Q1(a) key features

Good. You labelled (3,4), (2,0) and the asymptote x=1 on the right-hand side.

$(3,4)$, $x=1$

Q1(a) incomplete sketch

This sketch is incomplete for $y = f(|x|)$. You need the reflection in the $y$-axis, including $x = -1$, $(-2,0)$ and $(-3,4)$.

only right half

What to do next: Reflect the $x \ge 0$ part of $y = f(x)$ in the $y$-axis, and include both vertical asymptotes.

Q1(b) asymptote x=3

Correct. The zero of $f'(x)$ at $x=3$ becomes a vertical asymptote of $y = 1/f'(x)$.

$x=3$

Q1(b) branches through (1,0)

Correct. You have the intercept $(1,0)$ and the right branch shapes for $y = 1/f'(x)$.

rising left branch

Q1(b) max in 4th quadrant

Correct. For $x > 3$ the stationary point of $y = 1/f'(x)$ is below the $x$-axis.

4th quadrant max

Q1(c) shape

Good transformation to $y = -f(|x - 1|)$.

middle V at (1,0)

Q1(c) asymptotes

Correct asymptotes: x = 0, x = 2 and y = -2.

$x=0$, $x=2$, $y=-2$

Q1(c) key points

Correct turning points and axial intercepts.

(-2,-4),(4,-4)

Q2(a) intercept

Correct y-intercept (0, 1/4) and curve shape.

$(0,\tfrac{1}{4})$

Q2(a) asymptotes

Correct. Vertical asymptote x = 4 and horizontal asymptote y = 0.

$x=4$, $y=0$

Q2(b) existence

Correct. Since $R_f \subseteq D_g$, $gf$ exists.

Rf subset Dg

Q2(b) domain

Correct domain of gf.

R\{4}

Q2(b) range

Correct range (0, infinity), justified via the mapping.

$R_{gf}=(0,\infty)$

Q2(b) rule

Correct rule for gf(x).

$\ln\!\left(1+\dfrac{(x-4)^2}{4}\right)$

Q2(c) largest k

Correct. Largest k is 4.

$k=4$

Q2(d) plus-minus

Correct. You included both square-root branches.

±

Q2(d) reject branch

Correct rejection using the restricted domain $x < 4$.

rejt as x<4

Q2(d) inverse form

Correct inverse in similar form, with domain x > 0.

$4-\sqrt{4/x}$

Q3(a) height

Correct. You expressed $h$ using Pythagoras.

$h=\sqrt{a^2-x^2}$

Q3(a) shown result

Correct derivation of the given $V^2$ expression.

$V^2$ shown

Q3(b) differentiate

Good. You differentiated $V$ and set $dV/dx = 0$.

$\dfrac{\mathrm{d}V}{\mathrm{d}x}=0$

Q3(b) algebra error

When factorising $4a^2 x^3 - 6x^5 = 0$, the factor should be $2x^3(2a^2 - 3x^2) = 0$, so $x^2 = (2/3)a^2$, not $x = 2a$.

$x=2a$

What to do next: Factor out $2x^3$ carefully: $2x^3(2a^2 - 3x^2) = 0$.

Q3(b) max volume

The maximum volume must be in terms of $a$ only. With the correct $x$, max $V = a^3 / 3$.

left in terms of x,h

What to do next: Substitute $x^2 = (2/3)a^2$ into $V$ and simplify fully.

Q3(c) incomplete

Q3(c) is incomplete. You need the total surface area at the maximum volume: $\dfrac{5 + 3\sqrt{3}}{2} a^2$ (exact form from the scheme).

$x=2a$ only

What to do next: Find the lateral triangle area and hexagon area using the correct x, then add them.

Q4(a) reason

Correct. Without knowing p is real, the conjugate root theorem need not apply.

p may not be real

Q4(b) p=10

Correct substitution and result p = 10.

$p=10$

Q4(c) conjugate root

Correct use of the conjugate root theorem with real coefficients.

3-i root

Q4(c) quadratic factors

Correct quadratic factors.

$2z^{2}-2z+1$

Q4(c) all roots

Correct. All four roots found.

$\tfrac{1}{2} \pm \tfrac{i}{2}$

Q4(c) labelled axes

Good. Axes are labelled Re and Im.

$\mathrm{Re}/\mathrm{Im}$

Q4(c) plot 1/2±i/2

Correct plotting of 1/2 ± i/2.

$(1/2,±1/2)$

Q4(c) plot 3±i

Correct plotting of 3 ± i.

$(3,±1)$

Q4(d) wrong shape

The quadrilateral is a trapezium (two vertical parallel sides), not a parallelogram.

Parallelogram

What to do next: Name it as a trapezium (or trapezoid).

Q4(d) missing area

No clear area. Area $= \frac{1}{2}(1+2)(2.5) = 3.75$ (or $\frac{15}{4}$).

area scribbled

What to do next: Use (sum of parallel sides) × height / 2.

Q5(a) population not sample

These 20 members are a population: the coach surveyed every member of the club.

called sample

What to do next: Say population, because the questionnaire went to all club members.

Q5(b) advantage

The advantage should be about reducing bias / fairness / representativeness of a random sample in each category.

play well with

What to do next: Mention less bias, or a fairer / more representative selection.

Q5(c) number of teams

Correct. $\binom{5}{1}\binom{8}{2}\binom{7}{2} = 2940$.

2940

Q5(d) case idea

Good idea to split into cases for the flexible guard.

Case 1/2

Q5(d) wrong total

The correct total is 1925. Your Case 2 overcounts (use the remaining 7 forwards and 5 guards carefully, and avoid double-counting when the flexible guard is not selected).

2975

What to do next: Follow Method 2: 1575 + 350 = 1925, with no overlap.

Q6(a) incomplete Venn

Many regions are blank. Use independence of A and C to get n(A∩C)=10, then fill every region (and outside).

sparse Venn

What to do next: Compute P(A∩C)=P(A)P(C)=0.1 so n(A∩C)=10, then complete all 8 regions.

Q6(a) x and y placed

Good. You placed x in A∩B∩C and y in B∩C only.

x,y

Q6(b) missing

Q6(b) is missing. From independence of B and C, $y = 12 - x$, with greatest $y = 12$ and least $y = 2$.

blank

What to do next: Use n(B∩C)=12, then constraints x≥0, y≥0, 10-x≥0.

Q7(a) shown

Correct show that P(X=2)=1/n.

$\dfrac{1}{n}$

Q7(b) distribution

Correct probability distribution in terms of n.

table

Q7(c) standard deviation

Correct variance and standard deviation.

$\dfrac{26}{25}$

Q7(c) probability

Correct. $P(X \le \sigma) = P(X=0)+P(X=1)=1/5$.

$\dfrac{1}{5}$

Q7(d) incomplete

This expectation setup is incomplete and does not use the doll cost $10. The break-even value is n = 16.

expected losses

What to do next: Let Y be Happie's amount (or profit) per game, include the $10 cost, set E(Y)-10=0 (or E(profit)=0), solve n=16.

Q8(a)(i) independence

Correct independence assumption in context.

independent

Q8(a)(i) constant p

Correct constant probability assumption.

constant

Q8(a)(ii) range for p

Correct exact range for p.

$\dfrac{2}{17} < p < \dfrac{3}{17}$

Q8(b)(i) acceptance probability

Correct. P(accepted) ≈ 0.363.

0.363

Q8(b)(ii) conditional probability

Correct. Conditional probability ≈ 0.217.

0.217

Q9(a) alpha

Correct. $\alpha = 67.0$ using the mean point on the regression line.

67.0

Q9(b) scatter diagram

Clear scatter diagram with a decreasing trend.

scatter

Q9(c) k and c

Correct values of k and c for the better model.

k,c

Q9(d) calculation error

With $k = -0.0873$ and $c = 77.7$, $s \approx -0.0873(49) + 77.7 = 73.4$, not $77.4$.

77.4

What to do next: Recalculate -0.0873 × 49 + 77.7 carefully.

Q9(d) reliability

Good. You correctly noted that h = 7 is outside the data range, so the estimate is not reliable.

extrapolation

Q9(e) causation

Correlation is not causation. The statement is not valid; other factors may explain the trend.

said valid

What to do next: Say correlation does not imply causation.

Q10(a) sigma

Correct standardisation and σ ≈ 71.2.

71.2

Q10(b) probability

Correct. Using $S_1+S_2+S_3+L_1+L_2$ gives about $0.273$.

0.273

Q10(c) assumption wording

Say the amounts of fish caught by the vessels are independent of one another (not just that the vessels are independent).

vessels independent

What to do next: State independence of the catch amounts S1,S2,S3,L1,L2.

Q10(d) probability

Correct. Premium vessels give about 0.348.

0.348

Q10(e) sampling distribution

Correct distribution for the sample mean.

$110^{2}/n$

Q10(e) least n

Correct. Least n is 24.

$n=24$

Q11(a) wording

It is the sample mean (not the raw data) that is approximately normal by CLT, because n = 35 is large.

data follows normal

What to do next: Say: sample mean approx normal by CLT since n is large.

Q11(b) variance

250 is the sample variance. Unbiased estimate is $\dfrac{35}{34}\times 250 = 257.353$ (3 d.p.), not $250/34$.

7.353

What to do next: Use $s^2 = \dfrac{n}{n-1} \times$ (sample variance).

Q11(b) mean

Correct unbiased mean 47.143 (3 d.p.).

47.143

Q11(c) hypotheses

Correct hypotheses for a decrease in mean time.

H1: μ<50

Q11(c) p-value and conclusion

With the correct variance 257.353, p-value ≈ 0.146 > 0.1, so do not reject H0. There is insufficient evidence that the mean time has decreased.

reject H0

What to do next: Redo the test with $s^2 = 257.353$; conclude do not reject $H_0$.

Q11(d) missing

Q11(d) is blank. For the two-tailed 10% test, reject when y < 36.1 or y > 43.9.

blank d

What to do next: Use H0: μ=40 vs H1: μ≠40, critical z = ±1.6449.