WAEC SSCE - Further Mathematics - 2021 (Essay)

Detalhes da Resposta

Reading the polynomial as \(f(x)=2x^{3}+px^{2}+qx-5\) (the two linear terms in the printed stem are a typographical slip for a quadratic and a linear term).

Since \((x-1)\) is a factor, \(f(1)=0\):

\[2+p+q-5=0\Rightarrow p+q=3\quad(\text{i})\]

The remainder is \(27\) when divided by \((x+2)\), so \(f(-2)=27\):

\[2(-8)+p(4)+q(-2)-5=27\Rightarrow -16+4p-2q-5=27\]

\[4p-2q=48\Rightarrow 2p-q=24\quad(\text{ii})\]

Add (i) and (ii): \(3p=27\Rightarrow p=9\). Then \(q=3-9=-6\).

\[\boxed{p=9,\ q=-6}\]

Detalhes da Resposta

Detalhes da Resposta

Because selection is with replacement, each draw is independent with a constant probability of a bad mango:

\[p=\frac{6}{24}=\frac14,\qquad q=1-p=\frac34\]

Let \(X\) be the number of bad mangoes in \(n=6\) draws; \(X\sim\text{Bin}(6,\tfrac14)\). We need \(P(X\le3)\), which is easiest via the complement \(P(X\le3)=1-P(X\ge4)\).

\[P(4)=\binom{6}{4}\left(\tfrac14\right)^{4}\left(\tfrac34\right)^{2}=\frac{15\times9}{4096}=\frac{135}{4096}\]

\[P(5)=\binom{6}{5}\left(\tfrac14\right)^{5}\left(\tfrac34\right)=\frac{6\times3}{4096}=\frac{18}{4096}\]

\[P(6)=\left(\tfrac14\right)^{6}=\frac{1}{4096}\]

\[P(X\ge4)=\frac{135+18+1}{4096}=\frac{154}{4096}=\frac{77}{2048}\]

\[P(X\le3)=1-\frac{77}{2048}=\frac{1971}{2048}\approx0.9624\]

Detalhes da Resposta

Simplify the integrand first. Since \(\sqrt{x}=x^{1/2}\):

\[\frac{x(2x-3)}{\sqrt{x}}=\frac{2x^{2}-3x}{x^{1/2}}=2x^{3/2}-3x^{1/2}\]

Integrate:

\[\int_{1}^{9}\left(2x^{3/2}-3x^{1/2}\right)dx=\left[\frac{4}{5}x^{5/2}-2x^{3/2}\right]_{1}^{9}\]

At \(x=9\): \(9^{5/2}=3^{5}=243\) and \(9^{3/2}=27\), giving \(\tfrac45(243)-2(27)=194.4-54=140.4\).

At \(x=1\): \(\tfrac45-2=-1.2\).

\[140.4-(-1.2)=141.6=\frac{708}{5}\]

Detalhes da Resposta

(a) Horizontal projectile from a cliff

Horizontally the stone travels at a constant \(30\,\text{m/s}\); vertically it starts with zero vertical velocity and falls under gravity. First find the time of flight from the vertical motion, using \(h=\tfrac{1}{2}gt^{2}\):

\[50=\tfrac{1}{2}(10)t^{2}\;\Rightarrow\; 50=5t^{2}\;\Rightarrow\; t^{2}=10\;\Rightarrow\; t=\sqrt{10}\approx 3.16\,\text{s}.\]

The horizontal distance (range) is

\[R=\text{(horizontal speed)}\times t=30\sqrt{10}\approx 94.9\,\text{m}.\]

The stone lands about \(94.9\,\text{m}\) from the foot of the cliff.

(b) Body in equilibrium on two strings

The weight of the body is \(W=mg=2\times 10=20\,\text{N}\), acting vertically downward. String \(AP\) (tension \(T_1\)) makes \(56^{\circ}\) with the upward vertical, and string \(AR\) (tension \(T_2\)) is horizontal. Resolve the forces at \(A\).

Vertical equilibrium: only \(T_1\) has a vertical component, and it supports the weight:

\[T_1\cos 56^{\circ}=20\;\Rightarrow\; T_1=\frac{20}{\cos 56^{\circ}}=\frac{20}{0.5592}\approx 35.8\,\text{N}.\]

Horizontal equilibrium: the horizontal component of \(T_1\) is balanced by \(T_2\):

\[T_2=T_1\sin 56^{\circ}=20\tan 56^{\circ}=20(1.4826)\approx 29.7\,\text{N}.\]

Hence \(T_1\approx 35.8\,\text{N}\) and \(T_2\approx 29.7\,\text{N}\).

Detalhes da Resposta

Reading the expression as \(\left(p+\tfrac12\sqrt3\right)(1-\sqrt3)^{2}=3-\sqrt3\).

First expand the square:

\[(1-\sqrt3)^{2}=1-2\sqrt3+3=4-2\sqrt3\]

So:

\[\left(p+\tfrac{\sqrt3}{2}\right)(4-2\sqrt3)=3-\sqrt3\]

Note \(4-2\sqrt3=2(2-\sqrt3)\). Divide both sides:

\[p+\frac{\sqrt3}{2}=\frac{3-\sqrt3}{4-2\sqrt3}\]

Rationalise the right side by \(\times\dfrac{4+2\sqrt3}{4+2\sqrt3}\): denominator \(16-12=4\); numerator \((3-\sqrt3)(4+2\sqrt3)=12+2\sqrt3-6=6+2\sqrt3\). So the right side \(=\dfrac{6+2\sqrt3}{4}=\dfrac{3+\sqrt3}{2}\).

\[p=\frac{3+\sqrt3}{2}-\frac{\sqrt3}{2}=\frac{3}{2}\]

\[\boxed{p=\tfrac32}\]

Detalhes da Resposta

(a) The curve \(y=(x^{2}-x+1)(x-2)\) cuts the x-axis where \(y=0\). Since \(x^{2}-x+1=0\) has discriminant \(1-4<0\) (no real roots), the only x-intercept is \(x-2=0\), i.e. the point \((2,0)\).

Expand: \(y=x^{3}-3x^{2}+3x-2\), so \(\dfrac{dy}{dx}=3x^{2}-6x+3\). At \(x=2\): \(3(4)-12+3=3\). This is the tangent gradient, so the normal gradient is \(-\tfrac13\).

Normal at \((2,0)\): \(y-0=-\tfrac13(x-2)\), i.e.

\[x+3y-2=0\]

(b) \(P(-1,2),\ Q(5,1),\ R(3,-4)\). Midpoint of \(\overline{PR}=\left(\tfrac{-1+3}{2},\tfrac{2-4}{2}\right)=(1,-1)\).

Line through \(Q(5,1)\) and \((1,-1)\): gradient \(=\dfrac{1-(-1)}{5-1}=\dfrac{2}{4}=\tfrac12\).

\[y-1=\tfrac12(x-5)\Rightarrow x-2y-3=0\]

Detalhes da Resposta

(a) Reading the coefficients of \(x\) and \(y\) from each mapping:

\[P=\begin{pmatrix}-3&6\\4&1\end{pmatrix},\qquad Q=\begin{pmatrix}2&-3\\-4&-6\end{pmatrix}\]

(b) Image of \((-2,-3)\) under \(Q\): \((2(-2)-3(-3),\ -4(-2)-6(-3))=(-4+9,\ 8+18)=(5,26)\).

(c) "Q followed by P" means \(P\) acts second, so the single matrix is \(PQ\):

\[PQ=\begin{pmatrix}-3&6\\4&1\end{pmatrix}\begin{pmatrix}2&-3\\-4&-6\end{pmatrix}=\begin{pmatrix}-30&-27\\4&-18\end{pmatrix}\]

i.e. \((x,y)\to(-30x-27y,\ 4x-18y)\).

(d) Image of \((1,4)\): \((-30(1)-27(4),\ 4(1)-18(4))=(-30-108,\ 4-72)=(-138,-68)\).

(e) Anticlockwise rotation of \(225^{\circ}\): \(\begin{pmatrix}\cos225^{\circ}&-\sin225^{\circ}\\\sin225^{\circ}&\cos225^{\circ}\end{pmatrix}=\begin{pmatrix}-\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}&-\frac{\sqrt2}{2}\end{pmatrix}\).

Apply to \((-\sqrt2,\ 2\sqrt2)\):

\(x'=-\tfrac{\sqrt2}{2}(-\sqrt2)+\tfrac{\sqrt2}{2}(2\sqrt2)=1+2=3\)

\(y'=-\tfrac{\sqrt2}{2}(-\sqrt2)-\tfrac{\sqrt2}{2}(2\sqrt2)=1-2=-1\)

So \(P'=(3,-1)\).

Detalhes da Resposta

(a) Daily distances form an AP: first term \(a=500\text{m}\), common difference \(d=250\text{m}\).

(i) The daily run reaches \(15\text{km}=15000\text{m}\) when the \(n\)th term equals \(15000\):

\[a+(n-1)d=15000\Rightarrow 500+250(n-1)=15000\]

\[250(n-1)=14500\Rightarrow n-1=58\Rightarrow n=59\text{ days}\]

(ii) Total distance is the sum of 59 terms:

\[S_{59}=\frac{n}{2}(a+l)=\frac{59}{2}(500+15000)=\frac{59}{2}(15500)=456250\text{ m}=456.25\text{ km}\]

(b) GP: second term \(ar=-3\); sum to infinity \(\dfrac{a}{1-r}=\dfrac{25}{2}\).

From \(ar=-3\), \(a=-\dfrac{3}{r}\). Substitute:

\[\frac{-3/r}{1-r}=\frac{25}{2}\Rightarrow -6=25r(1-r)\Rightarrow 25r^{2}-25r-6=0\]

\[r=\frac{25\pm\sqrt{625+600}}{50}=\frac{25\pm35}{50}\Rightarrow r=\tfrac65\ \text{or}\ r=-\tfrac15\]

A sum to infinity requires \(|r|<1\), so \(\boxed{r=-\tfrac15}\) (reject \(\tfrac65\)).

Detalhes da Resposta

(a) Uniform retardation, so use \(v^{2}=u^{2}-2as\) with \(u=45,\ v=5,\ a=10\):

\[5^{2}=45^{2}-2(10)s\Rightarrow 25=2025-20s\]

\[20s=2000\Rightarrow s=100\text{ m}\]

(b) The bucket is pulled upward, so applying Newton's second law upward with \(m=16\text{kg},\ T=240\text{N},\ g=10\text{m/s}^{2}\):

\[T-mg=ma\Rightarrow 240-16(10)=16a\]

\[80=16a\Rightarrow a=5\text{ m/s}^{2}\ (\text{upward})\]

Detalhes da Resposta

Write the position vectors as coordinates: \(P(4,-5)\), \(Q(1,3)\), \(R(-5,2)\).

(a) Coordinates of M

In the parallelogram \(PQRM\) the vertices are taken in order \(P\to Q\to R\to M\), so the diagonals \(PR\) and \(QM\) bisect each other. Equating their midpoints:

\[\text{mid}(PR)=\left(\tfrac{4+(-5)}{2},\tfrac{-5+2}{2}\right)=\left(-\tfrac{1}{2},-\tfrac{3}{2}\right).\]

\[\text{mid}(QM)=\left(\tfrac{1+m_1}{2},\tfrac{3+m_2}{2}\right).\]

So \(1+m_1=-1\Rightarrow m_1=-2\) and \(3+m_2=-3\Rightarrow m_2=-6\).

\[\boxed{M(-2,-6)}.\]

(b) Acute angle between \(\overline{PM}\) and \(\overline{PQ}\)

\[\overline{PM}=M-P=(-2-4,\,-6-(-5))=(-6,-1),\]

\[\overline{PQ}=Q-P=(1-4,\,3-(-5))=(-3,8).\]

Using the scalar (dot) product,

\[\overline{PM}\cdot\overline{PQ}=(-6)(-3)+(-1)(8)=18-8=10,\]

\[|\overline{PM}|=\sqrt{(-6)^2+(-1)^2}=\sqrt{37},\qquad |\overline{PQ}|=\sqrt{(-3)^2+8^2}=\sqrt{73}.\]

\[\cos\theta=\frac{10}{\sqrt{37}\,\sqrt{73}}=\frac{10}{\sqrt{2701}}\approx 0.1924.\]

\[\theta=\cos^{-1}(0.1924)\approx 78.9^{\circ}\approx 79^{\circ}.\]

Since this value is already acute, the acute angle is \(79^{\circ}\).

Detalhes da Resposta

Use \({}^{n}C_{2}=\dfrac{n(n-1)}{2}\) with \(n=5y\):

\[{}^{5y}C_{2}=\frac{5y(5y-1)}{2}=190\]

So \(5y(5y-1)=380\). Let \(n=5y\):

\[n^{2}-n-380=0\Rightarrow n=\frac{1\pm\sqrt{1+1520}}{2}=\frac{1\pm39}{2}\]

Taking the positive root, \(n=20\). Since \(n=5y\):

\[5y=20\Rightarrow \boxed{y=4}\]

(The negative root \(n=-19\) is rejected.)

Detalhes da Resposta

Use \(\cos\theta=\dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|\,|\mathbf{y}|}\) with \(\mathbf{x}=\begin{pmatrix}-4\\3\end{pmatrix},\ \mathbf{y}=\begin{pmatrix}-9\\15\end{pmatrix}\).

Dot product: \(\mathbf{x}\cdot\mathbf{y}=(-4)(-9)+(3)(15)=36+45=81\).

Magnitudes: \(|\mathbf{x}|=\sqrt{(-4)^{2}+3^{2}}=\sqrt{25}=5\); \(|\mathbf{y}|=\sqrt{(-9)^{2}+15^{2}}=\sqrt{306}\approx17.49\).

\[\cos\theta=\frac{81}{5\times17.49}=\frac{81}{87.46}=0.9261\]

\[\theta=\cos^{-1}(0.9261)\approx22^{\circ}\ \text{(nearest degree)}\]

Detalhes da Resposta

(a)(i) Cumulative frequency table (cf against upper boundaries). \(N = 400\).

Height	Freq	Upper boundary	Cumulative freq
100 - 109	27	109.5	27
110 - 119	58	119.5	85
120 - 129	130	129.5	215
130 - 139	105	139.5	320
140 - 149	50	149.5	370
150 - 159	25	159.5	395
160 - 169	5	169.5	400

(a)(ii) Plot (109.5, 27), (119.5, 85), ... , (169.5, 400) and join with a smooth ogive.

(b)(i) Median at position \( \tfrac{400}{2} = 200\), in class 120 - 129 (\(L=119.5, \text{cf}=85, f=130\)):

\[ \text{Median} = 119.5 + \frac{200 - 85}{130}\times 10 = 119.5 + 8.85 \approx \mathbf{128.3 \text{ cm}} \]

(b)(ii) Interquartile range. \(Q_1\) at 100th (class 120 - 129):

\[ Q_1 = 119.5 + \frac{100 - 85}{130}\times 10 = 119.5 + 1.15 = 120.65 \]

\(Q_3\) at 300th (class 130 - 139, \(L=129.5, \text{cf}=215, f=105\)):

\[ Q_3 = 129.5 + \frac{300 - 215}{105}\times 10 = 129.5 + 8.10 = 137.60 \] \[ \text{IQR} = Q_3 - Q_1 = 137.60 - 120.65 \approx \mathbf{17.0 \text{ cm}} \]

(b)(iii) Percentage with height at most 130 cm. Read cf at 130 (class 130 - 139):

\[ \text{cf}(130) = 215 + \frac{130 - 129.5}{10}\times 105 = 215 + 5.25 = 220.25 \] \[ \text{Percentage} = \frac{220.25}{400}\times 100 \approx \mathbf{55\%} \]

Detalhes da Resposta

The box holds \(5\) red, \(7\) blue and \(4\) green bulbs, giving \(5+7+4=16\) bulbs. Two bulbs are drawn without replacement, so the number of equally likely selections is

\[\binom{16}{2}=\frac{16\times 15}{2}=120.\]

(a) Same colour

Count the same-colour pairs for each colour and add:

• Both red: \(\binom{5}{2}=10\)
• Both blue: \(\binom{7}{2}=21\)
• Both green: \(\binom{4}{2}=6\)

Favourable outcomes \(=10+21+6=37\), so

\[P(\text{same colour})=\frac{37}{120}.\]

(b) Different colours

Every draw is either the two bulbs matching or not, so "different colours" is the complement of "same colour":

\[P(\text{different})=1-\frac{37}{120}=\frac{83}{120}.\]

(c) At least one red

Use the complement "no red bulb". There are \(16-5=11\) non-red bulbs, giving \(\binom{11}{2}=\frac{11\times 10}{2}=55\) all-non-red pairs. Hence

\[P(\text{no red})=\frac{55}{120}=\frac{11}{24},\qquad P(\text{at least one red})=1-\frac{11}{24}=\frac{13}{24}.\]