NECO · 2022 · SSCE · General Maths · Paper 2

(a)

1. Multiply the top and bottom by $\sqrt3$ to clear the surd from the denominator: $\frac{5}{\sqrt3} = \frac{5\sqrt3}{3}$.
2. Put in $\sqrt3 = 1.73$: the numerator is $5 \times 1.73 = 8.65$.
3. Divide by 3: $8.65 \div 3 = 2.883\ldots$, which is $2.9$ to 2 significant figures.

(b)

1. Divide every term by 3 so that $x^2$ has coefficient 1: $x^2 - \frac{20}{3}x + 4 = 0$.
2. Move the constant to the right: $x^2 - \frac{20}{3}x = -4$.
3. Add the square of half the coefficient of $x$, $\left(\frac{10}{3}\right)^2 = \frac{100}{9}$, to both sides: $x^2 - \frac{20}{3}x + \frac{100}{9} = \frac{100}{9} - 4$.
4. Write the left side as a square and simplify the right: $\left(x - \frac{10}{3}\right)^2 = \frac{64}{9}$.
5. Take square roots, keeping both signs: $x - \frac{10}{3} = \pm\frac{8}{3}$.
6. So $x = \frac{10}{3} + \frac83 = 6$ or $x = \frac{10}{3} - \frac83 = \frac23$.

1. Let his income be ₦$x$.
2. The tax-free part is $\frac16 x$.
3. The taxable remainder is $x - \frac16 x = \frac56 x$.
4. The tax is 20% of the remainder: $\frac{20}{100} \times \frac56 x = \frac16 x$.
5. Set the tax equal to ₦2,400: $\frac16 x = 2400$.
6. Multiply both sides by 6: $x = 14\,400$.
7. The tax-free amount is $\frac16 \times 14\,400 = 2400$.
8. His income is ₦14,400.00 and the tax-free amount is ₦2,400.00.

(a)

1. Use the distance formula: $(5 - p)^2 + (9 - 5)^2 = 5^2$.
2. Simplify the known squares: $(5 - p)^2 + 16 = 25$.
3. Subtract 16: $(5 - p)^2 = 9$.
4. Take square roots: $5 - p = \pm 3$.
5. So $p = 2$ or $p = 8$.
6. Both points are 5 units from $(5, 9)$.

(b)

1. For slopes $m_1$ and $m_2$, the angle $\theta$ between the lines satisfies $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$.
2. Numerator: $12 - \frac25 = \frac{58}{5}$.
3. Denominator: $1 + 12 \times \frac25 = 1 + \frac{24}{5}$
   $= \frac{29}{5}$.
4. Divide $\frac{58}{5}$ by $\frac{29}{5}$: the fives cancel and $58 \div 29 = 2$, so $\tan\theta = 2$.
5. So $\theta = \tan^{-1} 2 \approx 63.43^\circ$.

(a)

1. The determinant is $|A| = 3 \times 2 - 4 \times 5$
   $= 6 - 20$
   $= -14$.
2. Swap the leading-diagonal entries and change the signs of the other two: the adjoint is $\begin{pmatrix} 2 & -4 \\ -5 & 3 \end{pmatrix}$.
3. Divide by the determinant: $A^{-1} = -\frac{1}{14}\begin{pmatrix} 2 & -4 \\ -5 & 3 \end{pmatrix}$
   $= \begin{pmatrix} -\frac17 & \frac27 \\ \frac{5}{14} & -\frac{3}{14} \end{pmatrix}$.

(b)

1. Write the equations as $A\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$, so $\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\begin{pmatrix} 1 \\ 3 \end{pmatrix}$.
2. Multiply the adjoint by the column: $\begin{pmatrix} 2(1) - 4(3) \\ -5(1) + 3(3) \end{pmatrix} = \begin{pmatrix} -10 \\ 4 \end{pmatrix}$.
3. Multiply by $-\frac{1}{14}$: $x = \frac{10}{14} = \frac57$ and $y = -\frac{4}{14} = -\frac27$.
4. Check in the first equation: $3\left(\frac57\right) + 4\left(-\frac27\right)$ is $\frac{15}{7} - \frac{8}{7}$, which equals 1.
5. So $x = \frac57$ and $y = -\frac27$.

1. $AD$ and $BC$ both pass through the centre $O$, so they are diameters.

(i)

1. $\angle AOB$ and $\angle COD$ are vertically opposite angles, so $\angle AOB = 85^\circ$.

(ii)

1. In triangle $OCD$, $OC = OD$ (radii), so the base angles are equal: $\angle OCD = \frac{180^\circ - 85^\circ}{2}$
   $= 47.5^\circ$.
2. $B$, $O$ and $C$ lie on one line, so $\angle BCD = \angle OCD = 47.5^\circ$.

(iii)

1. $\angle BAD$ and $\angle BCD$ both stand on the arc $BD$ (angles in the same segment), so $\angle BAD = 47.5^\circ$.

(iv)

1. $\angle BOD$ and $\angle COD$ lie on the straight line $BC$, so $\angle BOD = 180^\circ - 85^\circ$
   $= 95^\circ$.

(v)

1. In triangle $OBD$, $OB = OD$ (radii), so $\angle OBD = \frac{180^\circ - 95^\circ}{2}$
   $= 42.5^\circ$.

(a)

1. Read the logarithms from the tables: $\log 565 = 2.7520$, $\log 0.0536 = \bar{2}.7292$ and $\log 249.1 = 2.3964$.
2. Multiplying means adding logs: $2.7520 + \bar{2}.7292 = 1.4812$.
3. Dividing means subtracting logs: $1.4812 - 2.3964 = \bar{1}.0848$.
4. Squaring means doubling the log: $2 \times \bar{1}.0848 = \bar{2}.1696$.
5. The cube root means dividing by 3.
6. Write $\bar{2}.1696$ as $\bar{3} + 1.1696$ so the negative part divides exactly: $\bar{1} + 0.3899 = \bar{1}.3899$.
7. Take the antilog: $\text{antilog } \bar{1}.3899 \approx 0.2454$.

(b)

1. The area of a sector is $\frac{\theta}{360} \times \pi r^2$.
2. Substitute: $\frac{36}{360} \times \frac{22}{7} \times 7^2 = \frac{1}{10} \times 154$.
3. The minor sector has area $15.4\text{ cm}^2$.

(a)(i)

1. The first term is $a = 6$ and the common difference is $d = 9 - 6 = 3$.
2. The $n$th term of an A.P. is $T_n = a + (n - 1)d$, so $T_{99} = 6 + 98 \times 3$.
3. Work it out: $T_{99} = 6 + 294 = 300$.

(ii)

1. Substitute $a$ and $d$: $T_n = 6 + (n - 1) \times 3$.
2. Expand and simplify: $T_n = 3n + 3$.

(b)

1. Write $y = (1 - x^3)^{1/2}$ and let $u = 1 - x^3$, so $\frac{du}{dx} = -3x^2$.
2. By the chain rule, $\frac{dy}{dx} = \frac12 u^{-1/2} \times \frac{du}{dx}$
   $= \frac12(1 - x^3)^{-1/2} \times (-3x^2)$.
3. Simplify: $\frac{dy}{dx} = \frac{-3x^2}{2\sqrt{1 - x^3}}$.

(c)

1. Integrate the gradient: $y = \int (6x^2 + 10x - 7)\,dx$
   $= 2x^3 + 5x^2 - 7x + c$.
2. Put in the point $(-3, 5)$: $5 = 2(-27) + 5(9) - 7(-3) + c$.
3. Simplify: $5 = -54 + 45 + 21 + c = 12 + c$, so $c = -7$.
4. The curve is $y = 2x^3 + 5x^2 - 7x - 7$.

(a)

1. For a sphere of radius $r$: $x = 4\pi r^2$ and $y = \frac43 \pi r^3$.
2. Divide: $\frac{x}{y} = \frac{4\pi r^2}{\frac43 \pi r^3}$
   $= \frac{3}{r}$.
3. Set $\frac{3}{r} = \frac27$ and cross-multiply: $2r = 21$, so $r = 10.5$.
4. Surface area: $x = 4 \times \frac{22}{7} \times 10.5^2$
   $= 1386$.
5. Volume: $y = \frac43 \times \frac{22}{7} \times 10.5^3$
   $= 4851$.
6. So $x = 1386$ (square units) and $y = 4851$ (cubic units).

(b)

1. Rearrange $3x + 3y = 5$ as $y = -x + \frac53$: its gradient is $-1$.
2. Perpendicular gradients multiply to $-1$, so the new gradient is $1$.
3. Through $(3, 2)$: $y - 2 = 1(x - 3)$, so $y = x - 1$.
4. The $x$-intercept is where $y = 0$: $0 = x - 1$, so $x = 1$.
5. The line cuts the $x$-axis at $(1, 0)$.

(a)

1. Square both sides: $M^2 = \frac{N - M^2 f + 1}{d - N}$.
2. Multiply both sides by $(d - N)$: $M^2(d - N) = N - M^2 f + 1$.
3. Collect the $M^2$ terms on the left: $M^2(d - N) + M^2 f = N + 1$.
4. Factorise: $M^2(d - N + f) = N + 1$.
5. Divide and take the square root: $M = \sqrt{\frac{N + 1}{d - N + f}}$.
6. Substitute $N = 3$, $f = 4$, $d = 8$: $M = \sqrt{\frac{3 + 1}{8 - 3 + 4}}$
   $= \sqrt{\frac49}$
   $= \frac23$.

(b)

1. Inverse variation with the cube: $R = \frac{k}{S^3}$.
2. Use $R = 8$, $S = 4$: $8 = \frac{k}{64}$, so $k = 512$.
3. Put in $R = \frac{512}{216}$: $\frac{512}{216} = \frac{512}{S^3}$, so $S^3 = 216$.
4. Take the cube root: $S = 6$.

(c)

1. Write the root as a power: $y = x^{4/3}$.
2. Bring down the power and reduce it by 1: $\frac{dy}{dx} = \frac43 x^{1/3}$
   $= \frac43\sqrt[3]{x}$.

1. The cone sits on the cylinder, so they share the same radius $r$.
2. Find $r$ from the cone by Pythagoras: $r^2 = 10^2 - 8^2 = 36$, so $r = 6$ cm.
3. The height of the cylinder is $18 - 8 = 10$ cm.
4. Volume of the cylinder: $\pi r^2 h = \frac{22}{7} \times 36 \times 10$
   $\approx 1131.4\text{ cm}^3$.
5. Volume of the cone: $\frac13 \pi r^2 h = \frac13 \times \frac{22}{7} \times 36 \times 8$
   $\approx 301.7\text{ cm}^3$.
6. Add them: $1131.4 + 301.7 = 1433.1\text{ cm}^3$.
7. To 2 significant figures, the volume of the solid is $1400\text{ cm}^3$.

(a)

1. Substitute each $x$ into $y = 6 + 3x - 3x^2$.
2. For $x = -2$: $y = 6 - 6 - 12 = -12$.
3. For $x = 2$: $y = 6 + 6 - 12 = 0$.
4. For $x = 3$: $y = 6 + 9 - 27 = -12$.
5. For $x = 4$: $y = 6 + 12 - 48 = -30$.

(b)

1. Draw the axes to the given scales ($x$ from $-3$ to $4$, $y$ from $-30$ to $10$).
2. Plot the eight points $(-3, -30)$, $(-2, -12)$, $(-1, 0)$, $(0, 6)$, $(1, 6)$, $(2, 0)$, $(3, -12)$, $(4, -30)$ and join them with a smooth curve.

(c)

1. For $y = 2 - x$, plot $(-3, 5)$, $(0, 2)$ and $(4, -2)$ and draw a straight line through them.

(d)(i)

1. The top of the curve is halfway between the equal values at $x = 0$ and $x = 1$, so it is at $x = 0.5$.
2. There $y = 6 + 1.5 - 0.75 = 6.75$.
3. The maximum value of $y$ is $6.75$ at $x = 0.5$.

(ii)

1. The solutions are the $x$-values where the curve and the line cross: about $x = -0.7$ and $x = 2$.
2. Check by algebra: $6 + 3x - 3x^2 = 2 - x$ gives $3x^2 - 4x - 4 = 0$, which is $(3x + 2)(x - 2) = 0$.
3. So $x = -\frac23$ or $x = 2$.

(i)

1. The Econs sector is marked as a right angle, so $4x + 3y = 90$ (1).
2. The angles at the centre add up to $360^\circ$: $(4x + 3y) + 5x + (3x + 5y) + 10y = 360$.
3. Collect like terms: $12x + 18y = 360$.
4. Divide by 6: $2x + 3y = 60$ (2).
5. Subtract (2) from (1): $2x = 30$, so $x = 15$.
6. Put $x = 15$ into (2): $30 + 3y = 60$, so $y = 10$.
7. The sector angles are Econs $90^\circ$, Eng.
8. $75^\circ$, Maths $95^\circ$ and Civic Edu.
9. $100^\circ$, which add up to $360^\circ$.

(ii)

1. Maths has $3(15) + 5(10) = 95^\circ$.
2. As a percentage: $\frac{95}{360} \times 100 = 26.38\ldots$, so $26.4\%$ of the students offered Maths.

(iii)

1. The events are exclusive, so add the angles: Econs and Civic Edu. together take $90^\circ + 100^\circ = 190^\circ$.
2. The probability is $\frac{190}{360} = \frac{19}{36}$.