WAEC SSCE - Further Mathematics - 2016 (Essay)

Bayanin Amsa

(a) To find \(f(5)\) from \( f(x+2) = 6x^2 + 5x - 8 \), set \( x + 2 = 5 \), i.e. \( x = 3 \):

\[ f(5) = 6(3)^2 + 5(3) - 8 = 54 + 15 - 8 = 61. \]

(b) Rationalise by multiplying numerator and denominator by \( 4\sqrt2 + 2\sqrt3 \):

Denominator: \( (4\sqrt2 - 2\sqrt3)(4\sqrt2 + 2\sqrt3) = (4\sqrt2)^2 - (2\sqrt3)^2 = 32 - 12 = 20. \)

Numerator: \( (7\sqrt2 + 3\sqrt3)(4\sqrt2 + 2\sqrt3) = 56 + 14\sqrt6 + 12\sqrt6 + 18 = 74 + 26\sqrt6. \)

\[ \frac{74 + 26\sqrt6}{20} = \frac{37}{10} + \frac{13}{10}\sqrt6. \]

So \( p = \dfrac{37}{10},\ q = \dfrac{13}{10},\ r = 6 \).

Bayanin Amsa

Express \(r\) as a linear combination \(r = \alpha p + \beta q\):

\[\begin{pmatrix} 17 \\ 5 \end{pmatrix} = \alpha\begin{pmatrix} 5 \\ 3 \end{pmatrix} + \beta\begin{pmatrix} -1 \\ 2 \end{pmatrix}\]

This gives the equations:

\[5\alpha - \beta = 17,\qquad 3\alpha + 2\beta = 5\]

From the first, \(\beta = 5\alpha - 17\). Substitute:

\[3\alpha + 2(5\alpha - 17) = 5 \ \Rightarrow\ 13\alpha = 39 \ \Rightarrow\ \alpha = 3,\quad \beta = -2\]

Hence \(r = 3p - 2q\). Make \(q\) the subject:

\[2q = 3p - r \ \Rightarrow\ q = \tfrac{1}{2}(3p - r)\]

Check: \(\tfrac{1}{2}\left(3\begin{pmatrix}5\\3\end{pmatrix} - \begin{pmatrix}17\\5\end{pmatrix}\right) = \tfrac{1}{2}\begin{pmatrix}-2\\4\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix} = q\).

Bayanin Amsa

The operation is \( m * n = m + n + 2 \).

(a) Identity element. Let \(e\) be the identity, so \( m * e = m \) for all \(m\):

\[ m + e + 2 = m \Rightarrow e = -2. \] The identity element is \( e = -2 \).

(b) Inverse of n. Let \(n^{-1}\) be the inverse, so \( n * n^{-1} = e = -2 \):

\[ n + n^{-1} + 2 = -2 \Rightarrow n^{-1} = -4 - n. \] The inverse of \(n\) is \( n^{-1} = -(n+4) \).

Bayanin Amsa

(a) \(^{k}P_2 = k(k - 1) = 72\):

\[k^2 - k - 72 = 0 \ \Rightarrow\ (k - 9)(k + 8) = 0\]

Since \(k > 0\), \(k = 9\).

(b) \(2\cos^2\theta - 5\cos\theta = 3\). Let \(c = \cos\theta\):

\[2c^2 - 5c - 3 = 0 \ \Rightarrow\ (2c + 1)(c - 3) = 0\] \[c = -\tfrac{1}{2} \quad\text{or}\quad c = 3\ (\text{rejected, since } |\cos\theta| \le 1)\]

So \(\cos\theta = -\tfrac{1}{2}\). In \(0^\circ \le \theta \le 360^\circ\):

\[\theta = 120^\circ \quad\text{and}\quad \theta = 240^\circ\]

Bayanin Amsa

(a)(i) If books of the same subject must stand together, treat each subject as one block: 3 blocks arranged in \(3!\) ways, with the books inside each block also arranged.

\[3! \times 2! \times 5! \times 3! = 6 \times 2 \times 120 \times 6 = 8640\]

(ii) If only the Physics books must stand together, treat the 5 Physics books as one block. The items to arrange are: 2 Maths + 1 Physics block + 3 Chemistry \(= 6\) items.

\[6! \times 5! = 720 \times 120 = 86400\]

(b) \(p = \tfrac{13}{20} = 0.65,\ q = 0.35,\ n = 8\). Find \(P(\text{at least 3})\) via the complement:

\[P(0) = (0.35)^8 = 0.000225\] \[P(1) = \binom{8}{1}(0.65)(0.35)^7 = 0.003346\] \[P(2) = \binom{8}{2}(0.65)^2(0.35)^6 = 0.021747\] \[P(\ge 3) = 1 - (0.000225 + 0.003346 + 0.021747) = 1 - 0.025318 = 0.975\]

The probability that at least 3 speak English is \(0.975\) (3 s.f.).

Bayanin Amsa

Three points are collinear when the slope between any two pairs is the same.

Using \((5,2)\) and \((2,1)\): \[ m = \frac{2-1}{5-2} = \frac{1}{3}. \]

Using \((2,1)\) and \((-4,k)\) and equating the slope: \[ \frac{k-1}{-4-2} = \frac{1}{3} \Rightarrow \frac{k-1}{-6} = \frac{1}{3}. \]

\[ k - 1 = -2 \Rightarrow k = -1. \]

Bayanin Amsa

(a)(i) Cumulative frequency table

The cumulative frequency is obtained by adding each class frequency to the running total. The curve is plotted against the upper class boundary of each class, so these are tabulated as well. The total number of students is \(N = 550\).

Marks	Class boundaries	Frequency \(f\)	Cumulative frequency \(\sum f\)
1 - 10	0.5 - 10.5	3	3
11 - 20	10.5 - 20.5	17	20
21 - 30	20.5 - 30.5	41	61
31 - 40	30.5 - 40.5	85	146
41 - 50	40.5 - 50.5	97	243
51 - 60	50.5 - 60.5	115	358
61 - 70	60.5 - 70.5	101	459
71 - 80	70.5 - 80.5	64	523
81 - 90	80.5 - 90.5	21	544
91 - 100	90.5 - 100.5	6	550

(a)(ii) Cumulative frequency curve (ogive)

Plot the cumulative frequency (vertical axis) against the upper class boundary (horizontal axis), starting the curve from \((0.5,\,0)\), and join the points with a smooth S-shaped curve.

(b) Estimates read from the curve

(i) Number of students scoring between 32 and 74. Read the cumulative frequency at each mark from the ogive (dashed guide lines):

\[\text{CF}(32)\approx 61+\frac{32-30.5}{10}\times 85 \approx 74\]\[\text{CF}(74)\approx 459+\frac{74-70.5}{10}\times 64 \approx 481\]

Number scoring between 32 and 74:

\[481-74 = 407\ \text{students (approximately)}.\]

(ii) Pass mark, if 18% failed. The number who failed is

\[18\%\ \text{of}\ 550 = \frac{18}{100}\times 550 = 99.\]

The pass mark is the mark below which 99 students lie, i.e. the mark corresponding to \(\text{CF}=99\). Reading horizontally from \(99\) on the vertical axis to the curve and down to the mark axis:

\[\text{pass mark} \approx 30.5+\frac{99-61}{85}\times 10 \approx 35.\]

(iii) Lowest mark for distinction, if 8% passed with distinction. The distinction group is the top 8%, so the number below the distinction mark is

\[550-\Big(8\%\ \text{of}\ 550\Big)=550-44=506.\]

Read the mark corresponding to \(\text{CF}=506\):

\[\text{lowest distinction mark} \approx 70.5+\frac{506-459}{64}\times 10 \approx 78.\]

Bayanin Amsa

Momentum is conserved in each collision.

(a) Let the second mass be \(m\). Both bodies move in the same direction, then coalesce at \(6\text{ ms}^{-1}\):

\[ 3(8) + m(5) = (3 + m)(6). \]

\[ 24 + 5m = 18 + 6m \Rightarrow m = 6\text{ kg}. \]

(b) Now the \(6\text{ kg}\) body moves at \(5\text{ ms}^{-1}\) opposite to the \(3\text{ kg}\) body. Take the direction of the \(3\text{ kg}\) body as positive:

\[ 3(8) + 6(-5) = (3 + 6)v. \]

\[ 24 - 30 = 9v \Rightarrow v = \frac{-6}{9} = -0.67\text{ ms}^{-1}\ (\text{2 d.p.}). \]

The common velocity is \(0.67\text{ ms}^{-1}\) in the direction of the \(6\text{ kg}\) body's original motion.

Bayanin Amsa

(a)(i) By the binomial theorem:

\[(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7\]

(ii) The 5th, 6th and 7th coefficients are \(35,\ 21,\ 7\). They form an A.P. with common difference:

\[d = 21 - 35 = -14 \quad(\text{and } 7 - 21 = -14)\]

(b) Let \(f(x) = \sqrt{2x + 8x^2}\), with \(h = 1\). Compute the ordinates:

• \(f(1) = \sqrt{10} = 3.1623\)
• \(f(2) = \sqrt{36} = 6.0000\)
• \(f(3) = \sqrt{78} = 8.8318\)
• \(f(4) = \sqrt{136} = 11.6619\)
• \(f(5) = \sqrt{210} = 14.4914\)

Trapezium rule:

\[\int_1^5 f\,dx \approx \tfrac{h}{2}\big[(f_0 + f_4) + 2(f_1 + f_2 + f_3)\big]\] \[= \tfrac{1}{2}\big[(3.1623 + 14.4914) + 2(6 + 8.8318 + 11.6619)\big]\] \[= \tfrac{1}{2}\big[17.6537 + 52.9874\big] = \tfrac{1}{2}(70.6411) = 35.32\]

So the integral is approximately \(35.32\) (to 2 decimal places).

Bayanin Amsa

Probabilities of gaining admission: \( P(A)=\tfrac45,\ P(S)=\tfrac34,\ P(M)=\tfrac23 \). The events are independent.

Failure probabilities: \( P(A')=\tfrac15,\ P(S')=\tfrac14,\ P(M')=\tfrac13 \).

(a) None gains admission: \[ P(A')\,P(S')\,P(M') = \frac15 \cdot \frac14 \cdot \frac13 = \frac{1}{60}. \]

(b) Only Ago and Sulley gain admission (Musa fails): \[ P(A)\,P(S)\,P(M') = \frac45 \cdot \frac34 \cdot \frac13 = \frac{12}{60} = \frac{1}{5}. \]

Bayanin Amsa

By the division identity, dividend = divisor \(\times\) quotient + remainder:

\[ f(x) = (x^2 + 5x + 1)(2x - 5) + (30x + 16). \]

Expand the product:

\[ (x^2+5x+1)(2x-5) = 2x^3 - 5x^2 + 10x^2 - 25x + 2x - 5 = 2x^3 + 5x^2 - 23x - 5. \]

Add the remainder:

\[ f(x) = 2x^3 + 5x^2 - 23x - 5 + 30x + 16 = 2x^3 + 5x^2 + 7x + 11. \]

Compare with \( f(x) = 2x^3 + mx^2 + nx + 11 \):

\[ m = 5, \qquad n = 7. \]

Bayanin Amsa

(a) Work on the numerator:

\[\tfrac{3}{2}\log 27 = \tfrac{3}{2}\log 3^3 = \tfrac{9}{2}\log 3,\qquad 3\log 5\sqrt{5} = 3\log 5^{3/2} = \tfrac{9}{2}\log 5\] \[\text{Numerator} = \tfrac{9}{2}\log 3 - \tfrac{9}{2}\log 5 = \tfrac{9}{2}\log\tfrac{3}{5} = \tfrac{9}{2}\log 0.6\]

Dividing by \(\log 0.6\):

\[\frac{\tfrac{9}{2}\log 0.6}{\log 0.6} = \frac{9}{2} = 4.5\]

(b)(i) The matrices are:

\[A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix},\qquad B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\]

(ii) Apply \(A\) to \(P(-2, 2)\):

\[A(P) = (\,-2 + 2(2),\ -(-2) + 2\,) = (2,\ 4)\]

Then apply \(B\) to \((2, 4)\):

\[B(2,4) = (2(2) + 3(4),\ 2 + 2(4)) = (16,\ 10)\]

The image of \(P\) under \(A\) followed by \(B\) is \((16, 10)\).

Bayanin Amsa

Given \(u = 30\,\text{ms}^{-1},\ g = 10\,\text{ms}^{-2}\), upward taken positive.

(a) At maximum height \(v = 0\). Use \(v^2 = u^2 - 2gh\):

\[0 = 30^2 - 2(10)h \ \Rightarrow\ 20h = 900 \ \Rightarrow\ h = 45\,\text{m}\]

(b) Time to return to ground:

\[t = \frac{2u}{g} = \frac{2(30)}{10} = 6\,\text{s}\]

(c) Height \(40\,\text{m}\): use \(h = ut - \tfrac{1}{2}gt^2\):

\[40 = 30t - 5t^2 \ \Rightarrow\ 5t^2 - 30t + 40 = 0 \ \Rightarrow\ t^2 - 6t + 8 = 0\] \[(t - 2)(t - 4) = 0 \ \Rightarrow\ t = 2\,\text{s (going up)},\ t = 4\,\text{s (coming down)}\]

Bayanin Amsa

(a) Since \(r\) is parallel to \(m - n\):

\[m - n = (3i + 2j) - (i + j) = 2i + j\]

So \(r = \lambda(2i + j)\), giving \(\alpha = 2\lambda,\ \beta = \lambda\).

The magnitude of \(r\) is three times that of \(s = 2i - j\), where \(|s| = \sqrt{4 + 1} = \sqrt{5}\):

\[|r| = \sqrt{(2\lambda)^2 + \lambda^2} = |\lambda|\sqrt{5} = 3\sqrt{5} \ \Rightarrow\ |\lambda| = 3\]

Taking \(\lambda = 3\): \(\alpha = 6,\ \beta = 3\) (the other solution \(\lambda = -3\) gives \(\alpha = -6,\ \beta = -3\)).

(b) Using \(r = 6i + 3j\):

\[r - s = (6 - 2)i + (3 - (-1))j = 4i + 4j\] \[|r - s| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \approx 5.7\]

Direction: \(\tan\theta = \dfrac{4}{4} = 1\), so \(\theta = 45^\circ\) above the positive \(x\)-axis (bearing \(045^\circ\)).

Bayanin Amsa

(a) Value of k. The mean of \(1, 4, k, (k+4), 11\) is \((k+1)\):

\[ \frac{1 + 4 + k + (k+4) + 11}{5} = k + 1 \Rightarrow \frac{20 + 2k}{5} = k + 1. \]

\[ 20 + 2k = 5k + 5 \Rightarrow 15 = 3k \Rightarrow k = 5. \]

(b) Standard deviation. The numbers become \(1, 4, 5, 9, 11\) with mean \(k+1 = 6\).

Deviations from the mean: \(-5, -2, -1, 3, 5\); their squares: \(25, 4, 1, 9, 25\), summing to \(64\).

\[ \text{Variance} = \frac{64}{5} = 12.8, \qquad \text{S.D.} = \sqrt{12.8} \approx 3.58. \]