A Level H1 Mathematics Practice Paper 2

TuitionGoWhere Practice Paper - Maths H1 A-Level

Answer Key and Marking Scheme

Paper: Practice Paper 2 (Version 2 of 5) Total Marks: 100

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Section A: Pure Mathematics (40 marks)

Question 1 (8 marks)

(a) Find stationary points. [4 marks]

( y = x^3 - 6x^2 + 9x + 5 )

( \frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3) ) [M1 – correct differentiation and factorisation]

Set ( \frac{dy}{dx} = 0 ): ( x = 1 ) or ( x = 3 ) [A1]

When ( x = 1 ): ( y = 1 - 6 + 9 + 5 = 9 ) → (1, 9) [A1]

When ( x = 3 ): ( y = 27 - 54 + 27 + 5 = 5 ) → (3, 5) [A1]

(b) Determine nature. [2 marks]

( \frac{d^2y}{dx^2} = 6x - 12 ) [M1]

At ( x = 1 ): ( \frac{d^2y}{dx^2} = 6(1) - 12 = -6 < 0 ) → maximum point [A1]

At ( x = 3 ): ( \frac{d^2y}{dx^2} = 6(3) - 12 = 6 > 0 ) → minimum point [A1]

(c) Sketch. [2 marks]

• y-intercept: when ( x = 0 ), ( y = 5 ) → (0, 5) [B1]
• Correct shape with stationary points (1, 9) maximum and (3, 5) minimum clearly labelled [B1]

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Question 2 (8 marks)

(a) Solve ( 2x^2 + 5x - 12 \leq 0 ). [4 marks]

( 2x^2 + 5x - 12 = (2x - 3)(x + 4) = 0 ) [M1 – factorisation]

( x = \frac{3}{2} ) or ( x = -4 ) [A1]

Since coefficient of ( x^2 ) is positive (2 > 0), parabola opens upward. [M1 – sign analysis]

Solution: ( -4 \leq x \leq \frac{3}{2} ) [A1]

(b) Solve ( 2e^{2t} + 5e^t - 12 \leq 0 ). [4 marks]

Let ( u = e^t ), then ( 2u^2 + 5u - 12 \leq 0 ) [M1 – substitution]

From (a): ( -4 \leq u \leq \frac{3}{2} ) [M1]

Since ( u = e^t > 0 ) for all real ( t ), we have ( 0 < e^t \leq \frac{3}{2} ) [M1 – considering domain]

( t \leq \ln\left(\frac{3}{2}\right) ) [A1]

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Question 3 (8 marks)

(a) Find ( x ) that maximises profit. [5 marks]

( P(x) = 20x e^{-0.5x} )

Using product rule: ( P'(x) = 20e^{-0.5x} + 20x(-0.5)e^{-0.5x} = 20e^{-0.5x}(1 - 0.5x) ) [M1 – product rule; A1 – correct derivative]

Set ( P'(x) = 0 ): ( 20e^{-0.5x}(1 - 0.5x) = 0 ) [M1]

Since ( e^{-0.5x} \neq 0 ), ( 1 - 0.5x = 0 ) → ( x = 2 ) [A1]

( P''(x) = 20(-0.5)e^{-0.5x}(1 - 0.5x) + 20e^{-0.5x}(-0.5) = 10e^{-0.5x}(0.5x - 2) )

At ( x = 2 ): ( P''(2) = 10e^{-1}(1 - 2) = -10e^{-1} < 0 ) → maximum [M1 – verification; A1]

(b) Maximum weekly profit. [1 mark]

( P(2) = 20(2)e^{-1} = 40e^{-1} \approx 14.715 ) thousand = $14 715 [A1]

(c) Why profit decreases. [2 marks]

As ( x ) increases beyond 2, the exponential decay factor ( e^{-0.5x} ) dominates, causing the product ( 20x e^{-0.5x} ) to decrease. [B1] The marginal profit becomes negative (P'(x) < 0 for x > 2). [B1]

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Question 4 (8 marks)

(a) Differentiate ( y = \ln(3x^2 + 2) ). [2 marks]

( \frac{dy}{dx} = \frac{1}{3x^2 + 2} \cdot 6x = \frac{6x}{3x^2 + 2} ) [M1 – chain rule; A1]

(b) Tangent at ( x = 1 ). [4 marks]

At ( x = 1 ): ( y = \ln(3(1)^2 + 2) = \ln 5 ) [B1 – y-coordinate]

( \frac{dy}{dx}\bigg|_{x=1} = \frac{6(1)}{3(1)^2 + 2} = \frac{6}{5} ) [B1 – gradient]

Equation: ( y - \ln 5 = \frac{6}{5}(x - 1) ) [M1]

( y = \frac{6}{5}x - \frac{6}{5} + \ln 5 ) [A1 – correct form]

(c) Exact value of integral. [2 marks]

( \int_0^1 \frac{6x}{3x^2 + 2} , dx = \left[\ln(3x^2 + 2)\right]_0^1 ) [M1 – recognising antiderivative from (a)]

( = \ln 5 - \ln 2 = \ln\left(\frac{5}{2}\right) ) [A1]

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Question 5 (8 marks)

(a) Show ( A = 120x - 2x^2 ). [2 marks]

Perimeter of three sides: ( 2x + y = 120 ) → ( y = 120 - 2x ) [M1]

Area: ( A = xy = x(120 - 2x) = 120x - 2x^2 ) [A1]

(b) Find ( x ) for maximum area. [3 marks]

( \frac{dA}{dx} = 120 - 4x ) [M1]

Set ( \frac{dA}{dx} = 0 ): ( 120 - 4x = 0 ) → ( x = 30 ) [A1]

( \frac{d^2A}{dx^2} = -4 < 0 ) → maximum [A1]

(c) Maximum area. [1 mark]

( A = 120(30) - 2(30)^2 = 3600 - 1800 = 1800 ) m² [A1]

(d) Verify maximum. [2 marks]

( \frac{d^2A}{dx^2} = -4 ) [M1]

Since ( \frac{d^2A}{dx^2} < 0 ) for all ( x ), the stationary point at ( x = 30 ) is a maximum. [A1]

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Section B: Probability and Statistics (60 marks)

Question 6 (6 marks)

(a) Unbiased estimates. [3 marks]

Sample mean: ( \bar{x} = \frac{520}{10} = 52.0 ) minutes [A1]

Unbiased variance: ( s^2 = \frac{1}{n-1}\left[\sum x^2 - \frac{(\sum x)^2}{n}\right] = \frac{1}{9}\left[28,400 - \frac{520^2}{10}\right] ) [M1]

( = \frac{1}{9}[28,400 - 27,040] = \frac{1360}{9} \approx 151 ) minutes² (3 s.f.) [A1]

(b) Meaning of unbiased estimate. [1 mark]

An unbiased estimate is one whose expected value equals the true population parameter. The sample mean is an unbiased estimator of the population mean. [B1]

(c) Combined estimate. [2 marks]

Combined mean: ( \bar{x}_{\text{combined}} = \frac{10(52.0) + 15(54.0)}{25} = \frac{520 + 810}{25} = \frac{1330}{25} = 53.2 ) minutes [M1; A1]

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Question 7 (7 marks)

(a) Assumptions for binomial model. [2 marks]

1. Each light bulb is either defective or not defective (two possible outcomes). [B1]
2. The probability of a light bulb being defective (0.08) is constant for each bulb, and the bulbs are selected independently. [B1]

(b) Exactly 2 defective. [2 marks]

( X \sim B(20, 0.08) )

( P(X = 2) = \binom{20}{2}(0.08)^2(0.92)^{18} ) [M1]

( = 190 \times 0.0064 \times 0.92^{18} \approx 0.271 ) (3 s.f.) [A1]

(c) At most 3 defective. [2 marks]

( P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) ) [M1]

Using GC: ( P(X \leq 3) \approx 0.931 ) (3 s.f.) [A1]

(d) More than 1 defective. [1 mark]

( P(X > 1) = 1 - P(X \leq 1) = 1 - [P(X = 0) + P(X = 1)] \approx 1 - 0.516 = 0.484 ) (3 s.f.) [A1]

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Question 8 (7 marks)

(a) Mass less than 490 g. [2 marks]

( X \sim N(500, 8^2) )

( Z = \frac{490 - 500}{8} = -1.25 ) [M1]

( P(X < 490) = P(Z < -1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056 \approx 0.106 ) (3 s.f.) [A1]

(b) Mass between 495 g and 510 g. [3 marks]

( Z_1 = \frac{495 - 500}{8} = -0.625 ), ( Z_2 = \frac{510 - 500}{8} = 1.25 ) [M1]

( P(495 < X < 510) = P(-0.625 < Z < 1.25) ) [M1]

( = \Phi(1.25) - \Phi(-0.625) = 0.8944 - (1 - 0.7340) = 0.8944 - 0.2660 = 0.6284 \approx 0.628 ) (3 s.f.) [A1]

(c) New mean for 2% below 490 g. [2 marks]

Let new mean be ( \mu ). ( P(X < 490) = 0.02 )

( Z = \frac{490 - \mu}{8} = -2.0537 ) (from inverse normal) [M1]

( 490 - \mu = -2.0537 \times 8 = -16.43 )

( \mu = 490 + 16.43 = 506.43 \approx 506 ) g (3 s.f.) [A1]

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Question 9 (8 marks)

(a) Hypotheses. [2 marks]

( H_0: \mu = 50 ) (mean weight is 50 g) [B1]

( H_1: \mu < 50 ) (mean weight is less than 50 g) – one-tail test [B1]

(b) Test at 5% level. [4 marks]

Test statistic: ( Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{49.2 - 50}{2.5/\sqrt{40}} = \frac{-0.8}{0.3953} = -2.024 ) [M1; A1]

Critical value at 5% (one-tail, left): ( z_{\text{crit}} = -1.645 ) [B1]

Since ( -2.024 < -1.645 ), reject ( H_0 ). [M1]

There is sufficient evidence at the 5% significance level to conclude that the mean weight is less than 50 g. [A1]

(c) Significance at 1% level. [2 marks]

Critical value at 1% (one-tail, left): ( z_{\text{crit}} = -2.326 ) [B1]

Since ( -2.024 > -2.326 ), we do not reject ( H_0 ) at the 1% level. The result is not significant at the 1% level. [B1]

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Question 10 (8 marks)

(a) Product moment correlation coefficient. [2 marks]

( r = \frac{n\sum xy - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} ) [M1]

( = \frac{8(7120) - (96)(560)}{\sqrt{[8(1280) - 96^2][8(40,800) - 560^2]}} )

( = \frac{56,960 - 53,760}{\sqrt{[10,240 - 9216][326,400 - 313,600]}} )

( = \frac{3200}{\sqrt{1024 \times 12,800}} = \frac{3200}{\sqrt{13,107,200}} = \frac{3200}{3620.4} \approx 0.884 ) (3 s.f.) [A1]

(b) Interpretation. [1 mark]

There is a strong positive linear correlation between hours of revision and test score. [B1]

(c) Regression line of ( y ) on ( x ). [3 marks]

( b = \frac{n\sum xy - (\sum x)(\sum y)}{n\sum x^2 - (\sum x)^2} = \frac{3200}{1024} = 3.125 ) [M1; A1]

( a = \bar{y} - b\bar{x} = \frac{560}{8} - 3.125\left(\frac{96}{8}\right) = 70 - 3.125(12) = 70 - 37.5 = 32.5 ) [M1]

Equation: ( y = 32.5 + 3.13x ) (3 s.f.) [A1]

(d) Estimate for 15 hours. [2 marks]

When ( x = 15 ): ( y = 32.5 + 3.125(15) = 32.5 + 46.875 = 79.375 \approx 79.4 ) [A1]

This is interpolation since ( x = 15 ) lies within the range of the data (assuming data range covers 15 hours). The estimate is reliable because the correlation is strong (( r \approx 0.884 )) and it is interpolation. [B1]

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Question 11 (8 marks)

(a) Show ( \mu - 1.2816\sigma = 165 ). [2 marks]

( P(X < 165) = 0.10 )

Standardising: ( P\left(Z < \frac{165 - \mu}{\sigma}\right) = 0.10 ) [M1]

From normal tables, ( P(Z < -1.2816) = 0.10 )

So ( \frac{165 - \mu}{\sigma} = -1.2816 ) → ( \mu - 1.2816\sigma = 165 ) [A1]

(b) Find ( \mu ) and ( \sigma ). [4 marks]

( P(X > 185) = 0.05 ) → ( P(X < 185) = 0.95 )

( P\left(Z < \frac{185 - \mu}{\sigma}\right) = 0.95 ) → ( \frac{185 - \mu}{\sigma} = 1.6449 ) [M1]

So ( \mu + 1.6449\sigma = 185 ) [A1]

Subtracting equations: ( (185 - 165) = (1.6449 + 1.2816)\sigma ) [M1]

( 20 = 2.9265\sigma ) → ( \sigma = 6.834 ) cm

( \mu = 165 + 1.2816(6.834) = 165 + 8.758 = 173.8 ) cm [A1]

(c) Probability between 170 cm and 180 cm. [2 marks]

( Z_1 = \frac{170 - 173.8}{6.834} = -0.556 ), ( Z_2 = \frac{180 - 173.8}{6.834} = 0.907 ) [M1]

( P(170 < X < 180) = P(-0.556 < Z < 0.907) = \Phi(0.907) - \Phi(-0.556) )

( = 0.8179 - (1 - 0.7107) = 0.8179 - 0.2893 = 0.5286 \approx 0.529 ) (3 s.f.) [A1]

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Question 12 (8 marks)

(a) Hypotheses. [1 mark]

( H_0: \mu = 120 ) (mean lifetime is 120 hours)

( H_1: \mu < 120 ) (mean lifetime is less than 120 hours) [B1]

(b) Test statistic and ( p )-value. [3 marks]

( Z = \frac{117.5 - 120}{10/\sqrt{64}} = \frac{-2.5}{1.25} = -2.00 ) [M1; A1]

( p\text{-value} = P(Z < -2.00) = 1 - \Phi(2.00) = 1 - 0.9772 = 0.0228 ) [A1]

(c) Test at 5% level. [3 marks]

Since ( p\text{-value} = 0.0228 < 0.05 ), reject ( H_0 ). [M1]

There is sufficient evidence at the 5% significance level to reject the company's claim. [A1]

The data suggests the mean lifetime is less than 120 hours. [A1]

(d) Meaning of ( p )-value. [1 mark]

The ( p )-value is the probability of obtaining a sample mean of 117.5 hours or less, assuming the null hypothesis (that the true mean is 120 hours) is true. [B1]

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Question 13 (8 marks)

(a) Scatter diagram. [2 marks]

• Axes labelled: ( x ) (Advertising expenditure, 000) on horizontal axis, \( y \) (Sales revenue, 000) on vertical axis [B1]
• Appropriate scales, points plotted correctly [B1]

(b) Regression line. [3 marks]

( b = \frac{10(11,500) - (250)(420)}{10(7250) - 250^2} = \frac{115,000 - 105,000}{72,500 - 62,500} = \frac{10,000}{10,000} = 1.00 ) [M1; A1]

( a = \bar{y} - b\bar{x} = \frac{420}{10} - 1.00\left(\frac{250}{10}\right) = 42 - 25 = 17.0 ) [M1]

Equation: ( y = 17.0 + 1.00x ) [A1]

(c) Draw regression line. [1 mark]

Line passes through ( (\bar{x}, \bar{y}) = (25, 42) ) with gradient 1. Drawn correctly on scatter diagram. [B1]

(d) Estimate for $30 000. [2 marks]

When ( x = 30 ): ( y = 17.0 + 1.00(30) = 47.0 ) → $47 000 [A1]

This is interpolation if ( x = 30 ) is within the data range. The estimate is reliable if the linear model is a good fit (check ( r )). [B1]

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END OF ANSWER KEY

TuitionGoWhere Practice Paper (AI) – Version 2 of 5 – A-Level Maths H1