RI 2021 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student Method 1 3 2 2 0x x ax b    Since all the coefficients of the equation are real, then the conjugate of 11 i2 is also a root of the equation. That is, 11 i2 is a root of the equation. Let the third root be 3x .    3 2 3 2 3 1 12 1 i 1 i 2 2 52 4 x x ax b x x x x x x x x                       Compare coefficients of 2 3 3 3 3 : 2 2 4 5 27: 2 4 4 5constants: 5 4 x x x x a x a b x b            The other roots of the equations are 11 i2 and 4 . The values of a and b are 27 4 and 5 respectively. Method 2 Since 11 i2 is a root of 3 2 2 0x x ax b    , then 3 21 1 11 i 2 1 i 1 i 02 2 2 1 11 3 1i+2 i 1 i 04 8 4 2 7 27 1 i 04 8 2 a b a b a b a                                                    Comparing Imaginary parts: 27 1 2708 2 4a a    Real parts: 7 0 54 a b b     Since the question did not state that the use of GC is not allowed, we can use the GC to evaluate 311 i2     and 211 i2     . Raffles Institution H2 Mathematics (9758) Solution for 2021 A-Level Paper 2

2 | P a g e The equation becomes 3 2 272 5 04x x x    . Using GC to solve the equation, we get the other roots of the equation to be 11 i2 and 4 . We can use the GC to solve the equation.

3 | P a g e Question 2 No. Suggested Solution Remarks for Student (a) The value of h 5 1 4 5 5   . [ Note that   4 0 f (1 ) f (1) f (1 ) f (1 2 ) f (1 3 ) f (1 4 ) n nh h h h h h h h h h h             f (1)h represents the area of left-most rectangle in the diagram above. Similarly, f (1 )h h represents the area of second left- most rectangle in the diagram above, and f (1 4 )h h represents the area of right-most rectangle in the diagram above. So,   4 0 f (1 ) n nh h   represents the sum of the area of the 5 shaded rectangles. Clearly, the total area of the rectangles is less than the area of A. ] (b) A similar expression is   5 1 f (1 ) n nh h   or    4 0 f (1 1 ) n n h h    (c) 21f ( ) 1.20x x  Lower bound for area of A  4 0 24 0 f (1 ) 4 1 4 1 1 5.6085 20 5 n n nh h n                  Upper bound for area of A  5 1 25 1 f (1 ) 4 1 4 1 15 20 5 6.568 n n nh h n                  Note that you can use the GC to evaluate the sum 1+h 1+2h 1+3h 1+4h

4 | P a g e (d) x y g( )y x

5 | P a g e Question 3 No. Suggested Solution Remarks for Student (a)(i) 21h(2) (2) 3 52   So, 6 1gh(2) g(5)24 4   (a)(ii) g( ) 1.4 1 1.45 1 0.4 x x x x     (b)(i) 2 bk . The value of x has to be excluded from the domain of f as 2 x a x b   will not be defined at 2 bx . (b)(ii) Let f ( )y x . Then,   2 2 2 1 2 1 x ay x b xy by x a x y a by a byx y           1f ( ) 2 1 a bxx x    So, 1f( ) f ( ) 2 2 1 x x x a a bx x b x     Comparing, 1b and a. Note that if 1 2a and 1b , then 1 1 12f ( ) , \2 1 2 2 x x x x          This function f is not one- one and so, 1f would not exist. A more completely correct answer is , 1b and 1\ 2a      . However, Cambridge says that 1b and a is accepted as the above level of detail is not expected. (b) (iii) 1 4 4f ( 4) 9 9 a b a     Alternatively, you can apply 1f 4 f 4    from part (bii)

6 | P a g e Question 4 No. Suggested Solution Remarks for Student (a) Total time taken  1 10 40 25 325s2 2 n n u u     (b) Let r be the common ratio in the G.P. (Alfie’s programme). Then, 9 9 840 25 5r r   Total time taken by Suzie 10 9 1 9 825 15 325 25 10 894.797s 8 15                   Average speed 130 35 1.17 ms894.797   (c) 8 min 480 s At the 8th min, Suzie is swimming at the second phase of the programme (i.e. the constant speed for each lap). The time elapsed at the constant speed laps = 480 325 155s  Note that 6 25 150 s  So, at the 8th minute, Suzie has completed 6 laps and is 5 s into her 7th lap of the constant speed phase. Suzie is swimming away from her starting point.

7 | P a g e Question 5 No. Suggested Solution Remarks for Student (a) 2 2tan 5 d sec 5 1d 1tan55 x x x x x x C        Remember to put the arbitrary constant “+C” for indefinite integral (b) 0 0 0 1sin 2 sin 3 d cos5 cos d2 1 1sin 5 sin2 5 1 1sin sin 52 10 b b b x x x x x x x x b b            As this is a definite integral, there should not be “+C” (c)     1 d ln lnln ln ln ln ln ln ln ln ln , since 1 b b aa x xx x b a b a a b           Sinec the question stated 1 a b  , then ln 0a and ln 0b . You are expected to simplify your answer without the modulus. (d) 2 2 2 1 e d 2e d 2e dd x x x u u u xx          2 2 3 32 2 3 2 22 e 1 1d 2e d21 e 1 e 1 1 d2 1 4 1 4 1 e x x x x x x x uu Cu C           Remember to convert back to the original variable.

8 | P a g e Section B: Probability and Statistics Question 6 No. Suggested Solution Remarks for Student Total probability0.2 0.3 1 2 0.5 p p q p q          E 0.2 0.6 3 4 5 0.8 7 5X p p q p q        2E 0.2 1.2 9 16 25 1.4 25 25X p p q p q                 22 2 2 2 Var =E E 1.4 25 25 0.8 7 5 1.4 25 25 0.5 2 0.8 7 5 0.5 2 13.9 25 3.3 3 1.61 X X X p q p q p p p p p p                     Use GC to solve for p to give p = 0.2 Then q = 0.1 So, the mean score 0.8 7(0.2) 5(0.1) 2.7    Question 7 No. Suggested Solution Remarks for Student (a) ABRACADABRA: 5 A, 2 B, 2 R, 1 C, 1 D Total number of different arrangements11! 831605!2!2!  (b) _ BB _ RR _ CD _ The number of different arrangements 4 23! 2! 2! 144C     [BB, RR, CD can be arranged in 3! ways. 2! for CD can be reversed. Notice that we have to split the AAAA and the A since we can only have exactly 4 of the A’s next to each other. So, there are 4 2 2!C ways for AAAA and A to be placed.] (c) The number of arrangements where all 5 A’s are together7! 12602!2!  The required probability 1260 1 83160 66 

9 | P a g e Question 8 No. Suggested Solution Remarks for Student (a) Because the sales manager only wanted to test if the life span of the front tyre is greater than 20 000 miles, so 1-tail test should be used. 0 1 0 Null hypothesis, Alternative hypo 0 thesis H : 20 00 H : 2000,     where  is the population mean life span (in miles) of the front tyres. (b) ( 20) 9.4 9.420 20.18850 x x         2 2 ( 20)1 ( 20) 0.754955 0.75549 50 xx            The unbiased estimate of the population mean is 20 188 miles. The unbiased estimate of the population variance is 6 20.755 10 755000 miles  The life span x given in this part of the question is in thousand miles. Do read the question carefully. (c) Perform a one-tail test at 5% level of significance. Under 0H , 754955N 20 000,50X     approximately by Central Limit Theorem, since n = 50 is large. p-value P 20188X  = 0.063 > 0.05 Thus, we do not reject 0H and conclude that there is insufficient evidence at 5% level of significance to say that the population mean life span of the front tyres is more than 20 000 miles. (d) As the population distribution is not known, a sample size of 15 is too small for us to use the Central Limit Theorem. Thus, the test would not be appropriate.

10 | P a g e Question 9 No. Suggested Solution Remarks for Student (a) No, as according to the