021.08f.final

# 021.08f.final - HKUST MATH021 Concise Calculus Final...

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HKUST MATH021 Concise Calculus Final Examination (Version 1) Name: 13th Dec 2008 Student I.D.: 12:30-15:30 Lecture Section: Directions: DO NOT open the exam until instructed to do so. Please write your Name, ID number, and Lecture Section in the space provided above. You may use a HKEA approved calculator, but graphical calculators are NOT allowed. All mobile phones and electronic equipments other than calculators should be switched off during the examination. This is a closed book examination. When instructed to open the exam, please check that you have 9 pages of questions in addition to the cover page. Answer all questions. Show the working steps of your answers for full credit. Cheating is a serious offense. Students caught cheating are subject to a zero score as well as additional penalties. Trigonometric Identities cos 2 θ + sin 2 θ = 1 1 + tan 2 θ = sec 2 θ 1 + cot 2 θ = csc 2 θ sin 2 θ = 2 sin θ cos θ cos 2 θ = 2 cos 2 θ - 1 = 1 - 2 sin 2 θ tan 2 θ = 2 tan θ 1 - tan 2 θ sin( A + B ) = sin A cos B + cos A sin B sin( A - B ) = sin A cos B - cos A sin B cos( A + B ) = cos A cos B - sin A sin B cos( A - B ) = cos A cos B + sin A sin B sin A cos B = 1 2 sin( A + B ) + sin( A - B ) cos A cos B = 1 2 cos( A + B ) + cos( A - B ) sin A sin B = 1 2 cos( A - B ) - cos( A + B ) tan( A + B ) = tan A + tan B 1 - tan A tan B tan( A - B ) = tan A - tan B 1 + tan A tan B Question No. Points Out of Q. 1-10 30 Q. 11 20 Q. 12 10 Q. 13 10 Q. 14 10 Q. 15 10 Q. 16 10 Total Points 100

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1 Part I: Multiple choice questions. Only one answer in each question is correct. Each question is worth 3 point. Put your answers to the multiple choice questions in the boxes below. No credit will be given if you show your answers in other places. Question 1 2 3 4 5 6 7 8 9 10 Total Answer e a b b d a c a d c 1. Which of the following is the graph of a function y = f ( x ) in the xy plane? (a) The unit circle centered at the origin (b) The square with vertices (1 , 1), ( - 1 , 1), ( - 1 , - 1) and (1 , - 1) (c) The triangle with vertices ( - 1 , 0), (1 , 1), (0 , - 1) (d) All of the above (e) None of the above Solution None of the above. 2. Let a , b , c be positive numbers. Evaluate lim x 0 (1 - e - ax ) cos 2 bx xe cx 2 . (a) a (b) b (c) c (d) 0 (e) 1 Solution The limit is a , by L’Hˆ opital’s rule: lim x 0 (1 - e - ax ) cos 2 bx xe cx 2 = lim x 0 ae - ax cos 2 bx - 2 b (1 - e - ax ) cos bx sin bx e cx 2 + 2 cx 2 e cx 2 = a 3. How many local maximum should f ( x ) have if its derivative is f 0 ( x ) = ( x + 1)( x - 2) 3 ( x + 4) 2 ( x - 5) ? (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 Solution One local maximum f (2) at the critical point x = 2, by first derivative test.
2 4. Find Z 4 0 x 2 + 2 x - 3 dx . (a) 95 3 (b) 86 3 (c) 74 3 (d) 64 3 (e) 58 3 Solution x 2 + 2 x - 3 = ( x - 1)( x + 3) = 0 ⇐⇒ x = 1 or x = - 3. Z 4 0 x 2 + 2 x - 3 dx = Z 1 0 - ( x 2 + 2 x - 3) dx + Z 4 1 ( x 2 + 2 x - 3) dx = h - 1 3 x 3 - x 2 + 3 x i 1 0 + h 1 3 x 3 + x 2 - 3 x i 4 1 = 86 3 5. The value of Z 2 0 x 2 2 - x 2 dx is (a) 1 2 (b) 2 (c) 1 (d) π 2 (e) π Solution The answer is π 2 . Let x = 2 sin θ , dx = 2 cos θdθ . Then Z 2 0 x 2 2 - x 2 dx = Z π 2 0 2 sin 2 θ 2 cos θdθ 2 cos θ = Z π 2 0 2 sin 2 θdθ = Z π 2 0 (1 - cos 2 θ ) = h θ - 1 3 sin 2 θ i π 2 0 = π 2 6. If

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