WAEC SSCE - Further Mathematics - 2018

Given that \(y = x(x + 1)^{2}\), calculate the maximum value of y.

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Find the radius of the circle \(x^{2} + y^{2} - 8x - 2y + 1 = 0\).

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What is the probability of obtaining a head and a six when a fair coin and and a die are tossed together? 

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If \(\overrightarrow{OX} = \begin{pmatrix} -7 \\ 6 \end{pmatrix}\) and \(\overrightarrow{OY} = \begin{pmatrix} 16 \\ -11 \end{pmatrix}\), find \(\overrightarrow{YX}\).

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Find the angle between forces of magnitude 7N and 4N if their resultant has a magnitude of 9N.

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A body of mass 28g, initially at rest is acted upon by a force, F Newtons. If it attains a velocity of \(5.4ms^{-1}\) in 18 seconds, find the value of F.

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We can use the formula \(F = ma\) to solve the problem, where \(F\) is the force in Newtons, \(m\) is the mass in kilograms, and \(a\) is the acceleration in meters per second squared. We need to convert the mass from grams to kilograms and find the acceleration first. Given that the initial velocity, \(u\) = 0 \(ms^{-1}\), final velocity, \(v\) = \(5.4ms^{-1}\), and time, \(t\) = 18 seconds. We can use the formula \(v = u + at\) to find the acceleration: \(v = u + at\) \(5.4 = 0 + a\times 18\) \(a = \frac{5.4}{18} = 0.3ms^{-2}\) Now, we can substitute the values of mass and acceleration in the formula to find the force: \(F = ma\) \(F = 0.028\times 0.3\) \(F = 0.0084N\) Therefore, the value of force is 0.0084N.

Out of 70 schools, 42 of them can be attended by boys and 35 can be attended by girls. If a pupil is selected at random from these schools, find the probability that he/ she is from a mixed school.

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A car is moving at 120\(kmh^{-1}\). Find its speed in \(ms^{-1}\).

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To convert from kilometers per hour to meters per second, we need to use the following formula: 1 kilometer per hour = 0.277778 meters per second Therefore, to find the speed of the car in meters per second, we can multiply its speed in kilometers per hour by 0.277778. Speed in meters per second = 120 km/h x 0.277778 = 33.33336 m/s (rounded to 5 decimal places) Therefore, the correct answer is 33.3\(ms^{-1}\).

If the polynomial \(f(x) = 3x^{3} - 2x^{2} + 7x + 5\) is divided by (x - 1), find the remainder.

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Evaluate \(\int_{\frac{1}{2}}^{1} \frac{x^{3} - 4}{x^{3}} \mathrm {d} x\).

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If \(\sin\theta = \frac{3}{5}, 0° < \theta < 90°\), evaluate \(\cos(180 - \theta)\).

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Find the domain of \(g(x) = \frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}\)

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To find the domain of the given function, we need to identify any values of x that would make the denominator of the fraction equal to zero or negative, as these values would result in an undefined function. Additionally, we need to consider any other restrictions on x that may arise from the function's algebraic properties. In this case, the denominator of the function is the square root of \(9x^{2} + 1\). Since the square root of a negative number is not defined in the real number system, we know that the expression inside the square root must be non-negative. This means that: \(9x^{2} + 1 \geq 0\) Solving this inequality, we get: \(9x^{2} \geq -1\) Dividing both sides by 9, we get: \(x^{2} \geq -\frac{1}{9}\) Since the square of any real number is non-negative, we know that this inequality is satisfied for all real values of x. Therefore, there are no values of x that would make the denominator of the function equal to zero or negative, and the domain of the function is all real numbers. In other words, the correct answer is: \(x: x \in R\)

The general term of an infinite sequence 9, 4, -1, -6,... is \(u_{r} = ar + b\). Find the values of a and b.

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Two functions f and g are defined on the set of real numbers by \(f : x \to x^{2} + 1\) and \(g : x \to x - 2\). Find f o g.

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Find the equation to the circle \(x^{2} + y^{2} - 4x - 2y = 0\) at the point (1, 3).

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If \(\log_{3}a - 2 = 3\log_{3}b\), express a in terms of b.

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The given equation is \(\log_{3}a - 2 = 3\log_{3}b\). We can use the logarithmic property that \(\log_{a}b^{c} = c\log_{a}b\) to simplify the equation: \(\log_{3}a - 2 = \log_{3}b^{3}\) \(\log_{3}a = \log_{3}b^{3} + 2\) \(\log_{3}a = \log_{3}(b^{3}\cdot 3^{2})\) Using the property that if \(\log_{a}b = \log_{a}c\) then \(b = c\), we get: \(a = b^{3}\cdot 3^{2}\) \(a = 9b^{3}\) Therefore, the value of \(a\) in terms of \(b\) is \(a = 9b^{3}\).

Simplify \(\frac{\sqrt{3}}{\sqrt{3} -1} + \frac{\sqrt{3}}{\sqrt{3} + 1}\)

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Given that \(f(x) = 3x^{2} -  12x + 12\) and \(f(x) = 3\), find the values of x.

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To find the values of x where the function f(x) is equal to 3, we need to solve the equation:

3x^2 - 12x + 12 = 3

We can start by subtracting 3 from both sides:

3x^2 - 12x + 12 - 3 = 3 - 3
3x^2 - 12x + 9 = 0

Next, we can use the quadratic formula to find the solutions for x:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 3, b = -12, c = 9. Plugging in these values, we get:

x = (-(-12) ± √((-12)^2 - 4 * 3 * 9)) / 2 * 3
x = (12 ± √(144 - 108)) / 6
x = (12 ± √36) / 6
x = (12 ± 6) / 6

So, the two values of x that solve the equation are:

x = (12 + 6) / 6 = 18 / 6 = 3
x = (12 - 6) / 6 = 6 / 6 = 1

Therefore, the two values of x that make f(x) equal to 3 are 1 and 3.

Find the coefficient of \(x^{3}\) in the binomial expansion of \((x - \frac{3}{x^{2}})^{9}\).

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A particle starts from rest and moves through a distance \(S = 12t^{2} - 2t^{3}\) metres in time t seconds. Find its acceleration in 1 second.

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The distance moved by a particle is given by the equation \(S = 12t^{2} - 2t^{3}\), where S is the distance travelled in metres, and t is the time taken in seconds. To find the acceleration of the particle in 1 second, we need to differentiate the equation for distance with respect to time twice to obtain the equation for acceleration. First, we differentiate S with respect to time to get the velocity equation: $$\frac{dS}{dt} = \frac{d}{dt}(12t^{2} - 2t^{3}) = 24t - 6t^{2}$$ Next, we differentiate the velocity equation with respect to time to get the acceleration equation: $$\frac{d^2S}{dt^2} = \frac{d}{dt}(24t - 6t^{2}) = 24 - 12t$$ To find the acceleration in 1 second, we substitute t = 1 into the acceleration equation: $$\frac{d^2S}{dt^2} \bigg\rvert_{t=1} = 24 - 12(1) = 12ms^{-2}$$ Therefore, the acceleration of the particle in 1 second is 12\(ms^{-2}\).

Express \(\frac{13}{4}\pi\) radians in degrees.

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To convert radians to degrees, we use the following formula:

degrees = radians × 180°/π

where π is approximately equal to 3.14.

Substituting the given value of radians, we get:

degrees = (13/4)π × 180°/π

The π cancels out, leaving us with:

degrees = (13/4) × 180°

Simplifying the expression, we get:

degrees = 585°

Therefore, the answer is 585°.

Resolve \(\frac{3x - 1}{(x - 2)^{2}}, x \neq 2\) into partial fractions.

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How many numbers greater than 150 can be formed from the digits 1, 2, 3, 4, 5 without repetition?

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The marks scored by 4 students in Mathematics and Physics are ranked as shown in the table below

Calculate the Spearmann's rank correlation coefficient.

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Solve \(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\).

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

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The first term of a Geometric Progression (GP) is \(\frac{3}{4}\), If the product of the second and third terms of the sequence is 972, find its common ratio.

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A binary operation * is defined on the set of real numbers, by \(a * b = \frac{a}{b} + \frac{b}{a}\). If \((\sqrt{x} + 1) * (\sqrt{x} - 1) = 4\), find the value of x. 

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If the determinant of the matrix \(\begin{pmatrix} 2 & x \\ 3 & 5 \end{pmatrix} = 13\), find the value of x.

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Find the stationary point of the curve \(y = 3x^{2} - 2x^{3}\).

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The midpoint of M(4, -1) and N(x, y) is P(3, -4). Find the coordinates of N.

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We know that the midpoint of a line segment is the point that is exactly halfway between the endpoints of the segment. So, to find the coordinates of point N, we need to use the midpoint formula. The midpoint formula is: Midpoint = [(x1 + x2)/2, (y1 + y2)/2] where (x1, y1) and (x2, y2) are the coordinates of the two endpoints of the segment. In this problem, we are given that the midpoint of segment MN is P(3, -4), and one endpoint is M(4, -1). Let's call the coordinates of the other endpoint N(x, y). Using the midpoint formula, we can write: [(4 + x)/2, (-1 + y)/2] = (3, -4) Now, we can solve for x and y: (4 + x)/2 = 3 => 4 + x = 6 => x = 2 (-1 + y)/2 = -4 => -1 + y = -8 => y = -7 Therefore, the coordinates of point N are (2, -7). So, the correct option is (2, -7).

In how many ways can the letters of the word 'ELECTIVE' be arranged?

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Given that \(a = i - 3j\) and \(b = -2i + 5j\) and \(c = 3i - j\), calculate \(|a - b + c|\).

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If \(\alpha\) and \(\beta\) are the roots of \(2x^{2} - 5x + 6 = 0\), find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\).

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To find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\), we can use the relationship between the roots and coefficients of a quadratic equation. Let's start by finding the sum and product of the roots of the original equation, \(2x^{2} - 5x + 6 = 0\). We can use the formulae: Sum of roots, \(S = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}\) Product of roots, \(P = \frac{c}{a} = \frac{6}{2} = 3\) Now, let's add 1 to both roots: \(\alpha + 1\) and \(\beta + 1\) The sum of these roots would be: \((\alpha + 1) + (\beta + 1) = \alpha + \beta + 2\) And the product would be: \((\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1\) Now, we want to find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\). Let's call this equation \(ax^{2} + bx + c = 0\). According to the relationship between roots and coefficients, we know that: \(\frac{-b}{a} = \frac{5}{2}\) and \(\frac{c}{a} = 3\) Solving for \(b\) and \(c\) in terms of \(a\), we get: \(b = -\frac{5a}{2}\) and \(c = 3a\) Now, we can use the sum and product of the new roots to form two equations: \(\frac{-b}{a} = \frac{5}{2} \implies -\frac{5a}{2a} = \frac{5}{2} \implies -5 = 5\) This equation is clearly not true, which means that the first option, \(2x^{2} - 9x + 15 = 0\), is not the correct answer. Let's move on to the second option: \((\alpha + \beta) + 2 = \frac{5}{2} + 2 = \frac{9}{2}\) \((\alpha \beta + \alpha + \beta + 1) = 3 + \frac{5}{2} + \frac{5}{2} + 1 = 8\) Substituting the values of \(b\) and \(c\) in terms of \(a\), we get: \(a(x^{2} - \frac{9}{2}x + \frac{13}{2}) = 0\) Simplifying, we get: \(2x^{2} - 9x + 13 = 0\) Therefore, the answer is the second option, \(2x^{2} - 9x + 13 = 0\).

Find the constant term in the binomial expansion \((2x^{2} + \frac{1}{x})^{9}\)

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The constant term in a binomial expansion is the term that has no variable or a variable raised to the power of zero. To find the constant term in \((2x^{2} + \frac{1}{x})^{9}\), we need to look for the term that has no \(x\) or a \(x\) raised to the power of zero. To obtain the constant term, we need to choose the constant terms from each of the factors in the expansion, which are \(2x^2\) and \(\frac{1}{x}\), in such a way that their product is raised to a power that adds up to 9. In other words, we need to choose the constant terms that multiply to give a coefficient in the expansion of \((2x^2)^m (\frac{1}{x})^{9-m}\) that is independent of \(x\). The constant term in the expansion of \((2x^2)^m (\frac{1}{x})^{9-m}\) is \(\binom{9}{m}(2x^2)^m (\frac{1}{x})^{9-m}\), where \(\binom{9}{m}\) is the binomial coefficient that represents the number of ways to choose \(m\) items out of 9. For the constant term, we need to choose \(m\) such that the powers of \(x\) in the two factors add up to zero. That is, \begin{align*} 2m - (9-m) &= 0 \\ \Rightarrow m &= \frac{9}{3} \\ \Rightarrow m &= 3 \end{align*} Therefore, the constant term in the expansion of \((2x^2 + \frac{1}{x})^9\) is \(\binom{9}{3}(2x^2)^3 (\frac{1}{x})^6\). Plugging in the values, we get: \begin{align*} \binom{9}{3}(2x^2)^3 (\frac{1}{x})^6 &= \frac{9!}{3!6!}(2^3x^6)(\frac{1}{x^6}) \\ &= 84(8) \\ &= 672 \end{align*} Therefore, the constant term in the binomial expansion of \((2x^2 + \frac{1}{x})^9\) is \boxed{672}.

If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} + 5x + n = 0\), such that \(\alpha\beta = 2\), find the value of n.

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If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} + 5x + n = 0\), then by the quadratic formula, we have: \[\alpha, \beta = \frac{-5 \pm \sqrt{5^{2} - 4(2)(n)}}{2(2)}\] Simplifying, we get: \[\alpha, \beta = \frac{-5 \pm \sqrt{25 - 8n}}{4}\] We are also given that \(\alpha\beta = 2\). Therefore, we have: \[\alpha\beta = \frac{-5 + \sqrt{25 - 8n}}{4} \cdot \frac{-5 - \sqrt{25 - 8n}}{4} = 2\] Expanding the left-hand side, we get: \[\frac{25 - (25 - 8n)}{16} = 2\] Simplifying, we get: \[\frac{8n}{16} = 2\] \[n = 4\] Therefore, the answer is the fourth option, 4.

If \(\begin{vmatrix}  k & k \\ 4 & k \end{vmatrix} + \begin{vmatrix}  2 & 3 \\ -1 & k \end{vmatrix} = 6\), find the value of the constant k, where k > 0.

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If \(4x^{2} + 5kx + 10\) is a perfect square, find the value of k.

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We know that a perfect square trinomial has the form of \((ax + b)^2\), where a and b are constants. So, if we have a trinomial of the form \(4x^2 + 5kx + 10\), we can write it as \((2x + c)^2\), where c is another constant. Expanding the square, we get: \begin{align*} (2x + c)^2 &= 4x^2 + 4cx + c^2 \\ &= 4x^2 + (4c)x + c^2 \\ \end{align*} Comparing the coefficients of the two trinomials, we get: \begin{align*} 4 &= 4 \\ 5k &= 4c \\ 10 &= c^2 \end{align*} Dividing the second equation by 4, we get: \begin{align*} \frac{5k}{4} &= c \end{align*} Since c = \( \sqrt{10} \), we get: \begin{align*} \frac{5k}{4} &= \sqrt{10} \\ \Rightarrow k &= \frac{4\sqrt{10}}{5} \end{align*} So the value of k is \( \frac{4\sqrt{10}}{5} \).

\(P = {1, 3, 5, 7, 9}, Q = {2, 4, 6, 8, 10, 12}, R = {2, 3, 5, 7, 11}\) are subsets of \(U = {1, 2, 3, ... , 12}\). Which of the following statements is true?

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Calculate the standard deviation of 30, 29, 25, 28, 32 and 24.

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The sum of the first twelve terms of an Arithmetic Progression is 168. If the third term is 7, find the values of the common difference and the first term.

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(a) Solve : \(2^{3y + 2} - 7(2^{2y + 2}) - 31(2^{y}) - 8 = 0, y \in R\).

(b) Find \(\int (\sqrt{x^{2} + 1}) xdx\).

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A body of mass 20kg moving with a velocity of 80ms\(^{-1}\) collides with another body of mass 30kg moving with a velocity of 50ms\(^{-1}\). If they both moved in the same direction after collision, find their common velocity if they moved in the :

(a) same direction before collision ; (b) opposite direction before collision.

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The probabilities that Ali, Baba and Katty will gain admission to college are \(\frac{2}{3}, \frac{3}{4}\) and \(\frac{4}{5}\) respectively. Find the probability that:

(a) only Katty and Baba will gain admission ;

(b) none of them will gain admission ;

(c) at most two of them will gain admission.

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Ten coins were tossed together a number of times. The distribution of the number of heads obtained is given in the following table :

Calculate, correct to three decimal places, the :

(a) mean number of heads ;

(b) probability of getting an even head ;

(c) probability of getting an odd number.

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The position vectors of points A, B and C with respect to the origin are (8i - 2j), (2i + 6j) and (-10i + 4j) respectively. If ABCN is a parallelogram, find :

(a) the position vector of N;

(b) AN and AB ;

(c) correct to two decimal place, the acute angle between AN and AB.

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A circle is drawn through the points (3, 2), (-1, -2) and (5, -4). Find the :

(a) coordinates of the centre of the circle ;

(b) radius of the circle ;

(c) equation of the circle.

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(a)(i) Write down the binomial expansion of \((2 - \frac{1}{2}x)^{5}\) in ascending powers of x.

(ii) Using the expansion in (a)(i), find, correct to two decimal places, the value of \((1.99)^{5}\).

(b) The polynomial \(x^{3} + qx^{2} + rx + 9\), where q and r are constants, has (x + 1) as a factor and has a remainder -17 when divided by (x + 2). Find the values of q and r.

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Given that \(\log_{3} x - 3\log_{x} 3 + 2 = 0\), find the values of x.

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If \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24 \), find the values of x.

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Two panel of judges, X and Y, rank 8 brands of cooking oil as follows :

Calculate the Spearmann's rank correlation coefficient.

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(a) The probability that Kunle solves a particular question is \(\frac{1}{3}\) while that of Tayo is \(\frac{1}{5}\). If both of them attempt the question, find the probability that only one of them will solve the question.

(b) A committee of 8 is to be chosen from 10 persons. In how many ways can this be done if there is no restriction?

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(a) Using the substitution \(u = x - 2\), write \(\frac{x^{3} + 5}{(x - 2)^{4}}\) as an expression in terms of u.

(b) Using the answer in (a), express \(\frac{x^{3} + 5}{(x - 2)^{4}}\) in partial fractions.

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A uniform beam, XY, 4m long and weighing 350N rests on two pivots P and Q. It is kept in equilibrium by weights of 80N attached at X and 1000N attached at a point between P and Q such that it is 0.6m from Q. If XP = 0.8m and PQ = 2.2m.

(a) calculate the reactions at P and Q ;

(b) if the 1000N weight is replaced with a 1200N weight, at what point from Q should it be placed in order to maintain the equilibrium.

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Given that \(m = 3i - 2j ; n = 2i - 3j\) and \(p = -i + 6j\), find \(4m + 2n - 3p\).

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