RI Post+Prelim+Revision+P1+%28LT5%29 solutions

RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 4 Post-Prelim Revision Paper 1 (Source: A level questions) Total number of marks: 100 3 hour 1 9758/2024/01/Q1 The graph of 2 1xy ax bx c += ++ , where ,ab and c are non-zero constants, has an asymptote at 1 2x = − . The graph also has a turning point at 12, 9 −− . Find the values of ,ab and c . [4] Suggested Solution Remarks for Student [4] 2 1xy ax bx c += ++ Αsymptote at 2 11 1 022 2x a bc  = − ⇒ − + − +=   2 4 0 (1)abc⇒− + =  Graph passes through 11 12, 9 9 42 a bc − − ⇒− =− −+ 4 2 9 (2)a bc⇒ − +=  ( )( ) ( ) ( ) 2 22 2 22 12d d 2 ax bx c x ax by x ax bx c ax ax c b ax bx c + +− + += ++ − − +−= ++ 1Turning point at 2, 4 4 09 0 (3) a acb bc − − ⇒− + + − = ⇒− + =  Solving: 2, 1ab= = − and 1c = −

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ _____________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 2 of 20 2 9758/2023/01/Q1 Find the exact equation of the tangent to the curve ( ) 2 ln 11 5yx= − at the point where 2.x = [5] No. Suggested Solution Remarks for Student ( ) 2 ln 11 5yx= − Differentiate with respect to x, ( ) ( ) ( ) 1d 2 11 5 ( 5) 10 11 5d d 10 11 5d y xxyx y yxx = − −= − − = −− When 2x = , ( ) 2 ln 11 5(2) 1 eyy= −= ⇒= ( )d 10 11 5 10(e)(11 10) 10ed y yxx = − − = − −= − The equation of tangent is e 10e( 2) 10e 21e yx yx −= − − = −+ 3 9758/2024/01/Q4 (a) Without using a calculator, solve the inequality 43 .2 x xx −≥+ [4] (b) Hence, solve the inequality 34 .2 x xx −≥+ [2] No. Suggested Solution Remarks for Student (a) [4] 43 2 x xx −≥+ ( )( ) ( ) 4 23 02 xx x xx −+ −⇒≥ + ( )( ) ( ) 4 23 02 xx x xx ++ − + − ≤⇒ ( ) ( ) 22 5 6 0 and 0 and 2xx x x x x⇒ + − − ≤ ≠ ≠− ( )( )( )2 6 1 0 and 0 and 2xx x x x x⇒ + − + ≤ ≠ ≠− 2 1 or 0 6xx∴ −<≤ − <≤ No calculator is allowed so key algebraic steps need to be included in your working Multiplying by “−1” to convert to “ +x 2 ” term results in an easier algebraic manipulation that is less prone to careless mistakes −2 −1 0 6

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ _____________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 3 of 20 (b) [2] To solve 43 2 x xx −≥+ , we replace x by x in part (a): 2 1 or 0 6xx− <≤ − <≤ 0 2 1 has no solutionxx≥ ⇒ − < ≤− To solve 0 6, 0 and 6 0 and 6 6 60 o r 06 x xx xx xx <≤ < ≤ ⇒≠ −≤≤ ∴ − ≤< <≤ Alternatively, we can consider the graph of y = |x|: 4 9758/2024/0 2/Q2 A gardener designs a flower bed ABCDE in the shape of a rectangle with an equilateral triangle on one of the shorter sides. Side AE is of length a m and side ED is of length b m (see diagram). The total perimeter of the flower bed is 20 m. Find the maximum possible area of the flower bed, showing that it is a maximum value. Give your answer correct to 4 significant figures. [6]

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ _____________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 4 of 20 Suggested Solution Remarks for Student [6] Considering perimeter of flower bed: ( )32 3 20 10 * 2ba b a+ = ⇒= −  Area, 21 sin 602A ab a °= + Substitute (*): 2 2 31 310 222 36 104 Aa a a aa = −+    −= + d 36 10d2 A aa −= + d 36 0 10 0d2 20 63 A aa a −= ⇒ += ⇒= − 2 2 d 36 2.13 0d2 A a −= ≈− < By 2nd derivative test, maximum area occurs when 20 63 a = − . ∴Maximum area is 2 2 3 6 20 20 104 63 63 23.43 m (4sf ) −   + −−  = Since A contains 2a and a, it is more efficient to express A in terms of a only than in terms of b. You can verify the value obtained for maximum area by using a GC as follows: Interestingly, with the use of the phrase “shorter sides”, it is implied that ab< . However, when 20 4.6861 63 a = ≈ − , 3 2010 2.97092 63 b = −≈  − which suggests that ab> .

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ _____________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 5 of 20 5 9758/2023/01/Q3 Vectors a and b are such that 1= −a.b . It is also given that ( )×a b+a is perpendicular to ( )×a b+b . (a) Show that 1×=ab [3] (b) Hence find the angle between the direction of a and direction of b. [3] No. Suggested Solution Remarks for Student (a) ( ) ( )×+ ⊥ ×+aba abb ( ) ( ) ( ) ( ) ( ) ( ) .0 . . . .0 ×+ ×+ = × ×+× + ×+ = aba abb ab ab abbaab a b ( ) ( ) 2 1 0, since . . 0 and . × −= × = × = = − ab abb aab ab 1 2 1 1, since 0 ×= ×= ×≥ ab ab ab ( ) ( ).. 0× = ×=abb aab because ×ab is perpendicular to both a and b. The reason needs to be given. (b) Let θ be the angle between the direction of a and the direction of b. Then, . cos 1 θ= = −ab a b ---- (1) sin 1θ×= =a b ab ---- (2) sin(2) (1), 1cos tan 1 θ θ θ ÷= − = − ab ab So, 135θ = 

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ _____________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 6 of 20 6 9758/2023/01/Q4bcd (b) Given that 0n ≠ , show that 2 sin coscos d ,x nx nxx nx x cnn= ++∫ where c is an arbitrary constant. [3] (c) Using the result in part (b) show that, for all positive integers, n , the value of π 0 cos dx nx x∫ can be expressed as 2 k n , where the possible value(s) of k are to be determined [2] (d) Using the result in part (b) find the exact value of π 2 0 cos 2 d .x xx∫ [3] No. Suggested Solution Remarks for Student (b) 2 11cos d sin sin d 11 sin cos x nx x x nx nx xnn x nx nx cnn = − = ++ ∫∫ Integration by parts. d, cosd vu x nxx= = d sin1, d u nxvxn= = (c) Using the results in part (b), 0 2 0 22 22 cos d 11 sin cos 1 11 1sin cos (0)sin (0) cos (0) 11cos , since sin 0 for and cos 0 =1 x nx x x nx nxnn nn n nn nn n n nnnn π π ππ π ππ = + = +− + = −= ∈ ∫  When n is even, π 220 11 0cos π 1 and cos d 0n x nx x nn −= = = =∫ When n is odd, π 220 11 2cos π 1 and cos dn x nx x nn −− −= −= =∫ Therefore, π 20 cos d kx nx x n=∫ , where 0 when is even 2 when is odd nk n = −

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ _____________________________________ Y6 H2 Math Term 4 Post Prelim Revision: Paper 1 Page 7 of 20 (d) Since cos 2 0xx ≥ when π0 4x≤≤ and cos 2 0xx ≤ when ππ 42 x≤≤ , so in the interval π0 2x≤≤ , πcos 2 , when 0 ; 4cos 2 ππcos 2 , when . 42 xx x xx xx x  ≤≤=  − ≤≤ π 2 0 ππ 42 π0 4 ππ 42 π0 4 ππ 42 π0 4 cos 2 d cos 2 d cos 2 d cos 2 d cos 2 d sin 2 cos 2 sin 2 cos 2 24 24 x xx x xx x xx x xx x xx xx x xx x = +− = −   =+−+     ∫ ∫∫ ∫∫ ππ πsin cos 0 cos 042 2 2 424 π ππ πsin π sin coscos π2 42 2 242 4 π 1 1π00 0 08 4 48 π 4  = + −−    − +− −   = +−− −+ + + = Again, using your GC, sketc