WAEC SSCE - Further Mathematics - 2008 (Objective)

Question 1 Report

Marks	2	3	4	5	6	7	8
No of students	5	7	9	6	3	6	4

The table above shows the distribution of marks by some candidates in a test. If a student is selected at random, what is the probability that she scored at least 6 marks?

$\frac{3}{40}$

$\frac{1}{4}$

$\frac{13}{40}$

$\frac{27}{40}$

Probability

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Question 2 Report

Marks	2	3	4	5	6	7	8
No of students	5	7	9	6	3	6	4

The table above shows the distribution of marks by some candidates in a test. What is the median score?

3.5

4.0

4.5

5.0

Statistics

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Question 3 Report

If $2\log_{4} 2 = x + 1$, find the value of x.

-2

-1

0

1

Indices And Logarithmic Functions

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Question 4 Report

Given that $f '(x) = 3x^{2} - 6x + 1$ and f(3) = 5, find f(x).

$f(x) = x^{3} - 3x^{2} + x + 20$

$f(x) = x^{3} - 3x^{2} + x + 31$

$f(x) = x^{3} - 3x^{2} + x + 2$

$f(x) = x^{3} - 3x^{2} + x - 13$

Functions

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Question 5 Report

In the diagram above, forces P, Q and 50N are acting on a body at M. If the system is in equilibrium, calculate, in terms of $\theta$, the magnitude of P.

$\frac{50 \cos \theta}{\sin (\theta + 45)°}$

$\frac{50 \cos \theta}{\cos (\theta + 45)°}$

$\frac{50 \sin \theta}{\cos (\theta + 45)°}$

$\frac{50 \sin \theta}{\sin (\theta + 45)°}$

Vectors

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Question 6 Report

A test consists of 12 questions out of which candidates are to answer 10. If the first 6 are compulsory, in how many ways can each candidate select her questions?

40

25

15

10

Answer Details

Out of the 12 questions, the first 6 are compulsory and each candidate must answer them. Therefore, the candidate has to choose 4 questions out of the remaining 6 questions to answer. The number of ways to choose 4 questions out of 6 is given by the formula for combinations: \[\binom{6}{4} = \frac{6!}{4!2!} = \frac{6\times 5}{2\times 1} = 15\] Therefore, each candidate can select her questions in 15 ways. Hence, the answer is 15.

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Permutation And Combinations

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Question 7 Report

The probability of Jide, Atu and Obu solving a given problem are $\frac{1}{12}$, $\frac{1}{6}$ and $\frac{1}{8}$ respectively. Calculate the probability that only one solves the problem.

$\frac{1}{576}$

$\frac{55}{576}$

$\frac{77}{576}$

$\frac{167}{576}$

Answer Details

To calculate the probability that only one person solves the problem, we can consider the following cases: Case 1: Jide solves the problem, and Atu and Obu don't solve the problem. The probability of Jide solving the problem is $\frac{1}{12}$, the probability of Atu not solving the problem is $\frac{5}{6}$ (since the probability of Atu solving the problem is $\frac{1}{6}$), and the probability of Obu not solving the problem is $\frac{7}{8}$ (since the probability of Obu solving the problem is $\frac{1}{8}$). Therefore, the probability of this case is: \[\frac{1}{12} \times \frac{5}{6} \times \frac{7}{8} = \frac{35}{576}\] Case 2: Atu solves the problem, and Jide and Obu don't solve the problem. The probability of Atu solving the problem is $\frac{1}{6}$, the probability of Jide not solving the problem is $\frac{11}{12}$ (since the probability of Jide solving the problem is $\frac{1}{12}$), and the probability of Obu not solving the problem is $\frac{7}{8}$. Therefore, the probability of this case is: \[\frac{1}{6} \times \frac{11}{12} \times \frac{7}{8} = \frac{77}{576}\] Case 3: Obu solves the problem, and Jide and Atu don't solve the problem. The probability of Obu solving the problem is $\frac{1}{8}$, the probability of Jide not solving the problem is $\frac{11}{12}$, and the probability of Atu not solving the problem is $\frac{5}{6}$. Therefore, the probability of this case is: \[\frac{1}{8} \times \frac{11}{12} \times \frac{5}{6} = \frac{55}{576}\] So the total probability of only one person solving the problem is the sum of the probabilities of these three cases: \[\frac{35}{576} + \frac{77}{576} + \frac{55}{576} = \frac{167}{576}\] Therefore, the correct option is (d) $\frac{167}{576}$.

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Probability

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Question 8 Report

Given that $\overrightarrow{AB} = 5i + 3j$ and $\overrightarrow{AC} = 2i + 5j$, find $\overrightarrow{BC}$.

-7i - 8j

-3i + 2j

3i - 2j

3i + 8j

Answer Details

To find $\overrightarrow{BC}$, we need to subtract $\overrightarrow{AB}$ from $\overrightarrow{AC}$, as both of these vectors share the point $A$. So, \[\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}\] Substituting the given values, we get: \begin{align*} \overrightarrow{BC} &= \overrightarrow{AC} - \overrightarrow{AB} \\ &= (2i + 5j) - (5i + 3j) \\ &= 2i + 5j - 5i - 3j \\ &= \boxed{-3i + 2j} \end{align*} Therefore, the answer is option (B) -3i + 2j.

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Vectors

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Question 9 Report

A binary operation ,*, is defined on the set R, of real numbers by $a * b = a^{2} + b + ab$. Find the value of x for which $5 * x = 37$.

7

2

-2

-7

Binary Operations

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Question 10 Report

The coefficient of the 5th term in the binomial expansion of $(1 + kx)^{8}$, in ascending powers of x is $\frac{35}{8}$. Find the value of the constant k.

2

$\frac{1}{2}$

$-\frac{1}{2}$

-2

Answer Details

In the binomial expansion of $(1+kx)^8$, the coefficient of the 5th term in ascending powers of x is given by the expression: \[\binom{8}{4}(kx)^4(1)^{8-4}\] which simplifies to: \[\binom{8}{4}k^4x^4\] Using the formula for binomial coefficients, we can write: \[\binom{8}{4} = \frac{8!}{4!4!} = 70\] So, we can write the expression for the 5th term coefficient as: \[70k^4x^4\] We are given that this coefficient is $\frac{35}{8}$, so we can write: \[70k^4x^4 = \frac{35}{8}\] Simplifying this equation, we get: \[k^4x^4 = \frac{1}{16}\] Dividing both sides by $x^4$, we get: \[k^4 = \frac{1}{16}\] Taking the fourth root of both sides, we get: \[k = \pm \frac{1}{2}\] Since the expansion is of the form $(1+kx)^8$, the sign of k does not affect the coefficient of the 5th term, so we can take $k = \frac{1}{2}$. Therefore, the value of the constant k is $\frac{1}{2}$.

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Binomial Theorem

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Question 11 Report

Express $r = (12, 210°)$ in the form $a i + b j$.

$6(-i + \sqrt{3} j)$

$6(-\sqrt{3} i - j)$

$6(i - \sqrt{3} j)$

$6(i + \sqrt{3} j)$

Answer Details

We can express a vector in terms of its rectangular or Cartesian coordinates by using the formulas: \[x = r\cos\theta\] \[y = r\sin\theta\] where r is the magnitude of the vector and θ is the angle it makes with the positive x-axis. Using these formulas, we have: \[x = r\cos\theta = 12\cos 210° = 12 \times \frac{-\sqrt{3}}{2} = -6\sqrt{3}\] \[y = r\sin\theta = 12\sin 210° = 12 \times \frac{-1}{2} = -6\] Therefore, we can express vector r as: \[r = -6\sqrt{3}i - 6j = 6(-\sqrt{3}i - j)\] Hence, the correct option is: $6(-\sqrt{3}i - j)$.

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Vectors

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Question 12 Report

Evaluate $\int_{1}^{2} (2 + 2x - 3x^{2}) \mathrm {d} x$.

-2

2

8

10

Integration

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Question 13 Report

Express $\frac{7\pi}{6}$ radians in degrees.

315°

210°

105°

75°

Answer Details

To convert radians to degrees, we can use the conversion formula: $$\text{degrees} = \frac{\text{radians}}{\pi} \times 180^\circ$$ Substituting the value of $\frac{7\pi}{6}$ radians, we get: $$\text{degrees} = \frac{7\pi}{6} \div \pi \times 180^\circ$$ Simplifying the expression, we get: $$\text{degrees} = \frac{7}{6} \times 180^\circ$$ $$\text{degrees} = 210^\circ$$ Therefore, $\frac{7\pi}{6}$ radians is equal to 210 degrees.

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Trigonometry

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Question 14 Report

Two forces $F_{1} = (10N, 020°)$ and $F_{2} = (7N, 200°)$ act on a particle. Find the resultant force.

(3 N, 020°)

(3 N, 200°)

(17 N, 020°)

(17 N, 200°)

Vectors

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Question 15 Report

A rectangle has a perimeter of 24m. If its area is to be maximum, find its dimension.

12, 12

6, 6

4, 8

9, 3

Answer Details

To find the dimensions of a rectangle that has the maximum area with a given perimeter, we can start by expressing the perimeter in terms of the rectangle's dimensions. Let's call the length of the rectangle l and its width w. Then the perimeter P is given by: P = 2l + 2w We know that P = 24m, so we can substitute this into the above equation to get: 24 = 2l + 2w Simplifying this equation, we get: 12 = l + w Now we need to express the area of the rectangle in terms of l and w. The area A is given by: A = lw We want to find the values of l and w that will maximize A, subject to the constraint that 12 = l + w. One way to do this is to express l in terms of w using the equation 12 = l + w, and substitute this expression into the equation for A. We get: l = 12 - w A = lw = w(12 - w) = 12w - w^2 Now we have an expression for A in terms of just one variable, w. To find the maximum value of A, we can take the derivative of A with respect to w, set it equal to zero, and solve for w. We get: dA/dw = 12 - 2w = 0 w = 6 So the width of the rectangle that maximizes the area is 6m. To find the length, we can use the equation 12 = l + w, which gives us: l = 12 - w = 12 - 6 = 6 So the dimensions of the rectangle that has the maximum area with a perimeter of 24m are 6m by 6m. Therefore, the correct answer is 6, 6.

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Functions

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Question 16 Report

If $p = \begin{pmatrix} 2 \\ -2 \end{pmatrix} $ and $q = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, find $|q - \frac{1}{2}p|$.

$2\sqrt{2}$

$\sqrt{13}$

$5$

$\sqrt{29}$

Answer Details

To find the value of |q - 1/2p|, we first need to calculate q - 1/2p. q - 1/2p = $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ - 1/2$\begin{pmatrix} 2 \\ -2 \end{pmatrix}$ = $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ - $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ = $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ Now, we need to calculate the magnitude or length of the vector (2, 5). |q - 1/2p| = $\sqrt{(2)^2 + (5)^2}$ = $\sqrt{4 + 25}$ = $\sqrt{29}$ Therefore, the answer is option D: $\sqrt{29}$.

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Vectors

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Question 17 Report

If n items are arranged two at a time, the number obtained is 20. Find the value of n.

5

10

15

40

Answer Details

The number of ways to arrange n items taken two at a time is given by the formula: $_nP_2 = \frac{n!}{(n-2)!} = n(n-1)$ We are given that $_nP_2 = 20$, so we can set up the equation: \begin{aligned} n(n-1) &= 20 \\ n^2 - n - 20 &= 0 \end{aligned} We can solve for n using the quadratic formula: \begin{aligned} n &= \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-20)}}{2(1)} \\ &= \frac{1 \pm \sqrt{81}}{2} \\ &= \frac{1 \pm 9}{2} \end{aligned} We discard the negative solution, and the positive solution is: \begin{aligned} n &= \frac{1+9}{2} \\ &= 5 \end{aligned} Therefore, the value of n is 5.

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Permutation And Combinations

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Question 18 Report

The sum of the first n terms of a linear sequence is $S_{n} = n^{2} + 2n$. Find the common difference of the sequence.

5

4

3

2

Sequences And Series

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Question 19 Report

If $\begin{vmatrix} 4 & x \\ 5 & 3 \end{vmatrix} = 32$, find the value of x.

4

2

-2

-4

Answer Details

The determinant of a 2x2 matrix [a, b; c, d] is given by $ad-bc$. Here we are given the determinant of the matrix $\begin{vmatrix} 4 & x \\ 5 & 3 \end{vmatrix}$ to be 32. So, we have $(4 \times 3) - (x \times 5) = 32$. Simplifying this equation gives us $12 - 5x = 32$. Solving for x, we get $x = -4$. Therefore, the value of x is -4.

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Matrices And Linear Transformation

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Question 20 Report

The initial and final velocities of an object of mass 5 kg are $u = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ and $v = \begin{pmatrix} 4 \\ 7 \end{pmatrix}$ respectively. Find the magnitude of its change in momentum.

25

15

$3\sqrt{7}$

$\sqrt{10}$

Matrices And Linear Transformation

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Question 21 Report

The polynomial $2x^{3} + x^{2} - 3x + p$ has a remainder of 20 when divided by (x - 2). Find the value of constant p.

8

6

-6

-8

Answer Details

The problem tells us that when we divide the polynomial $2x^{3} + x^{2} - 3x + p$ by $(x - 2)$, the remainder is 20. We can use polynomial long division to find the quotient and remainder. \begin{array}{c|ccccc} \multicolumn{2}{r}{2x^2} & 5x & 7 \\ \cline{2-6} x-2 & 2x^3 & x^2 & -3x & +p & \\ \multicolumn{2}{r}{-2x^3} & +4x^2 & & &\\ \cline{2-3} \multicolumn{2}{r}{0} & 5x^2 & -3x & &\\ \multicolumn{2}{r}{} & -5x^2 & +10x & &\\ \cline{3-4} \multicolumn{2}{r}{} & 0 & 7x & +p & \\ \multicolumn{2}{r}{} & & -7x & +14 & \\ \cline{4-5} \multicolumn{2}{r}{} & & 0 & p+14 & \\ \end{array} The quotient is $2x^{2} + 5x + 7$, and the remainder is $p + 14$. We know that the remainder is equal to 20, so we can set up the equation: \[p + 14 = 20.\] Solving for $p$, we get: \[p = 20 - 14 = 6.\] Therefore, the value of the constant $p$ is 6, which is option (B).

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Polynomial Functions

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Question 22 Report

If $f(x) = 2x^{2} - 3x - 1$, find the value of x for which f(x) is minimum.

$\frac{3}{2}$

$\frac{4}{3}$

$\frac{3}{4}$

$\frac{2}{3}$

Answer Details

To find the value of x for which f(x) is minimum, we need to find the critical points of the function f(x) and determine whether they correspond to a minimum or maximum value. The critical points are the values of x where the derivative of f(x) is equal to zero or does not exist. Taking the derivative of f(x) with respect to x, we get: $$f'(x) = 4x - 3$$ Setting f'(x) equal to zero and solving for x, we get: $$4x - 3 = 0$$ $$x = \frac{3}{4}$$ Therefore, x = 3/4 is a critical point of f(x). To determine whether x = 3/4 corresponds to a minimum or maximum value of f(x), we can use the second derivative test. The second derivative of f(x) is: $$f''(x) = 4$$ Since f''(x) is positive for all values of x, we can conclude that x = 3/4 corresponds to a minimum value of f(x). Therefore, the value of x for which f(x) is minimum is x = 3/4.

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Polynomial Functions

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Question 23 Report

Two statements are represented by p and q as follows:

p : He is brilliant; q : He is regular in class

Which of the following symbols represent "He is regular in class but dull"?

$q \vee \sim p$

$q \edge \sim p$

$\sim q \edge \sim p$

$\sim q \vee \sim p$

Logical Reasoning

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Question 24 Report

The sum of the first n terms of a linear sequence is $S_{n} = n^{2} + 2n$. Determine the general term of the sequence.

n + 1

2n + 1

3n + 1

4n + 1

Answer Details

To find the general term of a linear sequence, we need to find the common difference between any two consecutive terms of the sequence. We know that the sum of the first n terms of the sequence is given by: $$S_{n} = n^{2} + 2n$$ If we subtract the sum of the first (n-1) terms from the sum of the first n terms, we get the nth term of the sequence: $$\begin{aligned} a_{n} &= S_{n} - S_{n-1} \\ &= (n^{2} + 2n) - [(n-1)^{2} + 2(n-1)] \\ &= 2n - 1 \end{aligned}$$ Therefore, the general term of the sequence is given by: $$a_{n} = 2n - 1$$ So, the correct option is \textbf{2n + 1}.

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Sequences And Series

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Question 25 Report

Which of the following quadratic curves will not intersect with the x- axis?

$y = 2 - 4x - x^{2}$

$y = x^{2} - 5x -1$

$y = 2x^{2} - x - 1$

$y = 3x^{2} - 2x + 4$

Answer Details

A quadratic curve is a polynomial of degree 2, which means that it can be written in the form of $y = ax^{2} + bx + c$, where a, b, and c are constants. For a quadratic curve to intersect with the x-axis, it must have at least one x-intercept. An x-intercept is a point on the curve where the value of y is equal to zero. In other words, it is where the curve crosses the x-axis. To find the x-intercepts of a quadratic curve, we need to solve the equation $y = ax^{2} + bx + c$ for x when y = 0. This gives us the quadratic formula, which is: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ If the discriminant ($b^{2} - 4ac$) is greater than or equal to zero, then the quadratic curve will intersect with the x-axis. If the discriminant is less than zero, then the quadratic curve will not intersect with the x-axis. Using this information, we can determine which of the given quadratic curves will not intersect with the x-axis: - $y = 2 - 4x - x^{2}$ The discriminant for this quadratic is $16 - 4(1)(2) = 8$, which is greater than zero. Therefore, this quadratic curve will intersect with the x-axis. - $y = x^{2} - 5x -1$ The discriminant for this quadratic is $25 + 4(1)(1) = 29$, which is greater than zero. Therefore, this quadratic curve will intersect with the x-axis. - $y = 2x^{2} - x - 1$ The discriminant for this quadratic is $(-1)^{2} - 4(2)(-1) = 9$, which is greater than zero. Therefore, this quadratic curve will intersect with the x-axis. - $y = 3x^{2} - 2x + 4$ The discriminant for this quadratic is $(-2)^{2} - 4(3)(4) = -44$, which is less than zero. Therefore, this quadratic curve will not intersect with the x-axis. Therefore, the quadratic curve that will not intersect with the x-axis is the one described by the equation $y = 3x^{2} - 2x + 4$.

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Polynomial Functions

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Question 26 Report

Find the value of the constant k for which $a = 4 i - k j$ and $b = 3 i + 8 j$ are perpendicular.

$\frac{2}{3}$

2

3

$\frac{3}{2}$

Answer Details

For two vectors to be perpendicular, their dot product must be zero. So, we can find the value of $k$ by taking the dot product of the given vectors and setting it equal to zero. The dot product of two vectors, $\vec{a} \cdot \vec{b}$, is given by: $\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z$ where $\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$ and $\vec{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k}$ are the components of the vectors along the x, y, and z axes. So, using the given vectors, we have: $\vec{a} \cdot \vec{b} = (4\hat{i} - k\hat{j}) \cdot (3\hat{i} + 8\hat{j}) = 4 \times 3 + (-k) \times 8 = 12 - 8k$ For the vectors to be perpendicular, this dot product must be zero, so: $12 - 8k = 0$ Solving for $k$, we get: $k = \frac{12}{8} = \frac{3}{2}$ Therefore, the answer is option (D) $\frac{3}{2}$. So, the constant $k$ must be equal to $\frac{3}{2}$ for the vectors $a$ and $b$ to be perpendicular.

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Vectors

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Question 27 Report

Find the derivative of $3x^{2} + \frac{1}{x^{2}}$

$6x + 2x^{2}$

$6x + \frac{1}{2x}$

$6x - \frac{2}{x^{3}}$

$6x - \frac{1}{2x}$

Answer Details

To find the derivative of $3x^{2} + \frac{1}{x^{2}}$, we will use the power rule and the chain rule of differentiation. First, using the power rule, we can differentiate the term $3x^{2}$ as follows: \[\frac{d}{dx}(3x^{2}) = 6x.\] Next, using the chain rule, we can differentiate the term $\frac{1}{x^{2}}$ as follows: \[\frac{d}{dx}\left(\frac{1}{x^{2}}\right) = -\frac{1}{x^{3}} \cdot \frac{d}{dx}(x^{-2}) = -\frac{1}{x^{3}} \cdot (-2x^{-3}) = \frac{2}{x^{5}}.\] Putting these together, we get: \[\frac{d}{dx}(3x^{2} + \frac{1}{x^{2}}) = 6x + \frac{2}{x^{5}}.\] Therefore, the answer is option (C) $6x - \frac{2}{x^{3}}$.

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Differentiation

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Question 28 Report

Simplify $\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}}$

$3\sqrt{2}$

$5\sqrt{2}$

$\frac{5\sqrt{2}}{2}$

$\frac{3\sqrt{2}}{2}$

Answer Details

First, we simplify the term inside the square root in the numerator: \begin{align*} \sqrt{3} + \sqrt{48} &= \sqrt{3} + \sqrt{16\cdot3} \\ &= \sqrt{3} + \sqrt{16}\sqrt{3} \\ &= \sqrt{3} + 4\sqrt{3} \\ &= 5\sqrt{3}. \end{align*} Now we substitute this simplified value back into the original expression: \begin{align*} \frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}} &= \frac{5\sqrt{3}}{\sqrt{6}} \\ &= \frac{5\sqrt{3}}{\sqrt{2}\sqrt{3}} \\ &= \frac{5}{\sqrt{2}} \\ &= \frac{5\sqrt{2}}{2}. \end{align*} Therefore, the simplified expression is $\frac{5\sqrt{2}}{2}$.

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Surds

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Question 29 Report

If $y = x^{2} - 6x + 11$ is written in the form $y = a(x - h)^{2} + k$, find the value of $(a + h + k)$.

-4

-3

0

6

Answer Details

We can start by completing the square to rewrite the quadratic equation in the form given. \begin{align*} y &= x^{2} - 6x + 11\\ &= (x^{2} - 6x + 9) + 2\\ &= (x - 3)^{2} + 2 \end{align*} Now we can see that the equation is in the desired form, with $a = 1$, $h = 3$, and $k = 2$. Therefore, \begin{align*} (a + h + k) &= (1 + 3 + 2)\\ &= 6 \end{align*} So the value of $(a + h + k)$ is 6.

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Polynomial Functions

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Question 30 Report

Find the locus of points which is equidistant from P(4, 5) and Q(-6, -1).

3x - 5y + 13 = 0

3x - 5y - 7 = 0

5x - 3y + 7 = 0

5x + 3y - 1 = 0

Answer Details

To find the locus of points equidistant from two fixed points, we need to find the perpendicular bisector of the line segment joining the two points. This perpendicular bisector will be the locus of all points that are equidistant from the two given points. So, let's first find the midpoint of the line segment PQ joining P(4, 5) and Q(-6, -1). The midpoint M is given by: \begin{align*} M &= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\\ &= \left(\frac{4 + (-6)}{2}, \frac{5 + (-1)}{2}\right)\\ &= (-1, 2) \end{align*} Now, let's find the slope of the line segment PQ: \begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1}\\ &= \frac{-1 - 5}{-6 - 4}\\ &= \frac{-6}{-10}\\ &= \frac{3}{5} \end{align*} Since the perpendicular bisector of PQ will be perpendicular to PQ, its slope will be the negative reciprocal of the slope of PQ: \begin{align*} m_\perp &= -\frac{1}{m}\\ &= -\frac{1}{\frac{3}{5}}\\ &= -\frac{5}{3} \end{align*} Now we have the slope of the perpendicular bisector and a point on it, which is the midpoint M. We can use point-slope form to find the equation of the perpendicular bisector: \begin{align*} y - y_1 &= m_\perp(x - x_1)\\ y - 2 &= -\frac{5}{3}(x + 1) \end{align*} Simplifying this equation, we get: \begin{align*} 3y - 6 &= -5x - 5\\ 5x + 3y - 1 &= 0 \end{align*} Therefore, the locus of points equidistant from P(4, 5) and Q(-6, -1) is the line 5x + 3y - 1 = 0. Hence, the correct option is (d) 5x + 3y - 1 = 0.

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Vectors

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Question 31 Report

What is the coordinate of the centre of the circle $5x^{2} + 5y^{2} - 15x + 25y - 3 = 0$?

$(\frac{15}{2}, -\frac{25}{2})$

$(\frac{3}{2}, -\frac{5}{2})$

$(-\frac{3}{2}, \frac{5}{2})$

$(-\frac{15}{2}, \frac{25}{2})$

Answer Details

The general equation of a circle in standard form is: $(x-h)^2 + (y-k)^2 = r^2$ where the center of the circle is located at (h, k) and has a radius of r. To get the equation of the circle into this form, we need to complete the square for both the x and y terms: \begin{aligned} 5x^{2} + 5y^{2} - 15x + 25y - 3 &= 0 \\ 5(x^{2} - 3x) + 5(y^{2} + 5y) &= 3 \\ 5(x^{2} - 3x + \frac{9}{4}) + 5(y^{2} + 5y + \frac{25}{4}) &= 3 + 5(\frac{9}{4} + \frac{25}{4}) \\ 5(x - \frac{3}{2})^{2} + 5(y + \frac{5}{2})^{2} &= 35 \\ (x - \frac{3}{2})^{2} + (y + \frac{5}{2})^{2} &= 7 \end{aligned} Comparing this to the standard form, we can see that the center of the circle is at $(\frac{3}{2}, -\frac{5}{2})$. Therefore, the answer is (B) $(\frac{3}{2}, -\frac{5}{2})$.

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Co-ordinate Geometry

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Question 32 Report

A function f is defined on R, the set of real numbers, by: $f : x \to \frac{x + 3}{x - 2}, x \neq 2$, find $f^{-1}$.

$f^{-1} : x \to \frac{2x + 3}{x - 1}, x \neq 1$

$f^{-1} : x \to \frac{x + 3}{x + 2}, x \neq -2$

$f^{-1} : x \to \frac{x - 1}{2x + 3}, x \neq -\frac{3}{2}$

$f^{-1}: x \to \frac{x - 2}{x + 3}, x \neq -3$

Answer Details

To find $f^{-1}$, we need to solve the equation: \[y = \frac{x + 3}{x - 2}\] for x in terms of y. To do this, we start by swapping the roles of x and y: \[x = \frac{y + 3}{y - 2}\] Now we solve for y in terms of x. We begin by multiplying both sides by $y-2$: \[x(y-2) = y + 3\] Expanding the left-hand side: \[xy - 2x = y + 3\] Rearranging the terms: \[xy - y = 2x + 3\] Factoring out y on the left-hand side: \[y(x-1) = 2x + 3\] Finally, dividing both sides by $x-1$: \[y = \frac{2x+3}{x-1}, x\neq 1\] Therefore, the inverse function of $f(x) = \frac{x+3}{x-2}, x\neq 2$ is: \[f^{-1}(x) = \frac{2x+3}{x-1}, x\neq 1\] Hence, the answer is $\boxed{\textbf{(A) }f^{-1} : x \to \frac{2x+3}{x-1}, x \neq 1}$.

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Functions

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Question 33 Report

Express $\frac{1}{1 - \sin 45°}$ in surd form.

$2 + \sqrt{2}$

$2 + \sqrt{3}$

$2 - \sqrt{2}$

$1 + 2\sqrt{2}$

Answer Details

We can start by using the identity $\sin 45° = \frac{\sqrt{2}}{2}$: \begin{align*} \frac{1}{1 - \sin 45°} &= \frac{1}{1 - \frac{\sqrt{2}}{2}} \\ &= \frac{1}{\frac{2 - \sqrt{2}}{2}} \\ &= \frac{2}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} \quad \text{(rationalizing the denominator)} \\ &= \frac{2(2 + \sqrt{2})}{2^2 - (\sqrt{2})^2} \\ &= \frac{2(2 + \sqrt{2})}{2} \\ &= \boxed{2 + \sqrt{2}} \end{align*} Therefore, the answer is option A: $2 + \sqrt{2}$.

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Surds

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Question 34 Report

If events A and B are independent and $P(A) = \frac{7}{12}$ and $P(A \cap B) = \frac{1}{4}$, find P(B).

$\frac{3}{7}$

$\frac{4}{7}$

$\frac{5}{7}$

$\frac{6}{7}$

Answer Details

If events A and B are independent, then the occurrence of event A does not affect the occurrence of event B, and vice versa. We can use the formula for the probability of the intersection of two independent events: $$P(A \cap B) = P(A) \cdot P(B)$$ Given that $P(A) = \frac{7}{12}$ and $P(A \cap B) = \frac{1}{4}$, we can substitute these values into the formula and solve for $P(B)$: $$\frac{1}{4} = \frac{7}{12} \cdot P(B)$$ Multiplying both sides by $\frac{12}{7}$, we get: $$P(B) = \frac{3}{7}$$ Therefore, the probability of event B is $\frac{3}{7}$. Option (a) is the correct answer.

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Probability

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Question 35 Report

A body starts from rest and moves in a straight line with uniform acceleration of $5 ms^{-2}$. How far, in metres, does it go in 10 seconds?

50 m

250 m

350 m

500 m

Answer Details

The problem gives us the initial velocity of the body, $u=0$ (because it starts from rest), the acceleration, $a=5 ms^{-2}$, and the time, $t=10s$, and asks us to find the distance, $s$, that the body travels in that time. We can use the following equation to solve the problem: $s = ut + \frac{1}{2}at^2$ where $u$ is the initial velocity, $a$ is the acceleration, $t$ is the time, and $s$ is the distance traveled. Plugging in the values we have: $s = 0 \times 10 + \frac{1}{2} \times 5 \times 10^2 = 250\text{ m}$ Therefore, the answer is option (B) 250 m. This equation gives us the distance traveled by an object with uniform acceleration starting from rest. It is derived by combining the equations of motion for constant acceleration.

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Dynamics

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Question 36 Report

The distance between P(x, 7) and Q(6, 19) is 13 units. Find the values of x.

1 or -7

1 or 7

1 or 11

5 or -5

Answer Details

To find the value of x, we can use the distance formula which states that the distance between two points P(x1, y1) and Q(x2, y2) is given by: d = sqrt((x2 - x1)^2 + (y2 - y1)^2) In this case, we are given the distance between P(x, 7) and Q(6, 19) as 13 units. Therefore, we can write: 13 = sqrt((6 - x)^2 + (19 - 7)^2) Simplifying this equation, we get: 169 = (6 - x)^2 + 144 25 = (6 - x)^2 Taking the square root of both sides, we get: 5 = 6 - x or 5 = x - 6 Solving for x in each case, we get: x = 1 or x = 11 Therefore, the possible values of x are 1 or 11. Note that we can also check that the distance between P(1, 7) and Q(6, 19) and between P(11, 7) and Q(6, 19) is indeed 13 units.

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Co-ordinate Geometry

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Question 37 Report

If $P = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}$ and $Q = \begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}$, find PQ.

$\begin{pmatrix} 4 & 1 \\ -2 & 9 \end{pmatrix}$

$\begin{pmatrix} -4 & 1 \\ 2 & 9 \end{pmatrix}$

$\begin{pmatrix} -4 & 3 \\ -2 & 13 \end{pmatrix}$

$\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}$

Answer Details

To find the product of matrices P and Q, we need to multiply the rows of the first matrix with the columns of the second matrix. Multiplying the first row of matrix P with the first column of matrix Q gives: \begin{pmatrix} 1 & -2 \end{pmatrix}\begin{pmatrix} -2 \\ 1 \end{pmatrix} = (-2\times 1) + (-2\times 1) = -4 Multiplying the first row of matrix P with the second column of matrix Q gives: \begin{pmatrix} 1 & -2 \end{pmatrix}\begin{pmatrix} 3 \\ 0 \end{pmatrix} = (1\times 3) + (-2\times 0) = 3 Similarly, multiplying the second row of matrix P with the first and second columns of matrix Q gives: \begin{pmatrix} 3 & 4 \end{pmatrix}\begin{pmatrix} -2 \\ 1 \end{pmatrix} = (-6) + 4 = -2 \begin{pmatrix} 3 & 4 \end{pmatrix}\begin{pmatrix} 3 \\ 0 \end{pmatrix} = 9 + 0 = 9 Hence, the product of matrices P and Q is: PQ = $\begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}$$\begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}$ = $\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}$ Therefore, the answer is option D: $\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}$.

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Matrices And Linear Transformation

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Question 38 Report