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∴ ==