JIM101_VC1_2010-2011_Solutions - PUSAT PENGAJIAN PENDIDIKAN...

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PUSAT PENGAJIAN PENDIDIKAN JARAK JAUH UNIVERSITI SAINS MALAYSIA SIDANG AKADEMIK 2010/2011 JIM101 – CALCULUS SOLUTIONS – VIDEO CONFERENCE 1 MUHAMAD FUAD ABDULLAH Answers 1. Solve the following inequalities (a) 3 2 + x x 0 2 1 0 2 3 0 3 2 0 3 2 2 2 - - + - - + - + x x x x x x x x x x x ) )( ( Critical points are x = 0 , x = 1 and x = 2 x < 0 0 < x < 1 1 < x < 2 x > 2 x + + + x – 1 + + x – 2 + x x x ) )( ( 2 1 - - + + 0 2 1 - - x x x ) )( ( must be 0 or negative, thus the answer is x < 0 and 1 ≤ x ≤ 2 or in the interval notation as ( 29 [ ] 2 1 0 , , -
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(b) 2 1 - x x 0 1 2 0 1 2 2 0 2 1 - - - + - - - x x x x x x x Critical points are x = 1 and x = 2 x < 1 1 < x < 2 x > 2 x – 1 + + 2 – x + + 1 2 - - x x + 0 1 2 - - x x must be 0 or negative, thus the answer is x <1 and x ≥ 2 or in the interval notation as ( 29 [ 29 - , , 2 1 2. Find all values of x for 0 1 2 3 2 = + - x x i x 3 2 3 1 6 1 2 2 6 2 6 1 2 2 2 6 1 8 2 6 8 2 6 12 4 2 3 2 1 3 4 2 2 2 ± = - ± = - ± = - ± = - ± = - ± = - - ± - - = ) ( ) )( ( ) ( ) ( where 1 - = i
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3. Given i z 3 2 1 + = and i z 2 1 2 + - = (a) If ( 29 1 1 1 1 θ + θ = sin cos i r z and ( 29 2 2 2 2 θ + θ = sin cos i r z , Find 2 1 2 1 θ θ and , , r r i z 3 2 1 + = , i z 2 1 2 + - = 13 3 2 2 2 1 = + = r , 5 2 1 2 2 2 = + - = ) ( r 0 1 1 1 3 56 2 3 . tan = θ = θ - , 0 0 2 1 1 2 6 116 4 63 180 2 1 2 . . tan tan = - = θ - = - = θ - - (b) ( 29 ( 29 i i z z 2 1 3 2 2 1 + - + = i i i - - = - - + - = 8 6
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  • Spring '10
  • saipon
  • Injective function, JAUH UNIVERSITI SAINS, PUSAT PENGAJIAN PENDIDIKAN, MUHAMAD FUAD ABDULLAH

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