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**Question 2**
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The equation of a circle is \(3x^{2} + 3y^{2} + 6x - 12y + 6 = 0\). Find its radius

**Answer Details**

To find the radius of the given circle, we need to rewrite the equation in standard form, which is of the form \((x - a)^2 + (y - b)^2 = r^2\), where \((a, b)\) is the center of the circle and \(r\) is its radius. Completing the square for the given equation, we have: \begin{align*} 3x^{2} + 3y^{2} + 6x - 12y + 6 &= 0\\ 3(x^{2} + 2x) + 3(y^{2} - 4y) &= -6\\ 3(x^{2} + 2x + 1) + 3(y^{2} - 4y + 4) &= -6 + 3 + 12\\ 3(x + 1)^{2} + 3(y - 2)^{2} &= 9 \end{align*} Dividing both sides by 3, we have: \[(x + 1)^{2} + (y - 2)^{2} = 3\] Comparing this with the standard form, we can see that the center of the circle is \((-1, 2)\), and the radius is \(\sqrt{3}\). Therefore, the answer is \boxed{\sqrt{3}}.

**Question 3**
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Find the standard deviation of the numbers 3,6,2,1,7 and 5.

**Answer Details**

To find the standard deviation of a set of numbers, we need to follow these steps: 1. Find the mean (average) of the numbers. 2. For each number, subtract the mean and square the result. 3. Find the mean of the squared differences. 4. Take the square root of the mean to get the standard deviation. So, let's apply these steps to the given set of numbers: 3, 6, 2, 1, 7, 5. 1. The mean is (3 + 6 + 2 + 1 + 7 + 5) / 6 = 4. 2. For each number, subtract the mean and square the result: \begin{align*} (3 - 4)^2 &= 1 \\ (6 - 4)^2 &= 4 \\ (2 - 4)^2 &= 4 \\ (1 - 4)^2 &= 9 \\ (7 - 4)^2 &= 9 \\ (5 - 4)^2 &= 1 \end{align*} 3. Find the mean of the squared differences: \begin{align*} \frac{1 + 4 + 4 + 9 + 9 + 1}{6} &= \frac{28}{6} \\ &= 4.67 \end{align*} 4. Take the square root of the mean to get the standard deviation: \begin{align*} \sqrt{4.67} &\approx 2.16 \end{align*} Therefore, the standard deviation of the given set of numbers is approximately 2.16. So, the correct answer is (b) 2.16.

**Question 4**
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A particle starts from rest and moves in a straight line such that its velocity, v, at time t seconds is given by \(v = (3t^{2} - 2t) ms^{-1}\). Calculate the distance covered in the first 2 seconds.

**Answer Details**

To find the distance covered in the first 2 seconds, we need to integrate the velocity function from 0 to 2 seconds: \begin{align*} \text{Distance} &= \int_{0}^{2} v\,dt \\ &= \int_{0}^{2} (3t^2 - 2t)\,dt \\ &= \left[\frac{3}{3}t^3 - \frac{2}{2}t^2\right]_{0}^{2} \\ &= \left(3\cdot 2^2 - 2\cdot 2^2\right) - \left(3\cdot 0^2 - 2\cdot 0^2\right) \\ &= 4 \text{ m}. \end{align*} Therefore, the distance covered in the first 2 seconds is 4 meters. Answer choice (b) is correct.

**Question 5**
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Simplify \(\sqrt[3]{\frac{8}{27}} - (\frac{4}{9})^{-\frac{1}{2}}\)

**Question 6**
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Evaluate \(\cos (\frac{\pi}{2} + \frac{\pi}{3})\)

**Answer Details**

To evaluate \(\cos (\frac{\pi}{2} + \frac{\pi}{3})\), we can use the formula for the cosine of the sum of two angles, which states that \[\cos(a+b) = \cos a \cos b - \sin a \sin b.\] Using this formula, we can simplify \(\cos (\frac{\pi}{2} + \frac{\pi}{3})\) as follows: \[\cos (\frac{\pi}{2} + \frac{\pi}{3}) = \cos \frac{\pi}{2} \cos \frac{\pi}{3} - \sin \frac{\pi}{2} \sin \frac{\pi}{3}\] Recall that \(\cos \frac{\pi}{2} = 0\) and \(\sin \frac{\pi}{2} = 1\), and we also know that \(\cos \frac{\pi}{3} = \frac{1}{2}\) and \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\) from the unit circle. Substituting these values into the equation above, we get: \[\cos (\frac{\pi}{2} + \frac{\pi}{3}) = 0 \cdot \frac{1}{2} - 1 \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2}\] Therefore, the value of \(\cos (\frac{\pi}{2} + \frac{\pi}{3})\) is \(\frac{-\sqrt{3}}{2}\)

**Question 7**
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A particle starts from rest and moves in a straight line such that its velocity, v, at time t seconds is given by \(v = (3t^{2} - 2t) ms^{-1}\). Determine the acceleration when t = 2 secs.

**Answer Details**

The acceleration of a particle is given by the derivative of its velocity with respect to time. So we differentiate the given equation for velocity to find the acceleration: $$a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t) = 6t - 2$$ When t = 2 seconds, the acceleration is: $$a = 6(2) - 2 = 10 \text{ ms}^{-2}$$ Therefore, the correct option is: \(\mathbf{10 ms^{-2}}\).

**Question 8**
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A binary operation * is defined on the set of real numbers R, by a* b = -1. Find the identity element under the operation *.

**Question 9**
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Determine the coefficient of \(x^{2}\) in the expansion of \((a + 3x)^{6}\).

**Question 10**
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Four doctors and two nurses are to sit round a circular table. In how many ways can this be done if the nurses are to sit together?

**Question 11**
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What percentage increase in the radius of a sphere will cause its volume to increase by 45%?

**Question 12**
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\(f(x) = (x^{2} + 3)^{2}\) is defines on the set of real numbers, R. Find the gradient of f(x) at x = \(\frac{1}{2}\).

**Answer Details**

To find the gradient of \(f(x) = (x^2+3)^2\) at \(x=\frac{1}{2}\), we need to differentiate f(x) with respect to x and then substitute x = \(\frac{1}{2}\). Using the chain rule, we have: \begin{align*} \frac{d}{dx}[(x^2+3)^2] &= 2(x^2+3) \cdot \frac{d}{dx}(x^2+3) \\ &= 2(x^2+3) \cdot 2x \\ &= 4x(x^2+3) \end{align*} So, the gradient of \(f(x)\) is \(4x(x^2+3)\). Substituting \(x=\frac{1}{2}\), we get: \begin{align*} \text{Gradient of } f(x) \text{ at } x = \frac{1}{2} &= 4\left(\frac{1}{2}\right)\left(\left(\frac{1}{2}\right)^2+3\right) \\ &= 4\left(\frac{1}{2}\right)\left(\frac{13}{4}\right) \\ &= \frac{13}{2} \\ &= 6.5 \end{align*} Therefore, the gradient of \(f(x)\) at \(x=\frac{1}{2}\) is 6.5.

**Question 13**
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The functions f and g are defined on the set, R, of real numbers by \(f : x \to x^{2} - x - 6\) and \(g : x \to x - 1\). Find \(f \circ g(3)\).

**Answer Details**

To find \(f \circ g(3)\), we first need to apply the function g to the input 3. Since \(g : x \to x - 1\), we have: \[g(3) = 3 - 1 = 2.\] Now we can use the output of g(3) as the input to the function f. Since \(f : x \to x^{2} - x - 6\), we have: \[f(g(3)) = f(2) = 2^{2} - 2 - 6 = -4.\] Therefore, the value of \(f \circ g(3)\) is -4

**Question 15**
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Evaluate \(\frac{1}{1 - \sin 60°}\), leaving your answer in surd form.

**Answer Details**

We know that \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\). Substituting this value in the given expression, we get: \[\frac{1}{1 - \sin 60^{\circ}} = \frac{1}{1 - \frac{\sqrt{3}}{2}}\] Rationalizing the denominator by multiplying both numerator and denominator by \(1 + \frac{\sqrt{3}}{2}\), we get: \begin{align*} \frac{1}{1 - \sin 60^{\circ}} &= \frac{1}{1 - \frac{\sqrt{3}}{2}} \cdot \frac{1 + \frac{\sqrt{3}}{2}}{1 + \frac{\sqrt{3}}{2}} \\ &= \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{3}{4}} \\ &= \frac{1 + \frac{\sqrt{3}}{2}}{\frac{1}{4}} \\ &= 4 + 2\sqrt{3} \end{align*} Therefore, the answer is \boxed{4 + 2\sqrt{3}}.

**Question 16**
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Find the equation of a circle with centre (-3, -8) and radius \(4\sqrt{6}\).

**Answer Details**

To find the equation of a circle with center \((a, b)\) and radius \(r\), we use the formula: \[(x - a)^2 + (y - b)^2 = r^2\] In this case, the center is \((-3, -8)\) and the radius is \(4\sqrt{6}\). So the equation of the circle is: \[(x - (-3))^2 + (y - (-8))^2 = (4\sqrt{6})^2\] which simplifies to: \[(x + 3)^2 + (y + 8)^2 = 96\] Expanding the left-hand side gives: \[x^2 + 6x + 9 + y^2 + 16y + 64 = 96\] which simplifies to: \[x^2 + y^2 + 6x + 16y - 23 = 0\] So the answer is (2) \(x^{2} + y^{2} + 6x + 16y - 23 = 0\).

**Question 17**
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If \(\log_{9} 3 + 2x = 1\), find x.

**Question 18**
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In how many ways can 3 prefects be chosen out of 8 prefects?

**Answer Details**

There are different ways to approach this problem, but one common method is to use the formula for combinations. In general, the number of ways to choose k items out of n distinct items (without repetition and without order) is given by the formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) where n! (read as "n factorial") means the product of all positive integers up to n, and 0! is defined as 1. In this specific problem, we want to choose 3 prefects out of 8, so we have: \(\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\) Therefore, there are 56 ways to choose 3 prefects out of 8 prefects. So the answer is: 56.

**Question 19**
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Find the remainder when \(5x^{3} + 2x^{2} - 7x - 5\) is divided by (x - 2).

**Answer Details**

To find the remainder when the polynomial \(5x^3 + 2x^2 - 7x - 5\) is divided by \((x-2)\), we can use the Remainder Theorem which states that the remainder of the polynomial division can be found by evaluating the polynomial at the root of the divisor. In this case, the root of the divisor \((x-2)\) is \(x=2\). So, we evaluate the polynomial at \(x=2\) as follows: \begin{align*} 5(2)^3 + 2(2)^2 - 7(2) - 5 &= 40 + 8 - 14 - 5 \\ &= 29 \end{align*} Therefore, the remainder is \(29\), and the correct option is (c).

**Question 20**
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The fourth term of a geometric sequence is 2 and the sixth term is 8. Find the common ratio.

**Question 21**
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\(f(x) = p + qx\), where p and q are constants. If f(1) = 7 and f(5) = 19, find f(3).

**Question 22**
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The sum and product of the roots of a quadratic equation are \(\frac{4}{7}\) and \(\frac{5}{7}\) respectively. Find its equation.

**Answer Details**

Let the quadratic equation be \(ax^2 + bx + c = 0\). According to the problem, we have: Sum of the roots: \(-\frac{b}{a} = \frac{4}{7}\) Product of the roots: \(\frac{c}{a} = \frac{5}{7}\) Using the formula for the sum and product of roots, we get: \(-\frac{b}{a} = \frac{4}{7} \implies b = -\frac{4}{7}a\) \(\frac{c}{a} = \frac{5}{7} \implies c = \frac{5}{7}a\) Substituting these values in the quadratic equation, we get: \(ax^2 - \frac{4}{7}ax + \frac{5}{7}a = 0\) Multiplying both sides by \(\frac{7}{a}\), we get: \(7x^2 - 4x + 5 = 0\) Therefore, the quadratic equation is \(7x^2 - 4x + 5 = 0\).

**Question 23**
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The probability that Kofi and Ama hit a target in a shooting competition are \(\frac{1}{6}\) and \(\frac{1}{9}\) respectively. What is the probability that only one of them hit the target?

**Question 24**
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A body of mass 25kg changes its speed from 15m/s to 35m/s in 5 seconds by the action of an applied force F. Find the value of F.

**Answer Details**

To find the value of force F, we need to use Newton's second law of motion which states that force is equal to mass multiplied by acceleration (F = ma). In this case, we know the mass of the body is 25kg and we can find the acceleration by using the formula: acceleration = change in velocity / time The change in velocity is 35m/s - 15m/s = 20m/s and the time is 5 seconds, so: acceleration = 20m/s / 5s = 4m/s^2 Now we can use Newton's second law to find the force: F = ma F = 25kg x 4m/s^2 F = 100N Therefore, the value of force F is 100N.

**Question 25**
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The mean of 2, 5, (x + 2), 7 and 9 is 6. Find the median.

**Answer Details**

Certainly, I can help with that! To find the median of a set of numbers, we need to first put them in order from smallest to largest. So, let's arrange the given numbers in ascending order: 2, 5, x+2, 7, 9 The next step is to find the middle value. If there are an odd number of values, the median is simply the middle value. If there are an even number of values, the median is the average of the two middle values. In this case, we have five numbers, which is an odd number. So, the median is simply the middle value of the ordered set. To find the middle value, we need to count from either end until we get to the middle. Since there are five numbers, the middle value will be the third one. 2, 5, x+2, 7, 9 The third number is x+2, so that is the median. Now, we are given that the mean of the five numbers is 6. To find the mean, we add up all the numbers and divide by the total number of numbers: (2 + 5 + x + 2 + 7 + 9)/5 = 6 (25 + x)/5 = 6 25 + x = 30 x = 5 So, the value of x that makes the mean of the five numbers equal to 6 is 5. Plugging this value of x back into the original set, we have: 2, 5, 7, 7, 9 The median of this set is the middle value, which is 7. Therefore, the median of the given set of numbers is 7. I hope this helps you understand how to find the median of a set of numbers, as well as how to use the mean to solve for missing values in the set!

**Question 26**
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Two forces 10N and 15N act on an object at an angle of 120° to each other. Find the magnitude of the resultant.

**Question 27**
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Express 75° in radians, leaving your answer in terms of \(\pi\).

**Answer Details**

To convert from degrees to radians, we use the conversion formula: radians = (pi/180) x degrees So to convert 75 degrees to radians, we have: radians = (pi/180) x 75 radians = (5pi/12) Therefore, the answer is (A) \(\frac{5\pi}{12}\).

**Question 28**
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Given that \(q = 9i + 6j\) and \(r = 4i - 6j\), which of the following statements is true?

**Answer Details**

To determine which of the statements is true, we can use the properties of vectors. We start by finding the magnitude of vector r using the formula: |magnitude of r| = sqrt(x^2 + y^2) where x and y are the coefficients of the i and j terms respectively. |magnitude of r| = sqrt(4^2 + (-6)^2) = sqrt(52) So the statement "The magnitude of r is 52 units" is true. Next, we can find the dot product of vectors q and r. If the dot product is equal to zero, then the vectors are perpendicular. If the dot product is non-zero, then the vectors are not perpendicular. q . r = (9 * 4) + (6 * -6) = 36 - 36 = 0 Since the dot product of q and r is zero, the statement "r and q are perpendicular" is true. Therefore, the correct answer is: - r and q are perpendicular.

**Question 29**
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A line is perpendicular to \(3x - y + 11 = 0\) and passes through the point (1, -5). Find its equation.

**Question 30**
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Find the unit vector in the direction of (-5i + 12j).

**Answer Details**

To find the unit vector in the direction of a given vector, we need to divide the vector by its magnitude. The magnitude of vector (-5i + 12j) can be found using the Pythagorean theorem as follows: $$\left|\begin{pmatrix}-5 \\ 12 \\\end{pmatrix}\right| = \sqrt{(-5)^2 + (12)^2} = 13$$ Therefore, the unit vector in the direction of (-5i + 12j) is: $$\frac{1}{13}\begin{pmatrix}-5 \\ 12 \\\end{pmatrix} = \frac{1}{13}(-5i + 12j)$$ So the correct option is \(\frac{1}{13}(-5i + 12j)\).

**Question 31**
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The inverse of a function is given by \(f^{-1} : x \to \frac{x + 1}{4}\).

**Answer Details**

To find the original function from its inverse, we can simply swap the roles of x and y in the equation for the inverse and then solve for y. Starting with the given inverse function: \[f^{-1} : x \to \frac{x + 1}{4}\] Swapping x and y: \[x = \frac{y + 1}{4}\] Solving for y: \[y+1 = 4x\] \[y = 4x - 1\] Therefore, the original function is: \[f : x \to 4x - 1\] So the correct answer is.

**Question 32**
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Marks | 5-7 | 8-10 | 11-13 | 14-16 | 17-19 | 20-22 |

No of students | 4 | 7 | 26 | 41 | 14 | 8 |

The table above shows the distribution of marks of students in a class. Find the upper class boundary of the modal class.

**Question 33**
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Find, correct to two decimal places, the acute angle between \(p = \begin{pmatrix} 13 \\ 14 \end{pmatrix}\) and \(q = \begin{pmatrix} 12 \\ 5 \end{pmatrix}\).

**Question 34**
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If \(^{3x}C_{2} = 15\), find the value of x?

**Answer Details**

The formula for finding the number of combinations of r items out of n distinct items is given by the formula: $$^{n}C_{r} = \frac{n!}{r!(n-r)!}$$ Where n! is the factorial of n, that is the product of all positive integers from 1 to n. In this question, we are given that $$^{3x}C_{2} = 15$$ Substituting the given values into the formula above, we have: $$^{3x}C_{2} = \frac{(3x)!}{2!(3x-2)!} = 15$$ Simplifying this equation, we have: $$\frac{(3x)(3x-1)(3x-2)!}{2!} = 15$$ Multiplying both sides by 2, we get: $$(3x)(3x-1)(3x-2)! = 30$$ We can observe that 3x-2! is the factorial of (3x-2) which is an integer, and 3x-1 and 3x are consecutive integers. Therefore, we can re-write the equation above as: $$(3x)(3x-1) = 10$$ Expanding the left-hand side, we have: $$9x^2 - 3x = 10$$ Bringing all the terms to one side, we get: $$9x^2 - 3x - 10 = 0$$ We can then factorize this quadratic equation as follows: $$(3x - 5)(3x + 2) = 0$$ Using the zero-product property, we get: $$3x - 5 = 0 \quad \text{or} \quad 3x + 2 = 0$$ Solving for x in each equation, we get: $$x = \frac{5}{3} \quad \text{or} \quad x = -\frac{2}{3}$$ Since x represents the number of items in the set from which we are selecting combinations, it must be a positive integer. Therefore, the only valid solution is: $$x = \frac{5}{3} = 1.67 \approx 2$$ Hence, the value of x is 2.

**Question 35**
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A basket contains 3 red and 1 white identical balls. A ball is drawn from the basket at random. Calculate the probability that it is either white or red.

**Answer Details**

The probability of an event happening is the number of ways that event can happen, divided by the total number of possible outcomes. In this case, the total number of balls in the basket is 4, so there are 4 possible outcomes. The number of ways to choose a red ball is 3, since there are 3 red balls in the basket. The number of ways to choose a white ball is 1, since there is only 1 white ball in the basket. Therefore, the probability of choosing a red ball or a white ball is: \[P(\text{red or white}) = \frac{\text{number of red or white balls}}{\text{total number of balls}} = \frac{3+1}{4} = \frac{4}{4} = 1\] So the answer is option D: 1, meaning that it is certain that the ball drawn will be either red or white.

**Question 36**
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A function is defined by \(f(x) = \frac{3x + 1}{x^{2} - 1}, x \neq \pm 1\). Find f(-3).

**Answer Details**

To find f(-3), we substitute -3 in the given function: \(f(-3) = \frac{3(-3) + 1}{(-3)^2 - 1} = \frac{-8}{8} = -1\) Therefore, the value of f(-3) is -1. So, the answer is (B) \(-1\).

**Question 38**
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A force of 200N acting on a body of mass 20kg initially at rest causes it to move a distance of 320m along a straight line for t secs. Find the value of t.

**Answer Details**

To solve this problem, we can use the kinematic equation: s = ut + (1/2)at^2 where s is the distance travelled, u is the initial velocity (0 in this case), a is the acceleration, and t is the time taken. We can rearrange this equation to solve for t: t = sqrt((2s)/a) where sqrt() means "square root of". In this case, the force of 200N causes an acceleration of: a = F/m = 200/20 = 10 m/s^2 And the distance travelled is given as 320m. Plugging in the values, we get: t = sqrt((2s)/a) = sqrt((2 * 320)/10) = sqrt(64) = 8s Therefore, the value of t is 8s, which is.

**Question 40**
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If \(\begin{pmatrix} 3 & 2 \\ 7 & x \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix} \), find x.

**Answer Details**

To solve this problem, we just need to perform the matrix multiplication on the left-hand side and equate it with the right-hand side, and then solve for the unknown variable x. \(\begin{pmatrix} 3 & 2 \\ 7 & x \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\) Performing the matrix multiplication on the left-hand side gives: \(\begin{pmatrix} 3(2) + 2(3) \\ 7(2) + x(3) \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\) Simplifying the left-hand side gives: \(\begin{pmatrix} 12 \\ 14 + 3x \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\) We can see that the first element of both matrices is already equal to 12, so we only need to equate the second elements: \(14 + 3x = 29\) Solving for x gives: \(3x = 15\) \(x = 5\) Therefore, the answer is x = 5.

**Question 41**
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If \(f(x) = 6x^{3} + 13x^{2} + 2x - 5\) and \(f(-1) = 0\), find the factors of f(x).

None

**Question 42**
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(a) Write the following as column vectors: \(r = (10N, 090°) ; q = (8N, 135°)\).

(b) Use your answer in (a) to find \((r + q)\).

None

**Question 43**
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If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} - 7x + 4 = 0\), find the equation whose roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\).

None

**Question 44**
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Solve \(2^{(2y + 2)} - 9(2^{y}) = -2\).

None

**Question 45**
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The table shows the frequency distribution of marks scored by some candidates in an examination.

Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |

Freq | 2 | 5 | 8 | 18 | 20 | 15 | 5 | 4 | 2 | 1 |

(a) Draw the cumulative frequency curve of the distribution.

(b) Use your graph to estimate the :

(i) semi-interquartile range of the distribution; (ii) percentage of candidates who passed with distinction if the least mark for distinction was 72.

None

**Question 46**
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(a) Find the equation of the tangent to curve \(\frac{x^{2}}{4} + y^{2} = 1\) at point \(1, \frac{\sqrt{3}}{2}\).

(b) Express \(\frac{3x + 2}{x^{2} + x - 2}\) in partial fractions.

None

**Answer Details**

None

**Question 47**
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(a) Find, from first principles, the derivative of \(f(x) = (2x + 3)^{2}\).

(b) Evaluate : \(\int_{1} ^{2} \frac{(x + 1)(x^{2} - 2x + 2)}{x^{2}} \mathrm {d} x\)

None

**Answer Details**

None

**Question 48**
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A tyre manufacturing company researched into the life span of one type of their motorcycle tyres. The results were as follows :

Distance (100km) |
10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 |

Number of tyres | 30 | 69 | 93 | 57 | 36 | 15 |

(a) Draw a histogram for the distribution.

(b) Use the histogram to estimate the mode.

None

**Question 49**
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(a) If \(A = \begin{pmatrix} -2 & 5 \\ 4 & 3 \end{pmatrix}\) and \(B = \begin{pmatrix} 3 & 1 \\ 2 & 3 \end{pmatrix}\), find the values of x and y such that \(BA = 2\begin{pmatrix} 3 & 7 \\ -2 & x \end{pmatrix} + \begin{pmatrix} y & 4 \\ 12 & -3 \end{pmatrix}\).

(b) Two functions, f and g are defined by \(f : x \to \frac{1}{2}x + 1\) and \(g : x \to \frac{5x - 1}{3}\). Find :

(i) \(g^{-1}\) ; (ii) \(g^{-1} \circ f\).

None

**Question 50**
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(a) Forces \(F_{1} = (3N, 210°)\) and \(F_{2} = (4N, 120°)\) act on a particle of mass 7kg which is at rest. Calculate the :

(i) acceleration of the particle ; (ii) velocity of the particle after 3 seconds.

(b) \(F_{1} = (2i + 3j)N, F_{2} = (-5j)N \) and \(F_{3} = (6i - 4j)N\) act on a body. Find the magnitude and direction of the fourth force that will keep the body in equilibrium.

None

**Question 51**
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(a) \(m \begin{pmatrix} 2 \\ 1 \end{pmatrix} + n \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}\) where m and n are scalars. Find the value of (m + n).

(b) A(-1, 3), B(2, -1) and C(5, 3) are the vertices of \(\Delta\) ABC.

(i) Express in column notation, the unit vectors parallel to AB and AC.

(ii) Use a dot product to calculate \(\stackrel \frown{BAC}\), correct to the nearest degree.

None

**Question 52**
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The images of points (2, -3) and (4, 5) under a linear transformation A are (3, 4) and (5, 6) respectively. Find the :

(a) matrix A ; (b) inverse of A ; (c) point whose image is (-1, 1).

None

**Question 53**
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(a) Write down the binomial expansion of \((2 - x)^{5}\) in ascending powers of x.

(b) Use your expansion in (a) to evaluate \((1.98)^{5}\) correct to four decimal places.

None

**Question 54**
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(a) An object P of mass 6.5kg is suspended by two light inextensible strings, AP and BP. The strings make angles 50° and 60° respectively with the downward vertical.

(i) Express the forces acting on P in component form; (ii) If P is at rest, write down the vector equation connecting all the forces; (iii) Calculate, correct to one decimal place, the tensions in the strings.

(b) A particle of mass 5 kg moves with initial velocity \(\frac{1}{2} m/s\) and final velocity \(\frac{3}{4} m/s \). Find the magnitude of its change in momentum.

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**Question 55**
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A bag contains 4 red, 6 blue and 8 green identical marbles.

(a) If three marbles are drawn at random, without replacement, calculate the probability that :

(i) all will be green ; (ii) all will have the same colour.

(b) If each marble is replaced before another is drawn, calculate the probability that all will have the same colour.

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**Question 56**
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A particle of mass 400g is moving under the action of two forces \(F_{1} = (35N, 210°), F_{2} = (35\sqrt{3} N, 300°)\) and a resistance of 40N. Find the magnitude of the

(a) resultant of \(F_{1}\) and \(F_{2}\).

(b) resultant force acting on the particle.

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**Question 57**
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A survey conducted revealed that four out of every twenty taxi drivers do not have a valid driving license. If 6 drivers are selected at random, calculate, correct to three decimal places, the probability that

(a) exactly 2 ;

(b) more than 3 ;

(c) at least 5; have valid driving license.

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**Question 58**
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(a) Evaluate \(\frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}}\) correct to two decimal places.

(b) A committee of 2 tutors and 5 pupils is to be formed among 6 tutors and 10 pupils. In how many ways can this be done if one particular tutor must be on the committee and two particular pupils must not be on the committee?

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**Answer Details**

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