November 2021 Paper 2 Higher

(a) The open circle at $-1$ means $-1$ is not included, and the arrow points right: $x > -1$.

(b) Filled circle at $-3$ (included) and open circle at 4 (not included), with a line joining them.

(a) $60 = 2^2 \times 3 \times 5$ and $84 = 2^2 \times 3 \times 7$.

HCF $= 2^2 \times 3 = 12$.

(b) $24 = 2^3 \times 3$ and $40 = 2^3 \times 5$.

LCM $= 2^3 \times 3 \times 5 = 120$.

(a) 20 km in 15 minutes. There are four lots of 15 minutes in an hour, so $20 \times 4 = 80$ km/h.

(b) Stop: flat line at 20 km from 10 15 to 10 25.

In 20 minutes at 75 km/h: $75 \times \frac{20}{60} = 25$ km.

Draw a line from (10 25, 20) to (10 45, 45).

(a) $x=-1$: $1 + 2 + 2 = 5$. $x=1$: $1 - 2 + 2 = 1$. $x=2$: $4 - 4 + 2 = 2$. $x=4$: $16 - 8 + 2 = 10$.

(b) Plot $(-2,10), (-1,5), (0,2), (1,1), (2,2), (3,5), (4,10)$ and join with a smooth curve.

(c) Draw the line $y = 4$. It crosses the curve at about $x = -0.7$ and $x = 2.7$.

Hypotenuse: $\sqrt{8^2 + 10^2} = \sqrt{164} = 12.806\ldots$ mm.

Outer edges: 8, 10, part of the first hypotenuse ($12.806 - 10 = 2.806$), the second hypotenuse $12.806$, and 8.

Perimeter $= 8 + 10 + 2.806 + 12.806 + 8 = 41.6$ mm (3 s.f.)

(a) $BC$ is opposite $56^\circ$ and 12 is adjacent, so use tan.

$BC = 12 \tan 56^\circ = 17.79\ldots = 17.8$ cm

(b) 15 is adjacent and 18 is the hypotenuse, so use cos.

$x = \cos^{-1}\left(\frac{15}{18}\right) = 33.557\ldots = 33.6^\circ$

Mass of A $= 1.8 \times 80 = 144$ g. Mass of B $= 1.2 \times 40 = 48$ g.

Total mass $= 192$ g, total volume $= 120$ cm³.

Density $= 192 \div 120 = 1.6$ g/cm³.

Cumulative frequency counts everyone up to the end of each class.

So 13 is the number with $t \le 20$, which covers $0 < t \le 20$, not just $10 < t \le 20$.

The running totals are right, but every interval after the first should start at 0.

(a) Upper quartile $= 188$, lower quartile $= 50$.

IQR $= 188 - 50 = 138$ minutes.

(b) The median is at 120 minutes, which is 2 hours, so half the people waited at least that long. Becky is right.

(c) The median on Tuesday is 100 minutes, lower than 120 on Monday, so on average people waited less on Tuesday.

Louise: $x \times 1.025^3 = 344\,605$, so $x = 344\,605 \div 1.025^3 = 320\,000$.

Sadiq: $320\,000 \times 1.02^2 \times 1.035 = 344\,580.48$.

Sadiq's investment is worth £344 580.48.

(a) From P to Q you go 6 right and 3 up. QR is $\frac{3}{2}$ of PQ, so from Q go $9$ right and $4.5$ up.

$R = (0 + 9, 3 + 4.5) = (9, 7.5)$.

(b) Gradient of L $= \frac{3}{6} = \frac{1}{2}$. Perpendicular gradient $= -2$.

The line passes through $(0, 3)$, so $y = -2x + 3$.

$(x - 2)(3x + 2) = 3x^2 + 2x - 6x - 4 = 3x^2 - 4x - 4$

$(3x^2 - 4x - 4)(2x + 3) = 6x^3 + 9x^2 - 8x^2 - 12x - 8x - 12$

$= 6x^3 + x^2 - 20x - 12$

Use the product rule: multiply the number of choices for each role.

$16 \times 120 \times 100 = 192\,000$

Opposite angles of a cyclic quadrilateral add to 180°, so $BCD = 180 \div 4 = 45°$ and $BAD = 135°$.

Alternate segment theorem: angle $SBD = BCD = 45°$.

$SBA = SBD - ABD = 45 - 20 = 25°$.

(a) $BC^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \cos 72° = 185 - 54.38 = 130.61...$

$BC = \sqrt{130.61...} = 11.4$ cm.

(b) Area $= \frac{1}{2} \times 8 \times 11 \times \sin 72° = 41.8$ cm².

(a) $x_1 = \sqrt[3]{10 - 2 \times 2} = \sqrt[3]{6} = 1.817...$

$x_2 = \sqrt[3]{10 - 2 \times 1.817...} = 1.853...$

$x_3 = \sqrt[3]{10 - 2 \times 1.853...} = 1.846...$

(b) $x = \sqrt[3]{10 - 2x}$ gives $x^3 = 10 - 2x$, so $x^3 + 2x - 10 = 0$: $a = 2$, $b = -10$.

Frequency = frequency density × class width:

$0.8 \times 5 = 4$, $1.6 \times 10 = 16$, $2.2 \times 10 = 22$, $1.2 \times 15 = 18$. Total $= 60$.

20% of 60 $= 12$ people.

In the top class, 12 people cover $12 \div 1.2 = 10$ marks, so the score is $40 - 10 = 30$.

(a) Draw $y = 0.6$ across the graph: it meets the curve at about $37°$ and $180 - 37 = 143°$.

The graph repeats every 360°, so also $37 + 360 = 397°$ and $143 + 360 = 503°$.

(b) Reflecting in the $x$-axis makes every $y$ negative: $y = -\sin x°$.

(c) $f(x - 2)$ shifts every point 2 to the right, e.g. a point at $x = 1$ moves to $x = 3$.

Volume ratio A : B $= 125 : 27$, so radius ratio $= \sqrt[3]{125} : \sqrt[3]{27} = 5 : 3$.

B : C $= 1 : 2 = 3 : 6$, so A : C radii $= 5 : 6$.

Surface areas go with the square: $25 : 36$.

Paths that end with rain on Wednesday (R = rain, N = no rain):

RRR: $0.7 \times 0.8 \times 0.8 = 0.448$

RNR: $0.7 \times 0.2 \times 0.6 = 0.084$

NRR: $0.3 \times 0.6 \times 0.8 = 0.144$

NNR: $0.3 \times 0.4 \times 0.6 = 0.072$

Total $= 0.748$

Rearranging: $g = \frac{4\pi^2 l}{T^2}$.

Bounds: $51.95 \le l < 52.05$ and $1.445 \le T < 1.455$.

Upper bound: $g = \frac{4 \times 3.142^2 \times 52.05}{1.445^2} = 984.4$ (to 1 d.p.).

Lower bound: $g = \frac{4 \times 3.142^2 \times 51.95}{1.455^2} = 969.0$ (to 1 d.p.).