WAEC SSCE - Further Mathematics - 2012
Tambaya 1 Rahoto
In computing the mean of 8 numbers, a boy mistakenly used 17 instead of 25 as one of the numbers and obtained 20 as the mean. Find the correct mean
Bayanin Amsa
Let's call the sum of the 8 correct numbers "S" and the incorrect number that was added "x". The mean of the 8 numbers is: S/8 But the boy used 17 instead of 25, so the sum he used was: S + 17 - 25 = S - 8 And he got a mean of 20, so we can set up the equation: (S - 8)/8 = 20 Solving for S, we get: S = 168 So the correct mean is: S/8 = 168/8 = 21 Therefore, the correct mean is 21, which is.
Tambaya 3 Rahoto
The angle of a sector of a circle is 0.9 radians. If the radius of the circle is 4cm, find the length of the arc of the sector.
Bayanin Amsa
The formula for finding the length of an arc of a sector is given by L = rθ, where L is the length of the arc, r is the radius of the circle, and θ is the angle in radians. Using this formula and the given values, we have: L = 4 x 0.9 L = 3.6 cm Therefore, the length of the arc of the sector is 3.6 cm. Answer: 3.6 cm.
Tambaya 4 Rahoto
Which of the following sets is equivalent to \((P \cup Q) \cap (P \cup Q')\)?
Bayanin Amsa
Tambaya 5 Rahoto
Solve the inequality \(x^{2} - 2x \geq 3\)
Tambaya 6 Rahoto
Find the unit vector in the direction of the vector \(-12i + 5j\).
Bayanin Amsa
To find the unit vector in the direction of the vector \(-12i + 5j\), we need to divide the vector by its magnitude. The magnitude of a vector with components \(a\) and \(b\) is given by the formula \(\sqrt{a^2+b^2}\). So, the magnitude of the vector \(-12i + 5j\) is \(\sqrt{(-12)^2+5^2} = 13\). Now, to get the unit vector, we divide each component of the vector by its magnitude: \[\frac{-12}{13}i + \frac{5}{13}j\] This is the unit vector in the direction of the vector \(-12i + 5j\). Therefore, the correct option is \(\frac{-12i}{13} + \frac{5j}{13}\).
Tambaya 7 Rahoto
The distance s in metres covered by a particle in t seconds is \(s = \frac{3}{2}t^{2} - 3t\). Find its acceleration.
Bayanin Amsa
To find the acceleration, we need to differentiate the distance formula with respect to time (t): \begin{align*} s &= \frac{3}{2}t^{2} - 3t \\ \frac{d}{dt}s &= \frac{d}{dt}\left(\frac{3}{2}t^{2}\right) - \frac{d}{dt}(3t) \\ \frac{d}{dt}s &= 3t - 3 \\ \end{align*} Therefore, the acceleration is the second derivative of the distance formula with respect to time: \begin{align*} \frac{d^{2}}{dt^{2}}s &= \frac{d}{dt}(3t - 3) \\ \frac{d^{2}}{dt^{2}}s &= 3 \\ \end{align*} Thus, the acceleration of the particle is a constant value of 3 \(ms^{-2}\). Therefore, the answer is \(3 ms^{-2}\).
Tambaya 8 Rahoto
The marks obtained by 10 students in a test are as follows: 3, 7, 6, 2, 8, 5, 9, 1, 4 and 10. Find the variance.
Bayanin Amsa
Tambaya 9 Rahoto
Which of the following is nor a measure of central tendency?
Bayanin Amsa
Variance is not a measure of central tendency. Measures of central tendency are used to describe the typical or central value of a set of data, while variance is a measure of how spread out the data is from the mean. Variance is a measure of variability, not centrality. Therefore, the answer is "Variance."
Tambaya 10 Rahoto
If (x + 1) is a factor of the polynomial \(x^{3} + px^{2} + x + 6\). Find the value of p.
Bayanin Amsa
If (x + 1) is a factor of the polynomial, it means that if we substitute -1 in the polynomial, it should give us zero. Therefore: \((-1)^3 + p(-1)^2 -1 + 6 = 0\) Simplifying the above equation, we get: \(-1 + p + 5 = 0\) \(p = -4\) Hence, the value of p is -4. So the correct answer is.
Tambaya 11 Rahoto
The marks obtained by 10 students in a test are as follows: 3, 7, 6, 2, 8, 5, 9, 1, 4 and 10. Find the mean mark.
Bayanin Amsa
To find the mean mark, we add up all the marks and divide by the number of students. Adding up all the marks, we get: 3 + 7 + 6 + 2 + 8 + 5 + 9 + 1 + 4 + 10 = 55 There are 10 students, so we divide the sum of the marks by 10: 55 / 10 = 5.50 Therefore, the mean mark is 5.50. Hence, the correct option is: 5.50.
Tambaya 12 Rahoto
A straight line makes intercepts of -3 and 2 on the x- and y- axes respectively. Find the equation of the line.
Tambaya 13 Rahoto
Simplify \(\frac{x^{3n + 1}}{x^{2n + \frac{5}{2}}(x^{2n - 3})^{\frac{1}{2}}}\)
Tambaya 14 Rahoto
Find the number of different arrangements of the word IKOTITINA.
Tambaya 15 Rahoto
If P(x - 3) + Q(x + 1) = 2x + 3, find the value of (P + Q).
Bayanin Amsa
To find the value of (P + Q), we need to first expand the left-hand side of the equation using distributive property. P(x - 3) + Q(x + 1) = Px - 3P + Qx + Q Then we can simplify it by combining the like terms. Px + Qx - 3P + Q + 2x + 3 = 0 Now, we can group the like terms together: (P + Q)x - 3P + Q + 3 = 2x + 3 Since the coefficients of x on both sides of the equation are equal, we can equate the corresponding coefficients of x: (P + Q) = 2 Therefore, the value of (P + Q) is 2. So the correct answer is: - 2
Tambaya 16 Rahoto
Differentiate \(\frac{x}{x + 1}\) with respect to x.
Bayanin Amsa
To differentiate \(\frac{x}{x + 1}\) with respect to x, we can use the quotient rule of differentiation, which states that for functions u(x) and v(x), the derivative of \(\frac{u(x)}{v(x)}\) is given by: \[\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}\] Applying this rule to the given function, we have: \[u(x) = x\] \[v(x) = x + 1\] So, we need to find u'(x) and v'(x): \[u'(x) = 1\] \[v'(x) = 1\] Substituting these values into the quotient rule formula, we get: \[\frac{d}{dx} \left( \frac{x}{x + 1} \right) = \frac{1(x + 1) - x(1)}{(x + 1)^2}\] Simplifying the numerator and denominator, we get: \[\frac{d}{dx} \left( \frac{x}{x + 1} \right) = \frac{1}{(x + 1)^2}\] Therefore, the correct answer is \(\frac{1}{(x + 1)^2}\).
Tambaya 17 Rahoto
Find the acute angle between the lines 2x + y = 4 and -3x + y + 7 = 0.
Bayanin Amsa
To find the acute angle between two lines, we need to find the angle between their respective direction vectors. The direction vector of the line 2x + y = 4 is v = i + 2j and the direction vector of the line -3x + y + 7 = 0 is w = -3i + j. The acute angle θ between two vectors a and b is given by the formula cos(θ) = (a.b) / (|a|.|b|), where a.b is the dot product of the vectors and |a| and |b| are their respective magnitudes. Using this formula, we can find cos(θ) for the given direction vectors as follows: cos(θ) = (v.w) / (|v|.|w|) = (-5) / (√5.√10) = -1/√2 Since we are looking for the acute angle, we need to take the inverse cosine of -1/√2 in the range [0, π/2]: θ = cos-1(-1/√2) ≈ 45° Therefore, the acute angle between the given lines is approximately 45°, which is.
Tambaya 19 Rahoto
If r denotes the correlation coefficient between two variables, which of the following is always true?
Bayanin Amsa
The correct answer is: \(-1 \leq r \leq 1\). The correlation coefficient (r) measures the degree of linear association between two variables. The value of r ranges from -1 to 1, where -1 indicates a perfect negative correlation, 0 indicates no correlation, and 1 indicates a perfect positive correlation. Since r can take on any value between -1 and 1 (including -1 and 1), the correct statement is \(-1 \leq r \leq 1\), which means that the correlation coefficient is always between -1 and 1, inclusive. Therefore, options (A), (B), and (C) are incorrect.
Tambaya 20 Rahoto
A box contains 4 red and 3 blue identical balls. If two are picked at random, one after the other without replacement, find the probability that one is red and the other is blue.
Bayanin Amsa
Tambaya 21 Rahoto
A binary operation, \(\Delta\), is defined on the set of real numbers by \(a \Delta b = a + b + 4\). Find the identity element.
Bayanin Amsa
An identity element in a binary operation is an element such that when it operates with any other element of the set, it does not change the other element. In this case, we need to find an element, say "x," such that for any real number "a," $$a \Delta x = a$$ Substituting the given definition of the operation, we get: $$a + x + 4 = a$$ Solving for x, we get: $$x = -4$$ Thus, -4 is the identity element for the given binary operation.
Tambaya 22 Rahoto
QRS is a triangle such that \(\overrightarrow{QR} = (3i + 2j)\) and \(\overrightarrow{SR} = (-5i + 3j)\), find \(\overrightarrow{SQ}\).
Bayanin Amsa
To find \(\overrightarrow{SQ}\), we can use the fact that \(\overrightarrow{SR} = \overrightarrow{SQ} + \overrightarrow{QR}\). Rearranging this equation to solve for \(\overrightarrow{SQ}\) gives us: $$\overrightarrow{SQ} = \overrightarrow{SR} - \overrightarrow{QR} = (-5i + 3j) - (3i + 2j) = -8i + j$$ Therefore, the value of \(\overrightarrow{SQ}\) is -8i + j. So, the correct option is (A) 8i + j.
Tambaya 23 Rahoto
Two forces 10N and 6N act in the directions 060° and 330° respectively. Find the x- component of their resultant.
Bayanin Amsa
Tambaya 24 Rahoto
If \(y = x^{3} - x^{2} - x + 6\), find the values of x at the turning point.
Tambaya 26 Rahoto
If the midpoint of the line joining (1 - k, -4) and (2, k + 1) is (-k, k), find the value of k.
Bayanin Amsa
We can start by using the midpoint formula which states that the midpoint of a line joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\). So, the midpoint of the line joining (1-k, -4) and (2, k+1) is: \begin{align*} &\left(\frac{(1-k)+2}{2},\frac{(-4)+(k+1)}{2}\right) \\ &\Rightarrow \left(\frac{3-k}{2},\frac{k-3}{2}\right) \end{align*} We are given that the midpoint is (-k, k), so we can equate the x and y coordinates: \begin{align*} \frac{3-k}{2} &=-k \\ \frac{k-3}{2} &=k \end{align*} Solving for k, we get: \begin{align*} k &= 2 \\ \end{align*} Therefore, the value of k is 2, which is the answer option labeled as "-2".
Tambaya 27 Rahoto
The equation of a circle is \(3x^{2} + 3y^{2} + 24x - 12y = 15\). Find its radius.
Bayanin Amsa
To find the radius of the circle, we need to use the standard form of the equation of a circle, which is \((x - a)^{2} + (y - b)^{2} = r^{2}\), where \((a, b)\) is the center of the circle and \(r\) is the radius. To convert the given equation to standard form, we can complete the square for both \(x\) and \(y\): \begin{align*} 3x^{2} + 3y^{2} + 24x - 12y &= 15 \\ 3(x^{2} + 8x) + 3(y^{2} - 4y) &= 15 \\ 3(x^{2} + 8x + 16) + 3(y^{2} - 4y + 4) &= 15 + 3(16) + 3(4) \\ 3(x + 4)^{2} + 3(y - 2)^{2} &= 72 \\ (x + 4)^{2} + (y - 2)^{2} &= 8^{2} \end{align*} Comparing this with the standard form, we see that the center of the circle is \((-4, 2)\) and the radius is \(8\). Therefore, the answer is (d) 5.
Tambaya 28 Rahoto
The diagram above is a velocity- time graph of a moving object. Calculate the distance travelled when the acceleration is zero.
Bayanin Amsa
Tambaya 29 Rahoto
A stone is dropped from a height of 45m. Find the time it takes to hit the ground. \([g = 10 ms^{-2}]\)
Bayanin Amsa
To solve this problem, we can use the formula: \[ s = ut + \frac{1}{2}at^2 \] where s is the distance, u is the initial velocity, t is the time, and a is the acceleration due to gravity. In this case, the initial velocity is zero because the stone is dropped from rest. We also know that the distance is 45m and the acceleration due to gravity is 10 \(ms^{-2}\). Thus, we have: \begin{align*} s &= ut + \frac{1}{2}at^2 \\ 45 &= 0t + \frac{1}{2}(10)t^2 \\ 45 &= 5t^2 \\ t^2 &= 9 \\ t &= 3 \end{align*} Therefore, the time it takes for the stone to hit the ground is 3 seconds. Hence, the answer is (a) 3.0 seconds.
Tambaya 30 Rahoto
A polynomial is defined by \(f(x + 1) = x^{3} + px^{2} - 4x + 2\), find f(2).
Tambaya 31 Rahoto
Evaluate \(\log_{10}(\frac{1}{3} + \frac{1}{4}) + 2\log_{10} 2 + \log_{10} (\frac{3}{7})\)
Bayanin Amsa
To simplify this expression, we can first use the identity: $$\log_{a}(b) + \log_{a}(c) = \log_{a}(bc)$$ Using this identity, we can simplify the given expression as follows: \begin{align*} \log_{10}\left(\frac{1}{3}+\frac{1}{4}\right) + 2\log_{10}(2) + \log_{10}\left(\frac{3}{7}\right) &= \log_{10}\left(\frac{7}{12}\right) + \log_{10}(2^2) + \log_{10}\left(\frac{3}{7}\right) \\ &= \log_{10}\left(\frac{7}{12} \cdot 2^2 \cdot \frac{3}{7}\right) \\ &= \log_{10}(1) \\ &= 0 \end{align*} Therefore, the answer is 0.
Tambaya 32 Rahoto
Given that \(^{n}P_{r} = 90\) and \(^{n}C_{r} = 15\), find the value of r.
Bayanin Amsa
We know that: $$^{n}P_{r} = \frac{n!}{(n-r)!} = 90$$ and $$^{n}C_{r} = \binom{n}{r} = \frac{n!}{r!(n-r)!} = 15$$ To find the value of r, we can use the formula: $$^{n}C_{r} = \frac{^{n}P_{r}}{r!}$$ Substituting the given values, we get: $$15 = \frac{90}{r!}$$ Simplifying the equation, we get: $$r! = 6$$ The only integer value of r that satisfies this equation is 3, since 3! = 6. Therefore, the answer is r = 3.
Tambaya 33 Rahoto
Find the values of x at the point of intersection of the curve \(y = x^{2} + 2x - 3\) and the lines \(y + x = 1\).
Bayanin Amsa
To find the point of intersection of the curve and the line, we need to solve the system of equations formed by equating the two equations: \begin{align*} y &= x^2 + 2x - 3 \\ y &= -x + 1 \end{align*} Setting the right-hand sides equal to each other, we get: \begin{align*} x^2 + 2x - 3 &= -x + 1 \\ x^2 + 3x - 4 &= 0 \\ (x + 4)(x - 1) &= 0 \end{align*} Thus, the values of x at the points of intersection are x = -4 and x = 1. To find the corresponding y-values, we substitute these values of x back into either equation. Using the equation y = x^2 + 2x - 3, we get: \begin{align*} y &= (-4)^2 + 2(-4) - 3 \\ &= 7 \end{align*} and \begin{align*} y &= (1)^2 + 2(1) - 3 \\ &= 0 \end{align*} Therefore, the points of intersection are (-4, 7) and (1, 0). So, the correct answer is (1, -4).
Tambaya 34 Rahoto
Find the constant term in the binomial expansion of \((2x - \frac{3}{x})^{8}\).
Bayanin Amsa
Tambaya 35 Rahoto
From the diagram above, which of the following represents the vector V in component form?
Bayanin Amsa
Tambaya 36 Rahoto
Given that \(\sqrt{6}, 3\sqrt{2}, 3\sqrt{6}, 9\sqrt{2},...\) are the first four terms of an exponential sequence (G.P), find in its simplest form the 8th term.
Bayanin Amsa
Tambaya 37 Rahoto
Given that \(\sin x = \frac{-\sqrt{3}}{2}\) and \(\cos x > 0\), find x.
Bayanin Amsa
Tambaya 39 Rahoto
A fair die is tossed twice. Find the probability of obtaining a 3 and a 5.
Bayanin Amsa
When a die is tossed, there are six possible outcomes, each with equal probability. Therefore, the probability of getting a 3 on the first toss is 1/6, and the probability of getting a 5 on the second toss is also 1/6. Since we want both events to occur, we need to multiply their probabilities: \[\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\] So, the probability of obtaining a 3 and a 5 is 1/36. Therefore, the correct option is: - \(\frac{1}{36}\)
Tambaya 40 Rahoto
Given that \(P = \begin{pmatrix} 2 & 1 \\ 5 & -3 \end{pmatrix}\) and \(Q = \begin{pmatrix} 4 & -8 \\ 1 & -2 \end{pmatrix}\), Find (2P - Q).
Bayanin Amsa
To find the value of (2P - Q), we first need to compute 2P and Q, and then subtract Q from 2P. To compute 2P, we multiply each element of matrix P by 2: 2P = \(\begin{pmatrix} 4 & 2 \\ 10 & -6 \end{pmatrix}\) To subtract Q from 2P, we subtract each corresponding element of matrix Q from matrix 2P: 2P - Q = \(\begin{pmatrix} 4-4 & 2+8 \\ 10-1 & -6+2 \end{pmatrix}\) = \(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\) Therefore, the answer is \(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\).
Tambaya 41 Rahoto
The table gives the distribution of heights in metres of 100 students.
| Height | 1.40-1.42 | 1.43-1.45 | 1.46-1.48 | 1.49-1.51 | 1.52-1.54 | 1.55-1.57 | 1.58-1.60 | 1.61-1.63 |
| Freq | 2 | 4 | 19 | 30 | 24 | 14 | 6 | 1 |
(a) Calculate the : (i) mean height ; (ii) mean deviation of the distribution.
(b) What is the probability that the height of a student selected at random is greater than the mean height of the distribution?
Tambaya 42 Rahoto
(a) A body P of mass 5kg is suspended by two light inextensible strings AP and BP attached to a ceiling. If the strings are inclined at angles 40° and 30° respectively to the downward vertical, find the tension in each of the strings. [Take \(g = 10 ms^{-2}\)].
(b) A constant force F acts on a toy car of mass 5 kg and increases its velocity from 5 ms\(^{-1}\) to 9 ms\(^{-1}\) in 2 seconds. Calculate :
(i) the magnitude of the force ; (ii) velocity of the toy car 3 seconds after attaining a velocity of 9 ms\(^{-1}\).
Tambaya 43 Rahoto
(a) Evaluate : \(\int_{1} ^{4} \frac{x(3x - 2)}{2\sqrt{x}} \mathrm {d} x\)
(b) The equation of a circle is given by \(2x^{2} + 2y^{2} - 8x + 5y - 10 = 0\). Find the :
(i) coordinates of the centre ; (ii) radius of the circle .
Tambaya 45 Rahoto
Three forces \(-63j , 32.14i + 38.3j\) and \(14i - 24.25j\) act on a body of mass 5kg. Find, correct to one decimal place, the :
(a) magnitude of the resultant force ;
(b) acceleration of the body.
Tambaya 46 Rahoto
The twenty-first term of an Arithmetic Progression is \(5\frac{1}{2}\) and the sum of the first twenty-one terms is \(94\frac{1}{2}\). Find the :
(a) first term ; (b) common difference ; (c) sum of the first thirty terms.
Tambaya 47 Rahoto
Express \(3x^{2} - 6x + 10\) in the form \(a(x - b)^{2} + c\), where a, b and c are integers. Hence state the minimum value of \(3x^{2} - 6x + 10\) and the value of x for which it occurs.
Tambaya 48 Rahoto
(a) Find the angle between the vectors \(a = \begin{pmatrix} -3 \\ 4 \end{pmatrix}\) and \(b = \begin{pmatrix} -8 \\ -15 \end{pmatrix}\).
(b) Given that \(a = (4N, 060°)\) and \(b = (3N, 120°)\), find, in component form, the unit vector along \(a - b\).
Bayanin Amsa
None
Tambaya 49 Rahoto
The marks scored by 35 students in a test are given in the table below.
| Marks | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 | 26-30 |
| Frequency | 2 | 7 | 12 | 8 | 5 | 1 |
Draw a histogram for the distribution.
Tambaya 50 Rahoto
(a) The polynomial \(f(x) = x^{3} + px^{2} - 10x + q\) is exactly divisible by \(x^{2} + x - 6\). Find the :
(i) values of p and q ; (ii) third factor.
(b) The volume of a cube is increasing at the rate of \(2\frac{1}{2} cm^{3} s^{-1}\). Find the rate of change of the side of the base when its length is 2cm.
Tambaya 51 Rahoto
(a) Given that \(p = (4i - 3j)\) and \(q = (-i + 5j)\), find r such that \(|r| = 15\) and is in the direction \((2p + 3q)\).
(b) 
Forces of magnitude 8N, 6N and 4N act at the point P, as shown in the above diagram. Find the : (i) magnitude ; (ii) direction of the resultant force.
Tambaya 52 Rahoto
Simplify \(^{n + 1}C_{4} - ^{n - 1}C_{4}\)
= \(\frac{(n + 1)!}{4! (n - 3)!} - \frac{(n - 1)!}{4! (n - 5)!}\)
= \(\frac{(n + 1)(n)(n - 1)(n - 2)(n - 3)!}{4! (n - 3)!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)!}{4! (n - 5)!}\)
= \(\frac{(n + 1)(n)(n - 1)(n - 2)}{4!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)}{4!}\)
= \(\frac{(n - 1)(n - 2) [n(n + 1) - (n - 3)(n - 4)]}{4!}\)
= \(\frac{(n - 1)(n - 2) [n^{2} + n - n^{2} + 7n - 12]}{24}\)
= \(\frac{(n - 1)(n - 2)[8n - 12]}{24}\)
= \(\frac{(n - 1)(n - 2)(2n - 3)}{6}\)
Tambaya 53 Rahoto
(a) Two items are selected at random from four items labelled (p, q, r, s).
(i) List the sample space if sampling is done (1) with replacement ; (2) without replacement.
(ii) Find the probability that r is at least one of the two objects selected : (1) in a(i)1 ; (2) in a(i)2.
(b) How many whole numbers from 100 to 999 are divisible by (i) 4 ; (ii) both 3 and 4?
Bayanin Amsa
None
Tambaya 54 Rahoto
(a) A fair die with six faces is thrown six times. Calculate, correct to three decimal places, the probability of obtaining :
(i) exactly three sixes ; (ii) at most three sixes.
(b) Eight percent of screws produced by a machine are defective. From a random sample of 10 screws produced by the machine, find the probability that :
(i) exactly two will be defective ; (ii) not more than two will be defective.
Tambaya 55 Rahoto
(a)(i) Find the sum of the series \(A(1 + r) + A(1 + r)^{2} + ... + A(1 + r)^{n}\).
(ii) Given that r = 8% and A = GH 40.00, find the sum of the 6th to 10th terms of the series in (i).
(b) Find the equation of the tangent to the curve \(y = \frac{1}{x}\) at the point on the curve when x = 2.
Tambaya 56 Rahoto
Write down the first three terms of the binomial expansion \((1 + ax)^{n}\) in ascending powers of x. If the coefficients of x and x\(^{2}\) are 2 and \(\frac{3}{2}\) respectively, find the values of a and n.
Tambaya 57 Rahoto
Two functions g and h are defined on the set R of real numbers by \(g : x \to x^{2} - 2\) and \(h : x \to \frac{1}{x + 2}\). Find :
(a) \(h^{-1}\), the inverse of h ;
(b) \(g \circ h\), when \(x = -\frac{1}{2}\).
Tambaya 58 Rahoto
(a) Write down the matrix A of the linear transformation \(A(x, y) \to (2x -y, -5x + 3y)\).
(b) If \(B = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\), find :
(i) \(A^{2} - B^{2}\) ; (ii) matrix \(C = B^{2} A\) ; (iii) the point \(M(x, y)\) whose image under the linear transformation \(C\) is \(M' (10, 18)\).
(c) What is the relationship between matrix A and matrix C?
