WAEC · 2019 · Private, 2nd series · Further Maths · Paper 2

1. A five-digit number from these digits is always greater than 5000.
2. A four-digit one must start with 5 or 6.
3. Every number must end in 2, 4 or 6.
4. Four digits starting with 5: 3 choices for the last digit, then $3 \times 2 = 6$ for the middle two: $18$.
5. Four digits starting with 6: 2 choices for the last digit (2 or 4), then $3 \times 2 = 6$: $12$.
6. Five digits: 3 choices for the last digit, then $4! = 24$ orders for the rest: $72$.
7. Total: $18 + 12 + 72 = 102$ even numbers.

1. The fourth term: $ar^3 = 54$.
2. The sixth term: $ar^5 = 486$.
3. Divide: $r^2 = \dfrac{486}{54} = 9$, so $r = 3$ (the ratio is positive).
4. Then $a = \dfrac{54}{27} = 2$.
5. The third term is $ar^2 = 2 \times 9 = 18$.

(a)

1. The identity $e$ satisfies $x * e = x$: $x + e - 3xe = x$.
2. Take $x$ from both sides: $e - 3xe = 0$, so $e(1 - 3x) = 0$.
3. This must hold for every $x$, so $e = 0$.
4. Check: $x * 0 = x + 0 - 0 = x$ ✓.

(b)

1. Make the $n^2$ term positive: multiply by $-1$ and turn the sign round: $5n^2 - 14n - 3 \ge 0$.
2. Factorise: two numbers that multiply to $5 \times (-3) = -15$ and add to $-14$ are $-15$ and 1.
3. So $5n^2 - 15n + n - 3 = 5n(n - 3) + 1(n - 3)$
   $= (5n + 1)(n - 3) \ge 0$.
4. The roots are $n = -\frac15$ and $n = 3$. "$\ge 0$" is outside them: $n \le -\frac15$ or $n \ge 3$.
5. Check with $n = 0$, between the roots: $3 + 0 - 0 = 3$, which is not $\le 0$ ✓.

1. $y + \delta y = 3(x + \delta x)^2 + 4(x + \delta x) - 1$
   $= 3x^2 + 6x\,\delta x + 3(\delta x)^2 + 4x + 4\,\delta x - 1$.
2. Take away $y = 3x^2 + 4x - 1$: $\delta y = 6x\,\delta x + 3(\delta x)^2 + 4\,\delta x$.
3. Divide by $\delta x$: $\dfrac{\delta y}{\delta x} = 6x + 3\,\delta x + 4$.
4. Let $\delta x \to 0$: $\dfrac{dy}{dx} = 6x + 4$.

1. Two throws give 36 equally likely outcomes.

(a)

1. Sums that are multiples of 3: 3 (2 ways), 6 (5 ways), 9 (4 ways) and 12 (1 way): 12 outcomes, so $P = \frac{12}{36} = \frac13$.

(b)

1. Between 3 and 6 means a sum of 4 (3 ways) or 5 (4 ways): 7 outcomes, so $P = \frac{7}{36}$.

1. Class marks $44.5, 54.5, \ldots, 114.5$ and $d = x - 74.5$: $-30, -20, -10, 0, 10, 20, 30, 40$.
2. $\sum fd = -120 - 240 - 140 + 0 + 70 + 100 + 60 + 40$
   $= -230$ and $\sum f = 56$.
3. Mean $= 74.5 + \dfrac{-230}{56}$
   $= 74.5 - 4.107$
   $\approx 70.39$, so the mean wage is $70.39.

1. Take the direction the ball is hit as positive: $u = -10\text{ m s}^{-1}$ and $v = 15\text{ m s}^{-1}$.
2. Change in momentum $= m(v - u)$
   $= 0.1(15 - (-10))$
   $= 2.5\text{ N s}$.
3. $F = \dfrac{\text{change in momentum}}{t}$
   $= \dfrac{2.5}{0.01}$
   $= 250\text{ N}$.

1. In $ABCD$, $\overrightarrow{AB} = \overrightarrow{DC}$, so $\mathbf d = \mathbf a + \mathbf c - \mathbf b$.
2. $\mathbf d = \begin{pmatrix} -3 + 0 - 1 \\ 1 - 1 - 2 \end{pmatrix}$
   $= \begin{pmatrix} -4 \\ -2 \end{pmatrix}$.
3. So $D(-4, -2)$.

(a)

1. Take 2 out of the $x$ terms only: $y = 2\left(x^2 - \frac52x\right) + 3$.
2. Complete the square inside: half of $-\frac52$ is $-\frac54$, so $x^2 - \frac52x = \left(x - \frac54\right)^2 - \frac{25}{16}$.
3. Multiply back by 2: $y = 2\left(x - \frac54\right)^2 - \frac{25}{8} + 3$
   $= 2\left(x - \frac54\right)^2 - \frac18$.
4. So $p = 2$, $q = -\frac54$ and $r = -\frac18$.
5. $q + r = -\frac{10}{8} - \frac18$
   $= -\frac{11}{8}$.

(b)

1. The curves meet where $x^2 - 3x + 2 = -x^2 + 3x + 2$, so $2x^2 - 6x = 0$.
2. Factorise: $2x(x - 3) = 0$, so $x = 0$ or $x = 3$.
3. Between them, $y = -x^2 + 3x + 2$ is on top (at $x = 1$ it is 4, the other is 0).
4. Area $= \displaystyle\int_0^3 \left[(-x^2 + 3x + 2) - (x^2 - 3x + 2)\right]dx$
   $= \int_0^3 (-2x^2 + 6x)\,dx$.
5. $= \left[-\frac{2x^3}{3} + 3x^2\right]_0^3$
   $= -18 + 27$
   $= 9$ square units.

(a)

1. Compare the sides: $3 + \sqrt2 \approx 4.41$, $2\sqrt3 \approx 3.46$ and $3\sqrt2 \approx 4.24$.
2. The largest angle is opposite the longest side, $3 + \sqrt2$.
3. By the cosine rule, $\cos\theta = \dfrac{(2\sqrt3)^2 + (3\sqrt2)^2 - (3 + \sqrt2)^2}{2(2\sqrt3)(3\sqrt2)}$.
4. Work out each square: $(2\sqrt3)^2 = 12$, $(3\sqrt2)^2 = 18$ and $(3 + \sqrt2)^2 = 11 + 6\sqrt2$.
5. The top is $19 - 6\sqrt2 \approx 10.515$.
6. The bottom is $12\sqrt6 \approx 29.394$, so $\cos\theta \approx 0.3577$.
7. So $\theta \approx 69.04^\circ$.

(b)

1. Factorise the bottom: $3x^2 + 4x + 1 = (3x + 1)(x + 1)$.
2. Write $\dfrac{2x - 1}{(3x + 1)(x + 1)} = \dfrac{P}{3x + 1} + \dfrac{Q}{x + 1}$.
3. Multiply through: $2x - 1 = P(x + 1) + Q(3x + 1)$.
4. Put $x = -1$: $-3 = -2Q$, so $Q = \frac32$.
5. Put $x = -\frac13$: $-\frac53 = \frac23P$, so $P = -\frac52$.
6. So $\dfrac{2x - 1}{3x^2 + 4x + 1} = -\dfrac{5}{2(3x + 1)} + \dfrac{3}{2(x + 1)}$.

(a)

1. $P = \begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix}$ and $Q = \begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}$.

(b)

1. Rearrange: $MQ = 2P + 3Q - 5I$
   $= \begin{pmatrix} 5 & 7 \\ -1 & 6 \end{pmatrix} - \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}$
   $= \begin{pmatrix} 0 & 7 \\ -1 & 1 \end{pmatrix}$.
2. $|Q| = 2 + 1 = 3$, so $Q^{-1} = \frac13\begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}$.
3. Multiply on the right by $Q^{-1}$: $M = \begin{pmatrix} 0 & 7 \\ -1 & 1 \end{pmatrix}Q^{-1}$
   $= \frac13\begin{pmatrix} 7 & 7 \\ -1 & 2 \end{pmatrix}$.

(a)

1. Upper boundaries $9.5, 19.5, \ldots, 99.5$ with cumulative frequencies $5, 10, 20, 38, 61, 84, 93, 97, 99, 100$.

(b)

1. Plot these points, starting from $(-0.5, 0)$, and join with a smooth curve.

(c)(i)

1. Read across from 50: median $\approx 39.5 + \dfrac{12}{23} \times 10$
   $\approx 44.7$.

(ii)

1. $Q_1 \approx 29.5 + \dfrac{5}{18} \times 10$
   $\approx 32.3$ and $Q_3 \approx 49.5 + \dfrac{14}{23} \times 10$
   $\approx 55.6$, so the semi-interquartile range is $\frac12(55.6 - 32.3) \approx 11.7$.

(d)

1. At least 60 marks: $9 + 4 + 2 + 1 = 16$ students, so $P = \dfrac{16}{100} = 0.16$.

(a)

1. $X \sim B\left(6, \frac12\right)$, so each outcome has probability $\left(\frac12\right)^6 = \frac{1}{64}$.
2. $P(X \le 3) = \dfrac{{}^6C_0 + {}^6C_1 + {}^6C_2 + {}^6C_3}{64}$
   $= \dfrac{1 + 6 + 15 + 20}{64}$
   $= \dfrac{42}{64}$
   $= \dfrac{21}{32}$.

(b)

1. There are 6 jurors and $\,{}^6C_2 = 15$ pairs.

(i)

1. 1 American and 1 African: $\,{}^3C_1 \times {}^2C_1 = 6$, so $P = \dfrac{6}{15} = \dfrac25$.

(ii)

1. Pairs of the same nationality: 2 Americans, $\,{}^3C_2 = 3$, or 2 Africans, $\,{}^2C_2 = 1$.
2. So $P(\text{different}) = 1 - \dfrac{4}{15}$
   $= \dfrac{11}{15}$.

(a)(i)

1. $T$: $8\sin30^\circ = 4$ east.
2. $Q$: $10\sin150^\circ = 5$ east.
3. The $x$-component of the resultant is $4 + 5 = 9\text{ N}$.

(ii)

1. North parts: $8\cos30^\circ + 10\cos150^\circ = 6.928 - 8.660$
   $= -1.732$.
2. $|\mathbf R| = \sqrt{81 + 3}$
   $= \sqrt{84}$
   $\approx 9.165\text{ N}$.
3. $a = \dfrac{F}{m}$
   $= \dfrac{9.165}{4}$
   $\approx 2.29\text{ m s}^{-2}$.

(b)

1. The weight, $150 \times 10 = 1500\text{ N}$, acts at the middle of the beam, 2 m from the end.
2. That is $1\text{ m}$ from the first support.
3. Let the second support be $x$ m from the first.
4. Moments about the first support: $588x = 1500 \times 1$, so $x \approx 2.55\text{ m}$.

(a)

1. The plane is smooth: $a = g\sin35^\circ$
   $= 10 \times 0.5736$
   $= 5.736\text{ m s}^{-2}$ down the plane.
2. Half the plane is $150\text{ m}$: $v^2 = 0 + 2(5.736)(150) \approx 1720.7$.
3. $v \approx 41\text{ m s}^{-1}$ (2 s.f.).

(b)(i)

1. $v^2 = u^2 + 2gs$: $30.5^2 = 5.5^2 + 20s$, so $s = \dfrac{900}{20} = 45\text{ m}$.

(ii)

1. $30.5 = 5.5 + 10t$, so $t = 2.5\text{ s}$.
2. Average speed $= \dfrac{45}{2.5} = 18\text{ m s}^{-1}$.