Final Examination F.5 Mathematics M2 Suggested Answers

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1 Final Eamination F.5 Mathematics M Suggested Answers. The (r + )-th term C 9 r ( ) 9 r r 9 C r r 7 7r For the 8 term, set 7 7r 8 r 5 Coefficient of 8 C d 8 ( ) set d if > slightly, d we have <, if < slightly, d >. Many students claim that when the maimum value occur, but they didn t test it. d attain it ma value when. Hence, (, 7) on y f (), and let y 8 + C, where C is a constant. sub (, 7) on y, we have C the equation of the curve is y a) Let f () 8 8 f ( ) ( ) 8( ) is a root of f () Q α is the only integral root of f (), α. b) f () ( + )( ) α, β + γ and βγ αβγ + αβ + αγ α(βγ + (β + γ)) ( + ).

2 . consider +, we have or. Now, + d + d + + d ( + ) d + ( + ) d Most of the students think that + d ± + d. 5. a) L.H.S. a b) Q a dv dt dv ds ds dt v dv ds dv v ds d ds (v ) R.H.S. s d ds (v ) s d ds (v ) s When v, s. v s ds v s + C. + C C v s + s v Not able to find v, from d ds (v ) s Given s, so a. Some of students thinks that 8 a d ds (v ), hence v 9 s. This is wrong, ce a is the function of s, so the integrand can t found like this.

3 . a) When t, T b) T 8 () 8() The initial temperature of the solution is 8 7 (85 ) C. 8 t 8t dt dt 8 (t 8) (t 8t + 5) (9 t) (t 8t + 5) When t 5, dt dt t 5 [9 (5)] [(5) (8)(5) + 5] 8 The temperature of the solution decreases at a rate of 8 C/min when the chemical has been just added for 5 min. 7 a) e d de e e d e e + C where C is a constant. b) by shell method, we have Volume y d e d Some students try to find the volume by ug dice method, but it is very difficult to do so, as it is hard to give in term of y de [ e ( e ) d ] [() e () ()e () + e() e() ] (by a)) A 5e A

4 8. a) ( + ) d lim ( + ) lim (( + ) ( + ) + ) (( + ) ( + ) + ) ( )( + ) lim lim ( ) ( + ) ( )( + ) ( )( + ) b) d (() ) When, y The equation of tangent : y + ( ) y + or + y 9. a) Let P(n) be the statement n n ( n) when n, LHS ; P() is true. RHS () ( ()) Assume P(n) is true for some positive integer k, i.e. k k ( k) when n k +, LHS k k + k ( k) k +

5 k ( k + ) ( k + ) k + ( k + ) RHS P(k) is true. By M.I., P(n) is true for all +ve integer n. k + b) lim n lim n lim n n ( n) ( n) n lim n ( n) n n Students overlook that only n as n, but is not a tend to, i.e. is not tend to.. a) ln y y lny y d (lny) y lny y y Not able to solve ln y. y lny d y y lny y + C b) i) For y e, when, y. The coordinates of A (, ) Equation of L: y ( )( ) y + ii) y e, ln y lny Rewrite y + as y. The required area ln y ( y) e e ln y e + y ( ) Fail to use the results of part a) to find the answer. The presentation on finding the area is not clear, it is better to identify which parts they are going to figure out.

6 e y [ y ln y y] + y (by (a)) e e [ ( )] + e e e+. a) Since f () is continuous at, and lim Hence, the vertical asymptote is. Let y m + c be the oblique asymptote. and lim +. + m lim ± lim lim ± ± c lim () lim ± ± Hence the oblique asymptote is y +. ( ) lim lim ± ± OR f () + ( + ) + ( ) b) ( ) ( ) 8 ( ) f f ( ) < < < < < > f () undefined 8 f () + undefined + f () undefined Maimum point (, ) Minimum point (, 8) No infleion point. Very careless to give is y. +,and claim that the asymptote They don t alert the degree of numerator and denominator are the same and need further division.

7 c) The graph of y f ( ) : y y f () min (, 8) y + ma (, ) O. a) f () f (a ) Let u a, a, u a du d a, u f () d [( a u) f (a u)]( du) a a a ( a f (a u) u f (a u))du a a f u a a f f () d a a ( ) du a u f (u)du ( ) d a f () d f ( ) d a a f ( ) d. b) Let u, u du d, u

8 + + d ( + u) du ( + u) + ( + u) (-u) ( u) + ( u) du d. + d d d d + + d d. c) Let f () f ( ) + ( ) ( ) + ( ) + f () Only a few of the students check the condition of the function before ug the results of part a). + d + d ( by a) )

9 . a) y...() + y 9...() Substituting () into (), + () 9 ( + 9)( ) or 9 (rejected) Substituting into (), y y ± The coordinates of P (, ) The coordinates of Q (, ) b) For y, y or For - + y 9, y 9 The required area or 9 For 9 d + d 9 Then d θ dθ. When, θ. When, θ. d, let θ. The required area 9 9 θ d+ θ dθ 9 + θ dθ Not familiar to solve d θ d θ θ θ +

10 9 + [ θ + θ θ ] 9 + [ θ + θ θ ] OR For 9 y d y, let y θ Then θ dθ When y, θ ) When y, θ Area ( 9 y y ) y ( θ θ) dθ θ dθ [θ + θ ( + θ )dθ ] [ 9 θ + 9 θ θ ] ( ) ( ) ( )( ) c) Rewrite + y 9 as y Not able to simplify this epression. The required volume d + 9 d

11 [ ] a) Let 5 θ + θ when, θ d θ θ dθ θ θ dθ when, θ 8θ θ dθ 5 d 5 θ θ 5 θ + θ 8θ θ dθ ( θ θ 8 θ θ )dθ 8 θ dθ 8 ( + θ)dθ θ + θ + b) Let t tan when, t dt sec d when, t d dt + t

12 + d d + ( + t + + t ) dt dt + t + dt + t Weak on handling complicated trigonometry and algebra simplification. Let t tanθ t, θ dt sec θ dθ t, θ + t dt sec θ dθ sec θ [θ] 5. a) Let f () p p + q then f () pp - p p Set f (), we have. > (slightly), f () > < (slightly), f () < f () attain its min. value when. least value of f () p + q. b) f () f () f () p + q p p + q p p + q

13 a sub b q/p, we have a p b q/p p a p p + b q q + q a b q/p a q/p b bq q ( /p) ab ab q (/q) ab. Not able to try a b q/p a p p + b q q ab