M2_Bk1_Sol_Ch00_E.pdf

# M2_Bk1_Sol_Ch00_E.pdf - 0 New Progress in Senior...

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© Hong Kong Educational Publishing Co. 2 New Progress in Senior Mathematics Module 2 Book 1 (Extended Part) Solution Guide 0 0 pp.6 – 8 p.6 1. 5 2 5 2 + = + 7 = 2. 10 7 10 7 = 3 = 3. 6 2 7 = + 6 = 4. 6 4 6 4 + = + 10 10 = = p.8 1. 1 2 3 4 5 ! 5 × × × × = 120 = 2. 1 8 9 10 ! 10 × × × × = 3628800 = 3. 1 2 3 4 1 7 8 9 ! 4 ! 9 × × × × × × × = 15120 5 6 7 8 9 = × × × × = 4. 1 6 7 8 1 10 11 12 ! 8 ! 12 × × × × × × × × = 11880 9 10 11 12 = × × × = 5. ) 1 2 3 4 ( ) 1 2 3 4 ( 1 6 7 8 ! 4 ! 4 ! 8 × × × × × × × × × × = 70 1 2 3 4 5 6 7 8 = × × × × × × = 6. ) 1 2 3 4 5 6 ( ) 1 2 3 4 5 ( 1 9 10 11 ! 6 ! 5 ! 11 × × × × × × × × × × × × × × = 462 1 2 3 4 5 7 8 9 10 11 = × × × × × × × × = pp.1 – 10 0.1 (a) 3 3 3 5 6 3 5 6 × = p.1 5 3 2 15 3 6 = = (b) 2 3 2 3 2 3 2 2 3 2 + + × = ) 2 3 ( 2 2 3 ) 2 3 ( 2 + = + = 0.2 3 2 2 2 1 2 2 2 1 + + + p.2 2 1 3 2 2 3 2 1 2 12 8 3 2 2 2 8 4 2 2 2 3 2 2 2 3 2 2 2 3 2 2 2 1 2 2 2 2 2 2 2 2 2 1 = + = + = + + + = 0.3 2 2 9 18 + x x p.2 ) 2 2 ( 9 2 ) 2 2 )( 2 ( 9 4 ) 2 ( ) 2 2 )( 9 18 ( 2 2 2 2 2 2 9 18 + + = + + = + + + = + + + + + = x x x x x x x x x x x 0.4 )) ( ( x g f p.4 x x x f 2 2 sin cos 1 ) (cos = = = 0.5 (a) x y 5 6 = p.5 5 6 6 5 y x y x = = 5 6 y x = is the inverse function of x y 5 6 = . Foundation Mathematics

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3 © Hong Kong Educational Publishing Co. Foundation Mathematics (b) 1 4 4 2 + + = x x y 2 1 1 2 ) 1 2 ( 2 ± = + = ± + = y x x y x y Each given value of y corresponds to two values of x . x is not a function of y . There is no inverse function for . 1 4 4 2 + + = x x y 0.6 (a) Let . sin ) ( x x f y = = p.6 ) sin( ) ( x x f = ) ( sin x f x = = x y sin = is an odd function. (b) Let . cos ) ( x x f y = = ) ( cos ) cos( ) ( x f x x x f = = = x y cos = is an even function. 0.7 (a) 6 4 3 = + x p.7 6 4 3 = + x or 6 4 3 = + x 3 4 = x or 9 4 = x 4 3 = x or 4 9 = x 4 3 or 4 9 = x (b) 0 3 2 = x 2 3 0 3 2 = = x x 0.8 (a) ! 4 ! 2 ! 6 6 4 = C p.9 15 = (b) ! 5 ! 3 ! 8 8 5 = C 56 = 0.9 165 2 3 = + n n C C p.10 0 990 990 3 3 2 3 165 ) 1 ( 2 ) 1 ( ) 1 )( 2 ( 3 ) 2 )( 1 ( 3 2 2 3 = = + + = + n n n n n n n n n n n n 0 ) 99 10 )( 10 ( 2 = + + n n n (by factor theorem) 0 10 = n or 0 99 10 2 = + + n n 10 = n or 2 296 10 ± = n (rejected) pp.1 – 10 Example 0.1T p.1 (a) 5 5 5 10 5 10 × = 5 2 5 5 10 = = (b) 2 3 2 2 3 2 2 3 2 5 2 3 2 5 + = + 2 2 3 2 10 ) 2 3 2 ( 5 ) 2 ( ) 3 2 ( ) 2 3 2 ( 5 2 2 = = = Example 0.2T p.2 2 5 1 2 3 5 2 1 + 2 2 3 4 2 4 5 2 2 3 5 2 ) 2 5 ( 2 2 3 5 2 2 ) 5 ( 2 5 ) 2 3 ( ) 5 2 ( 2 3 5 2 2 5 2 5 2 5 1 2 3 5 2 2 3 5 2 2 3 5 2 1 2 2 2 2 = = + = + = + + + =
© Hong Kong Educational Publishing Co. 4 New Progress in Senior Mathematics Module 2 Book 1 (Extended Part) Solution Guide 0 Example 0.3T p.2 ) 4 9 ( 3 25 ) 4 9 )( 25 ( 3 4 ) 9 ( ) 4 9 )( 75 3 ( 4 9 4 9 4 9 75 3 4 9 75 3 2 2 = = = + = + x x x x x x x x x x x x x Example 0.4T p.4 ) 3 2 tan( ) 3 2 ( )) ( ( = = x x f x g f Example 0.5T p.4 (a) x y 10 = y x log = y x log = is the inverse function of . 10 x y = (b) x y = 2 y x = 2 y x = is the inverse function of . x y = Example 0.6T p.6 (a) Let .

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