NECO · 2023 · SSCE · Further Maths · Paper 2

(a)

1. Draw two overlapping circles $X$ and $Y$ inside $U$: $g$ in the overlap, $e$ in $X$ only, $h$ in $Y$ only, and $f$ and $i$ outside both.

(b)(i)

1. $X'$ is everything not in $X$: $\{f, h, i\}$.

(ii)

1. $X \cup Y = \{e, g, h\}$, so $(X \cup Y)' = \{f, i\}$.

(iii)

1. $Y' = \{e, f, i\}$, so $X' \cap Y' = \{f, i\}$: the same as (ii), as De Morgan's law says.

(a)

1. The radius is the distance from the centre $(4, -5)$ to the point $(-3, 2)$: $r^2 = (4 - (-3))^2 + (-5 - 2)^2$.
2. So $r^2 = 7^2 + (-7)^2 = 49 + 49 = 98$.
3. The circle with centre $(a, b)$ and radius $r$ is $(x - a)^2 + (y - b)^2 = r^2$.
4. Here: $(x - 4)^2 + (y + 5)^2 = 98$.
5. Expand the brackets: $x^2 - 8x + 16 + y^2 + 10y + 25 = 98$.
6. Collect everything on one side: $x^2 + y^2 - 8x + 10y - 57 = 0$.

(b)

1. Write the root as a power: $\sqrt[3]{x^2} = x^{\frac23}$.
2. Dividing, $x^{-\frac23} \div x^{\frac23} = x^{-\frac23 - \frac23}$
   $= x^{-\frac43}$.
3. Dividing by $\frac{1}{x^2}$ is multiplying by $x^2$: $x^{-\frac43} \times x^2 = x^{-\frac43 + 2}$
   $= x^{\frac23}$.
4. Raise to the power $-3$ by multiplying the indices: $\left(x^{\frac23}\right)^{-3} = x^{-2}$, that is $\dfrac{1}{x^2}$.

(i)

1. $\mathbf x \cdot \mathbf y = (2)(4) + (-5)(3)$
   $= 8 - 15$
   $= -7$.
2. $|\mathbf x| = \sqrt{4 + 25} = \sqrt{29}$ and $|\mathbf y| = \sqrt{16 + 9} = 5$.
3. $\cos\theta = \dfrac{-7}{5\sqrt{29}}$
   $\approx -0.2600$, so $\theta \approx 105.1^\circ$.

(ii)

1. $5\mathbf x + 2\mathbf y = (10 + 8)\mathbf i + (-25 + 6)\mathbf j$
   $= 18\mathbf i - 19\mathbf j$.
2. Its length is $\sqrt{324 + 361} = \sqrt{685}$.
3. Unit vector: $\dfrac{1}{\sqrt{685}}(18\mathbf i - 19\mathbf j)$.

1. Seven ordinates means six strips, so $h = \dfrac{5 - 2}{6} = 0.5$.
2. The ordinates of $(2x + 3)^2$ at $x = 2, 2.5, \ldots, 5$ are $49, 64, 81, 100, 121, 144, 169$.
3. First and last: $49 + 169 = 218$.
4. Twice the rest: $2(64 + 81 + 100 + 121 + 144) = 1020$.
5. Trapezium rule: $\dfrac{0.5}{2}(218 + 1020) = 0.25 \times 1238$
   $= 309.5$.
6. (The exact value is $309\frac13$: the curve bends upwards, so the rule overestimates slightly.)

(a)

1. Choose 5 men from 10 and 3 women from 8, and multiply: $\,{}^{10}C_5 \times {}^8C_3$.
2. $= 252 \times 56 = 14\,112$ ways.

(b)

1. Put the principal on.
2. Choose the other 4 men from the remaining 9: $\,{}^9C_4 = 126$.
3. The women as before: $\,{}^8C_3 = 56$.
4. So $126 \times 56 = 7056$ ways.

1. Class marks: $5.5, 15.5, 25.5, 35.5, 45.5$.
2. Take $A = 25.5$, so $d = -20, -10, 0, 10, 20$.
3. $\sum fd = 9(-20) + 49(-10) + 0 + 8(10) + 2(20)$
   $= -550$.
4. $\sum fd^2 = 9(400) + 49(100) + 0 + 8(100) + 2(400)$
   $= 10\,100$.
5. $\sigma = \sqrt{\dfrac{10\,100}{100} - \left(\dfrac{-550}{100}\right)^2}$
   $= \sqrt{101 - 30.25}$
   $= \sqrt{70.75}$
   $\approx 8.41$.

1. East parts: $204\sin30^\circ + 300\sin90^\circ + 225\sin120^\circ = 102 + 300 + 194.86$
   $= 596.86$.
2. North parts: $204\cos30^\circ + 300\cos90^\circ + 225\cos120^\circ = 176.67 + 0 - 112.5$
   $= 64.17$.
3. $R = \sqrt{596.86^2 + 64.17^2}$
   $= \sqrt{360\,360}$
   $\approx 600\text{ N}$.

(i)

1. Start in the top-left cell and send as much as possible each time.
2. $P_1$ (8): 6 to $A_1$, which is then full, and its last 2 to $A_2$.
3. $P_2$ (10): 6 to $A_2$, which is then full, and 4 to $A_3$.
4. $P_3$ (19): 4 to $A_3$, which is then full, and 15 to $A_4$.

(ii)

1. Cost: $6(19) + 2(30) + 6(30) + 4(40) + 4(70) + 15(20)$.
2. $= 114 + 60 + 180 + 160 + 280 + 300 = 1094$ (hundreds of naira).

(a)

1. Divide by 3: $x^2 + y^2 + 4x - 2y - 15 = 0$, so $g = 2$ and $f = -1$.
2. The centre is $(-g, -f) = (-2, 1)$ and $r = \sqrt{4 + 1 + 15}$
   $= \sqrt{20}$
   $= 2\sqrt5$.

(b)

1. The radius to $(2, 3)$ has gradient $\dfrac{3 - 1}{2 + 2} = \dfrac12$.
2. The tangent is perpendicular to it, with gradient $-2$: $y - 3 = -2(x - 2)$, so $y = -2x + 7$.

(c)

1. On the circle, $x = -2 + 2\sqrt5\cos\theta$ and $y = 1 + 2\sqrt5\sin\theta$.
2. At $(2, 3)$: $\cos\theta = \dfrac{4}{2\sqrt5}$
   $= \dfrac{2}{\sqrt5}$ and $\sin\theta = \dfrac{2}{2\sqrt5}$
   $= \dfrac{1}{\sqrt5}$.
3. So $\tan\theta = \frac12$ with both positive: $\theta \approx 26.57^\circ$.

(a)

1. The top has degree 3 and the bottom, $(x - 1)(x + 2) = x^2 + x - 2$, has degree 2, so divide first.
2. $2x$ times the bottom is $2x^3 + 2x^2 - 4x$.
3. Take it away: $3x^2 - 2x + 4$ is left.
4. $3$ times the bottom is $3x^2 + 3x - 6$.
5. Take it away: $-5x + 10$ is left.
6. So the fraction is $2x + 3 + \dfrac{10 - 5x}{(x - 1)(x + 2)}$.
7. Split the remainder: $10 - 5x = A(x + 2) + B(x - 1)$ for $\dfrac{A}{x - 1} + \dfrac{B}{x + 2}$.
8. Put $x = 1$: $5 = 3A$, so $A = \frac53$.
9. Put $x = -2$: $20 = -3B$, so $B = -\frac{20}{3}$.
10. So the answer is $2x + 3 + \dfrac{5}{3(x - 1)} - \dfrac{20}{3(x + 2)}$.

(b)

1. Let $B$ play basketball and $T$ play table tennis.
2. The 3 who play both are counted in both, so $T + B - 3 = 60$, which gives $T + B = 63$.
3. Also $T = 3B + 7$.
4. Substitute: $3B + 7 + B = 63$, so $4B = 56$ and $B = 14$.
5. Then $T = 3(14) + 7 = 49$.
6. So 49 students play table tennis.

(c)

1. The identity $e$ satisfies $p * e = p$: $\dfrac{3p + 3e - 5}{3} = p$.
2. Multiply by 3: $3p + 3e - 5 = 3p$.
3. Take $3p$ from both sides: $3e = 5$, so $e = \frac53$.

(a)

1. Put the left side over one denominator: $\dfrac{(x - 1) + 2(x + 1)}{(x + 1)(x - 1)} = \dfrac{3x + 1}{x^2 - 1}$.
2. So $\dfrac{3x + 1}{x^2 - 1} = \dfrac{1}{x + 3}$.
3. Cross-multiply: $(3x + 1)(x + 3) = x^2 - 1$.
4. Expand the left side: $3x^2 + 10x + 3 = x^2 - 1$.
5. Take $x^2 - 1$ from both sides: $2x^2 + 10x + 4 = 0$.
6. Divide by 2: $x^2 + 5x + 2 = 0$.
7. Use the formula: $x = \dfrac{-5 \pm \sqrt{25 - 8}}{2}$
   $= \dfrac{-5 \pm \sqrt{17}}{2}$.
8. With $\sqrt{17} \approx 4.1231$: $x \approx -0.44$ or $x \approx -4.56$.
9. Neither is $-1$, $1$ or $-3$, so both are allowed.

(b)

1. Divide by $x - 2$ with synthetic division: write 2 on the left and the coefficients $2, -9, -21, 88, 48$.
2. Bring down 2.
3. Then $2 \times 2 = 4$ and $-9 + 4 = -5$.
4. $-5 \times 2 = -10$ and $-21 - 10 = -31$.
5. Next $-31 \times 2 = -62$ and $88 - 62 = 26$.
6. Then $26 \times 2 = 52$ and $48 + 52 = 100$.
7. So the quotient is $2x^3 - 5x^2 - 31x + 26$ and the remainder is 100.
8. Check: $f(2) = 32 - 72 - 84 + 176 + 48 = 100$ ✓.

(c)

1. Substitute $x = 2$ term by term: $p(2) = 32 + 5(16) + 9(8) + 11(4) + 12(2) + 13$.
2. So $p(2) = 32 + 80 + 72 + 44 + 24 + 13 = 265$.
3. Then $3p(2) = 3 \times 265 = 795$.

1. $X(10, 1)$ and $Y(-2, 7)$.
2. $\mathbf r = \mathbf p + 3\mathbf q$
   $= (10 - 6, 1 + 21)$, so $Z(4, 22)$.

(i)

1. Gradient of $XY$: $\dfrac{7 - 1}{-2 - 10} = -\dfrac12$.
2. So $y - 1 = -\frac12(x - 10)$, which gives $x + 2y = 12$.
3. Gradient of $YZ$: $\dfrac{22 - 7}{4 + 2} = \dfrac52$.
4. So $y - 7 = \frac52(x + 2)$, which gives $2y = 5x + 24$.

(ii)

1. $\overrightarrow{OK} = \alpha(4\mathbf i + 22\mathbf j)$, so $K = (4\alpha, 22\alpha)$.
2. $K$ is on $XY$: $4\alpha + 2(22\alpha) = 12$, so $48\alpha = 12$ and $\alpha = \frac14$.
3. $\overrightarrow{XK} = (1 - 10, 5.5 - 1)$
   $= (-9, 4.5)$ and $\overrightarrow{XY} = (-12, 6)$.
4. $(-9, 4.5) = \frac34(-12, 6)$, so $\beta = \frac34$.

(iii)

1. $K = \frac14(4, 22) = (1, 5\frac12)$.

1. Times in seconds: $4\text{ min} = 240\text{ s}$, $30\text{ min} = 1800\text{ s}$ and $3\text{ min} = 180\text{ s}$.

(a)

1. Distance = area under the velocity–time graph.
2. $\frac12(240)(35) + 1800(35) + \frac12(180)(35) = 4200 + 63\,000 + 3150$
   $= 70\,350\text{ m}$, which is $70.35\text{ km}$.

(b)

1. Average speed $= \dfrac{70\,350}{240 + 1800 + 180}$
   $= \dfrac{70\,350}{2220}$
   $\approx 31.69\text{ m s}^{-1}$.

(c)

1. Acceleration $= \dfrac{35}{240}$
   $\approx 0.1458\text{ m s}^{-2}$.
2. Retardation $= \dfrac{35}{180}$
   $\approx 0.1944\text{ m s}^{-2}$.

(d)

1. Halfway is $35\,175\text{ m}$.
2. The first $4200\text{ m}$ take $240\text{ s}$.
3. The rest, $30\,975\text{ m}$, is at $35\text{ m s}^{-1}$: $\dfrac{30\,975}{35} = 885\text{ s}$.
4. So $C$ is reached after $240 + 885 = 1125\text{ s}$ (18 min 45 s).

(a)(i)

1. At least one tail is everything except four heads: $1 - \left(\frac12\right)^4 = \frac{15}{16}$.

(ii)

1. Two heads and two tails: $\binom42\left(\frac12\right)^4 = \frac{6}{16}$
   $= \frac38$.

(b)

1. The holding cost per unit is $10\%$ of ₦2.00, which is ₦0.20.
2. The economic order quantity is $\sqrt{\dfrac{2 \times 800 \times 5}{0.2}} = \sqrt{40\,000}$
   $= 200$ units.
3. Ordering cost: $\dfrac{800}{200} \times 5 = ₦20$.
4. Holding cost: $\dfrac{200}{2} \times 0.20 = ₦20$.
5. Total variable cost: $₦20 + ₦20 = ₦40.00$ per annum.

(a)(i)

1. Row minima: 3 and 1, so the maximin value is 3.

(ii)

1. Column maxima: 5 and 4, so the minimax value is 4.
2. They differ, so there is no saddle point and mixed strategies are needed.

(iii)

1. Let $A$ play row 1 with probability $p$: $3p + 5(1 - p) = 4p + 1(1 - p)$, so $5 - 2p = 1 + 3p$ and $p = \frac45$.
2. $A$ plays $\left(\frac45, \frac15\right)$.
3. Let $B$ play column 1 with probability $q$: $3q + 4(1 - q) = 5q + 1(1 - q)$, so $4 - q = 1 + 4q$ and $q = \frac35$.
4. $B$ plays $\left(\frac35, \frac25\right)$.

(iv)

1. Value $= 3\left(\frac45\right) + 5\left(\frac15\right)$
   $= \frac{17}{5}$
   $= 3.4$.

(b)

1. $\sigma = \sqrt{25} = 5$.
2. $z_1 = \dfrac{80 - 72}{5} = 1.6$ and $z_2 = \dfrac{84 - 72}{5} = 2.4$.
3. $P = \Phi(2.4) - \Phi(1.6)$
   $= 0.9918 - 0.9452$
   $= 0.0466$, so about $0.0466 \times 5000 \approx 233$ students.