# Mock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =

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1 Mock Eam 7 Mock Eam 7 Section A. Reference: HKDSE Math M 0 Q (a) ( + k) n nn ( )( k) + nk ( ) + + nn ( ) k + nk A nk... () nn ( ) k... () From (), k...() n Substituting () into (), nn ( ) n 76n 76n 0n 7 nn ( ) 0 n or 0 (rejected) Substituting n into (), k A (b) Coefficient of C ( ). For n, L.H.S.! R.H.S. ( + )! - L.H.S. he proposition is true for n. Net, assume the proposition is true for n m, where m is a positive integer, that is, m k k! k (m + )! -. When n k +, L.H.S. m + k k! k m k k! + ( m + ) ( m + )! k [(m + )! ] + (m + ) (m + )! (by the assumption) (m + )!( + m + ) (m + )(m + )! (m + )! [(m + ) + ]! R.H.S. he proposition is also true for n m +. By the principle of mathematical induction, the proposition is true for all positive integers n. () A A (6) Hong Kong Educational Publishing Company

2 Mathematics: Mock Eam Papers Module (Etended Part) Second Edition Solution Guide. Reference: HKDSE Math M 0 Q d (csc ) d d d sin lim + h 0 h sin( h) sin sin sin( + h) lim h 0 hsin( + h)sin + ( + h) ( + h) cos sin lim h 0 hsin( + h)sin h cos + h sin lim h 0 sin( + h)sin h cos ( ) sin csccot (). Reference: HKCEE A. Math 00 Q (a) + tana tan B tanb + tan tan B tan + tan B tan B (b) + tan tan tan + tan tan tan + tan tan tan tan tan 0 (by (a)) < tan > < ± tan ( ) ( ) ()( ) () ± tan + A () Hong Kong Educational Publishing Company

3 Mock Eam 7. Reference: HKDSE Math M 0 Q9 (ln ) d (ln ) d[(ln ) ] (a) (ln ) ln d (ln ) ln d (ln ) ln d(ln ) (ln ) ln + d (ln ) ln + d (ln ) ln + + C, where C is a constant A e (b) Volume () ( e ) (ln y) dy e y(ln y) yln y + y ( e + ) 6. Reference: HKDSE Math M 0 Q e (by (a)) + A A (6) (a) Slope of the line - - Let (a, b) be the coordinates of P. dy d ( ab, ) + a + a a a Substituting (, b) into - y - 0, () - b - 0 b - he coordinates of P are (, -). A (b) dy + d y + d ln + + C Substituting (, -) into y ln + + C, ln + + C C he equation of the curve is y ln +. A (c) he equation of the normal is y ( ) ( ) + y + 0 A (6) Hong Kong Educational Publishing Company

4 Mathematics: Mock Eam Papers Module (Etended Part) Second Edition Solution Guide 7. (a) ( ) (ln )() f ( ) ln When f () 0, ln 0 ln e 0 < < e e > e f () When e, f () is maimum. he maimum value of f( ) f() e ln e e e A (b) By (a), f (e) f () for > 0. ln e eln ln e e e. Reference: HKDSE Math M 0 Q7 (a) Area OA OB i j k 0 i j + k ( 6) (6) A (b) Volume of the tetrahedron OABC OA OB OC 6 ( ) (i 6j + k) (i j + k) 6 ()() + ( 6)( ) + ()() 6 0 he required distance is the height of the tetrahedron OABC with respect to the base OAB. Hong Kong Educational Publishing Company

5 Mock Eam 7 Let h be the distance between point C and the plane OAB. h h he distance between point C and the plane OAB is. A () 9. (a) A + 60 k + 9k k 6 k k 7 7 A k k k 0 k k 6 k 6 + k 7 k k 0 6 k k + 6 k k 7 (b) A k 0 6 k k + 6 k k 7 k 0 k k + k k 7 ake k, we have A. he given system of linear equations can be rewritten as A y z 0 y A z he solution is, y, z -. A (7) + A Hong Kong Educational Publishing Company

6 Mathematics: Mock Eam Papers Module (Etended Part) Second Edition Solution Guide Section B 0. (a) DO DC (diag. of // gram) CD : DE : OE OC (6i + ) j i j A () (b) AC OB mi + nj BC OC OB (6i + ) j ( mi + nj) (6 m) i + ( n) j A AC BC ( in semicircle) AC BC 0 m(6 m) + n( n) 0 m + n 6m + n... () AB OB OA OB BC ( mi + nj) [(6 m) i + ( n)] j (m 6) i + (n ) j BE OE OB ( i ) j ( mi + nj) ( m) i + ( n) j AB BE (tangent radius) AB BE 0 (m 6)( m) + (n )( n) 0 m + n... () Substituting () into (), 6m + n n 6m... () Substituting () into (), 6m m + 6m m + 6m 600 m m 9 0 m ( ) ± ( ) ()( 9) () + - or (rejected) A 6 Hong Kong Educational Publishing Company

7 Mock Eam 7 Substituting m n + into (), (c) OB i + j + OB + + EB + i + + j 9 + i + j A (7) Note that ABE. If O is the incentre of DABE, then OB bisects ABE, i.e., OBE. EB OB EB 7 OB EB cos OBE OB OE 7 + A OBE 6 O is not the incentre of DABE. A () 7 Hong Kong Educational Publishing Company

8 Mathematics: Mock Eam Papers Module (Etended Part) Second Edition Solution Guide p q. (a) p q pq p q p q p q pq p q 0 p q 0 ( R R R ; R R R ) pq p q p q pq p q ( p )( p + ) ( q )( q + ) pq( p )( q ) p + q + pq( p)( q)( q p) (b) (i) If (E) does not have unique solution, then h h h ( h)()( h)( )( h) 0 (by (a)) h 0, or () A (ii) he augmented matri is a a b ~ 0 0 b a c 0 0 c 9a ( R R R ; R 9 R R ) a ~ 0 0 b a c b + a ( R R R ) A Since (E) is consistent, c - b + a 0. () (c) ake h, a, b, c -. hen we have + y + z ( E) : 9 + 9y + z 7 + 7y + z By (b)(ii), the augmented matri is Let y t, where t is any real number. hen t 6, y t and z. A Hong Kong Educational Publishing Company

9 Mock Eam 7 Substituting t 6, y t and z into + 6y - z 0, + t 6t 0 6 t t 6, y 6 6 6, z A (). Reference: HKDSE Math M 0 Q (a) (i) y ( ) dy (ii) + d dy When 0, d ( )( ) 0 or (rejected) A 0 < < dy d When Minimum length > - 0 +, then length of the road attains its minimum. A km km km + 7 km A (6) 9 Hong Kong Educational Publishing Company

10 Mathematics: Mock Eam Papers Module (Etended Part) Second Edition Solution Guide (b) (i) In DBCX, sinα BC 0 BX CX cosα BX XY CX sinβ sin( α β) XY sin β sin( α + β) sin β km sinαcosβ + cosαsin β km sinαcotβ + cosα β + km 0 cot (ii) Let L km XY. 0 cotβ + km dl β ( 0csc ) dt (0cot β + ) dl Note that 0. dt dβ dt When CY is the shortest, then CY XB, we have β csc β csc α sec α cotβ cot α tanα he rate of change of β is dβ 0 dt dβ dt α, radians per hour. A A A () 0 Hong Kong Educational Publishing Company

11 Mock Eam 7. (a) Let u -. hen du -d. When 0, u ; when, u f ( ) d ( u) f( u)( du) ( u) f( u) du ( ) f( ) d f( ) d f( ) d f ( ) d f ( ) d f( ) d 0 0 f( ) d (b) (i) Let u cos. hen du -sin d. When 0, u ; when p, u -. 0 sincos + d u cos + u du u + u du + u du + u du du + u du [ u] + u d + A () Hong Kong Educational Publishing Company

12 Mathematics: Mock Eam Papers Module (Etended Part) Second Edition Solution Guide (ii) Let tan q. hen d sec q dq. When -, θ ; when, θ. 0 sincos + d cos + d sec θdθ + tan θ [ θ] sec θdθ sec θ dθ sincos (c) Let f( ). + cos sin( )cos ( ) f( ) + cos ( ) sincos + cos f( ) sincos sincos d + + d (by (a)) cos 0 0 cos (by (b)) A (6) A () Before we use the result of (a), we must point out that the function fulfills the requirement. Hong Kong Educational Publishing Company

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