RVHS 2023 JC2 H2 MA Test 4 Solution

RVHS 2023 JC2 H2 Maths Lecture Test 4 1 Name : Index Number : Class : Date : Duration : 50 min Max. No. of Marks : 30 [Answer all the questions on writing papers. Up to 1 mark will be deducted for poor presentation.] 1. HCI MYE 9740/2023/Q5 (modified) It is given that 1 sin 2yx=+ . (i) Show that d cos 2d yyx x = . [2] (ii) By further differentiation of the result in part (i), find the Maclaurin series for y, up to and including the term in 3x . [4] (iii) Use the series found in part (ii) to find an approximate value for 1.8 0 1 sin 2 dxx+ . [1] (iv) Use your calculator to find 1.8 0 1 sin 2 dxx+ , giving your answer correct to 4 decimal places. [1] (v) Suggest one way to reduce the error of the approximation found in part (iii). [1] 2. Specimen Paper 9740/2017/P2/Q10 River Valley High School 2022 JC2 Mathematics (9758) Lecture Test 4 (Term 3)

RVHS 2023 JC2 H2 Maths Lecture Test 4 2 End of Paper 3. Specimen Paper 9740/2017/P2/Q6 (modified) (ii) Which of the formulae 2m ed f=+ and 3m gd h=+ , where e, f, g, h are constants, is the better model for the relationship between m and d? Explain fully how you decided, and find the constants for the better formula. [4] (iii) Use the formula you chose from part (ii) to estimate the mass of a giant pumpkin with over the top length 6m. Comment on the reliability of your estimate. [2]

RVHS 2023 JC2 H2 Maths Lecture Test 4 3 Solutions: 1 Solution [9] i 1 sin 2yx=+  2 1 sin 2yx=+ Differentiate with respect to x, d2 2cos 2d yyx x = d cos 2d yyx x = (Shown) ALT 1 sin 2yx=+ ( ) ( ) 1 2 d1 1 sin 2 2cos 2d2 d1 sin 2 cos 2d d cos 2 (shown)d y xxx yxx x yyx x − =+ += = ii Differentiate with respect to x, 22 2 dd 2sin 2dd yyyx xx + =− Differentiate with respect to x, 3 2 2 3 2 2 d d d d d 2 4cos 2d d d d d y y y y yyx x x x x x    + + =−       When x = 0, y = 1, d 1d y x = , 2 2 d 1d y x =− , 3 3 d 1d y x =− Maclaurin’s series expansion is y = 1 + x + 21 2! x−  + 31 3! x−  + … y  1 + x 21 2 x− 31 6 x− iii 1.8 1.8 23 00 111 sin 2 d 1 d 26 2.0106 x x x x x x+  + − − =  iv 1.8 0 1 sin 2 d 2.2010xx+= v To reduce the percentage error, we could include more terms of x with higher power in our series expansion.

RVHS 2023 JC2 H2 Maths Lecture Test 4 4 2 Solution [13] i The Production Manager should use a 2-tail test because he is looking for a change in either direction (whether the average time taken by the alternative process is greater than or less than that with the original process). ii This is because the sample size is large and thus by Central Limit Theorem can be applied to have the sample mean time T to follow a normal distribution for the test. iii Let  denote the population mean time to manufacture a certain type of electronic control panel. Unbiased estimate of population mean = 835.7 16.71450t== Unbiased estimate of population variance = s2 21 835.714067.17 2.026126531 2.03 (3 sf)49 50 = − = = To test H0 :  = 17 Against H1 :  ≠ 17 at 10 % sig level Under H0 , 2 17 N(0,1) 50 XZ s −= where 2 2.026126531s = p-value = 0.155389 > 0.1. Do not Reject H0. OR Critical region: reject H0 if 1.645calz  . zcal = −1.42075 We do not reject H0 and conclude that there is insufficient evidence at 10% level of significance that the average time taken for the manufacture of a control panel is not 17 hours using alternative process. iv Longer average time might lead to producing better quality panels (or might need fewer resources other than time) v To test H0 :  = 17 Against H1 :  < 17 at 10 % sig level To reject H0 at 10% level of significance:  2 16.7 17 1.28155 40  − −

RVHS 2023 JC2 H2 Maths Lecture Test 4 5 3 Solution [8] i The scatter diagram shows that the data can be better represented by a non-linear curve. m increases at a greater rate as d increases. ii Product moment correlation coefficient, r, between m and d2 is 0.988 Product moment correlation coefficient between m and d3 is 0.9995 Since the |r|-value for model m = gd3 + h is closer to 1, it is the better model for the relationship between m and d. m ≈ 0.571654 d3 + 3.743088675 ≈ 0.572 d3 + 3.74 iii When d = 6, m ≈ 127 The estimate is reliable since d = 6 is within the data range (2.31 to 9.17), hence it is an interpolation. In addition, 0.9995r = is close to 1 which indicate a strong positive linear correlation between x and y. 2 0.3 40 1.28155 1.28155 0.3 40 1.48052 2.19     − − − −   If 2 2.19  , there will be sufficient evidence to conclude at 10% level of significance that alternative process is shorter than 17 hours (and vice versa). 11 449 9.17 2.31 m kg d m