GCE A Level H Mathematics November 0 Paper i) f () f( f() ) f ( ),, 0 Let y y y y y y y f (). D f R f R\{0, } f () f (). Students must know that f () stands for f (f() ) and not [f()]. Students must also show that both the rule and domains of f and f are equal. (ii) f () f () f( f () ) f( f () ) f (),, 0 ) y + y + 5 0 dy dy d + y + y d + y 0 When dy d, + y y + y 0 y y Substitute y into original equation: + + 5 0 5 7 Substitute y into original equation: + + 5 0 No solution. Hence there is only one point where dy d. Coordinates (, ). Need to give the final answer in coordinates, instead of just, y.
i) a b 0 a b sin 0 a 0 or b 0 or a is parallel to b. (ii) n + + 9 (iii) cos 0 0 9 Hint to students: n 0 n is parallel to There is no need to work out the value of the angle. i) y (0, d ) ( a, 0) (b, 0) (c, 0) (0, d ) (ii) The tangents are parallel to the y ais. 5i) z ( + i) + (i) + (i) + i + i z ( + i)( + i) + i 6i 8 + i i + i + i + 5 + 5 i Teaching Pointers: Although the question says "without using a calculator", it is advisable to use a calculator to check your answers.
(ii) pz + q z p + pi 5 q + qi 5 p + q 5 0 q 5 p q 50p pz + q z p 5 ( 50p) p + p 9p is real 6a)(i) Let P(n) be the statement p n (7 n ). When n, LHS p P() is true. RHS (7 ) LHS Assume that P(k) is true for some k Z +, i.e., p k (7 k ). To prove that P(k+) is true, i.e., p k+ (7 k+ ). LHS p k+ p k 7 (7 k ) 7 8 k+ 7 7 k+ (7 k+ ) RHS P(k) is true P(k+) is true. Since P() is true and P(k) is true P(k+) is true, by the Principle of Mathematical Induction, P(n) is true for all n Z +. (ii) n p r n (7 r ) r r n 7 n r r r (7n) n 7n 9 (n ) (bi) As n, (n + )! 0. the series u r converges. Sum to infinity.
(ii) u n S n S n (n + )! [ n! ] + (n + ) (n + )! n (n + )! 7i) From GC,.885 to d.p. Let 6 7 7 6 0 ( ) 0 0 or since > 0. (ii) From GC, f() d 0.597 (iii) Area 0 f() + 7 d 0 6 d [ 7 7 5 5 ] 0 [ 7 7 5 5 7 5 (iv) f( ) ( ) 6 ( ) 7 6 7 f() f( ) f( ) 0 If is a root, then so is. Since the coefficients of f() are all real, all comple roots occur in conjugate pairs. If a + bi is a root, then so is a bi. From above, if a + bi is a root, then so is (a + bi). So the 6 roots of f() are and comple roots of the form a bi, where a, b R. 8i) 9 d sin + C ] Since the region is situated below the ais, the area is the negative of the integral. Or students can take the absolute value of the integral. Hint to Students: Students can use the PlySmlt App to get a hint of what the 6 roots look like. Remember to add an arbitrary constant for an indefinite integral.
(ii) 9 9 9 9 / [ 9 + [ + 8 + 8 8 + 5 6! 9 6 79 +... ] + 5 + 68 + 5 6 99 +... (iii) sin 9 d 5 + + 5 + 68 + 5 6 99 d + 6 + 5 0 + 5 7 9 + C When 0, sin 0 C 0. sin + 6 + 5 0 + 5 7 9 9i) 8 ( + 6) 5 0 5 5 + 0 cartesian equation of plane q is y z 0.! 9 +... ] (ii) From GC, equation of m is r 6 0 + /, R. (iii) Square of distance AB (6 + ) + ( + + ) + ( ) (5 + ) + ( + ) + ( ) 5
5 + 0 + + 6 + + + 6 + 9 + 8 + 50 ( + 7 ) + 50 [ ( + 7 ) 9 ] + 50 ( + 7 ) + 7 Distance is minimum when 7 Coordinates of nearest point (6 7 (), 7, 7 ) ( 8 7, 5 7, 7 ) Students can also use differentiation to find the minimum distance instead of completing the square. An "otherwise" method to do this question is to find the foot of the perpendicular from the point A to the line m. The final answer must be given as coordinates, instead of as a position vector. 0i) Substitute, d dt : k( + ) 5 k k 5 Shown (ii) + ( ) + [ ( ) ] + 5 ( ) d dt 5 [ 5 ( ) ] 5 ( d 5 dt ) 5 ln 5 + 5 5 t + C When t 0, C 5 ln 0 No need to have modulus signs for logarithm because it is given that 0. 5 ln 5 + 5 + 5 t t 5 + 5 ln 5 + 5 + 5 ln 5 + 6
(iii)(a) When, t 5 ln 5 + 5 5 ln 5 + 5 (b) (iv) / y 5 + When 0, t 5 ln 5.5 min 5 + ln 5 + 5 t 5 + 5 + e t/ 5 5 + ( 5 + )e t/ 5 ( + e t/ 5 ) ( 5 + )e t/ 5 + 5 ( ) + e t/ 5 + 5( e t/ 5 ) ( + e t/ 5 ) + 5 e t/ 5 + e t/ 5 Only need to draw the part of the curve for which 0 and t 0. The curve looks like a straight line, but it is actually curved..5 t i) Volume V r + r h r + r r dv dr r + 6r r 6 r + (6r r 6 ) / (6r 6r 5 ) r + r r 5 6r r 6 r r r + 6 r 7
When r r, dv dr r r r + 6 r 0 r r 6 r 6r 6 r r 6r (6 r ) 9r 9r + 0 5r 768r + 0 0 Shown (ii) From GC, r.07,.95 (iii) When r.07, dv dr.07 + (.07) (.07 ) 6.07 8. 0. Hence r.07 does not give a stationary value of V. r.95 h 6.950797 0.65 (iv) V y 9. When drawing the curve, students must draw the maimum point at r.95.95 r 8
GCE A Level H Mathematics November 0 Paper i) t d dt 6t y 6t dy dt 6 dy d 6 6t t 0. t.5 (ii) Equation of tangent at P is y 6p p ( p ) When 0, y 6p p p Coordinates of D is (0, p). Midpoint of PD is given by p p y 9, y 6p + p y 9 y 7 7 y 9p 9 + i) Let ( 5)( + 9) A 5 + B + C + 9 9 + A( + 9) + (B + C)( 5) Substitute 5 : 8 6 A A Comparing coefficients of : 9 + B B Comparing constants : 7 5C C 8 9 + EQ \i\in(0,, ) ( 5)( + 9) d 0 5 + + 8 + 9 d 0 5 d + 0 + 9 d + 0 8 + 9 d [ ln 5 + ln ( + 9) + 8 tan ] 0 ln + ln + 8 tan [ ln 5 + ln 9 ] ln 5 + 8 tan It is advisable to use the calculator to verify that the final answer is correct. i)(a) Distance + 8 + +... to 0 terms 0 [ 8 + (0 )8 ] 0 m 9
(b) Distance n [ 6 + (n )8 ] n [ 8 + 8n ] n(n + ) 5000 From GC, n n(n + ) 88 < 5000 760 < 5000 5 500 > 5000 least number of stages 5. (ii) Distance + 8 + 6 + +... to n terms 8 n 8( n ) From GC, n 8( n ) 9 088 < 0000 0 88 < 0000 676 > 0000 he has run eactly 0 km during the th stage. Distance from O 0000 88 86 m Since OA 8 9 096 > 86, he is running from O towards A. a)(i) z ( 5 + i) Im 5 Re (ii) z 6i z ( 0 i) Im 6 5 Re Students should use compasses and ruler to draw the loci as neatly as possible. Students who cannot draw neatly can use the graph paper provided. ( 0, ) By Pythagoras' Theorem, + 0
Values of z 5 + + i( ), 5 + i( + ). (b)(i) w 6 ( i) 6 (e i /6 ) 6 6e i 6e i since 0 < Even though the question says "without using a calculator", students can use a calculator to verify that their answer is correct. (ii) w n w* (e i /6 ) n e i /6 n e in 6 i 6 n e i(n+) 6 (n + ) 6 (n + ) 6 n [cos i sin w n w* is a real number (n + ) sin 6 0 (n + ) 6,,,... (n + ) 6,, 8,... n + 6,, 8,... n 5,, 7,... The smallest positive values of n are 5,, 7. 0000 5i) 500 0 Choose a random starting number from to 0, eg. 5. Select the 5th, 5th, 5th, 65th,... customers on the list for the survey. ] (ii) Advantage: The sample is evenly spread out over the list of customers. Disadvantage: It is biased if the list of customers has a periodic or cyclic pattern. Students must answer in contet of the question. 6i) No. of ways C 8 C 5 C 6 C 70 0 5 500 (ii) If the particular midfielder is in the team but not his brother, no. of ways C 8 C C 5 C 70 5 00 Students may also use the complementary method: Total number of ways no. of ways
If the particular attacker is in the team but not his brother, no. of ways C 8 C C 5 C 70 6 0 600 Total no. of ways 00 + 600 6800 (iii) If the particular remaining midfielder plays as a midfielder, no. of ways C 8 C C 5 C 70 5 50 If the particular midfielder plays as a defender, no. of ways C 8 C C 5 C 56 5 50 If the particular midfielder is not playing, no. of ways C 8 C C 5 C 70 5 50 Total no. of ways 50 + 50 + 50 880 where both brothers are in the team no. of ways where both brothers are not in the team Since the ways to choose goalkeepers and attackers are the same in all cases, students may write the working as C ( 8 C C + 8 C C + 8 C C ) 5 C (70 + 56 + 70 ) 5 This may help to simplify the working a little and avoid confusion. 7i) Let X no. of sies out of 0 rolls. X ~ B(0, 6 ). P(X ) 0.55 (ii) Let Y no. of sies out of 60 rolls. Y ~ B(60, 6 ). Since n 60 is large, np 0 > 5, nq 50 > 5, Y ~ N(0, 50 6 ) approimately. P(5 Y 8) P(.5 Y 8.5) by c.c. 0.7 Students must remember to do continuity correction. (iii) Let W no. of sies out of 60 rolls of the biased die. W ~ B(60, 5 ). Since n 60 > 50, np < 5, W ~ Po() approimately. P(5 W 8) P(W 8) P(W ) 0.50 8a)(i) y
(ii) y (b)(i) For model A, r 0.970 For model B, r 0.979 (ii) Since r is closer to for model B, model B is the better model. Regression line is P 659.78 ln m + 9569.559 700 ln m + 96000 (iii) Estimated price 659.78 ln 50 + 9569.559 $6000 9i) Let be the population mean no. of minutes the bus is late. H 0 :. H : <.
(ii) Unbiased estimate of population variance 0 9. 9 Under H 0, H 0 is not rejected T. ~ t(9). /9 0 t. > InvT(0., 9) /9 0 t. >.8087 /9 0 t >.8 Set of values { t R: t >.8 }.0. (iii) H 0 is rejected 0k.8087 /9 0 0. k/.8087 k 0.65076 k 0. Set of values { k R: 0 < k 0. } 0i)(a) Probability 0 0 0 0.00 Since sample size is small and population variance is unknown, the t test has to be used. Since H 0 is not rejected, the test statistic lies outside the critical region. InvT(0., 9) Since H 0 is rejected, the test statistic lies inside the critical region. 0. 0. t InvT(0., 9) t (b) Probability P(no stars) 9 8 9 0 0 0 0.5 (c) P(,, +) P( from set &, + from set ) + P( from set &, + from set ) + P( from set &, + from set ) 0 0 0 + 0 6 + 6 + 6 000 0.058 0 0 + 0 0 0 (ii) P(eactly ) P( from set, non from set & ) + P( from set, non from set & ) + P( from set, non from set & ) 8 9 0 0 0 + 9 9 0 0 0 + 9 8 0 0 0 7 + 6 + 7 000
0.06 P(+, o eactly ) P(+, o, ) P(eactly ) P(+,o, ) + P(+,,o) + P(o,+, ) + P(o,,+) + P(,+,o) + P(,o,+) P(eactly ) 0 0 0 + 0 0 0 + 0 6 + + 6 + 8 + 9 + 8 000 0.06 0.07 0.06 7 06 0 0 + 0 0.06 0 0 + 0 0 0 + 0 i) Let A, B no. of originals and prints sold in a week. A ~ Po() B ~ Po() (a) P(B > 8) P(B 8) 0.768 (b) A + B ~ Po( + ) Po() P(A + B < 5) P(A + B ) 0.675 (ii) Let W no. of originals sold in n weeks. W ~ Po(n) P(W < ) < 0.0 P(W 0) + P(W ) + P(W ) < 0.0 e n + e n n + e n (n)! < 0.0 e n ( + n + n ) < 0.0 From GC, n e n ( + n + n ) 0.0697 > 0.0 0.075 > 0.0 5 0.0077 < 0.0 smallest possible n 5. 0 0 (iii) Method : Since n 5 is large, by CLT, B + B +... + B 5 ~ N(5, 5 ) N(57, 57) approimately. P(B + B +... + B 5 > 550) 0.8 Method : Let Y no. of prints sold in year. Y ~ Po(5 ) Po(57) Since mean 57 > 0, Y ~ N(57, 57) approimately. P(Y > 550) Both methods are acceptable, even though the answers are different. If Central Limit Theorem is used, there is no need to do continuity correction. But if the Poisson Distribution is approimated by the Normal Distribution, then continuity correction is necessary 5
P(Y > 550.5) 0.86 (iv) The mean no. of paintings sold per week may not be constant, as more may be sold during festive seasons. The sale of paintings may not be independent, as a person who bought a painting will likely buy other paintings by the same artist. Students must give reasons in contet of the question. 6