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