Test1Sol(1804)(08-092nd)

Test1Sol(1804)(08-092nd) - T1Sol/MATH1804/YMC/2009 THE...

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Unformatted text preview: T1Sol/MATH1804/YMC/2009 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1804 University Mathematics A Outline Solution to Test 1 Name : U. No. : Group : Answer ALL 4 Questions • Write your answer in the space provided. • For full credits, show your steps clearly and give details to your solution. • No books and notes are allowed to use. 1. ( 6 points ) (a) Let A = (- 3 , 2] \ (0 , 1] , B = { x ∈ Z : | x +1 | ≤ 2 } and C = { x ∈ R : x 10 +6 x 8 + x 2- 1 > } . Find A ∩ B ∩ C . (b) Let f ( x ) =- √ 1- x 2 for ≤ x ≤ 1 . Find the inverse function of f ( x ) (with explanation) and specify where this inverse function is defined. Solution. (a) A = (- 3 , 0] ∪ (1 , 2] , B = {- 3 ,- 2 ,- 1 , , 1 } . Then A ∩ B = {- 2 ,- 1 , } , and hence A ∩ B ∩ C = {- 2 ,- 1 } (since x = 0 does not satisfy the inequality). (b) The function g ( x ) = √ 1- x 2 for- 1 ≤ x ≤ is the inverse function of f ( x ) because: For ≤ x ≤ 1 , g ◦ f ( x ) = g ( f ( x )) = g (- p 1- x 2 ) = p 1- (1- x 2 ) = √ x 2 = x and For- 1 ≤ x ≤ , f ◦ g ( x ) = f ( g ( x )) = f ( p 1- x 2 ) =- p 1- (1- x 2 ) =- √ x 2 =- (- x ) ( since x is negative ) = x Alternative method : Let y = f ( x ) =- √ 1- x 2 for ≤ x ≤ 1 . Then by projecting the graph onto the y-axis, we have- 1 ≤ y ≤ . Now, y 2 = 1- x 2 and hence x = p 1- y 2 for- 1 ≤ y ≤ . Then we replace x by f- 1 ( x ) and y by x , we have f- 1 ( x ) = p 1- x 2 for- 1 ≤ x ≤ which is the inverse function of f ( x ) . For your reference: 2. ( 8 points ) (a) Compute lim x → 1 √ 2 x- x 4- 3 √ x 1- 3 √ x 4 ....
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Test1Sol(1804)(08-092nd) - T1Sol/MATH1804/YMC/2009 THE...

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