2024 H2 MATH P1 with Solutions

1 1 The graph of 2 1xy ax bx c += ++ , where a, b 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 a, b and c. [4] [Solution] Asymptote at x = 1 2−  2 11 022a b c   − + − + =       2 4 0a b c− + = …Eq(1) When x = –2 and y = 1 9− , 1 2 1 9 4 2a b c −+−= −+ 4 2 9a b c− + = …Eq(2) y = 2 1x ax bx c + ++  ( ) 2 1ax bx c y x+ + = + Differentiating wrt x: ( ) ( ) 2 d 21d yax bx c ax b yx+ + + + = When x = –2, y = 1 9− and d d y x = 0, ( ) 10 4 1 9ab + − + − = 49ab− + =− …Eq(3) From GC, a = 2, b = –1 and c = –1 2 Two vectors a and b are such that |a  b| = 3. It is given that a is a unit vector and b.b = 9. Show that a and b are perpendicular. [3] [Solution] 3=ab  sin 3 =ab where  is the angle between vectors a and b. ( )( )1 3 sin 3 = since |a| = 1 and |b| = 93= |sin  | = 1  sin  = 1 or –1   = 90 or 270  a and b are perpendicular. (Shown)

2 3 The equation 32 0,z az bz c+ + + = where a, b and c are constants, has roots 3 and w is a complex number. (a) State a condition on a, b and c for the third root to be w*. [1] (b) Given that the condition in part (a) holds, and that w = – 1 + 2i, find the values of a, b and c. [3] [Solution] (a) All three numbers, i.e. a, b, and c, are real numbers. (b) Let f(z) = 32z az bz c+ + + f(3) = 27 + 9a + 3b + c = 0  9a + 3b + c = –27 …Eq(1) f(–1 + 2i) = ( ) ( ) ( ) 32 1 2i 1 2i 1 2ia b c− + + − + + − + + = 0 From GC, (11 – 2i) + a(–3 – 4i) – b + 2bi + c = 0 (11 – 3a – b + c) + i(–2 – 4a + 2b) = 0 By comparing real and imaginary parts, 3a + b – c = 11 …Eq(2) 4a – 2b = –2 …Eq(3) From GC, a = –1, b = –1, c = –15 4 (a) Without using a calculator, solve the inequality 43 2 x xx −+ . [4] (b) Hence, solve the inequality 34 2 x xx −+ . [2] [Solution] (a) 43 2 x xx −+ 43 02 x xx −−+ ( )( ) ( ) 4 2 3 02 x x x xx − + − + ( ) 246 02 x x x xx − + + +

3 ( ) 2 56 02 xx xx −− + ( )( )( )2 6 1 0x x x x+ − +   –2 < x  –1 or 0 < x  6 (b) Replacing x with |x|, we have –2 < |x|  –1 (reject since |x|  0) or 0 < |x|  6  –6 < x < 6 Since x  0, the solution is –6 < x < 0 or 0 < x < 6 5 A geometric series has first term a and common ratio r, where r < 0. The sum to infinity of the series is 18 and its third term is 8 .3 (a) Show that 32 27 27 4 0.rr− + = [3] (b) Find the values of r and a. [1] The nth term of the series is .nu (c) It is given that 11 80 . k rr r k r uu  = + = = Find the value of k. [3] [Solution] (a) S = 1 a r− = 18  a = 18 – 18r …Eq(1) ar2 = 8 3  3ar2 = 8 …Eq(2) Subst Eq(1) into Eq(2): 3(18 – 18r)r2 = 8 54r2 – 54r3 = 8 27r3 – 27r2 + 4 = 0 (Shown) (b) From GC, r = 1 3− , 2 3 , 2 3 (reject 2 3 since r < 0) From Eq(1): a = 18 – 18 1 3 − = 24 –2 –1 0 6

4 (c) 11 80 k rr r k r uu  = + = =  1 1 1 1 80 80 80 k k k r r r r r k r r r u u u u  = + = = = + = +     11 80 81 k rr rr uu  == =  ( ) 124 1 3 80 18 81 11 3 k −− = −−  4 1 1 1 3 81 3 k    − = = −        k = 4 6 A curve C has equation 3 2e x ya=+ , where a is a positive constant. The point T lies on C and has an x – coordinate of 1. (a) Use calculus to find the equation of tangent to C at T. Give the equation in the form e( )y px q ra= + + , where p, q and r are exact constants to be found. [5] (b) It is given that the tangent to C at T passes through the origin. (i) Find the exact value of a. [1] (ii) Find the area of the region bounded by C, the line x = 0 and the tangent at T. Give your answer correct to 1 decimal place. [2] [Solution] (a) y = 3 2ex a+ Differentiating wrt x: ( ) 3322d 2 3 e 6 ed xxy xxx == When x = 1, d 6ed y x = and y = 2e + a Hence the equation of the tangent at T is y – 2e – a = 6e(x – 1) y = e(6x – 6 + 2) + a y = e(6x – 4) + a  p = 6, q = –4 and r = 1

5 (b) (i) Since the y-intercept is 0, –4e + a = 0 a = 4e (ii) Area required = ( ) ( ) 31 0 2e 4e 6e 4e 4e dx xx+ − − + = 5.402 = 5.4 units2 (1 dp) 7 It is given that f ( ) cos( ).rr = (a) Show that f (2 1) f (2 1) 2sin sin(2 ).r r r − − + = [2] (b) Hence, given that sin 0  , show that ( ) 1 cos cos (2 1)sin(2 ) . 2sin n r nr  = −+= [3] (c) Hence find the three possible values of 1 sin cos .66 n r rr =            [3] [Solution] (a) f(2r – 1) – f(2r + 1) = ( ) ( )cos 2 1 cos 2 1rr − − +       = ( ) ( ) ( ) ( )2 1 2 1 2 1 2 12sin sin 22 r r r r   − − + − + +− = ( )2sin sin 2 r−− = 2sin sin 2 r (Shown) (b) ( ) ( ) ( ) 11 f 2 1 f 2 1sin 2 2sin nn rr rrr == − − += = ( ) ( ) 1 1 f 2 1 f 2 12sin n r rr = − − + C T y = e(6x – 4) + 1 x = 0 y x

6 = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) f 1 f 3 + f 3 f 5 + f 5 f 7 1 ...2sin + f 2 5 f 2 3 + f 2 3 f 2 1 + f 2 1 f 2 1 nn nn nn  −  −  −    − − − − − −  − − + = ( ) ( )f 1 f 2 1 2sin n  −+ = ( )cos cos 2 1 2sin n  −+ (Shown) (c) 1 ππsin cos66 n r rr =            = 1 1 ππ2sin cos2 6 6 n r rr =            = 1 1 πsin 226 n r r =     = ( )ππcos cos 2 11 66 π2 2sin 6 n −+    = ( )3 πcos 2 11 26 12 2 2 n −+     = ( )31 πcos 2 14 2 6 n−+ When n = 1, ( )31 πcos 2 14 2 6 n−+ = 31 πcos4 2 2− = 3 4 When n = 2, ( )31 πcos 2 14 2 6 n−+ = 3 1 5 πcos4 2 6− = 0 When n = 3, ( )31 πcos 2 14 2 6 n−+ = 3 1 7 πcos4 2 6− = 3 2 Hence the 3 possible values required are 0, 3 4 and 3 2 .

7 8 It is given that ( ) 2 cos 1 e . x y=− (a) Show that 2 4 2 dd 2edd xyy ky xx =−  , where k is a constant to be found. [3] (b) By differentiation of the result in part (a), find the first three non-zero terms of the Maclaurin expansion of cos (1 – e2x). [4] (c) The first two non-zero terms of the Maclaurin expansion of cos (1 – e2x) are equal to the first two non-zero terms of the series expansion of 2 1 a bx+ , where a and b are constants. Using standard series from the List of Formulae (MF26), find the values of a and b. [2] [Solution] (a) y = ( ) 2cos 1 e x− Differentiating wrt x: ( ) ( ) 22d 2e sin 1 ed xxy x = − − − = ( ) 222e sin 1 exx − Differentiating wrt x: ( ) ( ) ( ) 2 2 2 2 2 2 2 d 4e sin 1 e 2e 2e cos 1 ed x x x x xy x = − + − − ( ) ( ) 2 2 4 24e sin 1 e 4e cos 1 ex x x x= − − − 4d2 4ed xy yx=− 4d2 2 ed xy yx =−  ; k = 2 (Shown) (b) Differentiating wrt x: 3 2 2 4 4 4 4 3 2 2 d d d d d2 4 e 4 e 2 4 e 16 ed d d d d x x x xy y y y y yyx x x x x = − + = − − When x = 0, y = 1, d d y x = 0, 2 2 d d y x ( )2 0 2 4= − =− , ( ) 3 3 d 2 4 0 16 24d y x = − − − =− Hence the Maclaurin’s expansion required is y = 1 + 0 + 24 2! x− + 324 3! x− + … = 231 2 4 ...xx− − +

8 (c) 1 2 2 2 11 1 bx aaa bx − =+ +  = 211 1 ...2 bx aa  −+  (From (1 + x)n expansion in MF26)  212 x−= 22 3 2 1 1 11 2 2 bx bx aaa a  − = −   1 = 1 a and 3 2 2 2 b a − =−  a = 1 and ( ) 3 24 1 4b== 9 A curve C has parametric equations x = 3t2 + 2t, y = t2 + 2t3 for 0.t (a) Show that d d y ktx = , where k is a constant to be found. [2] (b) The tangent to C at a point P makes an angle of 60o with the x-axis. Find the exact coordinates of P.