WAEC SSCE - Further Mathematics - 2018 (Essay)

Detalles de la respuesta

Let \(x\) = number of heads and \(f\) = frequency. First find \(N = \sum f\).

\(x\)	0	1	2	3	4	5	6	7	8	9	10	Total
\(f\)	2	7	23	36	11	61	100	12	8	5	3	268
\(fx\)	0	7	46	108	44	305	600	84	64	45	30	1333

(a) Mean number of heads.

\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{1333}{268} = \mathbf{4.974} \]

(b) Probability of an even number of heads. Even outcomes are \(x = 0,2,4,6,8,10\) with frequencies \(2+23+11+100+8+3 = 147\).

\[ P(\text{even}) = \frac{147}{268} = \mathbf{0.549} \]

(c) Probability of an odd number of heads. Odd outcomes are \(x = 1,3,5,7,9\) with frequencies \(7+36+61+12+5 = 121\).

\[ P(\text{odd}) = \frac{121}{268} = \mathbf{0.451} \]

Check: \(0.549 + 0.451 = 1.000\).

Detalles de la respuesta

Expand the determinant along the first row.

\[\Delta=(x-3)\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}-(-4)\begin{vmatrix}5&2\\2&6-x\end{vmatrix}+3\begin{vmatrix}5&2\\2&-4\end{vmatrix}\]

The \(2\times2\) minors:

\(\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}=2(6-x)+8=20-2x\).

\(\begin{vmatrix}5&2\\2&6-x\end{vmatrix}=5(6-x)-4=26-5x\).

\(\begin{vmatrix}5&2\\2&-4\end{vmatrix}=-20-4=-24\).

So

\[\Delta=(x-3)(20-2x)+4(26-5x)+3(-24)\]

\[=(-2x^{2}+26x-60)+(104-20x)-72=-2x^{2}+6x-28\]

Set \(\Delta=-24\):

\[-2x^{2}+6x-28=-24\Rightarrow-2x^{2}+6x-4=0\Rightarrow x^{2}-3x+2=0\]

\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2\]

Detalles de la respuesta

By conservation of linear momentum, total momentum before = total momentum after. After collision they move together with common velocity \(v\); total mass \(=20+30=50\,\text{kg}\).

(a) Same direction before collision. Take both velocities as positive:

\[20(80)+30(50)=50v\]

\[1600+1500=50v\Rightarrow3100=50v\Rightarrow v=62\ \text{m/s}\]

(b) Opposite directions before collision. Take the second body's velocity as negative:

\[20(80)+30(-50)=50v\]

\[1600-1500=50v\Rightarrow100=50v\Rightarrow v=2\ \text{m/s}\]

The positive result shows the combined body moves in the direction of the \(20\,\text{kg}\) body.

Detalles de la respuesta

Use the general circle \(x^2+y^2+2gx+2fy+c=0\), with centre \((-g,-f)\) and radius \(\sqrt{g^2+f^2-c}\). Substitute the three points.

\((3,2)\): \(6g+4f+c=-13\)  (i)

\((-1,-2)\): \(-2g-4f+c=-5\)  (ii)

\((5,-4)\): \(10g-8f+c=-41\)  (iii)

(i)\(-\)(ii): \(8g+8f=-8\Rightarrow g+f=-1\). (iii)\(-\)(i): \(4g-12f=-28\Rightarrow g-3f=-7\).

Solving: \(f=\tfrac32,\ g=-\tfrac52\). From (ii): \(5-6+c=-5\Rightarrow c=-4\).

(a) Centre \((-g,-f)=\left(\tfrac52,\,-\tfrac32\right)\).

(b) Radius \(=\sqrt{g^2+f^2-c}=\sqrt{\tfrac{25}{4}+\tfrac94+4}=\sqrt{\tfrac{50}{4}}=\dfrac{5\sqrt2}{2}\approx3.54\).

(c) Equation:

\[x^2+y^2-5x+3y-4=0\]

(Check with \((3,2)\): \(9+4-15+6-4=0\).)

Detalles de la respuesta

Note that \(\log_{x}3=\dfrac{1}{\log_{3}x}\). Let \(y=\log_{3}x\). The equation becomes

\[y-\frac{3}{y}+2=0\]

Multiply through by \(y\):

\[y^{2}+2y-3=0\Rightarrow(y+3)(y-1)=0\Rightarrow y=-3\ \text{or}\ y=1\]

Convert back with \(y=\log_{3}x\):

\[y=1\Rightarrow x=3^{1}=3,\qquad y=-3\Rightarrow x=3^{-3}=\frac{1}{27}\]

\[x=3\quad\text{or}\quad x=\frac{1}{27}\]

Detalles de la respuesta

Given \(m=3i-2j,\ n=2i-3j,\ p=-i+6j\).

\[4m=12i-8j,\qquad 2n=4i-6j,\qquad 3p=-3i+18j\]

\[4m+2n-3p=(12i-8j)+(4i-6j)-(-3i+18j)\]

Collect \(i\) terms: \(12+4+3=19\). Collect \(j\) terms: \(-8-6-18=-32\).

\[4m+2n-3p=19i-32j\]

Detalles de la respuesta

The values given by X and Y are already ranks (1 to 8), so \(n = 8\). Form \(d = X - Y\) and \(d^2\).

Oil	X	Y	\(d = X - Y\)	\(d^2\)
A	8	6	2	4
B	5	3	2	4
C	1	4	-3	9
D	7	8	-1	1
E	2	5	-3	9
F	6	7	-1	1
G	3	1	2	4
H	4	2	2	4
Total \(\sum d^2\)				36

\[ r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 36}{8(64 - 1)} = 1 - \frac{216}{504} = 1 - 0.5714 \approx \mathbf{0.57} \]

\(r_s \approx 0.57\) indicates a moderate positive agreement between the two panels of judges in ranking the brands of cooking oil.

Detalles de la respuesta

Position vectors: \(A=(8,-2),\ B=(2,6),\ C=(-10,4)\).

(a) In parallelogram \(ABCN\) the diagonals \(AC\) and \(BN\) bisect each other, so they share a midpoint.

Midpoint of \(AC=\left(\tfrac{8-10}{2},\tfrac{-2+4}{2}\right)=(-1,1)\).

Set midpoint of \(BN=(-1,1)\): \(\tfrac{2+N_x}{2}=-1\Rightarrow N_x=-4\); \(\tfrac{6+N_y}{2}=1\Rightarrow N_y=-4\).

\[N=-4\mathbf{i}-4\mathbf{j}\]

(b) \(\overrightarrow{AN}=N-A=(-4-8,\,-4+2)=(-12,-2)\); \(\overrightarrow{AB}=B-A=(2-8,\,6+2)=(-6,8)\).

(c) \(\overrightarrow{AN}\cdot\overrightarrow{AB}=(-12)(-6)+(-2)(8)=72-16=56\).

\(|\overrightarrow{AN}|=\sqrt{144+4}=\sqrt{148}\approx12.17\); \(|\overrightarrow{AB}|=\sqrt{36+64}=10\).

\[\cos\theta=\frac{56}{10\sqrt{148}}=0.4603\Rightarrow \theta\approx62.59^{\circ}\]

Detalles de la respuesta

Let the first term be \(a\) and common difference \(d\).

Sum of first 12 terms is 168:

\[S_{12}=\frac{12}{2}\big(2a+11d\big)=168\Rightarrow6(2a+11d)=168\Rightarrow2a+11d=28\quad(1)\]

Third term is 7:

\[a+2d=7\quad(2)\]

From (2), \(a=7-2d\). Substitute into (1):

\[2(7-2d)+11d=28\Rightarrow14+7d=28\Rightarrow7d=14\Rightarrow d=2\]

Then \(a=7-2(2)=3\).

\[a=3,\qquad d=2\]

Detalles de la respuesta

(a) Let \(u=2^{y}\). Then \(2^{3y+2}=4u^{3}\) and \(2^{2y+2}=4u^{2}\), so

\[4u^{3}-7(4u^{2})-31u-8=0\Rightarrow 4u^{3}-28u^{2}-31u-8=0\]

Testing \(u=8\): \(4(512)-28(64)-31(8)-8=2048-1792-248-8=0\). So \((u-8)\) is a factor:

\[(u-8)(4u^{2}+4u+1)=0\Rightarrow (u-8)(2u+1)^{2}=0\]

Thus \(u=8\) or \(u=-\tfrac12\). Since \(u=2^{y}>0\), reject \(u=-\tfrac12\). Then \(2^{y}=8=2^{3}\), so

\[\boxed{y=3}\]

(b) \(\displaystyle\int x\sqrt{x^{2}+1}\,dx\). Let \(w=x^{2}+1\), so \(dw=2x\,dx\Rightarrow x\,dx=\tfrac12 dw\).

\[\int\sqrt{w}\cdot\tfrac12\,dw=\tfrac12\cdot\tfrac{2}{3}w^{3/2}+C=\tfrac13\left(x^{2}+1\right)^{3/2}+C\]

Detalles de la respuesta

Admission probabilities: Ali \(P(A)=\tfrac23\), Baba \(P(B)=\tfrac34\), Katty \(P(K)=\tfrac45\). Failures: \(P(A')=\tfrac13,\ P(B')=\tfrac14,\ P(K')=\tfrac15\). The three events are independent.

(a) Only Katty and Baba (so Ali fails):

\[P(A')\,P(B)\,P(K)=\tfrac13\times\tfrac34\times\tfrac45=\frac{12}{60}=\frac15\]

(b) None gains admission:

\[P(A')\,P(B')\,P(K')=\tfrac13\times\tfrac14\times\tfrac15=\frac{1}{60}\]

(c) At most two = not all three. So subtract the probability that all three gain admission:

\[1-P(A)\,P(B)\,P(K)=1-\left(\tfrac23\times\tfrac34\times\tfrac45\right)=1-\frac{24}{60}=1-\frac25=\frac35\]

Detalles de la respuesta

(a) \(P(K)=\tfrac{1}{3}\Rightarrow P(K')=\tfrac{2}{3}\); \(P(T)=\tfrac{1}{5}\Rightarrow P(T')=\tfrac{4}{5}\). The events are independent.

Only one solves = (Kunle solves, Tayo fails) OR (Kunle fails, Tayo solves):

\[P=\left(\tfrac{1}{3}\times\tfrac{4}{5}\right)+\left(\tfrac{2}{3}\times\tfrac{1}{5}\right)=\frac{4}{15}+\frac{2}{15}=\frac{6}{15}=\frac{2}{5}\]

(b) Choosing 8 from 10 with no restriction:

\[\binom{10}{8}=\binom{10}{2}=\frac{10\times9}{2}=45\ \text{ways}\]

Detalles de la respuesta

(a) With \(u=x-2\), we have \(x=u+2\), so

\[x^{3}+5=(u+2)^{3}+5=u^{3}+6u^{2}+12u+8+5=u^{3}+6u^{2}+12u+13\]

Therefore

\[\frac{x^{3}+5}{(x-2)^{4}}=\frac{u^{3}+6u^{2}+12u+13}{u^{4}}=\frac{1}{u}+\frac{6}{u^{2}}+\frac{12}{u^{3}}+\frac{13}{u^{4}}\]

(b) Replace \(u\) by \(x-2\):

\[\frac{x^{3}+5}{(x-2)^{4}}=\frac{1}{x-2}+\frac{6}{(x-2)^{2}}+\frac{12}{(x-2)^{3}}+\frac{13}{(x-2)^{4}}\]

Detalles de la respuesta

Measure positions from \(X\): \(X=0,\ P=0.8\text{m},\ Q=0.8+2.2=3.0\text{m},\ Y=4.0\text{m}\). The uniform weight \(350\text{N}\) acts at the centre, \(2.0\text{m}\). The \(80\text{N}\) acts at \(X=0\); the \(1000\text{N}\) is \(0.6\text{m}\) from \(Q\), i.e. at \(2.4\text{m}\).

(a) Vertical equilibrium: \(R_P+R_Q=80+350+1000=1430\).

Take moments about \(P\) (anticlockwise positive):

\[80(0.8)-350(1.2)-1000(1.6)+R_Q(2.2)=0\]

\[64-420-1600+2.2R_Q=0\Rightarrow R_Q=\frac{1956}{2.2}=889.09\text{N}\]

Then \(R_P=1430-889.09=540.91\text{N}\).

(b) Replace with \(1200\text{N}\) at distance \(s\) from \(Q\) (position \(3.0-s\)). Keeping the reaction at \(P\) unchanged at \(540.91\text{N}\), take moments about \(Q\):

\[R_P(2.2)=80(3.0)+350(1.0)+1200\,s\]

\[540.91(2.2)=240+350+1200s\Rightarrow 1190=590+1200s\Rightarrow s=0.5\text{m}\]

So the \(1200\text{N}\) weight should be placed \(0.5\text{m}\) from \(Q\) (between \(P\) and \(Q\)).

Detalles de la respuesta

(a)(i) Expand \(\left(2-\tfrac12 x\right)^{5}\) using \(\binom{5}{k}2^{5-k}\left(-\tfrac12 x\right)^{k}\):

\[32-40x+20x^{2}-5x^{3}+\tfrac58 x^{4}-\tfrac{1}{32}x^{5}\]

(ii) Put \(2-\tfrac12 x=1.99\Rightarrow \tfrac12 x=0.01\Rightarrow x=0.02\). Substitute:

\(32-40(0.02)+20(0.02)^{2}-5(0.02)^{3}+\cdots\)

\(=32-0.8+0.008-0.00004+\cdots\approx31.20796\).

\[(1.99)^{5}\approx31.21\ \text{(2 d.p.)}\]

(b) Let \(P(x)=x^{3}+qx^{2}+rx+9\).

\((x+1)\) is a factor \(\Rightarrow P(-1)=0\): \(-1+q-r+9=0\Rightarrow q-r=-8\).

Remainder \(-17\) on division by \((x+2)\Rightarrow P(-2)=-17\): \(-8+4q-2r+9=-17\Rightarrow 4q-2r=-18\Rightarrow 2q-r=-9\).

Subtracting: \((2q-r)-(q-r)=-9-(-8)\Rightarrow q=-1\), then \(r=q+8=7\).

\[\boxed{q=-1,\ r=7}\]