Mock Exam 1. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q1 (4 - x) 3

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1 Moc Exam Moc Exam Sectio A. Referece: HKDSE Math M 06 Q ( - x) + C () (-x) + C ()(-x) + (-x) 6 - x + x - x 6 ( x) + x (6 x + x x ) + C C C x + x + x + x (6 x + x x ) x x x x \ Costat term (6)() + (-)() + ()(6) + (-)(6). Referece: HKDSE Math M 0 Q 60 (5) d si( x + h) si x (si x) lim h 0 h x h h cos + si lim h 0 h h x h lim cos si + h 0 h (cos x)() cos x. Referece: HKDSE Math M 06 Q (a) Area ( u)( l u) square uits u l u square uits (b) Let v uits be the legth of PQ ad A square uits be the area of DOPQ. v l u dv dt u du dt Whe u e ad dv dt -, du - e dt du -e dt da du u + lu dt u dt ( + l e)( e) e \ The rate of chage of the area of DOPQ is -e square uits per secod. () (5) Hog Kog Educatioal Publishig Compay

2 Mathematics: Moc Exam Papers Module (Exteded Part) Secod Editio Solutio Guide. Referece: HKDSE Math M 0 Q Whe x 0, y ±. Differetiate both sides of the equatio of the curve with respect to x, y dy x l y y dy 0 y + + dy dy (0, ) (0, ) l9 ( )l9 l dy l l y x + y y yl y x + y + + x l \ At (0, ), the equatio of the taget is y +. x l At (0, -), the equatio of the taget is y. 5. Referece: HKDSE Math M 06 Q5 (a) Whe, L.H.S. (-)() - ( )[() + ] -- R.H.S. - \ The propositio is true for. Next, assume the propositio is true for m, where m is a positive iteger, that is, m m ( ) (m ) ( ) +, whe m +, m + L.H.S. ( ) m m+ ( ) + ( ) ( m + ) m ( ) (m + ) m + + ( ) ( m + ) m m+ ( ) (m + ) + ( ) ( m + ) m + ( ) ( m + m + ) m + ( ) [( m + ) + ] R.H.S. \ The propositio is true for m +. By the priciple of mathematical iductio, the propositio is true for all positive itegers. (7) (by the assumptio) Hog Kog Educatioal Publishig Compay

3 Moc Exam (b) ( )( + ) ( ) + ( ) ( ) ( )() ( ) () 07 ( ) [(07) ] (6) 6. (a) si x si(x + x) si x cos x + cos x si x si x ( - si x) + cos x ( si x cos x) si x - si x + si x cos x si x - si x + si x ( - si x) si x - si x + si x - si x si x - si x (b) si 0 si 7 si (6 ) si (6 ) si 6 - si 6 si 6 cos 6 (by (a)) - si 6 cos 6 (si 6 0) - ( - cos 6 ) cos 6 cos 6 - cos 6-0 \ cos 6 is a root of the equatio x - x - 0. \ cos 6 ( ) ± ( ) ()( ) () + 5 or - 5 (rejected) (7) 7. Referece: HKALE P. Math 0 Paper Q5 (a) Let x + si q. The cos q dq. siθ x \ cosθ x 6x x si θ \ \ cos θ x(6 x) 9 x(6 x) (cos θ)(cos θ) dθ 9 cos θdθ ( + cos θ) dθ 9 9si θ θ + + C 9 9siθcosθ θ + + C 9 x x x x + + C si x x + x x si 6 + C, where C is a costat. Hog Kog Educatioal Publishig Compay

4 Mathematics: Moc Exam Papers Module (Exteded Part) Secod Editio Solutio Guide (b) The equatio of the curve ca be rewritte as y ± x(6 x). I the first quadrat, y x(6 x). Whe y 0, x(6 x) 0 x(6 x) 0 x 0 or 6 The required volume 6 [ x(6 x)] 0 6 x(6 x) 0 9 x x + si 9 9. Referece: HKDSE Math M 06 Q 6x x 6 0 (by (a)) () (a) (i) (ii) (iii) M M Similarly, we have M 0. 0 T ( M ) T ( M ) 0 (b) (i) ( ) + It is the sum of a geometric sequece with first term ad commo ratio. Hog Kog Educatioal Publishig Compay

5 Moc Exam (ii) Similarly, () Sectio B 9. Referece: HKDSE Math M 06 Q9 (a) Sice P(-, 9) is a poit o C, we have (-) + p(-) + q(-) + 9 p - q 6... () dy x + px + q Sice P(-, 9) is a statioary poit o C, we have dy 0 x (-) + p(-) + q 0 p - q... () () - (): q - Substitutig q - ito (), p - (-) p (b) From (a), dy x + x d y 6x + d y 6( ) + < 0 x \ P is a maximum poit of C. i.e., P is ot a miimum poit of C. () () 5 Hog Kog Educatioal Publishig Compay

6 Mathematics: Moc Exam Papers Module (Exteded Part) Secod Editio Solutio Guide (c) Whe dy 0, d y x + x 0 (x )( x + ) 0 x or - x > f \ (-, 9) is a maximum poit., 7 is a miimum poit. () (d) Whe d y 0, x. x d y x < x x > f \ + +, 5 7 is the poit of iflexio. (e) The equatio of L is y 9. Whe y 9, x + x - x + 9 x + x - x - 0 (x + ) (x - ) 0 x or - Area [9 ( x + x x + )] ( x x + x + ) x x + x + x () () 6 Hog Kog Educatioal Publishig Compay

7 Moc Exam 0. (a) (i) Let u - x. The -du. Whe x, u. Whe x, u. (ii) By (a)(i), l si( x) l si udu l si udu l si x l si( x) lsi x 0 [l si( x) l si x] 0 x l si( ) 0 si x sicos x cossi x l 0 si x l(si cotx cos ) 0 (6) (b) (i) xcsc x cotx + x d(cot x + ) cotx + ( xd ) [l(cotx + )] [ xl(cotx + )] + l(cot x + ) l + l(cot x + ) 7 Hog Kog Educatioal Publishig Compay

8 Mathematics: Moc Exam Papers Module (Exteded Part) Secod Editio Solutio Guide (ii) Note that < <. Puttig i (a)(ii), we have l si x cot cos 0 l x cot + 0 [l(cotx + ) l ] 0 l(cot x + ) l xcsc x l l + cotx + l l. Referece: HKDSE Math M 06 Q (a) (i) () Let A be the coefficiet matrix. (by (b)(i)) (6) Choose a suitable value of i (a)(ii) so as to evaluate l(cot x + ). l x + cot x + l cot l(cot x + ) l A p p p 5 0 p p p p 5 ( R R R ) p 5 + ( p)( p) + ( p) + p( p) p p + ( p )( p ) (E) has a uique solutio if ad oly if A 0. \ p ad p. Hog Kog Educatioal Publishig Compay

9 Moc Exam () q p q p p 5 0 q p ( C + C C) q p ( p ) q p q - p q - p q q p ( C + C C ) p + q p ( p ) ( p )( q ) ( p )( p + q ) ( p )( p + q ) q(p - 5) + p + (q - ) - q - (p - 5) - p(q - ) pq - 7q + -(q - ) - q + ( - p) + + (q - ) - q( - p) pq - p - 5q + By the Cramer s rule, the solutio is ( p )( p + q ) ( p + q ) x ( p )( p ) p pq 7q + y ( p )( p ) pq p 5q + z p (ii) () The augmeted matrix is q 7 q q 0 0 q 6 ( R R R ; R R R ) 0 q ( R R R ) q Sice (E) is cosistet, -q - 0, i.e., q -. () By (a)(ii)(), the augmeted matrix is ( R + R R ) Let z t, where t is ay real umber. + The y -6 - t ad x - - t. \ The solutio is x - - t, y -6 - t ad z t, where t is ay real umber. (9) 9 Hog Kog Educatioal Publishig Compay

10 Mathematics: Moc Exam Papers Module (Exteded Part) Secod Editio Solutio Guide (b) The augmeted matrix is R R R, R \ The system of liear equatios is equivalet to (E) for p ad q -. By (a)(ii)(), x - - t, y -6 - t ad z t, where t is ay real umber. If xy z, the (- - t)(-6 - t) t + 6t + t t t + t (*) Note that - ()() - < 0. \ (*) does ot have real solutio. \ There is o solutio of the system of liear equatios satisfyig xy z. (). Referece: HKCEE A. Math 00 Q5 (a) OM a CM OM OC ( r) a sb Sice C is the circumcetre of DOPQ, OP CM 0 ( OP CM) a [( r) a sb] 0 ( r) a a sa b 0 ( r) s 0 6 r s 0 r + s 6... () (b) Let R be the mid-poit of OQ. OR b CR OR OC ra + ( s) b OQ CR 0 ( OQ CR) b [ ra + ( s) b] 0 rb a + ( s) b b 0 r + s 0 r + s 0 r + s... () \ \ (): r + 9s 9... () () - (): s s () Cosider OQ i CR, where R is the mid-poit of OQ. 0 Hog Kog Educatioal Publishig Compay

11 Moc Exam (c) Substitutig s + r 5 r \ r ito (), 5 ad s ON a + b a + b ( ) CN ON OC 5 a + b + a b 5 a + b Sice H is the orthocetre, OH PQ. Sice C is the circumcetre, CN PQ. \ OH // CN Sice OH // CN ad OH : CN :, OH + 5 b QH OH OQ 5 a + b QH OP 0 5 a + b ( a) 0 5 a a + b a 0 \ (5 6) 0 OH + a 5 b a + 5 b () (7) N is the mid-poit of PQ. Hog Kog Educatioal Publishig Compay

Mock Exam 4. 1 Hong Kong Educational Publishing Company. Since y = 6 is a horizontal. tangent, dy dx = 0 when y = 6. Section A

Mock Exam 4. 1 Hong Kong Educational Publishing Company. Since y = 6 is a horizontal. tangent, dy dx = 0 when y = 6. Section A Mock Eam Mock Eam Sectio A. Referece: HKDSE Math M Q d [( + h) + ( + h)] ( + ) ( + ) lim h h + h + h + h + + h lim h h h+ h + h lim h h + h lim ( + h + h + ) h + (). Whe, e + l y + 7()(y) l y y e + l y