H2 Math Paper 1 Solution

1 2012 H2 MH Prelim P1 Solutions Qn Solution 1 4 13 43 033 33 03 x x xx xx x x ≥− −⇒−≥−− +⇒≥ − 1 03 x x +⇒≥ − 1 or 3xx⇒≤ − > Replace x by x , 1 (N.A.)x⇒≤ − or 3x > 3 or 3xx⇒> < − Qn Solution 2 (i) Let nP be the statement “ 2 2 n nun −= for n +∈ ] ” LHS of 1P = () () 211 10 1214 1 22 2uu −− ⎡⎤=− − += =⎣⎦ RHS of 1P =() 2 1 112 2 − = = LHS of 1P 1P∴ is true. Assume that kP is true for some k +∈ ] , i.e. 2 2 k kuk −= We want to prove 1+kP , i.e. () ()2 1 1 12 k kuk − + + =+ () ()() 21 11LHS of 2 1 4 1 2 k kkkPu u k k −+ ++ ⎡⎤== − + −+ + ⎣⎦ () ()() 212 22 14 1 2 kkkk k −+− ⎡⎤=− + − + + ⎣⎦ () ()() () () () () 21 2 1 22 1 2 21 1 22 1 4 1 2 22 2 1 4 4 2 22 1 21 R H S o f k k k k k kk k kk k k kk kP −+ −+ −+ −+ + ⎡⎤=− + + + −⎣⎦ ⎡⎤=− − − + + −⎣⎦ ⎡⎤=+ + ⎣⎦ =+ = 1kP+∴ is true. Since 1P is true and kP is true 1kP+⇒ is true, by Mathematical Induction, nP is true for all n +∈ ] .

2 (ii) () () 2 1 11 10 21 32 1 24 2 NN n nn nn NN nn u u uu uu uu uu − − == − ⎡⎤−− + =−⎣⎦ =− +− +− +− ∑∑ # 2 2 N NuN −== (iii) 0S∞ = 3 (a) (i) 2 2 d 1d 1 xxx x −= − (ii) 1 1cos dx xx − ⎛⎞ ⎜⎟⎝⎠∫ = 2 1 2 11cos d22 1 xx xx x − ⎛⎞ −⎜⎟⎝⎠ −∫ = 2 12 11cos 122 x x cx − ⎛⎞ − −+⎜⎟⎝⎠

3 (b) Let xu 1= ⇒ ux 1= ⇒ d d x u = 2 1 u− when x = 3, u = 1 3 ; when x = 6, u = 1 6 6 2 3 1 d 9 x xx −∫ = 1 6 1 3 2 21 1 d 9u u uu ⎛⎞−⎜⎟− ⎝⎠∫ = 1 6 1 2 3 2 2 19 1 d u u u uu− ⎛⎞−⎜⎟⎝⎠∫ = 1 6 1 3 2 1 d 19 u u − −∫ = 1 6 1 3 11 sin 33 u−⎡⎤−⎢⎥⎣⎦ = 1 36 2 ππ⎡⎤−− ⎢⎥⎣⎦ = 9 π

4 Qn Solution 4 (i) Points of intersection: 4 x = x2 + 1 ⇒ (0.062997, 1.0040) and (2.2301, 5.9734) Area = 2 2.2301 0.062997 4( 1 ) dx xx−+∫ = 2.9747 ≈ 2.97 (ii) Let x = c such that 2 0.062997 4( 1 ) d c x xx−+∫ = 2 1 (2.9747) 3 3 2 0.062997 8 33 c xxx⎡⎤ −−⎢⎥ ⎣⎦ = 2 1 (2.9747) 3 3 28 33 ccc−− = 1.4664 ⇒ c = 1.07 (iii) Volume generated about y-axis = 4 5.9734 1.0040 (1 ) d 256 yπ yy−−∫ = 20.2 x y 0 y = 4 x y = x2 + 1

5 Qn Solution 5(i) Given ()1 ln 1 tan2y x=+ , 2e1 t a ny x=+ Differentiate throughout w.r.t x. 22 de2 s e cd y y xx ⎛⎞ =⎜⎟⎝⎠ 22 d2e secd y y xx = (shown) Differentiate throughout w.r.t x. () 222 2 dd d2e 4e 2sec sec tanddd yy yy y x xxxxx ⎛⎞+= ⎜⎟⎝⎠ 2222 2 2 dde2 e s e c t a n dd yy yy x xxx ⎛⎞+= ⎜⎟⎝⎠ When 0x= , 0y= , d1 d2 y x = , 2 2 d1 2d y x =− By Maclaurin’s series, 2 f ( ) f (0) f '(0) f ''(0) ...2! xxx=+ + + 211f( ) 0 . . . 22 ! 2 xxx ⎛⎞ ⎛ ⎞=+ + − +⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ 211 ... 24xx=− + (ii) () 1x x ab xab x −=++ 1 1xb xaa −⎛⎞=+ ⎜⎟⎝⎠ 1. . .xb xaa ⎛⎞=− +⎜⎟⎝⎠ 2 2 ...xb xa a =− + Given 22 2 11 24 xb x xxa a −= − , Comparing coefficient of ,x 11 22 aa =⇒= Comparing coefficient of 2,x 2 2 1 144 ba b a =⇒= = 2, 1ab∴ ==

6 Qn Solution 6 (i) (ii) ()cos e tx= , ( )sin e ty= , where πln 2t≤ ()d es i ned ttx t =− ()d ec o sed tty t = ()d cot ed ty x =− Gradient of normal ()tan e t= Equation of normal: () () ()s i ne t a ne c o sett tyx ⎡⎤−= − ⎣⎦ () () () () sin e tan e sin e tan e tt t t yx yx −= − ∴= Since y-intercept is 0, the normal passes through the origin. (iii) Given equation of normal is yx= , ()tan e 1 πe 4 πln 4 t t t = = = πln 4 π 2cos e cos 42x ⎛⎞== =⎜⎟ ⎝⎠ πln 4 π 2sin e sin 42y ⎛⎞== =⎜⎟ ⎝⎠ d 1d y x =− Equation of tangent: 22 22yx ⎡⎤−= − − ⎢⎥ ⎣⎦ 2yx=− + 1 x y 0 1

7 Qn Solution 7(a) Let a cm be the height of the shortest doll. Since the heights of the dolls are in A.P., sum of all their heights ()7 47 02 aa=+= Therefore, 20 45a== cm Height of the tallest doll 4( 71 ) 4 ( 4 )d=+ − = 12 26d∴ == cm 7(b) Let T1 be the time interval between 1st and 2nd bounces, T2 be the time interval between 2nd and 3rd bounces, and so on … Hence T1, T2, T3, …, Tn is a G.P. where 1 4, 0.9Tr= = . Given 0.4kT < () 1 40 . 9 0 . 4 k− ⇒< () ( )( )1l n 0 . 9 l n 0 . 1 22.854 k k ⇒− < ⇒> 23k∴ = Total time from 1st to kth bounce () 22 22 41 0 . 9 10 . 9S ⎡ ⎤−⎣ ⎦== − 36.061 36 (nearest sec.) ≈ =

8 Qn Solution 8 (a)(i) (a) (ii) (b) (i) 2 3 3ax x b bya x x x ++== + + Asymptotes are 3 and 0yx x⇒= + = Given that 3y x=+ is an oblique asymptote, a = 1 (ii) When y = 0, 2 30ax x b++ = C has no x-intercept ⇒ Discriminant < 0 ⇒ 94 ( 1 ) ( )0b−< ⇒ 9 4b> (Shown) y = − 2 2 0 y x x = 1 y = f(x) (4, −3) • y y = −2 0 x x = −1 f( )y x= • 2 (4,−3) 2− (− 4,− 3) • x = 1

9 (iii) 2 34 4 3xxyx x x ++== + + Asymptotes are 3 and 0yx x⇒= + = 2 3 1(3 ) ax x b xk x ++ =+ With b = 4 and a = 1, 2 34 3xx kxx ++ =+ From the graph, to have two real roots, k > 1. Qn Solution 9 (i) y 0 x f( )y x= Since any horizontal line y = k, k ∈ \ will cut the graph of f exactly once, hence f is one-one. Thus, f -1 exists. (ii) Let xxy 1−= ∴ 42 1 2 2 +±= yyx But x < 0, 21 422 yxy∴ =− + Hence, f -1 : x → 21 422 x x−+ , x ∈ (-∞ , ∞) (iii) y yx= x 3y x=+ ( 2, 1)−− x = 0 y • (2,7) •

10 f( )y x= (-1,0) 0 1f( )yx −= x (0,-1) 1ff ( )yx −= (iv) Df : (-∞,0) and Rg : [-1 , 1]. Since Rg ⊄ Df , therefore fg does not exist. (v) fh(x) = f(sin x) = xx sin 1sin − Hence, fh : x → xx sin 1sin − , π < x < 2π Range of fh = [0, )∞