WAEC SSCE - Further Mathematics - 2020 (Essay)

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(a) Given \(u = 10\,\text{ms}^{-1},\ a = 0.2\,\text{ms}^{-2},\ s = 100\,\text{m}\). Use \(s = ut + \tfrac{1}{2}at^2\):

\[100 = 10t + 0.1t^2 \ \Rightarrow\ t^2 + 100t - 1000 = 0\] \[t = \frac{-100 + \sqrt{100^2 + 4000}}{2} = \frac{-100 + \sqrt{14000}}{2} = \frac{-100 + 118.32}{2}\] \[t \approx 9.16\,\text{s}\]

(b) \(S = t^2 - 5t + 6\), so velocity \(v = \dfrac{dS}{dt} = 2t - 5\).

(i) Initial velocity at \(t = 0\): \(v = 2(0) - 5 = -5\,\text{ms}^{-1}\).

(ii) Momentarily at rest when \(v = 0\): \(2t - 5 = 0 \Rightarrow t = 2.5\,\text{s}\).

\[S = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25\,\text{m}\]

The particle is \(0.25\,\text{m}\) from \(O\) (on the negative side).

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

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

Add the components:

\[\sum x = 10 - 10 - 2 = -2,\qquad \sum y = 0 - 17.32 + 3.46 = -13.86\]

Magnitude of the resultant:

\[R = \sqrt{(-2)^2 + (-13.86)^2} = \sqrt{4 + 192.0} = \sqrt{196.0} = 14.0\,\text{N}\]

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

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(a) Evaluate each logarithm to base 2:

\[\log_2 8 = 3,\quad \log_2 16 = 4,\quad 4\log_2 2 = 4,\quad \log_4 16 = 2\] \[\frac{\log_2 8 + \log_2 16 - 4\log_2 2}{\log_4 16} = \frac{3 + 4 - 4}{2} = \frac{3}{2}\]

(b) Let the A.P. have first term \(a\) and common difference \(d\). The 1st, 3rd and 7th terms are \(a,\ a + 2d,\ a + 6d\). These are consecutive G.P. terms, so:

\[(a + 2d)^2 = a(a + 6d)\] \[a^2 + 4ad + 4d^2 = a^2 + 6ad \ \Rightarrow\ 4d^2 - 2ad = 0 \ \Rightarrow\ 2d(2d - a) = 0\]

Since \(d \neq 0\), \(a = 2d\). The sum of the first two A.P. terms is 6:

\[a + (a + d) = 2a + d = 6\]

Substituting \(a = 2d\): \(4d + d = 6 \Rightarrow d = \tfrac{6}{5} = 1.2\), and \(a = 2d = \tfrac{12}{5} = 2.4\).

(i) First term \(= 2.4\). (ii) Common difference \(= 1.2\).

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The operation is \( p * q = p + q - \dfrac{pq}{2} \), with identity element \(0\).

(a) Inverse of \(-1\). Let \(x\) be the inverse of \(-1\), so \( (-1) * x = 0 \):

\[ -1 + x - \frac{(-1)x}{2} = 0 \Rightarrow -1 + x + \frac{x}{2} = 0 \Rightarrow \frac{3x}{2} = 1 \Rightarrow x = \frac{2}{3}. \]

The inverse of \(-1\) is \( \dfrac{2}{3} \).

(b) Truth set of \( m * 7 = m * 5 \).

\( m * 7 = m + 7 - \dfrac{7m}{2} \) and \( m * 5 = m + 5 - \dfrac{5m}{2} \). Setting them equal:

\[ 7 - \frac{7m}{2} = 5 - \frac{5m}{2} \Rightarrow 7 - 5 = \frac{7m}{2} - \frac{5m}{2} \Rightarrow 2 = m. \]

The truth set is \( \{2\} \).

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(a) Since \((x + 2)\) is a factor, divide \(g(x) = 2x^3 + 11x^2 - x - 30\) by \((x + 2)\).

Using synthetic division with root \(x = -2\):

\[2x^3 + 11x^2 - x - 30 = (x + 2)(2x^2 + 7x - 15)\]

Factorise the quadratic:

\[2x^2 + 7x - 15 = (2x - 3)(x + 5)\]

So \(g(x) = (x + 2)(2x - 3)(x + 5)\). The zeros are:

\[x = -2,\qquad x = \tfrac{3}{2},\qquad x = -5\]

(b) Let \(a = 2^x\) and \(b = 3^y\). The equations become:

\[3a + \tfrac{b}{9} = 25 \ \Rightarrow\ 27a + b = 225\] \[a - 3b = -19\]

From the second equation \(a = 3b - 19\). Substitute:

\[27(3b - 19) + b = 225 \ \Rightarrow\ 82b - 513 = 225 \ \Rightarrow\ b = 9\]

So \(3^y = 9 \Rightarrow y = 2\), and \(a = 3(9) - 19 = 8\), so \(2^x = 8 \Rightarrow x = 3\).

Check: \(3(8) + 3^{0} = 25\) and \(8 - 3^{3} = -19\). Hence \(x = 3,\ y = 2\).

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Integrate \( (3x - 2)^5 \) by reversing the chain rule (raise the power, divide by the new power and by the derivative of the bracket, \(3\)):

\[ \int (3x - 2)^5\,dx = \frac{(3x - 2)^6}{6 \times 3} = \frac{(3x - 2)^6}{18}. \]

Evaluate between \(1\) and \(3\):

\[ \left[\frac{(3x - 2)^6}{18}\right]_1^3 = \frac{(7)^6}{18} - \frac{(1)^6}{18} = \frac{117649 - 1}{18} = \frac{117648}{18} = 6536. \]

The value of the integral is \( 6536 \).

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First find the direction vector \(w + x - 2y\):

\[w + x - 2y = (8i + 3j) + (6i - 5j) - 2(2i + 3j)\] \[= (8 + 6 - 4)i + (3 - 5 - 6)j = 10i - 10j\]

Its magnitude is:

\[|10i - 10j| = \sqrt{10^2 + (-10)^2} = \sqrt{200} = 10\sqrt{2}\]

The unit vector in this direction is:

\[\hat{u} = \frac{10i - 10j}{10\sqrt{2}} = \frac{1}{\sqrt{2}}i - \frac{1}{\sqrt{2}}j\]

Since \(|z| = 41\) and \(z\) points in this direction:

\[z = 41\hat{u} = \frac{41}{\sqrt{2}}i - \frac{41}{\sqrt{2}}j = \frac{41\sqrt{2}}{2}(i - j)\]

Therefore \(z \approx 29.0i - 29.0j\).

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(a) \(y = x^2(1 + x)^{3/2}\). Apply the product rule:

\[\frac{dy}{dx} = 2x(1 + x)^{3/2} + x^2 \cdot \tfrac{3}{2}(1 + x)^{1/2}\]

Factor out \((1 + x)^{1/2}\):

\[\frac{dy}{dx} = (1 + x)^{1/2}\left[2x(1 + x) + \tfrac{3}{2}x^2\right] = (1 + x)^{1/2}\left[2x + \tfrac{7}{2}x^2\right]\] \[\frac{dy}{dx} = \frac{x(4 + 7x)}{2}\sqrt{1 + x}\]

(b) Let the centre be \((h, k)\) with \(2k - h = 3\), so \(h = 2k - 3\). The centre is equidistant from \(P(2,3)\) and \(Q(6,7)\):

\[(h - 2)^2 + (k - 3)^2 = (h - 6)^2 + (k - 7)^2\]

Expanding and simplifying gives \(8h + 8k = 72\), i.e. \(h + k = 9\). With \(h = 2k - 3\):

\[2k - 3 + k = 9 \ \Rightarrow\ k = 4,\quad h = 5\]

Centre \((5, 4)\); radius squared \(= (5 - 2)^2 + (4 - 3)^2 = 10\).

\[(x - 5)^2 + (y - 4)^2 = 10 \quad\text{or}\quad x^2 + y^2 - 10x - 8y + 31 = 0\]

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Convert each ordinal rank to a number and apply \(r_s=1-\dfrac{6\sum d^2}{n(n^2-1)}\) with \(n=10\), so \(n(n^2-1)=990\).

Cand.	I	II	III
A	1	2	3
B	3	1	2
C	6	3	1
D	2	9	6
E	10	7	9
F	9	4	8
G	7	8	7
H	4	10	5
I	8	5	4
J	5	6	10

(a)(i) Examiners I and II. \(d=I-II:\ -1,2,3,-7,3,5,-1,-6,3,-1\); \(\sum d^2=1+4+9+49+9+25+1+36+9+1=144\).\[r_s=1-\frac{6(144)}{990}=1-\frac{864}{990}=0.127\]

(a)(ii) Examiners I and III. \(d=I-III:\ -2,1,5,-4,1,1,0,-1,4,-5\); \(\sum d^2=4+1+25+16+1+1+0+1+16+25=90\).\[r_s=1-\frac{6(90)}{990}=1-\frac{540}{990}=0.455\]

(a)(iii) Examiners II and III. \(d=II-III:\ -1,-1,2,3,-2,-4,1,5,1,-4\); \(\sum d^2=1+1+4+9+4+16+1+25+1+16=78\).\[r_s=1-\frac{6(78)}{990}=1-\frac{468}{990}=0.527\]

(b) Agreement. The largest coefficient is \(0.527\), so Examiners II and III agree most.

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The geometry of the string gives \(XO=4.2-2.4=1.8\,\text{m}\), \(OY=2.4\,\text{m}\), and \(XY=3.0\,\text{m}\).

(a) Angle \(XOY\)

\[ XO^2+OY^2=1.8^2+2.4^2=3.24+5.76=9=3^2=XY^2. \]

Therefore, by the converse of Pythagoras’ theorem,

\[ \angle XOY=90^\circ. \]

(b) Horizontal force \(P\), and (c) tension \(T\)

The two tensions have equal magnitude \(T\). From the right-angled triangle:

• For \(OX\), the horizontal and vertical components of tension are \(0.6T\) to the left and \(0.8T\) upwards.
• For \(OY\), the horizontal and vertical components of tension are \(0.8T\) to the right and \(0.6T\) upwards.

The mass is \(0.800\,\text{kg}\), so its weight is \(W=0.8g\).

Resolving vertically for equilibrium:

\[ 0.8T+0.6T=0.8g \]

\[ 1.4T=0.8g \]

\[ T=\frac{0.8g}{1.4}=\frac{4g}{7}\,\text{N}. \]

Taking \(g=10\,\text{m s}^{-2}\),

\[ T=\frac{40}{7}=5.71\,\text{N}. \]

Horizontally, the rightward component is \(0.8T\) and the leftward component is \(0.6T\). Their resultant is \(0.2T\) to the right, so \(P\) must act to the left:

\[ P=0.8T-0.6T=0.2T. \]

\[ P=0.2\left(\frac{40}{7}\right)=\frac{8}{7}=1.14\,\text{N}. \]

Thus:

• \(\angle XOY=90^\circ\)
• \(P=1.14\,\text{N}\), acting horizontally towards \(X\)
• \(T=5.71\,\text{N}\)

The supplied school reference answer is not consistent with equilibrium: it treats \(P\) as though it were found directly from a string length, and then combines \(P\) and the weight to obtain one tension. There are two equal tension forces, so their horizontal and vertical components must both be included when resolving forces.

Examination reminder: Draw the force directions at the hanging body, resolve both tensions into horizontal and vertical components, and apply equilibrium separately in each direction.

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(a) Total members \(= 6 + 8 = 14\). Ways to choose 4 \(= \binom{14}{4} = 1001\).

Use the complement of "fewer than 2 farmers":

\[\text{0 farmers} = \binom{6}{0}\binom{8}{4} = 70,\qquad \text{1 farmer} = \binom{6}{1}\binom{8}{3} = 6 \times 56 = 336\] \[P(\text{at least 2 farmers}) = \frac{1001 - 70 - 336}{1001} = \frac{595}{1001} = \frac{85}{143} \approx 0.594\]

(b) With \(P(X) = \tfrac{1}{5},\ P(Y) = \tfrac{1}{12},\ P(Z) = \tfrac{1}{6}\):

(i) None:

\[\left(\tfrac{4}{5}\right)\left(\tfrac{11}{12}\right)\left(\tfrac{5}{6}\right) = \frac{220}{360} = \frac{11}{18}\]

(ii) All:

\[\left(\tfrac{1}{5}\right)\left(\tfrac{1}{12}\right)\left(\tfrac{1}{6}\right) = \frac{1}{360}\]

(iii) At least one:

\[1 - \frac{11}{18} = \frac{7}{18}\]

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(a) \( p: y \to 2y + 3 \) and \( q: y \to y - 2 \). The composite \( q \circ p \) (q of p) means apply p first, then q:

\[ (q \circ p)(y) = q(p(y)) = q(2y + 3) = (2y + 3) - 2 = 2y + 1. \]

So \( q \circ p : y \to 2y + 1 \).

(b) Form four-digit odd numbers greater than 4000 from \( \{1, 7, 3, 8, 2\} \) with repetition allowed.

First (thousands) digit: must be \(\ge 4\) for the number to exceed 4000. Only \(7\) and \(8\) qualify \( \Rightarrow 2 \) choices.

Last (units) digit: must be odd, from \( \{1, 3, 7\} \Rightarrow 3 \) choices.

Middle two digits: any of the 5 digits, repetition allowed \( \Rightarrow 5 \times 5 = 25 \) choices.

\[ \text{Total} = 2 \times 5 \times 5 \times 3 = 150. \]

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The fraction \( \dfrac{3x^2 + 3x - 2}{(x-1)(x+1)} \) is improper (numerator and denominator both degree 2), so it resolves as \[ \frac{3x^2 + 3x - 2}{(x-1)(x+1)} = P + \frac{Q}{x-1} + \frac{R}{x+1}. \]

Find P. The denominator expands to \( x^2 - 1 \). Divide: \[ 3x^2 + 3x - 2 = 3(x^2 - 1) + (3x + 1), \] so \( P = 3 \) and the proper remainder is \( \dfrac{3x + 1}{(x-1)(x+1)} \).

Find Q and R from \( 3x + 1 = Q(x+1) + R(x-1) \).

At \( x = 1 \): \( 3(1) + 1 = Q(2) \Rightarrow 4 = 2Q \Rightarrow Q = 2 \).

At \( x = -1 \): \( 3(-1) + 1 = R(-2) \Rightarrow -2 = -2R \Rightarrow R = 1 \).

Hence \( Q = 2 \) and \( R = 1 \) (with \( P = 3 \)).

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(a) Bag: 10 red, 8 green (18 balls), two drawn without replacement. One red and one green can occur as (red then green) or (green then red):

\[ P = \frac{10}{18}\cdot\frac{8}{17} + \frac{8}{18}\cdot\frac{10}{17} = 2\cdot\frac{80}{306} = \frac{160}{306} = \frac{80}{153} \approx 0.523. \]

(b) Defective rate \( p = 0.2,\ q = 0.8,\ n = 12 \). Required \( P(2 \le X \le 5) = \sum_{k=2}^{5}\binom{12}{k}(0.2)^k(0.8)^{12-k} \):

• \( P(2) = \binom{12}{2}(0.2)^2(0.8)^{10} \approx 0.2835 \)
• \( P(3) = \binom{12}{3}(0.2)^3(0.8)^{9} \approx 0.2362 \)
• \( P(4) = \binom{12}{4}(0.2)^4(0.8)^{8} \approx 0.1329 \)
• \( P(5) = \binom{12}{5}(0.2)^5(0.8)^{7} \approx 0.0531 \)

\[ P(2 \le X \le 5) \approx 0.2835 + 0.2362 + 0.1329 + 0.0531 = 0.706. \]