June 2024 Paper 2 Higher

19 cm is the hypotenuse.

$CB^2 = 19^2 - 10^2 = 361 - 100 = 261$

$CB = \sqrt{261} = 16.155\ldots \approx 16.2$ cm

(a) $90 = 2 \times 45 = 2 \times 3 \times 15 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5$

(b) $A = 2^2 \times 3 = 12$, $B = 2 \times 3^2 = 18$.

LCM uses the highest power of each prime: $2^2 \times 3^2 = 36$

The formula is $H = \dfrac{72}{n}$.

Monday: $72 \div 6 = 12$ hours

Tuesday: $72 \div 9 = 8$ hours

Difference $= 12 - 8 = 4$ hours

Total discs $= 24 \div 0.16 = 150$

Red and blue $= 150 - 24 = 126$

One part $= 126 \div (5 + 4) = 14$

Red $= 5 \times 14 = 70$

(a) $x = -1$: $1 + 1 = 2$. $x = 1$: $1 - 1 = 0$. $x = 3$: $9 - 3 = 6$.

(b) Plot $(-2,6), (-1,2), (0,0), (1,0), (2,2), (3,6)$ and join with a smooth curve.

(c) Draw $y = 4$ and read where it crosses the curve: $x \approx -1.6$ and $x \approx 2.6$.

Take the shares as $1 : 6 : 14$ parts (21 parts).

Tina gives Andy $\frac{3}{7} \times 14 = 6$. Now Andy $= 7$, Tina $= 8$.

Tina gives Luke $12.5\%$ of $8 = 1$. Now Luke $= 7$, Tina $= 7$.

Final: Andy 7, Luke 7, Tina 7 parts, so Tina is correct.

Angles in a quadrilateral add to $360^\circ$:

$(2x+15) + (4x+15) + (4x+8) + (3x-3) = 360$

$13x + 35 = 360 \Rightarrow x = 25$

$A = 65^\circ$, $B = 115^\circ$, so $A + B = 180^\circ$.

Co-interior angles add to $180^\circ$, so $AD$ is parallel to $BC$ and $ABCD$ is a trapezium.

On the drawing, $QR = 40 \div 5 = 8$ cm.

Area $= \frac12 \times PQ \times 8 = 28$, so $PQ = 7$ cm on the drawing.

Real length $= 7 \times 5 = 35$ m.

(a) Numbers that round to 7.3 (2 s.f.) are from 7.25 up to (not including) 7.35, so the least value is $7.25$.

(b) No. Values such as 7.3495 are bigger than 7.349 but still round to 7.3; $N$ can be any value below 7.35.

$\sin a = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{7-2x}{8}$ and $\tan b = \dfrac{\text{opp}}{\text{adj}} = \dfrac{1+x}{4}$

$\dfrac{7-2x}{8} = \dfrac{1+x}{4} \Rightarrow 7-2x = 2+2x \Rightarrow 4x = 5 \Rightarrow x = 1.25$

(a) Running totals: $7, 7+21=28, 43, 54, 60$.

(b) Plot at the upper ends of each class, $(2,7),(4,28),(6,43),(8,54),(10,60)$, and join with a smooth curve.

(c) LQ at cf 15 is about 2.7 kg; UQ at cf 45 is about 6.3 kg. IQR $\approx 6.3 - 2.7 = 3.6$ kg.

(d) At 7.4 kg the cf is about 51, so $60 - 51 \approx 9$ parcels.

(a) $f = \dfrac{k}{d^2}$. With $f = 3.5, d = 8$: $k = 3.5 \times 64 = 224$, so $f = \dfrac{224}{d^2}$.

(b) $10 = \dfrac{224}{d^2} \Rightarrow d^2 = 22.4 \Rightarrow d = \sqrt{22.4} = 4.73$ (3 s.f.)

Draw the four lines: $x = 2$ (vertical), $y = -3$ (horizontal), $y = 2x + 1$ (through $(0,1)$, gradient 2) and $3x + 2y = 6$ (through $(0,3)$ and $(2,0)$).

The region is left of $x=2$, above $y=-3$, below $y=2x+1$ and below $3x+2y=6$.

The region is a quadrilateral with corners $(-2,-3)$, $(\tfrac47, \tfrac{15}{7})$, $(2, 0)$ and $(2,-3)$. Shade it and label it R.

(a) Draw a tangent to the curve at $t = 5$. For example it drops about 10 m/s over 4 seconds, so gradient $\approx -10 \div 4 = -2.5$ m/s².

(b) Readings: $v(0)=25, v(2)=16, v(4)=9, v(6)=4$. Three trapezia of width 2:

$\frac12 \times 2 \times (25+16) + \frac12 \times 2 \times (16+9) + \frac12 \times 2 \times (9+4) = 41 + 25 + 13 = 79$ m

Multiply top and bottom by $\sqrt a - 1$:

$\dfrac{1}{\sqrt a + 1} \times \dfrac{\sqrt a - 1}{\sqrt a - 1} = \dfrac{\sqrt a - 1}{a - 1}$

Critical values: $4x - 3 = 0 \Rightarrow x = \frac34$ and $x + 5 = 0 \Rightarrow x = -5$.

The product is negative between the roots (the graph is a U shape below the axis there).

So $-5 < x < \frac34$.

Mass is proportional to volume, so height ratio L : M $= \sqrt[3]{64} : \sqrt[3]{125} = 4 : 5$.

Area ratio M : P $= 144 : 16$, so height ratio $= \sqrt{144} : \sqrt{16} = 12 : 4 = 3 : 1$.

Make M the same: $4:5 = 12:15$ and $3:1 = 15:5$.

So L : M : P $= 12 : 15 : 5$.

Start: 4 red, 3 yellow, 1 green. More yellow than red remain only if Tony takes 3 red, or 2 red and the green.

$P(RRR) = \frac48 \times \frac37 \times \frac26 = \frac{24}{336}$

$P(\text{2 red, 1 green}) = 3 \times \frac48 \times \frac37 \times \frac16 = \frac{36}{336}$

Total $= \frac{60}{336} = \frac{5}{28}$

(a) $\overrightarrow{BC} = -\mathbf b + \mathbf a + k\mathbf b$, so $\overrightarrow{CN} = \frac59\overrightarrow{CB} = \frac59(\mathbf b - \mathbf a - k\mathbf b)$.

$\overrightarrow{MN} = \frac12\mathbf a + k\mathbf b + \frac59(\mathbf b - \mathbf a - k\mathbf b) = -\frac1{18}\mathbf a + \frac{4k+5}{9}\mathbf b = \frac1{18}(8k\mathbf b + 10\mathbf b - \mathbf a)$

(b) No: $\overrightarrow{MN}$ contains $-\frac1{18}\mathbf a$, so it is not a multiple of $\mathbf b = \overrightarrow{OB}$.

$y = 2x^2 - 12x + 7 = 2(x^2 - 6x) + 7 = 2[(x-3)^2 - 9] + 7 = 2(x-3)^2 - 11$

The minimum is when $x = 3$, giving $y = -11$.

Turning point: $(3, -11)$

$g(-x)$ reflects the curve in the $y$-axis; $+2$ then moves it up 2 units.

Key points: $(-1,1) \to (1,3)$, $(1,-2) \to (-1,0)$, $(3,1) \to (-3,3)$.

Draw the U-shaped curve through these points, minimum at $(-1, 0)$.

Join OA, OB and OP.

$\angle OAP = \angle OBP = 90^\circ$ (a tangent is perpendicular to the radius).

$OA = OB$ (both radii) and $OP$ is common.

So triangles $OAP$ and $OBP$ are congruent (RHS), hence $AP = BP$. (Or: $AP^2 = OP^2 - OA^2 = OP^2 - OB^2 = BP^2$.)