Pure Mathematics Unit 1 For Cape Examinations Answers

cape pure mathematics unit 2 module 1 practice questions

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CAPE PURE MATHEMATICS UNIT 2 MODULE 1 PRACTICE QUESTIONS

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1. CAPE Pure Mathematics Unit 2 Practice Questions By Carlon R.Baird MODULE 1: COMPLEX NUMBERS AND CALCULUS II 1. (a) Use de Moivres theorem to prove the trigonometric identity: 7 5 3 cos7 64cos 112cos 56cos 7cos (b) Use de Moivres theorem to evaluate 8 1 i (c) Express 2 cos3 sin3 cos sin q i q q i q in the form cos sinkq i kq where k is an integer to be determined. 2. If | 6| 2| 6 9 |z z i , (a) Use an algebraic method to show that the locus of z is a circle, stating its centre and its radius. (b) Sketch the locus of z on an Argand diagram. 3. Find dy dx in terms of x and y where 3 3 2 3 6 4x x y y x 4. (a) Find the derivative of the function 1 2ln( ) ( ) cot( ) sin( )cos( ) cos ( ) 9 ln(2 ) x h x x x x x x x (b) The curve C has equation 2 cos( )x y e x i. Show that the stationary points on C occur when tan( ) 2x ii. Find an equation of the tangent to C at the point where x=0 5. (a) Given that 8 ( , , ) 4 cos( ) sin(4 ) tan 0z f x y z xyz xy x e xz y i. Determine xf , yf , zf ii. Determine xyf , yxf , yzf (b) Given that 2 4 2 4 18 x p xv v x v
2. i. Determine p v and p x ii. Determine 2 p x v and 2 p v x 6. (a) Integrate with respect to x i. 2 10 1 x x ii. 2 15 1 x x iii. 2 2 8 1 x x (b) (i) Express the function 4 3 2 3 2 4 9 17 12 ( ) 4 4 x x x x h x x x x as partial fractions (ii) Hence, evaluate 4 4 3 2 3 2 3 4 9 17 12 4 4 x x x x dx x x x (c) Determine 1 2 1 tan 1 x dx x 7. Using the substitution secx ,find 2 2 1 1 1 x dx xx x 8. (a) Show that 4 3 1 (1 )x x x (b) Given that 1 3 0 (1 )n nI x x dx , show that 1 3 3 2 n n n I I n (c) Use your reduction formula to evaluate 4I . 9. Given that sin(2 1) sin( ) m m x J dx x , (a) Show that 1 sin2 m m mx J J m (b) Hence find 5J .
3. 10. Use the trapezium rule using 4 strips to estimate 3 0 1 tan( )x dx giving your answer to 3 significant figures.
4. By Carlon R. Baird
5. 1. (a) First lets consider 7 (cos sin )i Now, by de Moivres theorem 7 7 (cos sin ) cos7 sin7 cos7 sin7 (cos sin ) Using binomial expansion: i i i i 7 7 6 7 5 2 7 4 3 1 2 3 7 3 4 7 2 5 7 6 7 4 5 6 7 6 5 2 2 4 3 3 cos7 sin7 cos (cos )( sin ) (cos )( sin ) (cos )( sin ) (cos )( sin ) (cos )( sin ) (cos )( sin ) ( sin ) cos 7(cos )( sin ) 21(cos )( sin ) 35(cos )( sin ) i C i C i C i C i C i C i i i i i 3 4 4 2 5 5 6 6 7 7 7 6 5 2 4 3 3 4 2 5 6 7 35(cos )( sin ) 21(cos )( sin ) 7(cos )( sin ) sin cos 7cos sin 21cos sin 35cos sin 35cos sin 21cos sin 7cos sin sin i i i i i i i i Now equating real parts: 7 5 2 3 4 6 7 5 2 3 2 2 2 3 7 5 7 3 2 4 3 3 0 3 2 3 1 0 1 2 cos7 cos 21cos sin 35cos sin 7cos sin cos 21cos (1 cos ) 35cos (1 cos ) 7cos (1 cos ) cos 21cos 21cos 35cos 1 2cos cos 7cos (1) ( cos ) (1) ( cos ) (1) ( cC C C 2 3 0 3 3 7 5 7 3 5 7 2 4 6 7 5 7 3 5 7 3 5 7 7 7 7 os ) (1) ( cos ) cos 21cos 21cos 35cos 70cos 35cos 7cos 1 3cos 3cos cos cos 21cos 21cos 35cos 70cos 35cos 7cos 21cos 21cos 7cos cos 21cos 35cos C 7 5 5 5 3 3 7cos 21cos 70cos 21cos 35cos 21cos 7cos 7 5 3 cos7 64cos 112cos 56cos 7cos
6. 2 (cos3 sin3 ) cos(6 ) sin(6 ) cos sin cos7 sin7 q i q q q i q q q i q q i q (b) 8 2 2 1 Let ( 1 ) Let 1 ( 1) (1) 2 tan 1 tan (1) 4 z i p i r p arg 4 3 4 p 8 8 8 Rewriting in polar form: (cos sin ) 3 3 2 cos( ) sin( ) 4 4 3 3 2 cos( ) sin( ) 4 4 Now applying de Moivre's theorem: 3 3 ( 2) (cos(8 ) sin(8 )) 4 4 24 24 16(cos( ) sin( ) 4 4 p p r i p i z p z i z i z i ) 16(cos(6 ) sin(6 )) 16(1 (0)) 16. z i z i z (c) 2 (cos3 sin3 ) cos(2(3) ) sin(2(3) ) cos sin cos( ) sin( ) cos6 sin6 cos( ) sin( ) q i q q i q q i q q i q q i q q i q Recall that 1 1 1 2 1 2 2 2 (cos( ) sin( )) z r i z r Im z Re z arg p 1 1 Recall that cos( ) cos and sin( ) sin( )
7. C(-10,12) O 12 -10 y x 7k 2. (a) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 6 2 6 9 6 2 6 9 ( 6) 2 ( 6) ( 9) ( 6) 2 ( 6) ( 9) ( 6) 4 ( 6) ( 9) 12 36 4 12 36 18 81 12 36 4 48 144 4 72 324 3 60 3 72 432 0 ou z z i x iy x iy i x iy x y i x y x y x y x y x x y x x y y x x y x x y y x x y y 2 2 2 2 2 2 t by 3 20 24 144 0 By completing the square ( 10) 100 ( 12) 144 144 0 ( 10) ( 12) 100 The locus of z is a circle with radius 10 and centre (-10,12) x x y y x y x y (b)
8. 3. 3 3 2 2 2 2 2 2 2 3 6 4 :3 1 3 3 8 (3 3) 8 1 3 8 1 3 3 3 x x y y x d dy dy x y x dx dx dx dy y x x dx dy x x dx y 4. (a) 1 2 1 2 2 2 2 ln( ) ( ) cot( ) sin( )cos( ) cos ( ) 9 ln(2 ) ln( ) 1 ( ) sin( )cos( ) cos ( ) 9 ln(2 ) tan( ) 1 2 ln(2 ) ln( ) tan( ) 0 1 sec2 '( ) (ln(2 )) tan (cos )(cos ) (sin )( x h x x x x x x x x h x x x x x x x x x x xx x h x x x x x x 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 sin ) 18 1 1 (ln(2 ) ln( )) sec 1 '( ) cos sin 18 ln (2 ) tan 1 2 ln( ) sec 1 = cos sin 18 ln (2 ) tan 1 ln(2) sec 1 '( ) = cos sin 18 ln (2 ) tan 1 x x x x x xxh x x x x x x x x xx x x x x x x x x h x x x x x x x x
9. (b) i) 2 cosx y e x 2 2 2 2 2 2 2 cos( ) 2 sin( ) =2 cos( ) sin( ) = 2cos( ) sin( ) At stationary pts. 0 2cos( ) sin( ) 0 0 and 2cos( ) sin( ) 0 2cos( ) sin( ) 0 2cos( ) sin( ) x x x x x x x dy x e e x dx e x e x e x x dy dx e x x e x x x x x x sin( ) 2= cos( ) tan( ) 2 x x x ii) When 0x , 2(0) cos(0) 1y e We have co-ordinates (0,1) 2(0) 0 1 1 2cos(0) sin(0) 2 Gradient of tangent at x=0 is 2 So equation of tangent : ( ) 1 2( 0) 2 1 x dy e dx y y m x x y x y x 5. (a) 8 ( , , ) 4 cos( ) sin(4 ) tan( )z f x y z xyz xy x e xz y i) 8 8 4 (cos )( ) ( )( sin( )) (sin(4 )(0) ( )(4 cos(4 )) 0 4 cos( ) sin( ) 4 cos(4 ). z x z f yz x y xy x xz e z xz yz y x xy x ze xz 2 2 4 [( )(0) (cos )( )] 0 sec 4 cos( ) sec yf xz xy x x y xz x x y
10. 8 8 8 8 8 4 0 [ 4 cos(4 ) (sin(4 )(8 )] 0 4 4 cos(4 ) 8 sin(4 ) 4 4 ( cos(4 ) sin(4 )) z z z z z z f xy e x xz xz e xy xe xz e xz xy e x xz xz ii) 4 [ ][ sin( )] [cos( )][1] 0 4 sin( ) cos( ) xyf z x x x z x x x 4 cos( ) ( )(0) (sin( ))( ) 0 4 cos( ) sin( ) yxf z x xy x x z x x x 4 0 0 4 yxf x x (b) 2 4 2 4 18 x p xv v x v i) 2 2 2 2 4 0 4 2 2 p xv xv v x xv v 2 3 2 3 4 0 72 4 72 p v x x v v x v ii) 2 2 2 4 2 0 4 2 p v x v v v v 2 2 2 2 4 0 4 2 p v v v x v v 6 (a) i) 2 2 2 10 2 5 1 1 =5ln 1 x x dx dx x x x c
11. ii) 1 2 2 2 15 15 (1 ) 1 x dx x x dx x Recall that if some function 1 2 2 ( ) (1 )f x x 1 2 2 1 2 2 1 '( ) (2 )(1 ) 2 '( ) (1 ) f x x x f x x x So 1 1 2 22 2 1 2 2 15 (1 ) 15 (1 ) 15(1 ) x x dx x x dx x C iii) 2 2 2 2 8 2 8 1 1 1 x x dx dx dx x x x 2 2 1 2 1 2 2 4 1 1 2tan ( ) 4ln 1 x dx dx x x x x C (b) i) 4 3 2 3 2 4 9 17 12 ( ) 4 4 x x x x h x x x x This algebraic fraction is improper so we shall use algebraic long division: 3 2 4 3 2 4 3 2 2 4 4 4 9 17 12 4 4 0 5 17 12 x x x x x x x x x x x x x 2 3 2 5 17 12 ( ) 4 4 x x h x x x x x 2 2 2 2 5 17 12 ( 4 4) 5 17 12 ( 2) x x x x x x x x x x x Let 2 2 5 17 12 ( ) ( 2) x x q x x x 2 2 ( 2) A B C x x x Multiplying out both sides by 2 ( 2)x x gives 2 5 17 12x x 2 ( 2) ( 2)A x Bx x Cx
12. 2 2 Let 0; 5(0) 17(0) 12 (0 2) 4 12 3 x A A A 2 2 2 2 Comparing terms: 5 5 3 5 2 x Ax Bx x A B B B Comparing terms: 17 4 2 17 4 2 17 4(3) 2(2) 17 12 4 C= 17 16 1 x x Ax Bx Cx A B C C C 2 3 2 1 ( ) 2 ( 2) q x x x x So ( ) ( )h x x q x 2 3 2 1 ( ) 2 ( 2) h x x x x x ii) Hence, 4 44 3 2 3 2 2 3 3 4 9 17 12 3 2 1 4 4 2 ( 2) x x x x dx x dx x x x x x x
13. 4 4 4 4 2 3 3 3 3 1 1 3 2 ( 2) ( 2) x dx dx dx x dx x x 4 42 1 4 4 3 3 3 3 2 2 1 1 3 2 ( 2) 3 ln 2 ln 2 2 1 4 3 (4 2) (3 2) 3 ln(4) ln(3) 2 ln(4 2) ln(3 2) 2 2 1 1 9 4 1 8 3 ln( ) 2ln(2) 2