A Level H1 Mathematics Practice Paper 1

TuitionGoWhere Practice Paper - Maths H1 A-Level - MARKING SCHEME

Total Marks: 100

Section A: Pure Mathematics [40 marks]

Question 1 [8 marks]

(a) [3 marks] $y = 3(2x+1)^{-1/2} + e^{2x-1}$

$\frac{dy}{dx} = 3 \times (-\frac{1}{2})(2x+1)^{-3/2} \times 2 + e^{2x-1} \times 2$

$= -3(2x+1)^{-3/2} + 2e^{2x-1}$

$= -\frac{3}{(2x+1)^{3/2}} + 2e^{2x-1}$

Marking: 1 mark for differentiating first term, 1 mark for differentiating second term, 1 mark for correct final form

(b) [5 marks] $y = x^2 \ln x$, point where $x = e$

$\frac{dy}{dx} = 2x \ln x + x^2 \times \frac{1}{x} = 2x \ln x + x$

At $x = e$: $\frac{dy}{dx} = 2e \ln e + e = 2e + e = 3e$

At $x = e$: $y = e^2 \ln e = e^2$

Equation: $y - e^2 = 3e(x - e)$ $y = 3ex - 3e^2 + e^2 = 3ex - 2e^2$

Marking: 1 mark for product rule, 1 mark for gradient at x=e, 1 mark for y-coordinate, 1 mark for tangent formula, 1 mark for final form

Question 2 [7 marks]

(a) [3 marks] Volume = $x^2h = 500$, so $h = \frac{500}{x^2}$

Cost = Base cost + Side cost $C = 20x^2 + 4 \times 15 \times xh$ $C = 20x^2 + 60x \times \frac{500}{x^2}$ $C = 20x^2 + \frac{30000}{x}$

Marking: 1 mark for height expression, 1 mark for cost setup, 1 mark for final form

(b) [4 marks] $\frac{dC}{dx} = 40x - \frac{30000}{x^2}$

Setting $\frac{dC}{dx} = 0$: $40x = \frac{30000}{x^2}$ $40x^3 = 30000$ $x^3 = 750$ $x = \sqrt[3]{750} = 9.09$ m

Marking: 1 mark for differentiation, 1 mark for setting equal to zero, 1 mark for solving, 1 mark for final answer

Question 3 [10 marks]

(a) [3 marks] $2x^2 - 5x - 3 < 0$ $(2x + 1)(x - 3) < 0$ Critical points: $x = -\frac{1}{2}, x = 3$ Solution: $-\frac{1}{2} < x < 3$

Marking: 1 mark for factoring, 1 mark for critical points, 1 mark for correct inequality

(b)(i) [3 marks] $x = -\frac{b}{2a} = -\frac{-5}{2(2)} = \frac{5}{4}$ $y = 2(\frac{5}{4})^2 - 5(\frac{5}{4}) - 3 = \frac{25}{8} - \frac{25}{4} - 3 = -\frac{49}{8}$ Vertex: $(\frac{5}{4}, -\frac{49}{8})$

Marking: 1 mark for x-coordinate, 1 mark for y-coordinate calculation, 1 mark for final coordinates

(b)(ii) [2 marks] Sketch showing parabola opening upward, vertex at $(\frac{5}{4}, -\frac{49}{8})$, x-intercepts at $x = -\frac{1}{2}$ and $x = 3$

Marking: 1 mark for correct shape and vertex, 1 mark for correct intercepts

(c) [2 marks] Area = $\int_0^3 |2x^2 - 5x - 3| dx$ Since curve is negative between intercepts, split at $x = 3$: Area = $-\int_0^3 (2x^2 - 5x - 3) dx = -[\frac{2x^3}{3} - \frac{5x^2}{2} - 3x]_0^3$ $= -[18 - 22.5 - 9] = 13.5$ square units

Marking: 1 mark for setup with absolute value, 1 mark for correct calculation

Question 4 [8 marks]

(a) [4 marks] $\frac{5x-2}{(x-1)(2x+3)} = \frac{A}{x-1} + \frac{B}{2x+3}$

$5x - 2 = A(2x+3) + B(x-1)$

When $x = 1$: $3 = 5A$, so $A = \frac{3}{5}$ When $x = -\frac{3}{2}$: $-\frac{15}{2} - 2 = B(-\frac{5}{2})$, so $B = \frac{19}{5}$

$\frac{5x-2}{(x-1)(2x+3)} = \frac{3/5}{x-1} + \frac{19/5}{2x+3}$

Marking: 1 mark for correct form, 1 mark for finding A, 1 mark for finding B, 1 mark for final answer

(b) [4 marks] $\int \frac{5x-2}{(x-1)(2x+3)} dx = \frac{3}{5} \ln|x-1| + \frac{19}{5} \times \frac{1}{2} \ln|2x+3| + c$ $= \frac{3}{5} \ln|x-1| + \frac{19}{10} \ln|2x+3| + c$

Marking: 1 mark for integrating first term, 1 mark for integrating second term, 1 mark for correct coefficients, 1 mark for constant

Question 5 [7 marks]

(a) [4 marks] Let $u = e^x$, then $f(x) = 3u^2 - 12u + 9 = 0$ $3(u^2 - 4u + 3) = 0$ $3(u-1)(u-3) = 0$ So $u = 1$ or $u = 3$

Marking: 1 mark for substitution, 1 mark for factoring, 1 mark for solving quadratic, 1 mark for values of u

(b) [3 marks] $e^x = 1$ gives $x = 0$ $e^x = 3$ gives $x = \ln 3$

Marking: 1 mark for first solution, 2 marks for second solution

Section B: Probability and Statistics [60 marks]

Question 6 [12 marks]

(a) [2 marks]

1. Each component has the same probability (0.85) of lasting more than 1000 hours
2. The components are tested independently

Marking: 1 mark for each condition

(b) [2 marks] $X \sim B(20, 0.85)$ $P(X = 18) = \binom{20}{18}(0.85)^{18}(0.15)^2 = 0.229$

Marking: 1 mark for setup, 1 mark for answer

(c) [3 marks] $P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$ $= 1 - P(X \leq 14) = 1 - 0.196 = 0.804$

Marking: 1 mark for setup, 1 mark for calculation method, 1 mark for answer

(d) [2 marks] With p = 0.85, P(X ≤ 14) = 0.196, which is quite low (less than 20%). This suggests the result does not strongly support the manufacturer's claim.

Marking: 1 mark for calculation/reasoning, 1 mark for conclusion

(e) [3 marks] If p = 0.70, then X ~ B(20, 0.70) $P(X \geq 15) = 1 - P(X \leq 14) = 1 - 0.584 = 0.416$

Marking: 1 mark for new distribution, 1 mark for calculation, 1 mark for answer

Question 7 [10 marks]

(a) [2 marks] $X \sim N(150, 20^2)$ $P(X > 180) = P(Z > \frac{180-150}{20}) = P(Z > 1.5) = 0.0668$

Marking: 1 mark for standardization, 1 mark for answer

(b) [3 marks] Need $P(X > w) = 0.10$, so $P(X \leq w) = 0.90$ $P(Z \leq \frac{w-150}{20}) = 0.90$ $\frac{w-150}{20} = 1.282$ $w = 150 + 20(1.282) = 175.6$ g

Marking: 1 mark for setup, 1 mark for z-value, 1 mark for final answer

(c) [5 marks] Sample mean $\bar{X} \sim N(150, \frac{20^2}{25}) = N(150, 16)$ $P(145 < \bar{X} < 155) = P(\frac{145-150}{4} < Z < \frac{155-150}{4})$ $= P(-1.25 < Z < 1.25) = 0.8944 - 0.1056 = 0.789$

Marking: 1 mark for distribution of sample mean, 1 mark for variance, 1 mark for standardization, 1 mark for z-values, 1 mark for final answer

Question 8 [8 marks]

(a) [4 marks] $\bar{x} = \frac{12+15+18+14+16+20+13+17+19+16}{10} = \frac{160}{10} = 16.0$ thousand dollars

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1} = \frac{(12-16)^2 + (15-16)^2 + ... + (16-16)^2}{9}$ $= \frac{16 + 1 + 4 + 4 + 0 + 16 + 9 + 1 + 9 + 0}{9} = \frac{60}{9} = 6.67$ (thousand dollars)²

Marking: 2 marks for mean, 2 marks for unbiased variance

(b) [4 marks] 95% confidence interval: $\bar{x} \pm t_{0.025,9} \times \frac{s}{\sqrt{n}}$ $t_{0.025,9} = 2.262$, $s = \sqrt{6.67} = 2.583$ $16.0 \pm 2.262 \times \frac{2.583}{\sqrt{10}} = 16.0 \pm 1.85$ Interval: $(14.2, 17.8)$ thousand dollars

Marking: 1 mark for formula, 1 mark for t-value, 1 mark for calculation, 1 mark for final interval

Question 9 [15 marks]

(a) [2 marks] Scatter diagram with x-axis (Hours of sleep) from 3 to 11, y-axis (Test score) from 40 to 90, showing 7 points with clear positive relationship.

Marking: 1 mark for axes and labels, 1 mark for correctly plotted points

(b) [4 marks] $\bar{x} = 7$, $\bar{y} = 65$ $S_{xx} = \sum(x_i - \bar{x})^2 = 28$ $S_{yy} = \sum(y_i - \bar{y})^2 = 1008$ $S_{xy} = \sum(x_i - \bar{x})(y_i - \bar{y}) = 168$ $r = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} = \frac{168}{\sqrt{28 \times 1008}} = 0.999$

Marking: 1 mark for means, 1 mark for sums of squares, 1 mark for formula, 1 mark for answer

(c) [4 marks] $a = \frac{S_{xy}}{S_{xx}} = \frac{168}{28} = 6.00$ $b = \bar{y} - a\bar{x} = 65 - 6(7) = 23.0$ $y = 6.00x + 23.0$

Marking: 1 mark for gradient calculation, 1 mark for intercept calculation, 1 mark for equation form, 1 mark for correct values

(d) [1 mark] Line drawn correctly on scatter diagram passing through all points approximately.

Marking: 1 mark for correctly drawn line

(e) [2 marks] $y = 6.00(6.5) + 23.0 = 62.0$ This prediction is reliable as 6.5 hours is within the range of the data and close to existing data points.

Marking: 1 mark for calculation, 1 mark for comment on reliability

(f) [2 marks] 12 hours is well outside the range of the data (4-10 hours). Extrapolation this far beyond the data range is unreliable as the linear relationship may not hold.

Marking: 1 mark for identifying extrapolation, 1 mark for explaining why inappropriate

Question 10 [15 marks]

(a) [2 marks] $H_0: \mu = 70$ (new method has same mean as old method) $H_1: \mu > 70$ (new method has higher mean)

Marking: 1 mark for each hypothesis

(b) [8 marks] Test statistic: $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{74.5 - 70}{8/\sqrt{16}} = \frac{4.5}{2} = 2.25$

Critical value: $t_{0.05,15} = 1.753$ (one-tailed test)

Since $2.25 > 1.753$, we reject $H_0$ at the 5% significance level.

Conclusion: There is sufficient evidence to conclude that the new teaching method improves student performance.

Marking: 2 marks for test statistic, 1 mark for critical value, 1 mark for comparison, 2 marks for decision, 2 marks for conclusion in context

(c) [2 marks] A Type I error would be concluding that the new method improves performance when it actually doesn't (rejecting a true null hypothesis).

Marking: 2 marks for correct explanation in context

(d) [3 marks] If $\mu = 75$, we want $P(\text{accept } H_0) = P(t \leq 1.753)$ when $\mu = 75$ Under $H_1$: $t = \frac{\bar{X} - 70}{8/4} \sim t_{15}$ but with non-central parameter This requires calculation of $P(T \leq 1.753)$ where $T$ has mean $\frac{75-70}{2} = 2.5$ Type II error probability ≈ 0.067

Marking: 1 mark for setup, 1 mark for method, 1 mark for answer

TOTAL: 100 marks