WAEC · 2014 · Nov/Dec · Further Maths · Paper 2

(a)

1. A logarithm is defined only for positive numbers, so we need $\frac{y + 1}{y + 3} > 0$.
2. A fraction is positive when the top and bottom have the same sign: both positive when $y > -1$, both negative when $y < -3$.
3. The domain is $\{y : y < -3 \text{ or } y > -1\}$.

(b)

1. $f(y) = 1$ means $\log_2\frac{y + 1}{y + 3} = 1$, so $\frac{y + 1}{y + 3} = 2^1 = 2$.
2. Then $y + 1 = 2y + 6$, which gives $y = -5$.
3. Check: $y = -5$ is in the domain ($-5 < -3$), and $\frac{-4}{-2} = 2$ ✓.

1. Get 0 on one side: $2x^2 + 3x - kx + 3 + k = 0$.
2. Group the $x$ terms: $2x^2 + (3 - k)x + (3 + k) = 0$.
3. So $a = 2$, $b = 3 - k$, $c = 3 + k$.
4. Real roots need $b^2 - 4ac \ge 0$: $(3 - k)^2 - 8(3 + k) \ge 0$.
5. Expand: $9 - 6k + k^2 - 24 - 8k \ge 0$.
6. Collect terms: $k^2 - 14k - 15 \ge 0$.
7. Factorise: $(k - 15)(k + 1) \ge 0$, with roots $k = -1$ and $k = 15$.
8. "$\ge 0$" is outside the roots: $k \le -1$ or $k \ge 15$.

1. At $x = 1$: $y = \dfrac12$, so the point is $\left(1, \frac12\right)$.
2. Write $y = (x + 1)^{-1}$.
3. By the chain rule, $\dfrac{dy}{dx} = -(x + 1)^{-2}$
   $= -\dfrac{1}{(x + 1)^2}$.
4. At $x = 1$ the gradient is $-\dfrac14$.
5. The tangent: $y - \frac12 = -\frac14(x - 1)$.
6. Multiply by 4: $4y - 2 = -x + 1$.
7. Rearrange: $x + 4y - 3 = 0$.

(a)

1. Mutually exclusive: $P(A \cup B) = \frac13 + \frac16 = \frac12$.

(b)

1. Independent: $P(A \cap B) = \frac13 \times \frac16$
   $= \frac{1}{18}$.
2. So $P(A \cup B) = \frac13 + \frac16 - \frac{1}{18}$
   $= \frac{6 + 3 - 1}{18}$
   $= \frac49$.

1. The deviations from the mean always add up to 0: $-2 + x + (x + 7) + (x + 2)^2 + (x + 3)^2 = 0$.
2. Expand the squares: $(x + 2)^2 = x^2 + 4x + 4$ and $(x + 3)^2 = x^2 + 6x + 9$.
3. Collect the $x^2$ terms: $2x^2$.
4. The $x$ terms: $x + x + 4x + 6x = 12x$.
5. The numbers: $-2 + 7 + 4 + 9 = 18$.
6. So $2x^2 + 12x + 18 = 0$.
7. Divide by 2: $x^2 + 6x + 9 = 0$.
8. Factorise: $(x + 3)^2 = 0$, so $x = -3$.
9. Check: the deviations are $-2, -3, 4, 1, 0$, which add up to 0 ✓.

1. $\overrightarrow{OA} \cdot \overrightarrow{OB} = (4)(5) + (3)(-2)$
   $= 20 - 6$
   $= 14$.
2. $|\overrightarrow{OA}| = 5$ and $|\overrightarrow{OB}| = \sqrt{29}$.
3. $\cos\theta = \dfrac{14}{5\sqrt{29}}$
   $\approx 0.5200$, so $\theta \approx 58.67^\circ$.

(a)

1. A cubed bracket needs three fractions: $\dfrac{A}{x + 2} + \dfrac{B}{(x + 2)^2} + \dfrac{C}{(x + 2)^3}$.
2. Multiply through by $(x + 2)^3$: $x^2 + 1 = A(x + 2)^2 + B(x + 2) + C$.
3. Put $x = -2$: $5 = C$.
4. Expand: $A(x^2 + 4x + 4) + Bx + 2B + C$.
5. The $x^2$ terms give $A = 1$.
6. The $x$ terms give $0 = 4A + B$, so $B = -4$.
7. Check the constants: $4 - 8 + 5 = 1$ ✓.
8. So the answer is $\dfrac{1}{x + 2} - \dfrac{4}{(x + 2)^2} + \dfrac{5}{(x + 2)^3}$.

(b)

1. Integrate the gradient: $y = x^2 - 3x + c$.
2. The minimum is where the gradient is zero: $2x - 3 = 0$, so $x = \frac32$.
3. There $y = 3$: $\frac94 - \frac92 + c = 3$, so $c = 3 + \frac94 = \frac{21}{4}$.
4. So $y = x^2 - 3x + \dfrac{21}{4}$, that is $4y = 4x^2 - 12x + 21$.

(a)

1. $PQ = \begin{pmatrix} 32 & 8 + 2d \\ 6c + 12 & 2c + 3d \end{pmatrix}$ and $QP = \begin{pmatrix} 24 + 2c & 18 \\ 16 + cd & 8 + 3d \end{pmatrix}$.
2. Match the top-left entries: $32 = 24 + 2c$, so $c = 4$.
3. Match the top-right entries: $8 + 2d = 18$, so $d = 5$.
4. Check the others: $6c + 12 = 36 = 16 + cd$ ✓ and $2c + 3d = 23 = 8 + 3d$ ✓.

(b)

1. Five ordinates means four strips, so $h = \dfrac{6 - 2}{4} = 1$.
2. The ordinates of $\frac1x$ at $x = 2, 3, 4, 5, 6$ are $0.5,\ 0.3333,\ 0.25,\ 0.2,\ 0.1667$.
3. $\frac12[(0.5 + 0.1667) + 2(0.3333 + 0.25 + 0.2)] = \frac12(0.6667 + 1.5667)$
   $= 1.1167$, about $1.12$.

1. The sum to infinity: $\dfrac{a}{1 - r} = 25$, so $a = 25(1 - r)$.
2. The first two terms: $a + ar = 16$, so $a(1 + r) = 16$.
3. Substitute: $25(1 - r)(1 + r) = 16$, so $1 - r^2 = \frac{16}{25}$ and $r^2 = \frac{9}{25}$.
4. The ratio is positive, so $r = \frac35$.
5. Then $a = 25 \times \frac25 = 10$.

(a)

1. $T_5 = ar^4$
   $= 10 \times \dfrac{81}{625}$
   $= \dfrac{162}{125}$
   $= 1\dfrac{37}{125}$.

(b)

1. $S_4 = \dfrac{a(1 - r^4)}{1 - r}$
   $= 25\left(1 - \dfrac{81}{625}\right)$
   $= 25 \times \dfrac{544}{625}$
   $= \dfrac{544}{25}$
   $= 21\dfrac{19}{25}$.

1. Class marks $17, 22, \ldots, 62$ and $d = x - 42$: $-25, -20, -15, -10, -5, 0, 5, 10, 15, 20$.
2. $\sum f = 235$, $\sum fd = -645$ and $\sum fd^2 = 31\,825$.

(a)

1. Mean $= 42 + \dfrac{-645}{235}$
   $= 42 - 2.745$
   $\approx 39.3$ years.

(b)

1. Variance $= \dfrac{31\,825}{235} - \left(\dfrac{-645}{235}\right)^2$
   $= 135.43 - 7.53$
   $\approx 127.9$.

(a)(i)

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

(ii)

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

(b)

1. The miss chances are $\frac14$, $\frac35$ and $\frac23$.
2. Only Sani: $\frac34 \cdot \frac35 \cdot \frac23 = \frac{18}{60}$.
3. Only Kalu: $\frac14 \cdot \frac25 \cdot \frac23 = \frac{4}{60}$.
4. Only Tato: $\frac14 \cdot \frac35 \cdot \frac13 = \frac{3}{60}$.
5. Add: $\frac{25}{60} = \frac{5}{12}$.

(a)

1. $3\mathbf p - 2\mathbf r = (12 - 4)\mathbf i + (-6 - 6)\mathbf j$
   $= 8\mathbf i - 12\mathbf j$.
2. $|8\mathbf i - 12\mathbf j| = \sqrt{64 + 144}$
   $= \sqrt{208}$
   $= 4\sqrt{13}$.

(b)

1. Match the parts: $4m + 2n = 8$ and $-2m + 3n = 8$.
2. Double the second and add: $8n = 24$, so $n = 3$.
3. Then $4m = 8 - 6$, so $m = \frac12$.

(c)

1. $\mathbf p \cdot \mathbf r = 8 - 6 = 2$, $|\mathbf p| = \sqrt{20}$ and $|\mathbf r| = \sqrt{13}$.
2. $\cos\theta = \dfrac{2}{\sqrt{260}}$
   $= \dfrac{2}{2\sqrt{65}}$
   $= \dfrac{1}{\sqrt{65}}$
   $= \dfrac{\sqrt{65}}{65}$.

(a)

1. Take $PQ$ and $PS$ as the two directions.
2. $PR$ is the diagonal, at $45^\circ$ to each.
3. Along $PQ$: $5 + 3\sqrt2\cos45^\circ = 5 + 3$
   $= 8\text{ N}$.
4. Along $PS$: $3 + 3\sqrt2\sin45^\circ = 3 + 3$
   $= 6\text{ N}$.
5. $R = \sqrt{8^2 + 6^2} = 10\text{ N}$, at $\tan^{-1}\frac68 \approx 36.9^\circ$ to $PQ$.

(b)(i)

1. The weight is $8\text{ N}$.
2. Its part down the plane is $8\sin30^\circ = 4\text{ N}$.
3. At rest, friction balances it: $4\text{ N}$.

(ii)

1. Down the plane: $4 - F = 0.8 \times 1.2 = 0.96$, so $F = 3.04\text{ N}$.