio \(r\), with \(|r| < 1\).

Second term: \(ar = 3\), so \(a = \dfrac{3}{r}\).

Sum to infinity: \(\dfrac{a}{1 - r} = \dfrac{25}{2}\). Substitute \(a\):

Solve:

Both satisfy \(|r| < 1\), so the common ratio is \(r = \dfrac{3}{5}\) or \(r = \dfrac{2}{5}\).

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(a) Trapezium rule, seven ordinates, \(\displaystyle\int_{0}^{3}\frac{dx}{x^{2}+1}\).

Seven ordinates means six strips, so \(h=\dfrac{3-0}{6}=0.5\). With \(y=\dfrac{1}{x^{2}+1}\):

(b) Matrix method: \(-2x+y=3,\ -x+4y=1\).

Determinant \(=(-2)(4)-(1)(-1)=-8+1=-7\). Inverse:

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(a)(i) Scatter diagram. Using a scale of 2 cm to 10 marks on each axis, the nine pairs \((43,54),(46,53),(48,63),(39,30),(30,44),(60,75),(8,20),(45,33),(40,49)\) are plotted with Mathematics \((x)\) on the horizontal axis and Physics \((y)\) on the vertical axis. The line of best fit required in part (a)(iii) is drawn on the same diagram.

(a)(ii) The means.

So the mean point is \((\bar{x},\bar{y})=(39.9,\,46.8)\).

(a)(iii) Line of best fit. A single straight line is drawn through the mean point \((39.9,46.8)\), oriented so that the points are balanced roughly equally above and below it (shown as the straight line on the scatter diagram above).

(b) Equation of the line. Set out the sums needed for the least-squares gradient:

With \(n=9\), the gradient is

The line passes through \((\bar{x},\bar{y})\), so using \(y-\bar{y}=b(x-\bar{x})\):

(c) Physics mark when Mathematics \(=28\).

The predicted mark in Physics for a candidate who scored 28 in Mathematics is 35.5 (to one decimal place).

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(a) For \(x^2 + mx + 11 = 0\) with roots \(\alpha,\beta\): \(\alpha + \beta = -m\) and \(\alpha\beta = 11\).

Setting this equal to 27:

(b) From the line \(2x + 3y = 1\), \(x = \dfrac{1 - 3y}{2}\). Substitute into the circle \(2x^2 + 2y^2 + 4x + 9y - 9 = 0\) (dividing by 2 first: \(x^2 + y^2 + 2x + 4.5y - 4.5 = 0\)):

Multiply through by 4 and simplify:

For \(y = 1\): \(x = \dfrac{1 - 3}{2} = -1\), giving \((-1, 1)\).

For \(y = -1\): \(x = \dfrac{1 + 3}{2} = 2\), giving \((2, -1)\), which lies in the fourth quadrant.

Therefore \(P = (-1, 1)\) and \(Q = (2, -1)\).

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(a) If \({}^{18}C_r={}^{18}C_{r+2}\), find \({}^{r}C_5\).

Since \({}^{n}C_a={}^{n}C_b\) with \(a\ne b\) requires \(a+b=n\):

(b) Binomial with \(p=0.1,\ q=0.9,\ n=6\).

(i) Exactly 5:

(ii) None:

(iii) At most 2 (0, 1 or 2):