Math-2 Final Revision (1)

Mathematics – 02 Final Revision By: Mahmoud Osama & Mido Revised By : Akram Gamal Karem Moamen Edited By : Mahmoud Osama & Nourhan Osama Thanks to : Omar Atya Table of Contents Limits & Continuity ............................................................................................................. 002 Improper & Proper Fractions ......................................................................................... 010 Partial Derivatives ............................................................................................................... 033 Fourier Series ......................................................................................................................... 060 Maximum & Minimum Values ........................................................................................ 076 Taylor’s Theorem ................................................................................................................. 082 Improper Integrals .............................................................................................................. 094 Ellipse ........................................................................................................................................ 115 Parabola ..................................................................................................................................... 123 Series .......................................................................................................................................... 130 Page | 1 Choose The Correct Answer : 1] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟎,𝟎) 𝒙𝟐 −𝒚𝟐 𝒙𝟐 +𝒚𝟐 exists or not ? A- It exists C- 𝟏 B- No , it doesn’t exist D- ( −𝟏 ) Solution: By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(0,0) 𝑥 2 −𝑦 2 𝑥 2 +𝑦 2 = 𝑙𝑖𝑚(𝑥,𝑦)→(0,0) 0 0 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 There is no Simplification By Using Paths method 1] On X-axis , at 𝑦 = 0 , 𝑙𝑖𝑚𝑥→0 2] On Y-axis , at 𝑥 = 0 , 𝑙𝑖𝑚𝑦→0 𝑥 2 −0 𝑥 2 +0 0−𝑦 2 0+𝑦 2 = 𝑙𝑖𝑚𝑥→0 = 𝑙𝑖𝑚𝑦→0 𝑥2 𝑥2 = 𝑙𝑖𝑚𝑥→0 1 = 1 = 𝐿1 −𝑦 2 𝑦2 = 𝑙𝑖𝑚𝑦→0 − 1 = −1 = 𝐿2 Since 𝐿1 ≠ 𝐿2 , Then It doesn’t exist Choice B 2] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟎,𝟎) A- It exists at −𝟐 𝟑 B- No , it doesn’t exist 𝒙𝟐 −𝟐 𝟑+𝒚 exists or not ? C- 𝟏 D- ( −𝟏 ) Solution: By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(0,0) Page | 2 𝑥 2 −2 3+𝑦 = 𝑙𝑖𝑚(𝑥,𝑦)→(0,0) −2 3 = −2 3 Then It exist at −2 3 Choice A 3] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟎,𝟎) A- It exists at 𝒙𝟐 +𝟑𝒚𝟐 𝒙𝟐 +𝒚𝟐 exists or not ? C- 𝟏 B- No , it doesn’t exist D- 𝟑 Solution: By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(0,0) 𝑥 2 +3𝑦 2 0 = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑥 2 +𝑦 2 0 There is no Simplification By Using Paths method 1] On X-axis , at 𝑦 = 0 , 𝑙𝑖𝑚𝑥→0 2] On Y-axis , at 𝑥 = 0 , 𝑙𝑖𝑚𝑦→0 𝑥 2 +3𝑦 2 𝑥 2 +𝑦 2 𝑥 2 +3𝑦 2 𝑥 2 +𝑦 2 = 𝑙𝑖𝑚𝑥→0 = 𝑙𝑖𝑚𝑦→0 𝑥 2 +0 𝑥 2 +0 = 𝑙𝑖𝑚𝑥→0 1 = 1 = 𝐿1 0+3𝑦 2 0+𝑦 2 = 𝑙𝑖𝑚𝑦→0 3 = 3 = 𝐿2 Since 𝐿1 ≠ 𝐿2 , Then It doesn’t exist Choice B 4] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟏,𝟎) 𝒚 𝒙+𝒚−𝟏 A- It exists C- 𝟏 B- No , it doesn’t exist D- 𝟎 exists or not ? Solution: By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(1,0) Page | 3 𝑦 𝑥+𝑦−1 = 𝑙𝑖𝑚(𝑥,𝑦)→(1,0) 0 1+0−1 0 = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 0 There is no Simplification By Using Paths method 1] On X-axis , at 𝑦 = 0 , 𝑙𝑖𝑚𝑥→1 2] On Y-axis , at 𝑥 = 1 , 𝑙𝑖𝑚𝑦→0 0 𝑥+0−1 𝑦 1+𝑦−1 = 𝑙𝑖𝑚𝑥→1 0 = 0 = 𝐿1 = 𝑙𝑖𝑚𝑦→0 1 = 1 = 𝐿2 Since 𝐿1 ≠ 𝐿2 , Then It doesn’t exist Choice B 5] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟎,𝟎) 𝒙𝒚 𝑥2 +𝑦2 exists or not ? 𝟏 A- It exists C- 𝟐 B- No , it doesn’t exist D- 𝟎 Solution: By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(0,0) 𝑥𝑦 𝑥 2 +𝑦 2 0 = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 0 There is no Simplification By Using Paths method 𝑥∗0 1] On X-axis , at 𝑦 = 0 , 𝑙𝑖𝑚𝑥→0 𝑥 2 +02 2] On Y-axis , at 𝑥 = 0 , 𝑙𝑖𝑚𝑦→0 02 +𝑦 2 3] at 𝑥 = 𝑦 , 𝑙𝑖𝑚𝑦→0 𝑦∗𝑦 𝑦 2 +𝑦 2 0∗𝑦 = 𝑙𝑖𝑚𝑦→0 = 0 = 𝐿1 = 0 = 𝐿2 𝑦2 2𝑦 2 = 𝑙𝑖𝑚𝑦→0 Since 𝐿1 = 𝐿2 ≠ 𝐿3 , Then It doesn’t exist Choice B Page | 4 1 2 1 = = 𝐿3 2 6] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟎,𝟎) 𝒙𝒚𝟐 𝑥2 +𝑦4 exists or not ? 𝟏 A- It exists C- 𝟐 B- No , it doesn’t exist D- 𝟎 Solution: 𝑥𝑦 2 By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(0,0) 𝑥 2 +𝑦 4 0 = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 0 There is no Simplification By Using Paths method 1] On X-axis , at 𝑦 = 0 , 𝑙𝑖𝑚𝑥→0 2] On Y-axis , at 𝑥 = 0 , 𝑙𝑖𝑚𝑦→0 3] at 𝑥 = 𝑦 , 𝑙𝑖𝑚𝑦→0 2 4] at 𝑥 = 𝑦 , 𝑦∗𝑦 2 𝑦 2 +𝑦 4 = 𝑙𝑖𝑚𝑦→0 𝑥∗02 𝑥 2 +04 0∗𝑦 2 02 +𝑦 4 𝑦3 𝑦 2 +𝑦 4 𝑦 2 ∗𝑦 2 𝑙𝑖𝑚𝑦→0 (𝑦2 )2 4 +𝑦 = 0 = 𝐿1 = 0 = 𝐿2 = 𝑙𝑖𝑚𝑦→0 = 𝑙𝑖𝑚𝑦→0 𝑦 2 (𝑦) 𝑦 2 (1+𝑦2 ) 𝑦4 2𝑦 4 = 𝑙𝑖𝑚𝑦→0 = 𝑙𝑖𝑚𝑦→0 𝑦 1+𝑦 2 1 2 = 𝑙𝑖𝑚𝑦→0 0 = 0 = 𝐿3 1 = = 𝐿4 2 Since 𝐿1 = 𝐿2 = 𝐿3 ≠ 𝐿4 , Then It doesn’t exist Choice B 7] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟏,𝟐) (𝟓𝒙𝟑 − 𝒙𝟐 𝒚𝟐 ) exists or not ? A- It exists at 𝟏 C- It exists at −𝟏 B- No , it doesn’t exist D- 𝟏 Solution: By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(1,2) (5𝑥 3 − 𝑥 2 𝑦 2 ) = 5 ∗ 13 − 12 ∗ 22 = 1 Page | 5 Then it exists at 1 Choice A 8] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟎,𝟎) 𝑦 2 sin2 𝑥 𝑥 4 +𝑦 4 A- It exists C- 𝟎 B- No , it doesn’t exist D- 𝟏 exists or not ? Solution: By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(0,0) 𝑦 2 sin2 𝑥 𝑥 4 +𝑦4 0 = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 0 There is no Simplification By Using Paths method 1] On X-axis , at 𝑦 = 0 , 𝑙𝑖𝑚𝑥→0 2] On Y-axis , at 𝑥 = 0 , 𝑙𝑖𝑚𝑦→0 3] at 𝑥 = 𝑦 , 𝑙𝑖𝑚𝑦→0 𝑦 2 sin2 𝑦 𝑦 4 +𝑦 4 = 𝑙𝑖𝑚𝑦→0 02 sin2 𝑥 𝑥 4 +04 𝑦 2 sin2 0 04 +𝑦 4 sin2 𝑦 2𝑦 2 = 0 = 𝐿1 = 0 = 𝐿2 = 𝑙𝑖𝑚𝑦→0 1 2 ∗ 𝑙𝑖𝑚𝑦→0 sin2 𝑦 𝑦2 1 1 2 2 = ∗ 1 = = 𝐿3 Since 𝐿1 = 𝐿2 ≠ 𝐿3 , Then It doesn’t exist Choice B 9] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟏,−𝟏) 𝒆−𝒙𝒚 𝐜𝐨𝐬(𝒙 + 𝒚) exists or not ? A- It exists at 𝒆 C- It exists at 𝟎 B- No , it doesn’t exist D- 𝟏 Solution: By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(1,−1) 𝑒 −(1∗−1) 𝑐𝑜𝑠(1 + (−1)) = 𝑒 Then it exists at 𝑒 Choice A Page | 6 𝟒−𝒙𝒚 10] Is this limits 𝒍𝒊𝒎(𝒙,𝒚)→(𝟐,𝟏) 𝒙𝟐 +𝟑𝒚𝟐 exists or not ? 𝟐 A- It exists at 𝟕 C- It exists at 𝟎 B- No , it doesn’t exist D- 𝟏 Solution: By Substitution: 𝑙𝑖𝑚(𝑥,𝑦)→(2,1) Then it exists at 4−2∗1 22 +3∗12 = 2 7 2 7 Choice A 𝒙−𝒂𝒚 11] 𝒍𝒊𝒎(𝒙,𝒚)→(𝟎,𝟎) 𝒙+𝒂𝒚 along the path 𝒚 = 𝟎 𝒂 A- 𝟎 C- 𝒃 B- 𝟏 D- None of them Solution: By Substitution: 𝑙𝑖𝑚𝑥→0 𝑥−𝑎∗0 𝑥+𝑎∗0 = 𝑥 𝑥 =1 Choice B 𝒙𝟐 +𝒚𝟐 12] 𝒍𝒊𝒎 [𝒍𝒊𝒎 𝒙𝟐 −𝒚𝟐] = 𝒚→𝟎 𝒙→𝟎 A- 𝟎 C- 𝟏 B- −𝟏 D- None of them Page | 7 Solution: 𝑙𝑖𝑚 𝑥 2 +𝑦 2 𝑥→0 𝑥 2 −𝑦 2 𝑙𝑖𝑚 [𝑙𝑖𝑚 = 02 +𝑦 2 02 −𝑦 2 𝑥 2 +𝑦2 𝑦→0 𝑥→0 𝑥 2 −𝑦2 = 𝑦2 −𝑦 2 = −1 ] = 𝑙𝑖𝑚[−1] = −1 𝑦→0 Choice B 13] Determine the set of points at which the function is Continuous 𝒇(𝒙, 𝒚) = 𝒙𝒚 𝟏+𝒆𝒙−𝒚 A- It is continuous on 𝐑 C- It is continuous on 𝐑 − {𝟎} B- It is continuous on 𝐑 − {−𝟏} D- It isn’t continuous on 𝐑 Solution: We first find the Domain , then the function is continuous at its Domain Domain of the numerator ; the Domain of (𝑥𝑦) is R Domain of the Denominator ; the Domain of (1 + 𝑒 𝑥−𝑦 ) is R 1 + 𝑒 𝑥−𝑦 ≠ 0 , 𝑒 𝑥−𝑦 ≠ −1 So the function is continuous on R Choice A 14] Determine the set of points at which the function is Continuous 𝒇(𝒙, 𝒚) = 𝟏+𝒙𝟐 +𝒚𝟐 𝟏−𝒙𝟐 −𝒚𝟐 Solution: ➔ 1 − 𝑥 2 − 𝑦2 ≠ 0 , 𝑥 2 + 𝑦2 ≠ 1 , 𝑦2 ≠ 1 − 𝑥 2 𝑦 ≠ √1 − 𝑥 2 Page | 8 Then: function is continuous at: ∀(𝑥, 𝑦) ∈ 𝑅 | 𝑦 ≠ √1 − 𝑥 2 Domain = { 𝑦 ∈ 𝑅: 𝑦 ≠ √1 − 𝑥 2 } 15] Determine the set of points at which the function is Continuous 𝒇(𝒙, 𝒚) = 𝒍𝒏(𝒙𝟐 + 𝒚𝟐 − 𝟒) Solution: ➔ 𝑥 2 + 𝑦 2 − 4 > 0 , 𝑥 2 + 𝑦 2 > 4 , 𝑥 2 > 4 − 𝑦 2 , 𝑥 > √4 − 𝑦 2 Domain = { 𝑥 ∈ 𝑅: 𝑥 > √4 − 𝑦 2 } Page | 9 𝟕𝒙−𝟐𝟓 16] The partial Fraction decomposition of 𝒙𝟐 −𝟕𝒙+𝟏𝟐 is 𝟒 𝟑 𝟒 𝟑 A- 𝒙−𝟑 + 𝒙−𝟒 𝟒 𝟑 𝟒 𝟑 C- 𝒙−𝟑 − 𝒙−𝟒 B- 𝒙+𝟑 − 𝒙+𝟒 D- 𝒙+𝟑 + 𝒙−𝟒 Solution: 7𝑥−25 7𝑥−25 𝑥 2 −7𝑥+12 = (𝑥−3)(𝑥−4) = 7𝑥−25 (𝑥−3)(𝑥−4) = A 𝑥−3 + B (Proper Fraction) 𝑥−4 𝐴(𝑥−4)+𝐵(𝑥−3) (𝑥−3)(𝑥−4) Then: 7𝑥 − 25 = 𝐴(𝑥 − 4) + 𝐵(𝑥 − 3) At x = 3 ➔ 21 − 25 = A(3 − 4) + 0 , −4 = −A , then: 𝐀 = 𝟒 At x = 4 ➔ 28 − 25 = 0 + B(4 − 3) , 3 = B , then: 𝐁 = 𝟑 Then: 7𝑥−25 𝑥 2 −7𝑥+12 = 4 𝑥−3 + 3 𝑥−4 17] The partial Fraction decomposition of 𝟐 𝟏 𝟏 𝟐 𝟏 𝟏 A- 𝒙−𝟏 − (𝒙−𝟏)𝟐 − 𝒙−𝟐 𝒙𝟐 −𝟑𝒙+𝟏 is (𝒙−𝟏)𝟐 (𝒙−𝟐) 𝟐 𝟏 𝟑 𝟐 𝟏 𝟏 C- 𝒙−𝟏 + (𝒙−𝟏)𝟐 − 𝒙−𝟐 B- 𝒙−𝟏 − (𝒙−𝟏)𝟐 + 𝒙−𝟐 D- 𝒙−𝟏 + (𝒙−𝟏)𝟐 − 𝒙−𝟐 Solution: 𝑥 2 −3𝑥+1 (𝑥−1)2 (𝑥−2) = A 𝑥−1 B + (𝑥−1)2 + C 𝑥−2 (Proper Fraction) By multiplying in two sides by : (𝑥 − 1)2 (𝑥 − 2) ➔ 𝑥 2 − 3𝑥 + 1 = A(𝑥 − 1)(𝑥 − 2) + B(𝑥 − 2) + C(𝑥 − 1)2 At x = 1 ➔ 1 − 3 + 1 = 0 + B(1 − 2) + 0 , −1 = −B , then: 𝐁 = 𝟏 Page | 10 At x = 2 ➔ 4 − 6 + 1 = 0 + 0 + C(2 − 1)2 , −1 = C , then: 𝐂 = −𝟏 At x = 0 ➔ 1 = A(0 − 1)(0 − 2) + B(0 − 2) + C(0 − 1)2 , 1 = A(−1)(−2) + 1(−2) + (−1)(−1)2 = 2A − 3 = 1 , then: 𝐀 = 𝟐 𝑥 2 −3𝑥+1 2 Then: (𝑥−1)2 (𝑥−2) = 𝑥−1 1 + (𝑥−1)2 − 1 𝑥−2 Choice D 𝟏 18] The partial Fraction decomposition of 𝒙𝟒 (𝒙+𝟐)is 𝟏 𝟏 𝟏 𝟏 𝟏 C- 𝟏𝟔𝒙 + 𝟖𝒙𝟐 − 𝟒𝒙𝟑 + 𝟐𝒙𝟒 + 𝟏𝟔(𝒙+𝟐) 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 D- − 𝟏𝟔𝒙 + 𝟏𝟔𝒙𝟐 + 𝟒𝒙𝟑 + 𝟐𝒙𝟒 + 𝟖(𝒙+𝟐) A- − 𝟏𝟔𝒙 − 𝟖𝒙𝟐 − 𝟒𝒙𝟑 + 𝟐𝒙𝟒 − 𝟏𝟔(𝒙+𝟐) 𝟏 𝟏 𝟏 B- − 𝟏𝟔𝒙 + 𝟖𝒙𝟐 − 𝟒𝒙𝟑 + 𝟐𝒙𝟒 + 𝟏𝟔(𝒙+𝟐) 𝟏 𝟏 𝟏 𝟏 𝟏 Solution: 1 𝑥 4 (𝑥+2) = A 𝑥 + B C D E 𝑥 𝑥 𝑥 𝑥+2 2 + 3 + 4 + (Proper Fraction) By multiplying in two sides by : 𝑥 4 (𝑥 + 2) ➔ 1 = A𝑥 3 (𝑥 + 2) + B𝑥 2 (𝑥 + 2) + C𝑥 (𝑥 + 2) + D(𝑥 + 2) + E𝑥 4 1 = A𝑥 4 + 2A𝑥 3 + B𝑥 3 + 2B𝑥 2 + C𝑥 2 + 2C𝑥 + D𝑥 + 2D + E𝑥 4 At x = 0 ➔ 1 = 2D , then: 𝐃 = 𝟏 𝟐 At x = −𝟐 ➔ 1 = E(−2)4 , then: 𝐄 = 𝟏 𝟏𝟔 Equating Coefficient (𝒙𝟒 ) ➔ 0 = A + E ; A = −E , then: 𝐀 = Equating Coefficient (𝒙𝟑 ) ➔ 0 = 2A + B ; then: 𝐁 = 𝟏 Equating Coefficient (𝒙𝟐 ) ➔ 0 = 2B + C ; then: 𝐂 = −𝟏 Page | 11 𝟖 𝟒 −𝟏 𝟏𝟔 𝟏 Then: 1 𝑥 4 (𝑥+2) =− 1 16𝑥 1 + 8𝑥 2 − 1 4𝑥 3 + 1 2𝑥 4 + 1 16(𝑥+2) Choice B 𝟗𝒙−𝟕 19] The partial Fraction decomposition of (𝒙+𝟑)(𝒙𝟐 +𝟏) is 𝟏𝟕 𝟏𝟕𝒙−𝟔 A- − 𝟓(𝒙+𝟑) + 𝟓(𝒙𝟐 +𝟏) 𝟏𝟕 C- 𝟏𝟕𝒙−𝟔 B- (𝒙+𝟑) + 𝟓(𝒙𝟐 −𝟏) D- 𝟏𝟕 𝟏𝟕𝒙−𝟔 𝟏𝟕 𝟏𝟕𝒙−𝟔 − 𝟓(𝒙𝟐 +𝟏) 𝟓(𝒙+𝟑) + (𝒙𝟐 +𝟏) 𝟓(𝒙+𝟑) Solution: 9𝑥−7 (𝑥+3)(𝑥 2 +1) = A 𝑥+3 + B𝑥+C (Proper Fraction) 𝑥 2 +1 By multiplying in two sides by : (𝑥 + 3)(𝑥 2 + 1) ➔ 9𝑥 − 7 = A(𝑥 2 + 1) + (B𝑥 + C)(𝑥 + 3) 9𝑥 − 7 = A𝑥 2 + A + B𝑥 2 + 3B𝑥 + C𝑥 + 3C At 𝒙 = −𝟑 ➔ −27 − 7 = A((−3)2 + 1) , then: 𝐀 = − Equating Coefficient (𝒙𝟐 ) ➔ 0 = A + B , then: 𝐁 = 𝟏𝟕 𝟏𝟕 𝟓 Equating Coefficient (𝒙𝟏 ) ➔ 9 = 3B + C , then: 𝐂 = − Then: 9𝑥−7 (𝑥+3)(𝑥 2 +1) =− 9𝑥−7 (𝑥+3)(𝑥 2 +1) Choice A Page | 12 17 5(𝑥+3) =− + 17 5(𝑥+3) 17 6 𝑥−5 5 𝑥 2 +1 + 𝟓 𝟔 𝟓 multiply by 𝟓 in numerator & denominator 17𝑥−6 5(𝑥 2 +1) 20] The partial Fraction decomposition of 𝟏 𝒙+𝟏 𝒙+𝟏 𝟏 𝒙+𝟏 𝒙+𝟏 A- 𝟒(𝟏−𝒙) + 𝟐(𝟏+𝒙𝟐 ) − 𝟒(𝟏+𝒙𝟐 )𝟐 C- B- 𝟒(𝟏−𝒙) − 𝟐(𝟏+𝒙𝟐 ) − 𝟒(𝟏+𝒙𝟐)𝟐 D- 𝒙𝟐 is (𝟏−𝒙)(𝟏+𝒙𝟐 )𝟐 𝟏 𝒙+𝟏 𝒙+𝟏 𝟏 𝒙+𝟏 𝒙+𝟏 + 𝟒(𝟏+𝒙𝟐 ) − 𝟐(𝟏+𝒙𝟐 )𝟐 𝟒(𝟏−𝒙) + 𝟐(𝟏+𝒙𝟐 ) − 𝟒(𝟏+𝒙𝟐 )𝟐 𝟐(𝟏−𝒙) Solution: 𝑥2 (1−𝑥)(1+𝑥 2 )2 A B𝑥+C D𝑥+E = (1−𝑥) + (1+𝑥 2 ) + (1+𝑥 2 )2 (Proper Fraction) By multiplying in two sides by : (1 − 𝑥 )(1 + 𝑥 2 )2 𝑥 2 = A(1 + 𝑥 2 )2 + (B𝑥 + C)(1 − 𝑥 )(1 + 𝑥 2 ) + (D𝑥 + E)(1 − 𝑥 ) 𝑥 2 = A + A𝑥 4 + 2A𝑥 2 + B𝑥 − B𝑥 2 + C − C𝑥 + B𝑥 3 − B𝑥 4 + C𝑥 2 −C𝑥 3 + D𝑥 + E − D𝑥 2 − E𝑥 At 𝒙 = 1 ➔ 1 = A(1 + 12 )2 , then: 𝐀 = 𝟏 𝟒 Equating Coefficient (𝒙𝟒 ) ➔ 0 = A − B , then: 𝐁 = 𝟏 Equating Coefficient (𝒙𝟑 ) ➔ 0 = B − C , then: 𝐂 = 𝟏 𝟒 𝟒 Equating Coefficient (𝒙𝟐 ) ➔ 1 = 2A − B + C − D , then: 𝐃 = − Equating Coefficient (𝒙𝟏 ) ➔ 0 = B − C + D − E , then: 𝐄 = − Then: 𝑥2 (1−𝑥)(1+𝑥 2 )2 = Choice C Page | 13 1 4(1−𝑥) + = 1 4(1−𝑥) 𝑥+1 4(1+𝑥 2 ) − + 1 1 𝑥+4 4 (1+𝑥 2 ) 𝑥+1 2(1+𝑥 2 )2 1 1 −2𝑥−2 + (1+𝑥 2 )2 𝟏 𝟐 𝟏 𝟐 𝟐𝒙−𝟏 21] The partial Fraction decomposition of (𝒙+𝟐)(𝒙−𝟑)is 𝟏 𝟏 𝟏 𝟏 A- 𝒙+𝟐 + 𝒙+𝟑 C- B- 𝒙−𝟐 − 𝒙−𝟑 D- 𝟏 𝟏 𝟏 𝟏 − 𝒙−𝟑 𝒙+𝟐 + 𝒙−𝟑 𝒙+𝟐 Solution: 2𝑥−1 (𝑥+2)(𝑥−3) = A 𝑥+2 + B 𝑥−3 (Proper Fraction) By multiplying in two sides by : (𝑥 + 2)(𝑥 − 3) ➔ 2𝑥 − 1 = A(𝑥 − 3) + B(𝑥 + 2) At 𝒙 = 𝟑 , 2 ∗ 3 − 1 = 0 + B(5) , 5 = 5B , then: 𝐁 = 𝟏 At 𝒙 = −𝟐 , 2 ∗ −2 − 1 = A(−2 − 3) + 0 , −5 = A(−5) , then: 𝐀 = 𝟏 2𝑥−1 Then: (𝑥+2)(𝑥−3) = 1 𝑥+2 + 1 𝑥−3 Choice D 𝟐𝒙+𝟓 22] The partial Fraction decomposition of (𝒙−𝟐)(𝒙+𝟏) is 𝟑 𝟏 𝟑 𝟏 A- 𝒙−𝟐 − 𝒙+𝟏 C- B- 𝒙+𝟑 − 𝒙+𝟏 D- 𝟑 𝟏 𝟑 𝟐 − 𝒙+𝟏 𝒙+𝟐 − 𝒙+𝟏 𝒙−𝟐 Solution: 2𝑥+5 (𝑥−2)(𝑥+1) = A 𝑥−2 + B 𝑥+1 (Proper Fraction) By multiplying in two sides by : (𝑥 − 2)(𝑥 + 1) ➔ 2𝑥 + 5 = A(𝑥 + 1) + B(𝑥 − 2) At 𝒙 = −𝟏 , 2 ∗ −1 + 5 = 0 + B(−1 − 2) , 3 = −3B , then: 𝐁 = −𝟏 Page | 14 At 𝒙 = 𝟐 , 2 ∗ 2 + 5 = A(2 + 1) + 0 , 9 = 3A , then: 𝐀 = 𝟑 2𝑥+5 Then: (𝑥−2)(𝑥+1) = 3 𝑥−2 − 1 𝑥+1 Choice A 𝟑 23] The partial Fraction decomposition of (𝒙−𝟏)(𝟐𝒙−𝟏) is 𝟑 𝟔 𝟑 𝟔 A- 𝒙−𝟏 + 𝟐𝒙−𝟏 C- B- 𝒙−𝟏 − 𝟐𝒙−𝟏 D- 𝟑 𝟔 𝟖 𝟔 − 𝟐𝒙+𝟏 𝒙−𝟏 − 𝟐𝒙−𝟏 𝒙−𝟏 Solution: 3 (𝑥−1)(2𝑥−1) = A 𝑥−1 + B 2𝑥−1 (Proper Fraction) By multiplying in two sides by : (𝑥 − 1)(2𝑥 − 1) ➔ 3 = A(2𝑥 − 1) + B(𝑥 − 1) At 𝒙 = 𝟏 , 3 = A(2 ∗ 1 − 1) + 0 , 3 = A , then: 𝐀 = 𝟑 𝟏 1 1 𝟐 2 2 At 𝒙 = , 3 = B ( − 1) , 3 = − B , then: 𝐁 = −𝟔 3 Then: (𝑥−1)(2𝑥−1) = 3 𝑥−1 − 6 2𝑥−1 Choice B 𝟏 24] The partial Fraction decomposition of (𝒙+𝟒)(𝒙−𝟐) is 𝟏 𝟏 𝟏 𝟏 A- − 𝟔(𝒙+𝟒) + 𝟔(𝒙+𝟐) B- − 𝟔(𝒙+𝟒) + 𝟕(𝒙−𝟐) Page | 15 𝟏 𝟏 C- − 𝟔(𝒙+𝟒) + 𝟔(𝒙−𝟐) D- 𝟏 𝟏 + 𝟔(𝒙−𝟐) 𝟔(𝒙+𝟒) Solution: 1 (𝑥+4)(𝑥−2) = A 𝑥+4 + B (Proper Fraction) 𝑥−2 By multiplying in two sides by : (𝑥 + 4)(𝑥 − 2) ➔ 1 = A(𝑥 − 2) + B(𝑥 + 4) At 𝒙 = 𝟐 , 1 = 0 + B ∗ 6 , then: 𝐁 = 𝟏 𝟔 At 𝒙 = −𝟒 , 1 = A ∗ −6 + 0 , then: 𝐀 = − 1 Then: (𝑥+4)(𝑥−2) = − 1 6(𝑥+4) + 𝟏 𝟔 1 6(𝑥−2) Choice C 25] To resolve a combined fraction into its parts is called A- rational fraction C- partial fraction B- combined fraction D- None of them Solution: Choice C 26] The partial Fraction decomposition of 𝟐 𝟏 𝟑 𝟐 𝟏 𝟑 A- (𝒙+𝟐) − (𝒙+𝟐)𝟐 + (𝒙+𝟏) C- B- (𝒙+𝟐) + (𝒙+𝟐)𝟐 + (𝒙+𝟏) D- 𝟓𝒙𝟐 +𝟏𝟕𝒙+𝟏𝟓 (𝒙+𝟐)𝟐 (𝒙+𝟏) 𝟐 𝟏 𝟑 𝟒 𝟏 𝟑 − (𝒙+𝟐)𝟐 − (𝒙+𝟏) (𝒙+𝟐) − (𝒙+𝟐)𝟐 + (𝒙+𝟏) (𝒙+𝟐) Solution: 5𝑥 2 +17𝑥+15 (𝑥+2)2 (𝑥+1) Page | 16 A B C = (𝑥+2) + (𝑥+2)2 + (𝑥+1) is (Proper Fraction) By multiplying in two sides by : (𝑥 + 2)2 (𝑥 + 1) ➔ 5𝑥 2 + 17𝑥 + 15 = A(𝑥 + 2)(𝑥 + 1) + B(𝑥 + 1) + C(𝑥 + 2)2 = A(𝑥 2 + 3𝑥 + 2) + B𝑥 + B + C(𝑥 2 + 4𝑥 + 4) At 𝒙 = −𝟐 , 5 ∗ (−2)2 + 17 ∗ −2 + 15 = B(−2 + 1) ➔ 1 = B(−1) , then: 𝐁 = −𝟏 At 𝒙 = −𝟏 , 5 ∗ (−1)2 + 17 ∗ −1 + 15 = C(−1 + 2)2 ➔ 3 = C , then: 𝐂 = 𝟑 Equating Coefficient (𝒙𝟐 ) ➔ 5 = A + C , 5 = A + 3 , then: 𝐀 = 𝟐 5𝑥 2 +17𝑥+15 (𝑥+2)2 (𝑥+1) Then: 2 1 3 = (𝑥+2) − (𝑥+2)2 + (𝑥+1) Choice A 27] The partial Fraction decomposition of 𝒙 (𝒙−𝟑)𝟑 (𝟐𝒙+𝟏) is 𝟐 𝟏 𝟑 𝟒 A- − 𝟑𝟒𝟑(𝒙−𝟑) + + + 𝟒𝟗(𝒙−𝟑)𝟐 𝟕(𝒙−𝟑)𝟑 𝟑𝟒𝟑(𝟐𝒙+𝟏) 𝟐 𝟏 𝟑 𝟒 B- − 𝟑𝟒𝟑(𝒙−𝟑) − + + 𝟒𝟗(𝒙−𝟑)𝟐 𝟕(𝒙−𝟑)𝟑 𝟑𝟒𝟑(𝟐𝒙+𝟏) C- 𝟐 𝟑𝟒𝟑(𝒙−𝟑) + 𝟏 𝟒𝟗(𝒙−𝟑)𝟐 + 𝟑 𝟕(𝒙−𝟑)𝟑 + 𝟒 𝟑𝟒𝟑(𝟐𝒙+𝟏) 𝟐 𝟏 𝟑 𝟒 D- − 𝟑𝟒𝟑(𝒙−𝟑) + + + 𝟐 𝟑 𝟒𝟗(𝒙−𝟑) 𝟕(𝒙+𝟑) 𝟑𝟒𝟑(𝟐𝒙+𝟏) Solution: 𝑥 (𝑥−3)3 (2𝑥+1) A B C D = (𝑥−3) + (𝑥−3)2 + (𝑥−3)3 + (2𝑥+1) (Proper Fraction) By multiplying in two sides by : (x − 3)3 (2x + 1) ➔ 𝑥 = A(𝑥 − 3)2 (2𝑥 + 1) + B(𝑥 − 3)(2𝑥 + 1) + C(2𝑥 + 1) + D(𝑥 − 3)3 = A(2𝑥 3 − 11𝑥 2 + 12𝑥 + 9) + B(2𝑥 2 − 5𝑥 − 3) + C(2𝑥 + 1) +D(𝑥 3 − 9𝑥 2 + 27𝑥 − 27) Page | 17 At 𝒙 = 𝟑 , 3 = C(2 ∗ 3 + 1) , 3 = C ∗ 7 , then: 𝐂 = 𝟏 1 1 𝟐 2 2 3 1 343 2 8 At 𝒙 = − , − = D (− − 3) , − = D (− 𝟑 𝟕 𝟒 ) , then: 𝐃 = 𝟑𝟒𝟑 Equating Coefficient (𝒙𝟑 ) ➔ 0 = 2A + D , 0 = 2A + 4 343 , then: 𝐀 = − 𝟐 𝟑𝟒𝟑 Equating Coefficient (𝒙𝟐 ) ➔ 0 = −11A + 2B − 9D , 0 = −11 ∗ − 𝑥 Then: (𝑥−3)3 (2𝑥+1) = − 2 343(𝑥−3) + 2 343 1 49(𝑥−3)2 + 2B − 9 ∗ + 3 7(𝑥−3)3 + 4 343 , then: 𝐁 = 𝟏 𝟒𝟗 4 343(2𝑥+1) Choice A 28] The partial Fraction decomposition of 𝟏 𝟏 𝟏 𝟏 A- 𝟐(𝒙−𝟏) + (𝒙−𝟏)𝟐 C𝟏 B- 𝟐(𝒙−𝟏) + (𝒙+𝟏)𝟐 + 𝟐(𝒙+𝟏) 𝒙𝟐 +𝟏 (𝒙−𝟏)𝟐 (𝒙+𝟏) 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 − (𝒙−𝟏)𝟐 + 𝟐(𝒙+𝟏) 𝟐(𝒙−𝟏) D- 𝟐(𝒙−𝟏) + (𝒙−𝟏)𝟐 + 𝟐(𝒙+𝟏) Solution: 𝑥 2 +1 (𝑥−1)2 (𝑥+1) A B C = (𝑥−1) + (𝑥−1)2 + (𝑥+1) (Proper Fraction) By multiplying in two sides by : (𝑥 − 1)2 (𝑥 + 1) ➔ 𝑥 2 + 1 = A(𝑥 − 1)(𝑥 + 1) + B(𝑥 + 1) + C(𝑥 − 1)2 = A(𝑥 2 − 1) + B(𝑥 + 1) + C(𝑥 2 − 2𝑥 + 1) At 𝒙 = 𝟏 , 2 = B(𝑥 + 1) , then: 𝐁 = 𝟏 At 𝒙 = −𝟏 , 2 = C(𝑥 − 1)2 , 2 = 4C , then: 𝐂 = Page | 18 is 𝟏 𝟐 Equating Coefficient (𝒙𝟐 ) 1 𝟏 2 𝟐 ➔ 1 = A + C , 1 = A + , then: 𝐀 = 𝑥 2 +1 Then: (𝑥−1)2 (𝑥+1) = 1 1 2(𝑥−1) + (𝑥−1)2 + 1 2(𝑥+1) Choice D 29] The partial Fraction decomposition of 𝟐𝒙+𝟏 𝟑 𝟐𝒙+𝟏 𝟑 A- (𝒙𝟐 +𝒙+𝟐) + (𝟐𝒙−𝟏) C- B- (𝒙𝟐 +𝒙+𝟐) − (𝟐𝒙−𝟏) 𝒙𝟐 −𝟑𝒙−𝟕 (𝒙𝟐 +𝒙+𝟐)(𝟐𝒙−𝟏) 𝟐𝒙−𝟏 𝟑 𝟐𝒙+𝟏 𝟑 − (𝟐𝒙−𝟏) (𝒙𝟐 +𝒙+𝟐) D- (𝒙𝟐 +𝒙+𝟐) − (𝟐𝒙+𝟏) Solution: 𝑥 2 −3𝑥−7 A𝑥+B (𝑥 2 +𝑥+2)(2𝑥−1) = (𝑥 2 C +𝑥+2) + (2𝑥−1) (Proper Fraction) By multiplying in two sides by : (𝑥 2 + 𝑥 + 2)(2𝑥 − 1) 𝑥 2 − 3𝑥 − 7 = (A𝑥 + B)(2𝑥 − 1) + C(𝑥 2 + 𝑥 + 2) = 2A𝑥 2 − A𝑥 + 2B𝑥 − B + C(𝑥 2 + 𝑥 + 2) 𝟏 33 𝟐 4 At 𝒙 = , − =C∗ 11 4 , then: 𝐂 = −𝟑 Equating Coefficient (𝒙𝟐 ) 1 = 2A + C , 1 = 2A − 3 , then: 𝐀 = 𝟐 Equating Coefficient (𝒙𝟏 ) −3 = −A + 2B + C , −3 = −2 + 2B − 3 , then: 𝐁 = 𝟏 Then: (𝑥 2 Choice B Page | 19 𝑥 2 −3𝑥−7 +𝑥+2)(2𝑥−1) 2𝑥+1 = (𝑥 2 +𝑥+2) 3 − (2𝑥−1) is 30] The partial Fraction decomposition of 𝟏𝟔 𝟖𝒙−𝟏𝟐 𝟐𝒙−𝟑 𝟏𝟔 −𝟖𝒙+𝟏𝟐 𝟐𝒙+𝟑 𝟏𝟔 −𝟖𝒙+𝟏𝟐 𝟐𝒙+𝟑 𝟏𝟑 (𝟐𝒙+𝟑)(𝒙𝟐 +𝟏)𝟐 is A- 𝟏𝟑(𝟐𝒙+𝟑) − 𝟏𝟑(𝒙𝟐 +𝟏) − (𝒙𝟐 +𝟏)𝟐 C- + 𝟏𝟑(𝒙𝟐 +𝟏) + (𝒙𝟐 +𝟏)𝟐 𝟏𝟑(𝟐𝒙+𝟑) B- 𝟏𝟑(𝟐𝒙+𝟑) − 𝟏𝟑(𝒙𝟐 +𝟏) − (𝒙𝟐 +𝟏)𝟐 𝟏𝟔 −𝟖𝒙+𝟏𝟐 𝟐𝒙+𝟑 D- − 𝟏𝟑(𝟐𝒙+𝟑) + 𝟏𝟑(𝒙𝟐 +𝟏) − (𝒙𝟐 +𝟏)𝟐 Solution: 13 (2𝑥+3)(𝑥 2 +1)2 A B𝑥+C = (2𝑥+3) + (𝑥 2 +1) D𝑥+E + (𝑥 2 (Proper Fraction) +1)2 By multiplying in two sides by : (2𝑥 + 3)(𝑥 2 + 1)2 13 = A(𝑥 2 + 1)2 + (B𝑥 + C)(2𝑥 + 3)(𝑥 2 + 1) + (D𝑥 + E)(2𝑥 + 3) = A(𝑥 4 + 2𝑥 2 + 1) + 2B𝑥 4 + 3B𝑥 3 + 2B𝑥 2 + 3B𝑥 + 2C𝑥 3 +3C𝑥 2 + 2C𝑥 + 3C + 2D𝑥 2 + 3D𝑥 + 2E𝑥 + 3E 3 2 𝟑 2 At 𝒙 = − , 13 = A ((− ) + 1) , 13 = A ∗ 𝟐 2 169 16 , then: 𝐀 = 𝟏𝟔 𝟏𝟑 Equating Coefficient (𝒙𝟒 ) 0 = A + 2B , 0 = 16 13 + 2B , then: 𝐁 = − 𝟖 𝟏𝟑 Equating Coefficient (𝒙𝟑 ) 0 = 3B + 2C , 0 = 3 ∗ − 8 13 + 2C , then: 𝐂 = 𝟏𝟐 𝟏𝟑 Equating Coefficient (𝒙𝟐 ) 0 = 2A + 2B + 3C + 2D , 0 = 2 ∗ ➔ 0 = 4 + 2D , then: 𝐃 = −𝟐 Page | 20 16 13 +2∗− 8 13 +3∗ 12 13 + 2D Equating Coefficient (𝒙𝟏 ) 0 = 3B + 2C + 3D + 2E , 0 = 3 ∗ − 8 13 +2∗ 12 13 + 3 ∗ −2 + 2E ➔ 0 = 6 + 2E , then: 𝐄 = 𝟑 Then: 13 (2𝑥+3)(𝑥 2 +1)2 = 16 13(2𝑥+3) = 8 12 13 13 2 (𝑥 +1) − 𝑥+ + 16 13(2𝑥+3) + −2𝑥+3 + (𝑥 2 −8𝑥+12 13(𝑥 2 +1) +1)2 2𝑥−3 − (𝑥 2 = +1)2 16 13(2𝑥+3) − 8𝑥−12 13(𝑥 2 +1) Choice A 31] The partial Fraction decomposition of −𝟐𝒙+𝟑 𝟔 −𝟐𝒙+𝟑 𝟔 A- 𝟕(𝒙𝟐 −𝒙+𝟏) − 𝟕(𝟑𝒙−𝟐) C- B- 𝟕(𝒙𝟐 −𝒙+𝟏) + 𝟕(𝟑𝒙−𝟐) 𝒙 (𝒙𝟐 −𝒙+𝟏)(𝟑𝒙−𝟐) 𝟐𝒙+𝟑 𝟔 −𝟐𝒙−𝟑 𝟔 + 𝟕(𝟑𝒙−𝟐) 𝟕(𝒙𝟐 −𝒙+𝟏) D- 𝟕(𝒙𝟐 −𝒙+𝟏) + 𝟕(𝟑𝒙−𝟐) Solution: 𝑥 A𝑥+B (𝑥 2 −𝑥+1)(3𝑥−2) = (𝑥 2 −𝑥+1) C + (3𝑥−2) (Proper Fraction) By multiplying in two sides by : (𝑥 2 − 𝑥 + 1)(3𝑥 − 2) ➔ 𝑥 = (A𝑥 + B)(3𝑥 − 2) + C(𝑥 2 − 𝑥 + 1) = 3A𝑥 2 − 2A𝑥 + 3B𝑥 − 2B + C(𝑥 2 − 𝑥 + 1) At 𝒙 = 𝟐 𝟑 , 2 3 =C∗ 7 9 , then: 𝐂 = 𝟔 𝟕 Equating Coefficient (𝒙𝟐 ) 6 𝟐 7 𝟕 0 = 3A + C , 0 = 3A + , then: 𝐀 = − Equating Coefficient (𝒙𝟏 ) 2 6 𝟑 7 7 𝟕 1 = −2A + 3B − C , 1 = −2 ∗ − + 3B − , then: 𝐁 = Page | 21 is 2𝑥−3 − (𝑥 2 +1)2 Then: (𝑥 2 2 3 −7 𝑥+7 𝑥 = (𝑥 2 −𝑥+1)(3𝑥−2) = −𝑥+1) + −2𝑥+3 7(𝑥 2 −𝑥+1) 6 7(3𝑥−2) + 6 7(3𝑥−2) Choice B 32] The partial Fraction decomposition of 𝟏 𝟏 A- − 𝒙𝟐 − (𝟏+𝒙𝟐 ) 𝟏 C- 𝟏 𝟏 𝒙𝟐 +𝒙𝟒 is 𝟏 + (𝟏+𝒙𝟐) 𝒙𝟐 𝟏 B- 𝒙𝟐 − (𝟏−𝒙𝟐 ) 𝟏 𝟏 D- 𝒙𝟐 − (𝟏+𝒙𝟐 ) Solution: 1 𝑥 2 +𝑥 4 = 1 𝑥 2 (1+𝑥 2 ) = A 𝑥 + B 𝑥2 C𝑥+D + (1+𝑥 2 ) (Proper Fraction) By multiplying in two sides by : 𝑥 2 (1 + 𝑥 2 ) ➔ 1 = A(𝑥 )(1 + 𝑥 2 ) + B(1 + 𝑥 2 ) + (C𝑥 + D)(𝑥 2 ) = A(𝑥 + 𝑥 3 ) + B(1 + 𝑥 2 ) + C𝑥 3 + D𝑥 2 At 𝒙 = 𝟎 ➔ 1 = B(1) ➔ then: 𝐁 = 𝟏 Equating Coefficient (𝒙𝟑 ) 0 = A + C ➔ A = −C Equating Coefficient (𝒙𝟐 ) 0 = B + D ➔ D = −B ➔ then: 𝐃 = −𝟏 Equating Coefficient (𝒙𝟏 ) 0 = A ➔ then: 𝐀 = 𝟎 then: 𝐂 = 𝟎 Then: 1 𝑥 2 (1+𝑥 2 ) Choice D Page | 22 0 1 𝑥 𝑥 = + 0∗𝑥−1 2 + (1+𝑥 2 ) = 1 𝑥 1 2 − (1+𝑥 2 ) 33] The partial Fraction decomposition of 𝟐 𝒙+𝟏 𝟐 𝒙+𝟏 A- 𝟑(𝒙+𝟏) + 𝟑(𝒙𝟐 +𝒙+𝟏) C- B- 𝟑(𝒙+𝟏) + 𝟑(𝒙𝟐 −𝒙−𝟏) 𝒙𝟐 +𝟏 𝒙𝟑 +𝟏 is 𝟐 𝒙+𝟏 𝟐 𝒙+𝟏 + 𝟑(𝒙𝟐 −𝒙+𝟏) 𝟑(𝒙+𝟏) D- 𝟑(𝒙+𝟏) − 𝟑(𝒙𝟐 −𝒙+𝟏) Solution: 𝑥 2 +1 𝑥 3 +1 A B𝑥+C = (𝑥+1) + (𝑥 2 −𝑥+1) (Proper Fraction) By multiplying in two sides by : 𝑥 3 + 1 ➔ 𝑥 2 + 1 = A(𝑥 2 − 𝑥 + 1) + (B𝑥 + C)(𝑥 + 1) = A𝑥 2 − Ax + A + B𝑥 2 + B𝑥 + C𝑥 + C At 𝒙 = −𝟏 ➔ 𝟐 = A((−1)2 − (−1) + 1) ➔ 2 = 3A , then: 𝐀 = 𝟐 𝟑 Equating Coefficient (𝒙𝟐 ) 2 𝟏 3 𝟑 1 = A + B ➔ 1 = + B , then: 𝐁 = Equating Coefficient (𝒙𝟏 ) 2 1 𝟏 3 3 𝟑 0 = −A + B + C ➔ 0 = − + + C , then: 𝐂 = Then: 𝑥 2 +1 𝑥 3 +1 = 2 3(𝑥+1) + 1 1 𝑥+3 3 (𝑥 2 −𝑥+1) = 2 3(𝑥+1) + x+1 3(𝑥 2 −𝑥+1) Choice C 34] The partial Fraction decomposition of 𝟏 (𝒙+𝟏)(𝒙𝟐 −𝟏)(𝒙𝟐 +𝟓) is 𝟓 𝟏 𝟏 𝒙−𝟏 𝟓 𝟏 𝟏 𝒙−𝟏 A- − 𝟕𝟐(𝒙+𝟏) − − − C- − 𝟕𝟐(𝒙+𝟏) − + + 𝟏𝟐(𝒙+𝟏)𝟐 𝟐𝟒(𝒙−𝟏) 𝟑𝟔(𝒙𝟐 +𝟓) 𝟏𝟐(𝒙+𝟏)𝟐 𝟐𝟒(𝒙−𝟏) 𝟑𝟔(𝒙𝟐 +𝟓) 𝟓 𝟏 𝟏 𝒙−𝟏 𝟓 𝟏 𝟏 𝒙−𝟏 B- − 𝟕𝟐(𝒙+𝟏) + + + D- 𝟕𝟐(𝒙+𝟏) − + + 𝟏𝟐(𝒙+𝟏)𝟐 𝟐𝟒(𝒙−𝟏) 𝟑𝟔(𝒙𝟐 +𝟓) 𝟏𝟐(𝒙+𝟏)𝟐 𝟐𝟒(𝒙−𝟏) 𝟑𝟔(𝒙𝟐 +𝟓) Page | 23 Solution: 1 1 (𝑥+1)(𝑥 2 −1)(𝑥 2 +5) = (𝑥+1)(𝑥−1)(𝑥+1)(𝑥 2 A 1 +5) B = (𝑥+1)2 (𝑥−1)(𝑥 2 C +5) (Proper Fraction) D𝑥+E = (𝑥+1) + (𝑥+1)2 + (𝑥−1) + (𝑥 2 +5) By multiplying in two sides by : (𝑥 + 1)2 (𝑥 − 1)(𝑥 2 + 5) ➔ 1 = A(𝑥 + 1)(𝑥 − 1)(𝑥 2 + 5) + B(𝑥 − 1)(𝑥 2 + 5) + C(𝑥 + 1)2 (𝑥 2 + 5) +(D𝑥 + E)(𝑥 + 1)2 (𝑥 − 1) ➔ 1 = A(𝑥 4 + 4𝑥 2 − 5) + B(𝑥 3 − 𝑥 2 + 5𝑥 − 5) + C(𝑥 4 + 2𝑥 3 + 6𝑥 2 + 10𝑥 + 5) +D𝑥 4 + D𝑥 3 − D𝑥 2 − D𝑥 + E𝑥 3 + E𝑥 2 − E𝑥 − E At 𝒙 = −𝟏 ➔ 1 = B((−1) − 1)((−1)2 + 5) ➔ 1 = B(−12) , then: 𝐁 = − 𝟏 𝟏𝟐 𝟏 At 𝒙 = 𝟏 ➔ 1 = C(𝑥 + 1)2 (𝑥 2 + 5) ➔ 1 = C(2)2 (12 + 5) = C(24) , then: 𝐂 = 𝟐𝟒 Equating Coefficient (𝒙𝟒 ) 0=A+C+D➔0= A+ 1 24 +D 1 Equating Coefficient (𝒙𝟑 ) 0 = B + 2C + D + E ➔ 0 = − 1 12 +2∗ 1 24 + D + E ➔ D + E = 0 ➔ D = −E Equating Coefficient (𝒙𝟐 ) 0 = 4A − B + 6C − D + E ➔ 0 = 4A − (− = 4A + 1 12 1 1 4 3 + + 2E = 4A + + 2E = 0 From equations 1 & 2 0=A+ 1 24 + D && D = −E ➔0=A+ Page | 24 1 24 − E , then: E = A + 1 24 1 1 ) + 6 ∗ 24 − (−E) + E 12 3 2 From equation 3 1 1 1 3 3 24 4A + + 2E = 0 ➔ 4A + + 2 ∗ (A + 1 1 3 12 4A + + 2A + E=A+ 1 24 5 5 𝟓 = 0 ➔ 6A + 12 = 0 ➔ 6A = − 12 , then: 𝐀 = − 𝟕𝟐 ➔E=− D = −E ➔ D = − (− Then: )=0 5 72 + 1 24 1 =− 1 36 1 ) = 36 36 1 (𝑥+1)2 (𝑥−1)(𝑥 2 +5) =− 5 72(𝑥+1) =− − 5 72(𝑥+1) 1 12(𝑥+1)2 − + 1 12(𝑥+1) 1 24(𝑥−1) 2 + + 1 24(𝑥−1) 1 1 𝑥−36 36 (𝑥 2 +5) + 𝑥−1 36(𝑥 2 +5) Choice C 35] The partial Fraction decomposition of 𝒙𝟐 +𝟐𝒙+𝟑 is (𝒙−𝟏)𝟐 (𝒙−𝟐)(𝒙𝟐 +𝟒) −𝟑𝟖 𝟔 𝟏𝟏 𝟐𝟗𝒙−𝟔 −𝟑𝟖 𝟔 𝟏𝟏 𝟐𝟗𝒙−𝟔 𝟑𝟖 𝟔 𝟏𝟏 𝟐𝟗𝒙−𝟔 −𝟑𝟖 𝟔 𝟏𝟏 𝟐𝟗𝒙−𝟔 A- 𝟐𝟓(𝒙−𝟏) − 𝟓(𝒙−𝟏)𝟐 + 𝟖(𝒙−𝟐) + (𝒙𝟐 +𝟒) C- 𝟐𝟓(𝒙−𝟏) + 𝟓(𝒙−𝟏)𝟐 + 𝟖(𝒙−𝟐) + (𝒙𝟐 +𝟒) B- 𝟐𝟓(𝒙−𝟏) − 𝟓(𝒙−𝟏)𝟐 + 𝟖(𝒙−𝟐) + (𝒙𝟐+𝟒) D- 𝟐𝟓(𝒙−𝟏) − 𝟓(𝒙−𝟏)𝟐 + 𝟖(𝒙−𝟐) − (𝒙𝟐 +𝟒) Solution: 𝑥 2 +2𝑥+3 (𝑥−1)2 (𝑥−2)(𝑥 2 +4) A B C D𝑥+E = (𝑥−1) + (𝑥−1)2 + (𝑥−2) + (𝑥 2 +4) (Proper Fraction) By multiplying in two sides by : (𝑥 − 1)2 (𝑥 − 2)(𝑥 2 + 4) ➔ 𝑥 2 + 2𝑥 + 3 = A(𝑥 − 1)(𝑥 − 2)(𝑥 2 + 4) + B(𝑥 − 2)(𝑥 2 + 4) +C(𝑥 − 1)2 (𝑥 2 + 4) + (D𝑥 + E)(𝑥 − 1)2 (𝑥 − 2) = A(𝑥 4 − 3𝑥 3 + 6𝑥 2 − 12𝑥 + 8) + B(𝑥 3 − 2𝑥 2 + 4𝑥 − 8) + C(𝑥 4 − 2𝑥 3 + 5𝑥 2 − 8𝑥 + 4) +𝐷𝑥 4 − 4D𝑥 3 + 5D𝑥 2 − 2D𝑥 + E𝑥 3 − 4E𝑥 2 + 5E𝑥 − 2E Page | 25 At 𝒙 = 𝟏 ➔ 12 + 2 ∗ 1 + 3 = B(−1) ∗ 5 ➔ 6 = −5B , then: 𝐁 = − 𝟔 𝟓 At 𝒙 = 𝟐 ➔ 22 + 2 ∗ 2 + 3 = C(2 − 1)2 (22 + 4) ➔ 11 = C(8) , then: 𝐂 = Equating Coefficient (𝒙𝟒 ) ➔0=A+C+D=A+ 11 8 +D=0➔A+D=− 11 1 8 Equating Coefficient (𝒙𝟑 ) 6 22 5 8 ➔ 0 = −3A + B − 2C − 4D + E ➔ 0 = −3A − − ➔ 0 = −3A − 79 20 − 4D + E ➔ 3A + 4D − E = − − 4D + E 79 2 20 Equating Coefficient (𝒙𝟐 ) ➔ 1 = 6A − 2B + 5C + 5D − 4E ➔ 1 = 6A + ➔ 1 = 6A + 371 40 12 5 + 55 8 + 5D − 4E ➔ 6A + 5D − 4E = − + 5D − 4E 331 3 40 Equating Coefficient (𝒙𝟏 ) ➔ 2 = −12A + 4B − 8C − 2D + 5E ➔ 2 = −12A − 79 5 − 2D + 5E ➔ 12A + 2D − 5E = − 89 5 4 By Solving Equations 2 , 3 , 4 together 3A + 4D − E = − 79 20 6A + 5D − 4E = − ➔ E = 3A + 4D + 331 40 79 )=− 20 ➔ 6A + 5D − 12A − 16D − 79 Page | 26 301 40 20 2 ➔ By substitute in E ➔ 6A + 5D − 4 (3A + 4D + ➔ −6A − 11D = 79 5 =− 3 331 40 331 40 ➔ 6A + 11D = − 301 40 5 𝟏𝟏 𝟖 12A + 2D − 5E = − 89 5 ➔ By substitute in E 79 ➔ 12A + 2D − 5 (3A + 4D + )=− 20 ➔ 12A + 2D − 15A − 20D − 79 ➔ −3A − 18D = 39 20 =− 4 ➔ A + 6D = − 4 89 5 89 5 13 6 20 Solve Equation 5 , 6 together 6A + 11D = − A=− 13 20 301 40 13 , A + 6D = − − 6D ➔ 6 (− ➔− 13 20 39 10 20 − 6D) + 11D = − − 36D + 11D = − 29 ➔ −25D = − Then: 𝐀 = − 8 ➔𝐃= 301 40 301 40 𝟐𝟗 𝟐𝟎𝟎 𝟑𝟖 𝟐𝟓 By Substitution in equation 2 : 3A + 4D − E = − Then: 79 20 ➔3∗− 𝑥 2 +2𝑥+3 (𝑥−1)2 (𝑥−2)(𝑥 2 +4) 𝑥 2 +2𝑥+3 = (𝑥−1)2 (𝑥−2)(𝑥 2 +4) = = 38 25 +4∗ −38 25(𝑥−1) −38 25(𝑥−1) − − 29 200 −E=− 6 5(𝑥−1)2 6 5(𝑥−1)2 + + 79 20 11 + 8(𝑥−2) 11 ➔𝐄=− 29 3 𝑥−100 200 (𝑥 2 +4) 29𝑥−6 8(𝑥−2) + (𝑥 2 +4) Choice A 36] The partial Fraction decomposition of 𝟕 𝟏𝟑 A- 𝟏 − 𝒙−𝟐 + 𝒙−𝟑 𝟕 𝟏𝟑 B- 𝟐 − 𝒙−𝟐 + 𝒙−𝟑 Page | 27 𝒙𝟐 +𝒙+𝟏 𝒙𝟐 −𝟓𝒙+𝟔 𝟕 𝟏𝟑 𝟕 𝟏𝟑 C- 𝟏 + 𝒙−𝟐 + 𝒙−𝟑 D- 𝟏 − 𝒙−𝟐 − 𝒙−𝟑 is 𝟑 𝟏𝟎𝟎 Solution: 𝑥 2 +𝑥+1 1 (Improper Fraction) 𝑥 2 −5𝑥+6 𝑥 2 +𝑥+1 𝑥 2 −5𝑥+6 =1+ 6𝑥−5 6𝑥−5 𝑥 2 − 5𝑥 + 6 𝑥 2 −5𝑥+6 𝑥2 + 𝑥 + 1 − 𝑥 − 5𝑥 + 6 2 (Proper Fraction) 𝑥 2 −5𝑥+6 6𝑥−5 6𝑥−5 𝑥 2 −5𝑥+6 A B = (𝑥−2)(𝑥−3) = (𝑥−2) + (𝑥−3) 6𝑥 − 5 By multiplying in two sides by : (𝑥 − 2)(𝑥 − 3) ➔ 6𝑥 − 5 = A(𝑥 − 3) + B(𝑥 − 2) At 𝒙 = 𝟑 ➔ 6 ∗ 3 − 5 = B(1) , then: 𝐁 = 𝟏𝟑 At 𝒙 = 𝟐 ➔ 6 ∗ 2 − 5 = A(2 − 3) , 7 = A(−1) , then: 𝐀 = −𝟕 6𝑥−5 7 𝑥 2 −5𝑥+6 Then: 13 = − (𝑥−2) + (𝑥−3) 𝑥 2 +𝑥+1 𝑥 2 −5𝑥+6 7 13 = 1 − (𝑥−2) + (𝑥−3) Choice A 37] The partial Fraction decomposition of 𝟓 𝟏 𝟓 𝟏 A- 𝟐𝒙 + 𝟑 + 𝟑𝒙+𝟏 + 𝒙−𝟏 B- 𝟐𝒙 − 𝟑 + 𝟑𝒙+𝟏 + 𝒙−𝟏 𝟔𝒙𝟑 +𝟓𝒙𝟐 −𝟕 𝟑𝒙𝟐 −𝟐𝒙−𝟏 𝟓 3𝑥 2 −2𝑥−1 6𝑥 3 +5𝑥 2 −7 3𝑥 2 −2𝑥−1 𝟏 C- 𝟐𝒙 + 𝟑 − 𝟑𝒙+𝟏 + 𝒙−𝟏 𝟓 𝟏 D- 𝟐𝒙 + 𝟑𝒙+𝟏 + 𝒙−𝟏 Solution: 6𝑥 3 +5𝑥 2 −7 is 2𝑥 + 3 (Improper Fraction) = 2𝑥 + 3 + 8𝑥−4 3𝑥 2 − 2𝑥 − 1 6𝑥 3 + 5𝑥 2 − 7 − 3 2 6𝑥 − 4𝑥 − 2𝑥 3𝑥 2 −2𝑥−1 9𝑥 2 + 2𝑥 − 7 2 9𝑥 − 6𝑥 − 3 Page | 28 8𝑥 − 4 − 4𝑥−4 (Proper Fraction) 3𝑥 2 −2𝑥−1 8𝑥−4 8𝑥−4 3𝑥 2 −2𝑥−1 A B = (3𝑥+1)(𝑥−1) = (3𝑥+1) + (𝑥−1) By multiplying in two sides by : (3𝑥 + 1)(𝑥 − 1) ➔ 8𝑥 − 4 = A(𝑥 − 1) + B(3𝑥 + 1) 𝟏 1 1 At 𝒙 = − ➔ 8 ∗ (− ) − 4 = A (− − 1) ➔ − 𝟑 3 3 20 3 4 = A (− ) , then: 𝐀 = 𝟓 3 At 𝒙 = 𝟏 ➔ 8 ∗ 1 − 4 = B(3 ∗ 1 + 1) ➔ 4 = B(4) , then: 𝐁 = 𝟏 8𝑥−4 A 3𝑥 2 −2𝑥−1 Then: B 5 1 = (3𝑥+1) + (𝑥−1) = (3𝑥+1) + (𝑥−1) 6𝑥 3 +5𝑥 2 −7 3𝑥 2 −2𝑥−1 5 1 = 2𝑥 + 3 + (3𝑥+1) + (𝑥−1) Choice A 38] The partial Fraction decomposition of 𝟐𝒙𝟒 +𝟑𝒙𝟐 +𝟏 𝒙𝟐 +𝟑𝒙+𝟐 −𝟒𝟓 𝟔 C- 𝟐𝒙𝟐 + 𝟔𝒙 + 𝟏𝟕 + (𝒙+𝟐) − (𝒙+𝟏) −𝟒𝟓 𝟔 D- 𝟐𝒙𝟐 − 𝟔𝒙 + 𝟏𝟕 + (𝒙+𝟐) + (𝒙+𝟏) A- 𝟐𝒙𝟐 − 𝟔𝒙 + 𝟏𝟕 − (𝒙+𝟐) + (𝒙+𝟏) B- 𝟐𝒙𝟐 − 𝟔𝒙 + 𝟏𝟕 + (𝒙+𝟐) + (𝒙+𝟏) Solution: 2𝑥 4 +3𝑥 2 +1 𝑥 2 +3𝑥+2 2𝑥 4 +3𝑥 2 +1 𝑥 2 +3𝑥+2 −39𝑥−33 −39𝑥−33 𝑥 2 +3𝑥+2 = −𝟒𝟓 𝟔 𝟒𝟓 𝟔 2𝑥 2 − 6𝑥 + 17 (Improper Fraction) = 2𝑥 2 − 6𝑥 + 17 + 𝑥 2 + 3𝑥 + 2 −39𝑥−33 𝑥 2 +3𝑥+2 (Proper Fraction) 𝑥 2 +3𝑥+2 is −39𝑥−33 (𝑥+2)(𝑥+1) = A B + (𝑥+2) (𝑥+1) By multiplying in two sides by : (𝑥 + 2)(𝑥 + 1) ➔ −39𝑥 − 33 = A(𝑥 + 1) + B(𝑥 + 2) 2𝑥 4 + 3𝑥 2 + 1 4 3 2𝑥 + 6𝑥 + 4𝑥 −6𝑥 3 − 𝑥 2 + 1 3 2 − − −6𝑥 − 18𝑥 − 12𝑥 17𝑥 2 + 12𝑥 + 1 2 17𝑥 + 51𝑥 + 34 −39𝑥 − 33 Page | 29 2 − At 𝒙 = −𝟐 ➔ −39 ∗ (−2) − 33 = A(−2 + 1) ➔ 45 = −A , then: 𝐀 = −𝟒𝟓 At 𝒙 = −𝟏 ➔ −39 ∗ (−1) − 33 = B(−1 + 2) ➔ 6 = B , then: 𝐁 = 𝟔 −39𝑥−33 𝑥 2 +3𝑥+2 Then: A B −45 6 = (𝑥+2) + (𝑥+1) = (𝑥+2) + (𝑥+1) 2𝑥 4 +3𝑥 2 +1 𝑥 2 +3𝑥+2 45 6 = 2𝑥 2 − 6𝑥 + 17 − (𝑥+2) + (𝑥+1) Choice B 𝒙𝟑 +𝟏 39] The partial Fraction decomposition of 𝟏+𝒙 C- 𝒙 − 𝒙𝟐 +𝟏 𝟏−𝒙 𝟏−𝒙 B- 𝟐𝒙 + 𝒙𝟐 +𝟏 D- 𝒙 + 𝒙𝟐 +𝟏 Solution: 𝑥 2 +1 𝑥 3 +1 𝑥 2 +1 𝑥 (Improper Fraction) =𝑥+ 𝑥2 + 1 1−𝑥 𝑥3 + 1 3 𝑥 +𝑥 𝑥 2 +1 40] The partial Fraction decomposition of 𝟔𝟒 𝟏 𝟑𝟏𝟏 A- 𝒙 + 𝟐 + 𝟐𝟓(𝒙−𝟏) + 𝟓(𝒙−𝟏)𝟐 + 𝟐𝟓(𝒙+𝟒) 𝟔𝟒 𝟏 𝟑𝟏𝟏 B- 𝒙 − 𝟐𝟓(𝒙−𝟏) + 𝟓(𝒙−𝟏)𝟐 + 𝟐𝟓(𝒙+𝟒) 𝟔𝟒 𝟏 𝟑𝟏𝟏 C- 𝒙 − 𝟐 + 𝟐𝟓(𝒙−𝟏) + 𝟓(𝒙−𝟏)𝟐 + 𝟐𝟓(𝒙+𝟒) 𝟏 𝟑𝟏𝟏 D- 𝟐𝟓(𝒙+𝟏) + 𝟓(𝒙−𝟏)𝟐 + 𝟐𝟓(𝒙+𝟒) Page | 30 − 1−𝑥 Choice D 𝟔𝟒 is 𝟏−𝒙 A- 𝒙 + 𝒙𝟐 +𝟏 𝑥 3 +1 𝒙𝟐 +𝟏 𝒙𝟒 +𝟒𝒙𝟐 +𝒙−𝟓 𝒙𝟑 +𝟐𝒙𝟐 −𝟕𝒙+𝟒 is Solution: 𝑥 4 +4𝑥 2 +𝑥−5 (Improper Fraction) 𝑥 3 +2𝑥 2 −7𝑥+4 𝑥 4 +4𝑥 2 +𝑥−5 𝑥 3 +2𝑥 2 −7𝑥+4 =𝑥−2+ 15𝑥 2 −17𝑥+3 𝑥 3 +2𝑥 2 −7𝑥+4 𝑥 3 +2𝑥 2 −7𝑥+4 (Proper Fraction) 𝑥 3 +2𝑥 2 −7𝑥+4 15𝑥 2 −17𝑥+3 15𝑥 2 −17𝑥+3 = 15𝑥 2 −17𝑥+3 𝑥 3 +2𝑥 2 −3𝑥−4𝑥+4 = 15𝑥 2 −17𝑥+3 15𝑥 2 −17𝑥+3 𝑥(𝑥 2 +2𝑥−3)−4(𝑥−1) 15𝑥 2 −17𝑥+3 = (𝑥−1)(𝑥(𝑥+3)−4) = (𝑥−1)(𝑥 2 ➔ 15𝑥 2 −17𝑥+3 (𝑥−1)2 (𝑥+4) A B +3𝑥−4) = 15𝑥 2 −17𝑥+3 𝑥(𝑥+3)(𝑥−1)−4(𝑥−1) 15𝑥 2 −17𝑥+3 = (𝑥−1)(𝑥+4)(𝑥−1) = 15𝑥 2 −17𝑥+3 (𝑥−1)2 (𝑥+4) C = (𝑥−1) + (𝑥−1)2 + (𝑥+4) By multiplying in two sides by : (𝑥 − 1)2 (𝑥 + 4) ➔ 15𝑥 2 − 17𝑥 + 3 = A(𝑥 − 1)(𝑥 + 4) + B(𝑥 + 4) + C(𝑥 − 1)2 ➔ 15𝑥 2 − 17𝑥 + 3 = A(𝑥 − 1)(𝑥 + 4) + B(𝑥 + 4) + C(𝑥 − 1)2 = A𝑥 2 + 3A𝑥 − 4A + B𝑥 + 4B + C𝑥 2 − 2𝑥 + 1 At 𝒙 = 𝟏 ➔ 15 ∗ (1)2 − 17 ∗ 1 + 3 = B(1 + 4) , then: 𝐁 = 𝟏 𝟓 At 𝒙 = −𝟒 ➔ 15 ∗ (−4)2 − 17 ∗ (−4) + 3 = C(−4 − 1)2 ➔ 311 = 25C , then: 𝐂 = Equating Coefficient (𝒙𝟐 ) 15 = A + C ➔ A = 15 − 15𝑥 2 −17𝑥+3 (𝑥−1)2 (𝑥+4) Then: 64 25(𝑥−1) 𝑥 4 +4𝑥 2 +𝑥−5 𝑥 3 +2𝑥 2 −7𝑥+4 Choice C Page | 31 = + 311 25 , then: 𝐀 = 1 5(𝑥−1) 2 + =𝑥−2+ 𝟔𝟒 𝟐𝟓 311 25(𝑥+4) 64 25(𝑥−1) + 1 5(𝑥−1)2 + 311 25(𝑥+4) 𝟑𝟏𝟏 𝟐𝟓 41] The partial Fraction decomposition of 𝟐 𝟏 𝟐 𝟏 𝟗𝒙𝟐 +𝟐𝟎𝒙−𝟏𝟎 is (𝒙+𝟐)(𝟑𝒙−𝟏) 𝟐 A- 𝟐 + (𝒙+𝟐) − (𝟑𝒙−𝟏) 𝟏 C- (𝒙+𝟐) − (𝟑𝒙−𝟏) 𝟐 B- 𝟑 + (𝒙+𝟐) + (𝟑𝒙−𝟏) 𝟏 D-𝟑 + (𝒙+𝟐) − (𝟑𝒙−𝟏) Solution: 9𝑥 2 +20𝑥−10 (𝑥+2)(3𝑥−1) = 9𝑥 2 +20𝑥−10 (𝑥+2)(3𝑥−1) = 3 + (𝑥+2)(3𝑥−1) 3𝑥 2 +5𝑥−2 3 (Improper Fraction) 5𝑥−4 3𝑥 2 + 5𝑥 − 2 9𝑥 2 + 20𝑥 − 10 2 5𝑥−4 (𝑥+2)(3𝑥−1) 5𝑥−4 (𝑥+2)(3𝑥−1) 9𝑥 2 +20𝑥−10 (Proper Fraction) = A (𝑥+2) + 9𝑥 + 15𝑥 − 6 5𝑥 − 4 B (3𝑥−1) By multiplying in two sides by : (𝑥 + 2)(3𝑥 − 1) ➔ 5𝑥 − 4 = A(3𝑥 − 1) + B(𝑥 + 2) At 𝒙 = −𝟐 ➔ 5 ∗ (−2) − 4 = A(3 ∗ (−2) − 1) , then: A = 2 𝟏 1 1 𝟑 3 3 At 𝒙 = ➔ 5 ∗ − 4 = B ( + 2) , then: B = −1 5𝑥−4 (𝑥+2)(3𝑥−1) Then: 9𝑥 2 +20𝑥−10 (𝑥+2)(3𝑥−1) Choice D Page | 32 2 1 = (𝑥+2) − (3𝑥−1) 2 1 = 3 + (𝑥+2) − (3𝑥−1) − 42] If 𝒇(𝒙, 𝒚) = 𝒙𝟑 + 𝒙𝟐 𝒚𝟑 − 𝟐𝒚𝟐 then 𝒇𝒙 (𝟐, 𝟏) , 𝒇𝒚 (𝟏, 𝟐) respectively equal A- 𝟏𝟔 , 𝟒 C- 𝟒 , 𝟏𝟔 B- 𝟎 , 𝟏𝟔 D- None of them Solution: 𝑓𝑥 = 𝜕𝑓 𝜕𝑥 = 3𝑥 2 + 2𝑥𝑦 3 − 0 , At (𝟐, 𝟏) ➔ 3 ∗ 22 + 2 ∗ 2 ∗ 13 = 12 + 4 = 16 Then: 𝑓𝑥 (2,1) = 16 𝑓𝑦 = 𝜕𝑓 𝜕𝑦 = 0 + 3𝑥 2 𝑦 2 − 4𝑦 , At (𝟏, 𝟐) ➔ 3 ∗ 12 ∗ 22 − 4 ∗ 2 = 12 − 8 = 4 Then: 𝑓𝑦 (1,2) = 4 Choice A Page | 33 𝒙 43] If 𝒇(𝒙, 𝒚) = 𝒔𝒊𝒏 (𝟏+𝒚) then 𝒇𝒙 , 𝒇𝒚 respectively equal AB- 𝐜𝐨𝐬( 𝒙 ) 𝟏+𝒚 𝟏+𝒚 𝐜𝐨𝐬( 𝒙 ) 𝟏+𝒚 𝟏+𝒚 , , 𝒙 ) 𝟏+𝒚 (𝟏+𝒚)𝟐 C- 𝒙 ) 𝟏+𝒚 (𝟏+𝒚)𝟐 D- None of them −𝒙 𝐜𝐨𝐬( −𝒙 𝐬𝐢𝐧( 𝒙 ) 𝟏+𝒚 (𝟏+𝒚)𝟐 −𝒙 𝐜𝐨𝐬( , 𝐜𝐨𝐬( 𝒙 ) 𝟏+𝒚 𝟏+𝒚 Solution: 𝑥 𝑓𝑥 = 𝑓𝑦 = 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦 𝑥 = cos ( 𝜕 1+𝑦 = cos ( 𝑥 𝑥 𝑥 cos(1+𝑦) 1 ) ⋅ 𝜕𝑥 (1+𝑦) = cos (1+𝑦) ⋅ 1+𝑦 = 𝜕 1+𝑦 𝑥 ) ⋅ 𝜕𝑦 (1+𝑦) 1+𝑦 = cos ( 𝑥 1+𝑦 )⋅ ((1+𝑦)⋅0)−(𝑥⋅1) (1+𝑦)2 = cos ( 𝑥 1+𝑦 −𝑥 ) ⋅ (1+𝑦)2 = 𝑥 ) 1+𝑦 (1+𝑦)2 −𝑥 cos( Choice A 𝜕𝑧 𝜕𝑧 44] Find 𝜕𝑥 , 𝜕𝑦 respectively if z is defined by (𝑥, 𝑦) where 𝑥 3 + 𝑦 3 + 𝑧 3 + 6𝑥𝑦𝑧 = 1 𝒙𝟐 +𝟐𝒚𝒛 𝒚𝟐 +𝟐𝒙𝒛 𝒚𝟐 +𝟐𝒙𝒛 A- −𝒛𝟐−𝟐𝒙𝒚 , − −𝒛𝟐 −𝟐𝒙𝒚 𝒙𝟐 +𝟐𝒚𝒛 B- −𝒛𝟐−𝟐𝒙𝒚 , C- −𝒛𝟐 −𝟐𝒙𝒚 , −𝒛𝟐 −𝟐𝒙𝒚 𝒚𝟐 −𝟐𝒙𝒛 𝒙𝟐 +𝟐𝒚𝒛 D- −𝒛𝟐 −𝟐𝒙𝒚 , −𝒛𝟐 −𝟐𝒙𝒚 Solution: 𝑓𝑥 = 𝜕𝑓 𝜕𝑥 = 3𝑥 2 + 0 + 3𝑧 2 ⋅ 3𝑥 2 + 6𝑦𝑧 + 𝜕𝑧 𝜕𝑥 3𝑥 2 + 6𝑦𝑧 = − Page | 34 𝜕𝑧 𝜕𝑥 + 6𝑦𝑧 + 6𝑥𝑦 ⋅ (3𝑧 2 + 6𝑥𝑦) = 0 𝜕𝑧 𝜕𝑥 𝒙𝟐 +𝟐𝒚𝒛 (3𝑧 2 + 6𝑥𝑦) 𝜕𝑧 𝜕𝑥 =0 𝒚𝟐 +𝟐𝒙𝒛 −𝒛𝟐 −𝟐𝒙𝒚 𝜕𝑧 = 𝜕𝑥 𝑓𝑦 = 3𝑥 2 +6𝑦𝑧 −(3𝑧 2 +6𝑥𝑦) 𝜕𝑓 𝜕𝑦 𝜕𝑧 𝜕𝑦 3𝑦 2 + 6𝑥𝑧 = − 𝜕𝑦 = 3(𝑥 2 +2𝑦𝑧) −3(𝑧 2 +2𝑥𝑦) = 0 + 3𝑦 2 + 3𝑧 2 ⋅ 3𝑦 2 + 6𝑥𝑧 + 𝜕𝑧 = 3𝑦 2 +6𝑥𝑧 −(3𝑧 2 +6𝑥𝑦) 𝜕𝑧 𝜕𝑦 = 𝑥 2 +2𝑦𝑧 −𝑧 2 −2𝑥𝑦 + 6𝑥𝑧 + 6𝑥𝑦 ⋅ 𝜕𝑧 𝜕𝑦 =0 (3𝑧 2 + 6𝑥𝑦) = 0 𝜕𝑧 𝜕𝑦 = (3𝑧 2 + 6𝑥𝑦) 3(𝑦 2 +2𝑥𝑧) −3(𝑧 2 +2𝑥𝑦) = 𝑦 2 +2𝑥𝑧 −𝑧 2 −2𝑥𝑦 Choice D 45] Find the second partial derivative of 𝒇(𝒙, 𝒚) = 𝒙𝟑 + 𝒙𝟐 𝒚𝟑 − 𝟐𝒚𝟐 [𝒇𝒙𝒙 , 𝒇𝒙𝒚 , 𝒇𝒚𝒙 , 𝒇𝒚𝒚 ] respectively A- 𝟔𝒙 + 𝟐𝒚𝟑 / 𝟔𝒙𝒚𝟐 / 𝟔𝒙𝒚𝟐 / 𝟔𝒙𝟐 𝒚 − 𝟒 B- 𝟔𝒙𝟐 𝒚 − 𝟒 / 𝟔𝒙𝒚𝟐 / 𝟔𝒙𝒚𝟐 / 𝟔𝒙 − 𝟐𝒚𝟑 C- 𝟔𝒙𝟐 𝒚 − 𝟒 / 𝟔𝒙𝒚𝟐 / 𝟔𝒙 + 𝟐𝒚𝟑 / 𝟔𝒙𝒚𝟐 D- None of them Solution: 𝑓𝑥 = 3𝑥 2 + 2𝑥𝑦 3 , 𝑓𝑦 = 3𝑥 2 𝑦 2 − 4𝑦 𝑓𝑥𝑥 = 6𝑥 + 2𝑦 3 𝑓𝑥𝑦 = 6𝑥𝑦 2 𝑓𝑦𝑥 = 6𝑥𝑦 2 𝑓𝑦𝑦 = 6𝑥 2 𝑦 − 4 Note Choice A Page | 35 Since 𝑓𝑥𝑦 = 𝑓𝑦𝑥 , then: the Function is continuous. 46] Find 𝒇𝒙𝒙𝒚𝒛 of 𝒇(𝒙, 𝒚, 𝒛) = 𝐬𝐢𝐧(𝟑𝒙 + 𝒚𝒛) A- −𝟗 𝐜𝐨𝐬(𝟑𝒙 + 𝒚𝒛) − 𝟗𝒚𝒛 𝐬𝐢𝐧(𝟑𝒙 + 𝒚𝒛) B- −𝟗 𝐜𝐨𝐬(𝟑𝒙 + 𝒚𝒛) + 𝟗𝒚𝒛 𝐬𝐢𝐧(𝟑𝒙 + 𝒚𝒛) C- −𝟗 𝐜𝐨𝐬(𝟑𝒙 + 𝒚𝒛) + 𝟗𝒚𝒛 𝐬𝐢𝐧(𝟑𝒙 − 𝒚𝒛) D- 𝟗 𝐜𝐨𝐬(𝟑𝒙 + 𝒚𝒛) + 𝟗𝒚𝒛 𝐬𝐢𝐧(𝟑𝒙 + 𝒚𝒛) Solution: 𝑓𝑥 = cos(3𝑥 + 𝑦𝑧) ∗ 3 = 3 cos(3𝑥 + 𝑦𝑧) 𝑓𝑥𝑥 = −3 sin(3𝑥 + 𝑦𝑧) ∗ 3 = − 9 sin(3𝑥 + 𝑦𝑧) 𝑓𝑥𝑥𝑦 = −9 cos(3𝑥 + 𝑦𝑧) ∗ 𝑧 = −9𝑧 cos(3𝑥 + 𝑦𝑧) 𝑓𝑥𝑥𝑦𝑧 = −9 cos(3𝑥 + 𝑦𝑧) + 9𝑦𝑧 sin(3𝑥 + 𝑦𝑧) Choice B 47] Find the first partial derivative of the following functions : ( I ) 𝒇(𝒙, 𝒚) = 𝒙𝟒 𝒚𝟑 + 𝟖𝒙𝟐 𝒚 [𝒇𝒙 , 𝒇𝒚 ] respectively A- 𝟒𝒙𝟑 𝒚𝟑 + 𝟏𝟔𝒙𝒚 , 𝟑𝐱 𝟒 𝒚𝟐 + 𝟖𝒙𝟐 B- 𝟑𝒙𝟒 𝒚𝟐 + 𝟖𝒙𝟐 , 𝟒𝒙𝟑 𝒚𝟑 + 𝟏𝟔𝒙𝒚 C- 𝟒𝒙𝟑 𝒚𝟑 + 𝟏𝟔𝒙𝒚 , 𝟑𝒙𝟒 𝒚𝟐 − 𝟖𝒙𝟐 D- None of them Solution: 𝑓𝑥 = 4𝑥 3 𝑦 3 + 16𝑥𝑦 , 𝑓𝑦 = 3𝑥 4 𝑦 2 + 8𝑥 2 Choice A Page | 36 𝟓 ( II ) 𝒈(𝒖, 𝒗) = (𝒖𝟐 𝒗 − 𝒗𝟑 ) [𝒈𝒖 , 𝒈𝒗 ] respectively 𝟒 𝟒 A- 𝟏𝟎𝒖𝒗(𝒖𝟐 𝒗 + 𝒗𝟑 ) , 𝟓(𝒖𝟐 𝒗 − 𝒗𝟑 ) (𝒖𝟐 − 𝟑𝒗𝟐 ) 𝟒 B- 𝟓(𝒖𝟐 𝒗 − 𝒗𝟑 ) (𝒖𝟐 − 𝟑𝒗𝟐 ) , 𝟏𝟎𝒖𝒗(𝒖𝟐 𝒗 − 𝒗𝟑 ) 𝟒 𝟒 𝟒 C- 𝟏𝟎𝒖𝒗(𝒖𝟐 𝒗 − 𝒗𝟑 ) , 𝟓(𝒖𝟐 𝒗 − 𝒗𝟑 ) (𝒖𝟐 − 𝟑𝒗𝟐 ) D- None of them Solution: 𝑔𝑢 = 5(𝑢2 𝑣 − 𝑣 3 )4 ∗ 2𝑢𝑣 = 10𝑢𝑣 (𝑢2 𝑣 − 𝑣 3 )4 𝑔𝑣 = 5(𝑢2 𝑣 − 𝑣 3 )4 ∗ (𝑢2 − 3𝑣 2 ) Choice C ( III ) 𝒇(𝒙, 𝒕) = 𝒆−𝒕 𝒄𝒐𝒔 𝝅𝒙 [𝒇𝒙 , 𝒇𝒕 ] respectively A- −𝒆−𝒕 𝒄𝒐𝒔 𝝅𝒙 , 𝝅𝒆−𝒕 𝒔𝒊𝒏 𝝅𝒙 B- 𝝅𝒆−𝒕 𝒔𝒊𝒏 𝝅𝒙 , −𝒆−𝒕 𝒄𝒐𝒔 𝝅𝒙 C- −𝝅𝒆−𝒕 𝒔𝒊𝒏 𝝅𝒙 , −𝒆−𝒕 𝒄𝒐𝒔 𝝅𝒙 D- None of them Solution: 𝑓𝑥 = (−𝑒 −𝑡 𝑠𝑖𝑛 𝜋𝑥 ) ∗ 𝜋 = −𝜋𝑒 −𝑡 𝑠𝑖𝑛 𝜋𝑥 𝑓𝑡 = −𝑒 −𝑡 𝑐𝑜𝑠 𝜋𝑥 Choice C Page | 37 ( IV ) 𝒇(𝒙, 𝒚) = √𝒙 𝒍𝒏 𝒕 [𝒇𝒙 , 𝒇𝒕 ] respectively 𝒍𝒏 𝒕 A- 𝟐 C- √ √𝒙 𝒕 , 𝒙 √𝒙 𝒍𝒏 𝒕 , 𝒕 𝟐 √𝒙 B- 𝒍𝒏 𝒕 √𝒙 , √𝒙 𝒕 D- None of them Solution: 𝑙𝑛 𝑡 𝑓𝑥 = 2√𝑥 𝑓𝑡 = √𝑥 𝑡 Choice A 𝒙 ( V ) 𝒇(𝒙, 𝒚) = 𝒚 [𝒇𝒙 , 𝒇𝒚 ] respectively 𝟏 𝒙 A- 𝒚 ,− 𝒚 𝟏 𝒙 C- 𝒚 , 𝒚𝟐 B- 𝟏 𝟓 ,− 𝒚𝟐 𝒚 D- None of them Solution: 𝑓𝑥 = 𝑥 𝑦 1 𝑦 = 𝑥𝑦 −1 ➔ 𝑓𝑦 = −𝑥𝑦 −2 = − Choice D Page | 38 𝑥 𝑦2 48] If 𝒇(𝒙, 𝒚) is a differentiable function such that 𝒙 = 𝒙(𝒓, 𝒔) and 𝒚 = 𝒚(𝒓, 𝒔) , then 𝝏𝒇 𝝏𝒇 𝝏𝒇 𝝏𝒙 𝝏𝒚 𝝏𝒇 𝝏𝒇 𝒅𝒙 𝝏𝒇 𝒅𝒚 A- 𝝏𝒓 = 𝝏𝒙 𝝏𝒚 + 𝝏𝒓 𝝏𝒓 C- 𝝏𝒓 = 𝝏𝒙 𝒅𝒓 + 𝝏𝒚 𝒅𝒓 𝝏𝒇 𝒅𝒇 𝝏𝒙 𝒅𝒇 𝝏𝒚 𝝏𝒇 𝝏𝒇 𝝏𝒙 𝝏𝒇 𝝏𝒚 B- 𝝏𝒓 = 𝒅𝒙 𝝏𝒓 + 𝒅𝒚 𝝏𝒓 D- 𝝏𝒓 = 𝝏𝒙 𝝏𝒓 + 𝝏𝒚 𝝏𝒓 Solution: Since 𝑓(𝑥, 𝑦) has 2 variables && 𝑥 and 𝑦 have 2 variables too Then: 𝜕𝑓 𝜕𝑟 = 𝜕𝑓 𝜕𝑥 𝜕𝑥 𝜕𝑟 + 𝜕𝑓 𝜕𝑦 𝜕𝑦 𝜕𝑟 Choice D 49] If 𝒇(𝒙, 𝒚) is differentiable at (𝒂, 𝒃) , then A- 𝒇(𝒙, 𝒚) is continuous at (𝒂, 𝒃) B- 𝒇(𝒙, 𝒚) is not continuous at (𝒂, 𝒃) C- 𝒇(𝒙, 𝒚) is not defined at (𝒂, 𝒃) D- None of them Solution: Since 𝑓 (𝑥, 𝑦) is differentiable at (𝑎, 𝑏) ➔ 𝑓(𝑥, 𝑦) is continuous at (𝑎, 𝑏) Choice A 𝝏𝒖 50] If 𝒖(𝒙, 𝒚) = 𝐥𝐧(𝒙𝟐 + 𝒚 + 𝟐) , then 𝝏𝒙 at (𝟏, 𝟑) is 𝟏 A- 𝟔 𝟏 C- 𝟑 Page | 39 𝟏 B- 𝟐 D- None of them Solution: 𝜕𝑢 𝜕𝑥 = 2𝑥 𝑥 2 +𝑦+2 At (1,3) ➔ 2𝑥 𝑥 2 +𝑦+2 = 2∗1 12 +3+2 2 1 6 3 = = Choice B 51] If 𝒇(𝒙, 𝒚) is a differentiable function such that 𝒙 = 𝒙(𝒓) and 𝒚 = 𝒚(𝒓) , then 𝝏𝒇 𝝏𝒇 𝝏𝒇 𝝏𝒙 𝝏𝒚 𝝏𝒇 𝝏𝒇 𝒅𝒙 𝝏𝒇 𝒅𝒚 A- 𝝏𝒓 = 𝝏𝒙 𝝏𝒚 + 𝝏𝒓 𝝏𝒓 𝝏𝒇 𝒅𝒇 𝝏𝒙 𝒅𝒇 𝝏𝒚 𝝏𝒇 𝝏𝒇 𝝏𝒙 𝝏𝒇 𝝏𝒚 B- 𝝏𝒓 = 𝒅𝒙 𝝏𝒓 + 𝒅𝒚 𝝏𝒓 C- 𝝏𝒓 = 𝝏𝒙 𝒅𝒓 + 𝝏𝒚 𝒅𝒓 D- 𝝏𝒓 = 𝝏𝒙 𝝏𝒓 + 𝝏𝒚 𝝏𝒓 Solution: Since 𝑓(𝑥, 𝑦) has 2 variables && 𝑥 and 𝑦 have 1 variable each Then: 𝜕𝑓 𝜕𝑟 = 𝜕𝑓 𝑑𝑥 𝜕𝑥 𝑑𝑟 + 𝜕𝑓 𝑑𝑦 𝜕𝑦 𝑑𝑟 Choice C 𝝏𝟐 𝒖 𝟐 52] If 𝒖(𝒙, 𝒚) = 𝐥𝐧(𝒙 + 𝒚 + 𝟐) , then 𝝏𝒚𝟐 at (𝟏, 𝟑) is 𝟏 𝟏 A- 𝟑𝟔 B- 𝟔 𝟏 𝟏 C- − 𝟑𝟔 D- − 𝟔 Solution: 𝜕𝑢 𝜕𝑦 = 1 𝑥 2 +𝑦+2 Page | 40 𝜕2 𝑢 𝜕𝑦 2 = (𝑥 2 +𝑦+2)∗0−1∗(1) (𝑥 2 +𝑦+2)2 = (12 −1 −1 +3+2)2 = (6)2 = −1 36 Choice C 53] Determine whether each of the following functions is a solution of the Laplace’s equation I- 𝒖(𝒙, 𝒕) = 𝐬𝐢𝐧(𝒙 − 𝒂𝒕) III- 𝒖(𝒙, 𝒕) = 𝒆𝒙−𝒕 II- 𝒖(𝒙, 𝒕) = 𝐬𝐢𝐧(𝒙 + 𝒕) IV- 𝒖(𝒙, 𝒕) = 𝒆𝒙 𝐬𝐢𝐧 𝒕 A] I and II B] I , II and III C] IV only D] I and IV Solution: I ] 𝑢(𝑥, 𝑡) = sin(𝑥 − 𝑎𝑡) 𝑢𝑥 = cos(𝑥 − 𝑎𝑡) 𝑢𝑥𝑥 = − sin(𝑥 − 𝑎𝑡) 𝑢𝑡 = cos(𝑥 − 𝑎𝑡) ∗ −𝑎 = −𝑎 cos(𝑥 − 𝑎𝑡) 𝑢𝑡𝑡 = −𝑎 ∗ − sin(𝑥 − 𝑎𝑡) ∗ −𝑎 = −𝑎2 sin(𝑥 − 𝑎𝑡) 𝑢𝑥𝑥 + 𝑢𝑡𝑡 = − sin(𝑥 − 𝑎𝑡) + (−𝑎2 sin(𝑥 − 𝑎𝑡)) ≠ 0 Then 𝒖 doesn’t satisfy Laplace’s equation II ] 𝑢(𝑥, 𝑡) = sin(𝑥 + 𝑡) 𝑢𝑥 = cos(𝑥 + 𝑡) , 𝑢𝑥𝑥 = − sin(𝑥 + 𝑡) 𝑢𝑡 = cos(𝑥 + 𝑡) , 𝑢𝑡𝑡 = − sin(𝑥 + 𝑡) 𝑢𝑥𝑥 + 𝑢𝑡𝑡 = − sin(𝑥 + 𝑡) + (− sin(𝑥 + 𝑡)) = −2 sin(𝑥 + 𝑡) ≠ 0 Then 𝒖 doesn’t satisfy Laplace’s equation Page | 41 III ] 𝑢(𝑥, 𝑡) = 𝑒 𝑥−𝑡 𝑢𝑥 = 𝑒 𝑥−𝑡 , 𝑢𝑥𝑥 = 𝑒 𝑥−𝑡 𝑢𝑡 = 𝑒 𝑥−𝑡 ∗ −1 = −𝑒 𝑥−𝑡 𝑢𝑡𝑡 = −𝑒 𝑥−𝑡 ∗ −1 = 𝑒 𝑥−𝑡 𝑢𝑥𝑥 + 𝑢𝑡𝑡 = 𝑒 𝑥−𝑡 + 𝑒 𝑥−𝑡 = 2𝑒 𝑥−𝑡 ≠ 0 Then 𝒖 doesn’t satisfy Laplace’s equation IV ] 𝑢(𝑥, 𝑡) = 𝑒𝑥 sin 𝑡 𝑢𝑥 = 𝑒 𝑥 sin 𝑡 , 𝑢𝑥𝑥 = 𝑒 𝑥 sin 𝑡 𝑢𝑡 = 𝑒 𝑥 cos 𝑡 , 𝑢𝑡𝑡 = −𝑒 𝑥 sin 𝑡 𝑢𝑥𝑥 + 𝑢𝑡𝑡 = 𝑒 𝑥 sin 𝑡 + (−𝑒 𝑥 sin 𝑡) = 0 Then 𝒖 satisfies Laplace’s equation Choice C 54] Determine whether each of the following functions is a solution of the Wave equation I- 𝒖(𝒙, 𝒕) = (𝒙 − 𝒂𝒕)𝟔 + (𝒙 + 𝒂𝒕)𝟔 II- 𝒖(𝒙, 𝒕) = 𝐬𝐢𝐧(𝒙 − 𝒂𝒕) + 𝒍𝒏(𝒙 + 𝒂𝒕) III- 𝒖(𝒙, 𝒕) = 𝐬𝐢𝐧(𝒙 − 𝒂𝒕) A] I and II B] I , II and III C] I D] II and III Page | 42 Solution: I ] 𝑢(𝑥, 𝑡) = (𝑥 − 𝑎𝑡)6 + (𝑥 + 𝑎𝑡)6 𝑢𝑥 = 6(𝑥 − 𝑎𝑡)5 + 6(𝑥 + 𝑎𝑡)5 𝑢𝑥𝑥 = 30(𝑥 − 𝑎𝑡)4 + 30(𝑥 + 𝑎𝑡)4 𝑢𝑡 = 6(𝑥 − 𝑎𝑡)5 ∗ (−𝑎) + 6(𝑥 + 𝑎𝑡)5 ∗ 𝑎 = −6𝑎(𝑥 − 𝑎𝑡)5 + 6𝑎(𝑥 + 𝑎𝑡)5 𝑢𝑡𝑡 = −6𝑎(𝑥 − 𝑎𝑡)4 ∗ 5 ∗ (−𝑎) + 6𝑎(𝑥 + 𝑎𝑡)4 ∗ 5 ∗ 𝑎 = 30𝑎2 (𝑥 − 𝑎𝑡)4 + 30𝑎2 (𝑥 + 𝑎𝑡)4 𝑢𝑡𝑡 = 𝑎2 𝑢𝑥𝑥 ➔ 30𝑎2 (𝑥 − 𝑎𝑡)4 + 30𝑎2 (𝑥 + 𝑎𝑡)4 = 𝑎2 (30(𝑥 − 𝑎𝑡)4 + 30(𝑥 + 𝑎𝑡)4 ) Then 𝒖 satisfies Wave equation II ] 𝑢(𝑥, 𝑡) = 𝑠𝑖𝑛(𝑥 − 𝑎𝑡) + 𝑙𝑛(𝑥 + 𝑎𝑡) 𝑢𝑥 = cos(𝑥 − 𝑎𝑡) + 1 𝑥+𝑎𝑡 𝑢𝑡 = cos(𝑥 − 𝑎𝑡) ∗ −𝑎 + 1 , 𝑢𝑥𝑥 = − sin(𝑥 − 𝑎𝑡) − (𝑥+𝑎𝑡)2 𝑎 𝑥+𝑎𝑡 𝑎2 𝑢𝑡𝑡 = −𝑎2 sin(𝑥 − 𝑎𝑡) − (𝑥+𝑎𝑡)2 𝑎2 1 𝑢𝑡𝑡 = 𝑎2 𝑢𝑥𝑥 ➔ −𝑎2 sin(𝑥 − 𝑎𝑡) − (𝑥+𝑎𝑡)2 = 𝑎2 (− sin(𝑥 − 𝑎𝑡) − (𝑥+𝑎𝑡)2 ) Then 𝒖 satisfies Wave equation III ] 𝑢(𝑥, 𝑡) = sin(𝑥 − 𝑎𝑡) 𝑢𝑥 = cos(𝑥 − 𝑎𝑡) , 𝑢𝑡 = −𝑎 cos(𝑥 − 𝑎𝑡) 𝑢𝑥𝑥 = − sin(𝑥 − 𝑎𝑡) , 𝑢𝑡𝑡 = −𝑎2 sin(𝑥 − 𝑎𝑡) 𝑢𝑡𝑡 = 𝑎2 𝑢𝑥𝑥 ➔ −𝑎2 sin(𝑥 − 𝑎𝑡) = 𝑎2 (− sin(𝑥 − 𝑎𝑡)) Then 𝒖 satisfies Wave equation Page | 43 Choice B 𝒅𝒛 55] If 𝒛 = 𝒙𝟐 𝒚 + 𝟑𝒙𝒚𝟒 , 𝒙 = 𝐬𝐢𝐧 𝟐𝒕 , 𝒚 = 𝐜𝐨𝐬 𝒕 , find A- 𝟔 B- −𝟔 C- 𝟏𝟐 D- 𝟎 𝒅𝒕 when 𝒕 = 𝟎 Solution: 𝜕𝑧 𝑑𝑥 = 2𝑥𝑦 + 3𝑦 4 , 𝜕𝑥 𝜕𝑧 𝑑𝑦 = 𝑥 2 + 12𝑥𝑦 3 , 𝜕𝑦 ➔ 𝑑𝑧 = 𝑑𝑡 𝜕𝑧 𝜕𝑥 ∗ 𝑑𝑥 𝑑𝑡 = 2 cos 2𝑡 𝑑𝑡 + 𝑑𝑡 𝜕𝑧 𝜕𝑦 ∗ = − sin 𝑡 𝑑𝑦 𝑑𝑡 = (2𝑥𝑦 + 3𝑦 4 ) ∗ (2 cos 2𝑡) + (𝑥 2 + 12𝑥𝑦 3 ) ∗ (− sin 𝑡) At 𝒕 = 𝟎 ➔ 2 ∗ (2𝑥𝑦 + 3𝑦 4 ) + 0 = 4𝑥𝑦 + 6𝑦 4 𝑥 = sin 2𝑡 , 𝑦 = cos 𝑡 , then: 𝒙 = 𝟎 , 𝒚 = 𝟏 Then: 4𝑥𝑦 + 6𝑦 4 = 4 ∗ 0 ∗ 1 + 6 ∗ 14 = 6 Choice A 56] If 𝒛 = 𝒆𝒙 𝐬𝐢𝐧 𝒚 , 𝒙 = 𝒔𝒕𝟐 , 𝒚 = 𝒔𝟐 𝒕 , find 𝝏𝒛 𝝏𝒔 , 𝝏𝒛 𝝏𝒕 respectively A- 𝒕𝒆𝒙 (𝒕 𝐬𝐢𝐧 𝒚 + 𝟐𝒔 𝐜𝐨𝐬 𝒚) , 𝒔𝒆𝒙 (𝟐𝒕 𝐬𝐢𝐧 𝒚 + 𝒔 𝐜𝐨𝐬 𝒚) B- 𝒔𝒆𝒙 (𝟐𝒕𝒆𝒙 𝐬𝐢𝐧 𝒚 + 𝒔 𝐜𝐨𝐬 𝒚) , 𝒕𝒆𝒙 (𝒕 𝐬𝐢𝐧 𝒚 + 𝟐𝒔 𝐜𝐨𝐬 𝒚) C- 𝒕𝒆𝒙 (𝒕 𝐬𝐢𝐧 𝒚 + 𝟐𝒔 𝐜𝐨𝐬 𝒚) , 𝒆𝒙 (𝟐𝒕𝒆𝒙 𝐬𝐢𝐧 𝒚 + 𝒔 𝐜𝐨𝐬 𝒚) D- None of them Solution: 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑠 = 𝑒 𝑥 sin 𝑦 , = 𝜕𝑧 𝜕𝑥 ∗ 𝜕𝑥 𝜕𝑠 + 𝜕𝑥 = 𝑡2 , 𝜕𝑠 𝜕𝑧 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑠 𝜕𝑥 𝜕𝑡 = 2𝑠𝑡 , 𝜕𝑧 𝜕𝑦 = 𝑒 𝑥 cos 𝑦 , 𝜕𝑦 𝜕𝑠 = 2𝑠𝑡 , 𝜕𝑦 𝜕𝑡 = 𝑠2 = (𝑒 𝑥 sin 𝑦 ) ∗ (𝑡 2 ) + (𝑒 𝑥 cos 𝑦) ∗ (2𝑠𝑡) = 𝑡 2 𝑒 𝑥 sin 𝑦 + 2𝑠𝑡𝑒 𝑥 cos 𝑦 = 𝒕𝒆𝒙 (𝒕 𝐬𝐢𝐧 𝒚 + 𝟐𝒔 𝐜𝐨𝐬 𝒚) Page | 44 𝜕𝑧 = 𝜕𝑡 𝜕𝑧 𝜕𝑥 ∗ 𝜕𝑥 𝜕𝑡 + 𝜕𝑧 ∗ 𝜕𝑦 𝜕𝑦 𝜕𝑡 = (𝑒 𝑥 sin 𝑦 ) ∗ (2𝑠𝑡) + (𝑒 𝑥 cos 𝑦) ∗ (𝑠 2 ) = 2𝑠𝑡𝑒 𝑥 sin 𝑦 + 𝑠 2 𝑒 𝑥 cos 𝑦 = 𝒔𝒆𝒙 (𝟐𝒕 𝐬𝐢𝐧 𝒚 + 𝒔 𝐜𝐨𝐬 𝒚) Choice A 57] If 𝒖 = 𝒙𝟒 𝒚 + 𝒚𝟐 𝒛𝟑 , 𝒛 = 𝒔𝒓𝟐 𝐬𝐢𝐧 𝒕 , 𝒙 = 𝒓𝒔𝒆𝒕 , 𝒚 = 𝒓𝒔𝟐 𝒆−𝒕 find 𝝏𝒖 , 𝝏𝒓 𝝏𝒖 𝝏𝒔 , 𝝏𝒖 𝝏𝒕 , when 𝒓 = 𝟐 , 𝒔 = 𝟏 , 𝒕 = 𝟎 respectively A- 𝟖𝟎 , 𝟏𝟗𝟎 , 𝟗𝟔 B- 𝟖𝟎 , 𝟏𝟗𝟐 , 𝟗 C- 𝟖𝟎 , 𝟏𝟗𝟐 , 𝟗𝟔 D- None of them Solution: 𝜕𝑢 𝜕𝑧 = 3𝑦 2 𝑧 2 , 𝜕𝑧 𝜕𝑟 = 2𝑠𝑟 sin 𝑡 , 𝜕𝑢 𝜕𝑥 = 4𝑥 3 𝑦 , 𝜕𝑥 𝜕𝑟 = 𝑠𝑒 𝑡 , 𝜕𝑢 𝜕𝑦 = 𝑥 4 + 2𝑦𝑧 3 , 𝑠 2 𝑒 −𝑡 𝜕𝑢 𝜕𝑟 = 𝜕𝑢 𝜕𝑧 ∗ 𝜕𝑧 𝜕𝑟 + 𝜕𝑢 𝜕𝑥 ∗ 𝜕𝑥 𝜕𝑟 + 𝜕𝑢 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑟 = (3𝑦 2 𝑧 2 ) ∗ (2𝑠𝑟 sin 𝑡) + (4𝑥 3 𝑦 ) ∗ (𝑠𝑒 𝑡 ) + (𝑥 4 + 2𝑦𝑧 3 ) ∗ (𝑠 2 𝑒 −𝑡 ) At 𝑟 = 2 , 𝑠 = 1 , 𝑡 = 0 ➔ (3𝑦 2 𝑧 2 ) ∗ (2 ∗ 1 ∗ 2 ∗ sin 0) + (4𝑥 3 𝑦 ) ∗ (1 ∗ 𝑒 0 ) + (𝑥 4 + 2𝑦𝑧 3 ) ∗ (12 𝑒 −0 ) = 0 + (4𝑥 3 𝑦 ) ∗ (1 ∗ 1) + (𝑥 4 + 2𝑦𝑧 3 ) ∗ (1 ∗ 1) = (4𝑥 3 𝑦 ) + (𝑥 4 + 2𝑦𝑧 3 ) Page | 45 𝜕𝑦 𝜕𝑟 = ➔ 𝑧 = 𝑠𝑟 2 sin 𝑡 , At 𝑟 = 2 , 𝑠 = 1 , 𝑡 = 0 Then: 𝑧 = 1 ∗ 22 ∗ sin 0 = 0 ➔ 𝑥 = 𝑟𝑠𝑒 𝑡 , At 𝑟 = 2 , 𝑠 = 1 , 𝑡 = 0 Then: 𝑥 = 2 ∗ 1 ∗ 𝑒 0 = 2 ➔ 𝑦 = 𝑟𝑠 2 𝑒 −𝑡 , At 𝑟 = 2 , 𝑠 = 1 , 𝑡 = 0 Then: 𝑦 = 2 ∗ 1 ∗ 1 = 2 By Substitution in the equation : (4𝑥 3 𝑦 ) + (𝑥 4 + 2𝑦𝑧 3 ) = (4 ∗ 23 ∗ 2) + (24 + 2 ∗ 2 ∗ 03 ) = 64 + 16 = 80 𝑢 = 𝑥 4 𝑦 + 𝑦 2 𝑧 3 , 𝑧 = 𝑠𝑟 2 sin 𝑡 , 𝑥 = 𝑟𝑠𝑒 𝑡 , 𝑦 = 𝑟𝑠 2 𝑒 −𝑡 𝜕𝑢 ➔ 𝜕𝑧 = 3𝑦 2 𝑧 2 , 𝜕𝑧 𝜕𝑠 = 𝑟 2 sin 𝑡 , 𝜕𝑢 𝜕𝑥 = 4𝑥 3 𝑦 , 𝜕𝑥 𝜕𝑠 = 𝑟𝑒 𝑡 , 𝜕𝑢 𝜕𝑦 = 𝑥 4 + 2𝑦𝑧 3 , 𝜕𝑦 𝜕𝑠 = 2𝑟𝑠𝑒 −𝑡 𝜕𝑢 𝜕𝑠 = 𝜕𝑢 𝜕𝑧 ∗ 𝜕𝑧 𝜕𝑠 + 𝜕𝑢 𝜕𝑥 ∗ 𝜕𝑥 𝜕𝑠 + 𝜕𝑢 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑠 = (3𝑦 2 𝑧 2 ) ∗ (𝑟 2 sin 𝑡 ) + (4𝑥 3 𝑦 ) ∗ (𝑟𝑒 𝑡 ) + (𝑥 4 + 2𝑦𝑧 3 ) ∗ (2𝑟𝑠𝑒 −𝑡 ) At 𝑟 = 2 , 𝑠 = 1 , 𝑡 = 0 ➔ (3𝑦 2 𝑧 2 ) ∗ (22 sin 0 ) + (4𝑥 3 𝑦 ) ∗ (2 ∗ 𝑒 0 ) + (𝑥 4 + 2𝑦𝑧 3 ) ∗ (2 ∗ 2 ∗ 1 ∗ 𝑒 −0 ) = 0 + (4𝑥 3 𝑦 ) ∗ (2) + (𝑥 4 + 2𝑦𝑧 3 ) ∗ (4) = 8𝑥 3 𝑦 + 4𝑥 4 + 8𝑦𝑧 3 At 𝑟 = 2 , 𝑠 = 1 , 𝑡 = 0 ➔𝑧=0 , 𝑥 =2 , 𝑦 =2 Then: 8 ∗ 23 ∗ 2 + 4 ∗ 24 + 8 ∗ 2 ∗ 03 = 128 + 64 + 0 = 192 Page | 46 𝑢 = 𝑥 4 𝑦 + 𝑦 2 𝑧 3 , 𝑧 = 𝑠𝑟 2 sin 𝑡 , 𝑥 = 𝑟𝑠𝑒 𝑡 , 𝑦 = 𝑟𝑠 2 𝑒 −𝑡 𝜕𝑢 𝜕𝑧 ➔ 𝜕𝑧 = 3𝑦 2 𝑧 2 , = 𝑠𝑟 2 cos 𝑡 , 𝜕𝑡 𝜕𝑢 𝜕𝑥 = 4𝑥 3 𝑦 , 𝜕𝑥 𝜕𝑡 = 𝑟𝑠𝑒 𝑡 , 𝜕𝑢 𝜕𝑦 = 𝑥 4 + 2𝑦𝑧 3 , 𝜕𝑦 𝜕𝑡 = −𝑟𝑠 2 𝑒 −𝑡 𝜕𝑢 𝜕𝑡 = 𝜕𝑢 𝜕𝑧 ∗ 𝜕𝑧 𝜕𝑡 + 𝜕𝑢 𝜕𝑥 ∗ 𝜕𝑥 𝜕𝑡 + 𝜕𝑢 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑡 = (3𝑦 2 𝑧 2 ) ∗ (𝑠𝑟 2 cos 𝑡) + (4𝑥 3 𝑦) ∗ (𝑟𝑠𝑒 𝑡 ) + (𝑥 4 + 2𝑦𝑧 3 ) ∗ (−𝑟𝑠 2 𝑒 −𝑡 ) At 𝑟 = 2 , 𝑠 = 1 , 𝑡 = 0 (3𝑦 2 𝑧 2 ) ∗ (1 ∗ 22 ∗ 𝑐𝑜𝑠 0) + (4𝑥 3 𝑦) ∗ (2 ∗ 1 ∗ 𝑒 0 ) + (𝑥 4 + 2𝑦𝑧 3 ) ∗ (−2 ∗ 12 ∗ 𝑒 −0 ) = 12𝑦 2 𝑧 2 + 8𝑥 3 𝑦 − 2𝑥 4 − 4𝑦𝑧 3 At 𝑟 = 2 , 𝑠 = 1 , 𝑡 = 0 ➔𝑧=0 , 𝑥 =2 , 𝑦 =2 12 ∗ 22 ∗ 02 + 8 ∗ 23 ∗ 2 − 2 ∗ 24 − 4 ∗ 2 ∗ 03 = 0 + 128 − 32 + 0 = 96 Choice C 𝝏𝒛 𝝏𝒛 58] Use Implicit Derivative to find 𝝏𝒙 and 𝝏𝒚 respectively 1] 𝒚𝒛 + 𝒙 𝒍𝒏 𝒚 = 𝒛𝟐 𝒍𝒏 𝒚 𝒛𝒚+𝒙 𝒍𝒏 𝒚 𝟖𝒛𝒚+𝒙 𝒛𝒚+𝒙 A- 𝟐𝒛−𝒚 , 𝟐𝒛𝒚−𝒚𝟐 𝒍𝒏 𝒚 B- 𝟐𝒛𝒚−𝒚𝟐 , 𝟐𝒛−𝒚 C- 𝟐𝒛+𝟗𝒚 , 𝟐𝒛𝒚−𝒚𝟐 D- None of them Solution: ➔𝒚∗ 𝝏𝒛 𝝏𝒙 + 𝒍𝒏 𝒚 = 𝟐𝒛 ∗ ➔ 𝒍𝒏 𝒚 = 𝟐𝒛 ∗ ➔ 𝝏𝒛 𝝏𝒙 = Page | 47 𝒍𝒏 𝒚 𝟐𝒛−𝒚 𝝏𝒛 𝝏𝒙 −𝒚∗ 𝝏𝒛 𝝏𝒙 𝝏𝒛 𝝏𝒙 → 𝒍𝒏 𝒚 = 𝝏𝒛 𝝏𝒙 (𝟐𝒛 − 𝒚) 𝝏𝒛 ➔𝒛+𝒚∗ 𝝏𝒚 𝒙 𝝏𝒛 𝒚 𝝏𝒚 + = 𝟐𝒛 ∗ 𝒙 𝝏𝒛 𝒚 𝝏𝒚 ➔ 𝒛 + = 𝟐𝒛 ∗ −𝒚∗ 𝝏𝒛 𝒙 𝝏𝒛 𝒚 𝝏𝒚 →𝒛+ = 𝝏𝒚 (𝟐𝒛 − 𝒚) 𝒙 ➔ 𝝏𝒛 𝝏𝒚 = 𝒛+𝒚 𝟐𝒛−𝒚 𝒛𝒚+𝒙 = 𝟐𝒛𝒚−𝒚𝟐 Choice A 2] 𝒙𝟐 − 𝒚𝟐 + 𝒛𝟐 − 𝟐𝒛 = 𝟒 𝒚 −𝒙 −𝒙 𝒚 𝒙 A- 𝒛−𝟏 , 𝒛−𝟏 𝒚 B- 𝒛−𝟏 , 𝒛−𝟏 C- 𝒛−𝟏 , 𝒛−𝟏 D- None of them Solution: ➔ 2𝑥 + 2𝑧 ∗ ➔ 2𝑧 ∗ ➔ 𝜕𝑧 𝜕𝑥 = 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑥 −2∗ −2𝑥 2𝑧−2 = ➔ −2𝑦 + 2𝑧 ∗ ➔ 2𝑧 ∗ ➔ 𝜕𝑧 𝜕𝑦 = 𝜕𝑧 𝜕𝑦 2𝑧−2 Page | 48 = 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑥 =0 = −2𝑥 → 𝜕𝑧 𝜕𝑥 (2𝑧 − 2) = −2𝑥 −𝑥 𝑧−1 𝜕𝑧 𝜕𝑦 −2∗ 2𝑦 Choice C −2∗ −2∗ 𝜕𝑧 𝜕𝑦 𝑦 𝑧−1 𝜕𝑧 𝜕𝑦 =0 = 2𝑦 → 𝜕𝑧 𝜕𝑦 (2𝑧 − 2) = 2𝑦 59] Find 𝒚′ of 𝒙𝟑 + 𝒚𝟑 = 𝟔𝒙𝒚 A- 𝟑𝒙𝟐 −𝟔𝒚 B- − 𝟑𝒚𝟐 −𝟔𝒙 C- − 𝟑𝒙𝟐 +𝟔𝒚 𝟑𝒙𝟐 −𝟔𝒚 𝟑𝒚𝟐 −𝟔𝒙 D- None of them 𝟑𝒚𝟐 −𝟔𝒙 Solution: 𝑥 3 + 𝑦 3 − 6𝑥𝑦 = 0 ➔ 𝑓𝑥 = 3𝑥 2 − 6𝑦 , 𝑓𝑦 = 3𝑦 2 − 6𝑥 𝑦′ = − 𝑓𝑥 𝑓𝑦 =− 3𝑥 2 −6𝑦 3𝑦 2 −6𝑥 Choice B 60] Find Second Partial Derivative to : I] 𝒇(𝒙, 𝒚) = 𝒙𝟑 𝒚𝟓 + 𝟐𝒙𝟒 𝒚 Solution: 𝑓𝑥 = 3𝑥 2 𝑦 5 + 8𝑥 3 𝑦 ➔ 𝑓𝑥𝑥 = 6𝑥𝑦 5 + 24𝑥 2 𝑦 𝑓𝑦 = 5𝑥 3 𝑦 4 + 2𝑥 4 ➔ 𝑓𝑦𝑦 = 20𝑥 3 𝑦 3 𝑓𝑥𝑦 = 15𝑥 2 𝑦 4 + 8𝑥 3 , 𝑓𝑦𝑥 = 15𝑥 2 𝑦 4 + 8𝑥 3 Since 𝑓𝑥𝑦 = 𝑓𝑦𝑥 , Then The Function Is Continuous II] 𝒇(𝒙, 𝒚) = 𝐬𝐢𝐧𝟐 (𝒎𝒙 + 𝒏𝒚) Solution: 𝑓𝑥 = 2 sin(𝑚𝑥 + 𝑛𝑦) cos(𝑚𝑥 + 𝑛𝑦) ∗ 𝑚 = 𝑚 sin 2(𝑚𝑥 + 𝑛𝑦) 𝑓𝑥𝑥 = 𝑚 ∗ cos 2(𝑚𝑥 + 𝑛𝑦) ∗ 2 ∗ 𝑚 = 2𝑚2 cos 2(𝑚𝑥 + 𝑛𝑦) 𝑓𝑦 = 2 sin(𝑚𝑥 + 𝑛𝑦) cos(𝑚𝑥 + 𝑛𝑦) ∗ 𝑛 = 𝑛 sin 2(𝑚𝑥 + 𝑛𝑦) 𝑓𝑦𝑦 = 𝑛 ∗ cos 2(𝑚𝑥 + 𝑛𝑦) ∗ 2 ∗ 𝑛 = 2𝑛2 cos 2(𝑚𝑥 + 𝑛𝑦) 𝑓𝑥𝑦 = 2𝑚𝑛 cos 2(𝑚𝑥 + 𝑛𝑦) = 𝑓𝑦𝑥 , then the function is continuous Page | 49 61] Use the chain rule to find 𝒅𝒛 𝒅𝒕 𝒅𝒘 or 𝒅𝒕 I] 𝒛 = 𝒙𝟐 + 𝒚𝟐 + 𝒙𝒚 , 𝒙 = 𝐬𝐢𝐧 𝒕 , 𝒚 = 𝒆𝒕 Solution: ➔ 𝜕𝑧 𝜕𝑥 𝑑𝑥 𝑑𝑡 ➔ 𝑑𝑧 = 𝑑𝑡 𝜕𝑧 𝜕𝑥 ∗ 𝑑𝑥 𝑑𝑡 = 2𝑥 + 𝑦 , = 𝑐𝑜𝑠 𝑡 , 𝜕𝑧 𝜕𝑥 ∗ 𝑑𝑥 𝑑𝑡 + + 𝜕𝑧 𝜕𝑦 𝑑𝑦 𝑑𝑡 𝜕𝑧 𝜕𝑦 𝜕𝑧 𝜕𝑦 ∗ 𝑑𝑦 𝑑𝑡 = 2𝑦 + 𝑥 = 𝑒𝑡 ∗ 𝑑𝑦 = (2𝑥 + 𝑦 ) ∗ (𝑐𝑜𝑠 𝑡) + (2𝑦 + 𝑥 ) ∗ (𝑒 𝑡 ) 𝑑𝑡 = 2𝑥 𝑐𝑜𝑠 𝑡 + 𝑦 𝑐𝑜𝑠 𝑡 + 2𝑦𝑒 𝑡 + 𝑥𝑒 𝑡 = 2 ∗ sin 𝑡 𝑐𝑜𝑠 𝑡 + 𝑒 𝑡 𝑐𝑜𝑠 𝑡 + 2 ∗ 𝑒 𝑡 ∗ 𝑒 𝑡 + sin 𝑡 ∗ 𝑒 𝑡 = 2 sin 𝑡 𝑐𝑜𝑠 𝑡 + 𝑒 𝑡 𝑐𝑜𝑠 𝑡 + 2𝑒 2𝑡 + 𝑒 𝑡 sin 𝑡 II] 𝒛 = 𝐜𝐨𝐬(𝒙 + 𝟒𝒚) , 𝒙 = 𝟓𝒕𝟒 , 𝒚 = 𝟏 𝒕 Solution: ➔ 𝜕𝑧 𝜕𝑥 𝑑𝑥 𝑑𝑡 ➔ 𝑑𝑧 = 𝑑𝑡 𝜕𝑧 𝜕𝑥 ∗ 𝑑𝑥 + 𝑑𝑡 𝜕𝑧 𝜕𝑦 ∗ 𝑑𝑦 𝑑𝑡 = − 𝑠𝑖𝑛(𝑥 + 4𝑦) , = 20𝑡 3 , 𝜕𝑧 𝜕𝑥 ∗ 𝑑𝑥 𝑑𝑡 + 𝑑𝑦 𝑑𝑡 𝜕𝑧 𝜕𝑦 =− ∗ 𝑑𝑦 𝑑𝑡 𝜕𝑧 𝜕𝑦 = −4 𝑠𝑖𝑛(𝑥 + 4𝑦) 1 𝑡2 1 = (− 𝑠𝑖𝑛(𝑥 + 4𝑦)) ∗ (20𝑡 3 ) + (−4 𝑠𝑖𝑛(𝑥 + 4𝑦)) ∗ (− 2 ) 𝑡 = −20𝑡 3 𝑠𝑖𝑛(𝑥 + 4𝑦) + 3 4 4 𝑠𝑖𝑛(𝑥+4𝑦) 𝑡2 1 = −20𝑡 𝑠𝑖𝑛 (5𝑡 + 4 ∗ ) + 𝑡 Page | 50 1 4 𝑠𝑖𝑛(5𝑡 4 +4∗ 𝑡 ) 𝑡2 III] 𝒘 = 𝒍𝒏 √𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 , 𝒙 = 𝒔𝒊𝒏 𝒕 , 𝒚 = 𝒄𝒐𝒔 𝒕 , 𝒛 = 𝒕𝒂𝒏 𝒕 Solution: 𝑑𝑤 ➔ 𝑑𝑡 = 𝜕𝑤 𝜕𝑥 ∗ 𝑑𝑥 + 𝑑𝑡 𝜕𝑤 𝜕𝑦 ∗ 𝑑𝑦 𝑑𝑡 + 𝜕𝑤 𝜕𝑧 𝑑𝑧 ∗ 𝑑𝑡 2𝑥 𝜕𝑤 = 𝜕𝑥 2√𝑥2 +𝑦2 +𝑧2 √𝑥 2 +𝑦 2 +𝑧 2 = 𝑥 𝑥 2 +𝑦2 +𝑧 2 , 𝑑𝑥 𝑑𝑡 = 𝑐𝑜𝑠 𝑡 2𝑦 𝜕𝑤 = 𝜕𝑦 2√𝑥2 +𝑦2 +𝑧2 √𝑥 2 +𝑦 2 +𝑧 2 = 𝑦 𝑥 2 +𝑦2 +𝑧 2 , 𝑑𝑦 𝑑𝑡 = − 𝑠𝑖𝑛 𝑡 2𝑧 𝜕𝑤 = 𝜕𝑧 ➔ 𝜕𝑤 𝜕𝑥 =( √𝑥 2 +𝑦 2 +𝑧 2 ∗ 𝑑𝑥 𝑑𝑡 + 𝑥 𝜕𝑦 ∗ 𝑧 𝑥 2 +𝑦2 +𝑧 2 𝑑𝑦 𝑑𝑡 + 𝜕𝑤 𝜕𝑧 ∗ , 𝑑𝑧 = 𝑠𝑒𝑐 2 𝑡 𝑑𝑡 𝑑𝑧 𝑑𝑡 𝑡)2 +(𝑐𝑜𝑠 𝑡)2 +(𝑡𝑎𝑛 𝑡)2 1+(𝑡𝑎𝑛 𝑡)2 𝑧 ) ∗ (𝑐𝑜𝑠 𝑡) + (𝑥 2 +𝑦2 +𝑧 2 ) ∗ (− 𝑠𝑖𝑛 𝑡) + (𝑥 2 +𝑦2 +𝑧 2 ) ∗ (𝑠𝑒𝑐 2 𝑡) 𝑠𝑖𝑛 𝑡∗𝑐𝑜𝑠 𝑡 𝑡𝑎𝑛 𝑡∗𝑠𝑒𝑐 2 𝑡 Page | 51 𝜕𝑤 = 𝑦 𝑥 2 +𝑦 2 +𝑧 2 = (𝑠𝑖𝑛 = 2√𝑥2 +𝑦2 +𝑧2 = − (𝑠𝑖𝑛 𝑡𝑎𝑛 𝑡∗𝑠𝑒𝑐 2 𝑡 𝑠𝑒𝑐 2 𝑡 𝑐𝑜𝑠 𝑡∗𝑠𝑖𝑛 𝑡 𝑡)2 +(𝑐𝑜𝑠 𝑡)2 +(𝑡𝑎𝑛 𝑡)2 = 𝑡𝑎𝑛 𝑡 + (𝑠𝑖𝑛 𝑡𝑎𝑛 𝑡∗𝑠𝑒𝑐 2 𝑡 𝑡)2 +(𝑐𝑜𝑠 𝑡)2 +(𝑡𝑎𝑛 𝑡)2 62] Use the chain rule to find 𝝏𝒛 𝝏𝒔 or 𝝏𝒛 𝝏𝒕 I] 𝒛 = 𝒙𝟐 𝒚𝟑 , 𝒙 = 𝒔 𝒄𝒐𝒔 𝒕 , 𝒚 = 𝒔 𝒔𝒊𝒏 𝒕 Solution: ➔ 𝜕𝑧 𝜕𝑥 ➔ 𝜕𝑧 𝜕𝑠 = 𝜕𝑧 𝜕𝑥 𝜕𝑥 ∗ 𝜕𝑠 = 2𝑥𝑦 3 , 𝜕𝑧 𝜕𝑠 + 𝜕𝑥 𝜕𝑠 𝜕𝑧 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑠 = 𝑐𝑜𝑠 𝑡 , 𝜕𝑧 𝜕𝑦 = 3𝑥 2 𝑦 2 , 𝜕𝑦 𝜕𝑠 = 𝑠𝑖𝑛 𝑡 = (2𝑥𝑦 3 ) ∗ (𝑐𝑜𝑠 𝑡) + (3𝑥 2 𝑦 2 ) ∗ (𝑠𝑖𝑛 𝑡) = (2 ∗ 𝑠 𝑐𝑜𝑠 𝑡 ∗ (𝑠 𝑠𝑖𝑛 𝑡)3 ) ∗ (𝑐𝑜𝑠 𝑡) + (3 ∗ (𝑠 𝑐𝑜𝑠 𝑡)2 ∗ (𝑠 𝑠𝑖𝑛 𝑡)2 ) ∗ (𝑠𝑖𝑛 𝑡) = 2𝑠 4 𝑐𝑜𝑠 2 𝑡 𝑠𝑖𝑛3 𝑡 + 3𝑠 4 𝑐𝑜𝑠 2 𝑡 𝑠𝑖𝑛3 𝑡 = 5𝑠 4 𝑠𝑖𝑛3 𝑡 𝑐𝑜𝑠 2 𝑡 ➔ 𝜕𝑧 𝜕𝑡 𝜕𝑧 𝜕𝑥 ➔ = 𝜕𝑧 𝜕𝑥 ∗ 𝜕𝑥 𝜕𝑡 = 2𝑥𝑦 3 , 𝜕𝑧 𝜕𝑡 + 𝜕𝑥 𝜕𝑡 𝜕𝑧 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑡 = −𝑠 𝑠𝑖𝑛 𝑡 , 𝜕𝑧 𝜕𝑦 = 3𝑥 2 𝑦 2 , 𝜕𝑦 𝜕𝑡 = 𝑠 𝑐𝑜𝑠 𝑡 = (2𝑥𝑦 3 ) ∗ (−𝑠 𝑠𝑖𝑛 𝑡) + (3𝑥 2 𝑦 2 ) ∗ (𝑠 𝑐𝑜𝑠 𝑡) = (2 ∗ 𝑠 𝑐𝑜𝑠 𝑡 ∗ (𝑠 𝑠𝑖𝑛 𝑡)3 ) ∗ (−𝑠 𝑠𝑖𝑛 𝑡) + (3(𝑠 𝑐𝑜𝑠 𝑡)2 (𝑠 𝑠𝑖𝑛 𝑡)2 ) ∗ (𝑠 𝑐𝑜𝑠 𝑡) = −2𝑠 5 𝑐𝑜𝑠 𝑡 𝑠𝑖𝑛4 𝑡 + 3𝑠 5 𝑐𝑜𝑠 3 𝑡 𝑠𝑖𝑛2 𝑡 II] 𝒛 = 𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔 ∅ , 𝜽 = 𝒔𝒕𝟐 , ∅ = 𝒔𝟐 𝒕 Solution: ➔ ∂z ∂θ ∂z ∂s = ∂z ∂θ ∗ ∂θ ∂s + ∂z ∂∅ = cos θ cos ∅ , Page | 52 ∗ ∂θ ∂s ∂∅ ∂s = t2 , ∂z ∂∅ = − sin θ sin ∅ , ∂∅ ∂s = 2st ∂z ➔ = (cos θ cos ∅ ) ∗ (t 2 ) + (− sin θ sin ∅) ∗ (2st) ∂s = (cos 𝑠𝑡 2 cos 𝑠 2 𝑡 ) ∗ (t 2 ) + (− sin 𝑠𝑡 2 sin 𝑠 2 𝑡) ∗ (2st) = 𝒕𝟐 𝒄𝒐𝒔 𝒔𝒕𝟐 𝒄𝒐𝒔 𝒔𝟐 𝒕 − 𝟐𝒔𝒕 𝒔𝒊𝒏 𝒔𝒕𝟐 𝒔𝒊𝒏 𝒔𝟐 𝒕 ∂z ➔ ∂z ∂θ = ∂t ∂z ∂θ ∗ ∂θ ∂t + ∂z ∂∅ ∂θ = cos θ cos ∅ , ➔ ∂z ∂∅ ∗ ∂t ∂t = 2st , ∂z ∂∅ ∂∅ = − sin θ sin ∅ , ∂t = s2 = (cos θ cos ∅ ) ∗ (2st) + (− sin θ sin ∅) ∗ (s2 ) ∂t = (𝑐𝑜𝑠 𝑠𝑡 2 𝑐𝑜𝑠 𝑠 2 𝑡 ) ∗ (2𝑠𝑡) + (− 𝑠𝑖𝑛 𝑠𝑡 2 𝑠𝑖𝑛 𝑠 2 𝑡) ∗ (𝑠 2 ) = 𝟐𝒔𝒕 𝒄𝒐𝒔 𝒔𝒕𝟐 𝒄𝒐𝒔 𝒔𝟐 𝒕 − 𝒔𝟐 𝒔𝒊𝒏 𝒔𝒕𝟐 𝒔𝒊𝒏 𝒔𝟐 𝒕 III] 𝝋 = 𝒆𝜶+𝟐𝜷 , 𝜶 = 𝒔 𝒕 𝒕 , 𝜷=𝒔 Solution: ➔ 𝜕𝜑 𝜕𝛼 𝜕𝜑 𝜕𝑠 = 𝜕𝜑 𝜕𝛼 ∗ 𝜕𝛼 𝜕𝑠 = 𝑒 𝛼+2𝛽 , + 𝜕𝛼 𝜕𝑠 𝜕𝜑 𝜕𝛽 = ∗ 1 𝜕𝛽 𝜕𝑠 𝜕𝜑 , 𝑡 𝜕𝛽 = 2𝑒 𝛼+2𝛽 , 𝜕𝛽 =− 𝜕𝑠 𝑡 𝑠2 𝑠 ➔ 𝜕𝜑 ➔ 𝜕𝜑 𝜕𝜑 = 𝑒 𝛼+2𝛽 , 𝜕𝛼 ➔ 𝜕𝑠 𝜕𝑡 𝜕𝜑 𝜕𝑡 Page | 53 1 𝑡 = (𝑒 𝛼+2𝛽 ) ∗ ( ) + (2𝑒 𝛼+2𝛽 ) ∗ (− 2 ) = 𝑡 𝑠 = 𝜕𝜑 𝜕𝛼 ∗ 𝜕𝛼 𝜕𝑡 + 𝜕𝛼 𝜕𝑡 𝜕𝜑 𝜕𝛽 ∗ =− 𝑒𝑡 𝑡 +2 𝑠 𝑡 − 𝑠 𝑡 +2 𝑠 2𝑡𝑒 𝑡 𝑠2 𝜕𝛽 𝜕𝑡 𝑠 𝑡2 𝑠 , 𝜕𝜑 𝜕𝛽 = 2𝑒 𝛼+2𝛽 , 𝜕𝛽 𝜕𝑡 1 = 1 𝑠 = (𝑒 𝛼+2𝛽 ) ∗ (− 2 ) + (2𝑒 𝛼+2𝛽 ) ∗ ( ) = − 𝑡 𝑠 𝑠 𝑡 +2 𝑠 𝑠𝑒 𝑡 𝑡2 + 𝑠 𝑡 +2 𝑠 2𝑒 𝑡 𝑠 63] Find 𝒅𝒚 𝒅𝒙 Using Partial derivatives I] 𝒚 𝒄𝒐𝒔 𝒙 = 𝒙𝟐 + 𝒚𝟐 Solution: ➔ 𝑑𝑦 =− 𝑑𝑥 𝐹𝑥 = 𝜕𝐹 𝜕𝑥 𝜕𝐹 𝐹𝑦 = 𝑑𝑦 𝑑𝑥 𝜕𝑦 =− 𝐹𝑥 𝐹𝑦 = −𝑦 𝑠𝑖𝑛 𝑥 = 2𝑥 ➔ −𝑦 𝑠𝑖𝑛 𝑥 − 2𝑥 = 0 = 𝑐𝑜𝑠 𝑥 = 2𝑦 ➔ 𝑐𝑜𝑠 𝑥 − 2𝑦 = 0 (−𝑦 𝑠𝑖𝑛 𝑥−2𝑥) (𝑐𝑜𝑠 𝑥−2𝑦) =− −(𝑦 𝑠𝑖𝑛 𝑥+2𝑥) (𝑐𝑜𝑠 𝑥−2𝑦) = 𝑦 𝑠𝑖𝑛 𝑥+2𝑥 𝑐𝑜𝑠 𝑥−2𝑦 II] 𝒕𝒂𝒏−𝟏 (𝒙𝟐 𝒚) = 𝒙 + 𝒙𝒚𝟐 Solution: ➔ 𝑑𝑦 =− 𝑑𝑥 𝐹𝑥 = 𝜕𝐹 𝐹𝑦 = 𝑑𝑦 𝑑𝑥 𝜕𝑥 𝜕𝐹 𝜕𝑦 = = 𝐹𝑥 𝐹𝑦 1 1+(𝑥 2 𝑦) 2 2 ∗ 2𝑥𝑦 = 1 + 𝑦 ➔ 1 1+(𝑥 2 𝑦) 2 2 ∗ 𝑥 = 2𝑥𝑦 ➔ 2𝑥𝑦 2 2 −𝑦 −1) 2 1+(𝑥 𝑦) ( =− 𝑥2 2 −2𝑥𝑦) 1+(𝑥2 𝑦) ( III] 𝒆𝒚 𝒔𝒊𝒏 𝒙 = 𝒙 + 𝒙𝒚 Solution: ➔ 𝑑𝑦 𝑑𝑥 Page | 54 =− 𝐹𝑥 𝐹𝑦 2𝑥𝑦 1+(𝑥 2 𝑦)2 𝑥2 1+(𝑥 2 𝑦)2 − 𝑦2 − 1 = 0 − 2𝑥𝑦 = 0 𝐹𝑥 = 𝐹𝑦 = 𝑑𝑦 𝑑𝑥 𝜕𝐹 𝜕𝑥 𝜕𝐹 𝜕𝑦 =− = 𝑒 𝑦 𝑐𝑜𝑠 𝑥 = 1 + 𝑦 ➔ 𝑒 𝑦 𝑐𝑜𝑠 𝑥 − 1 − 𝑦 = 0 = 𝑒 𝑦 𝑠𝑖𝑛 𝑥 = 𝑥 ➔ 𝑒 𝑦 𝑠𝑖𝑛 𝑥 − 𝑥 = 0 (𝑒 𝑦 𝑐𝑜𝑠 𝑥−1−𝑦) (𝑒 𝑦 𝑠𝑖𝑛 𝑥−𝑥) 64] If 𝒇(𝒙, 𝒚) where 𝒙 = 𝒆𝒖 𝒄𝒐𝒔 𝒗 & 𝒚 = 𝒆𝒖 𝒔𝒊𝒏 𝒗 , Show that 𝝏𝒇 𝝏𝒇 𝝏𝒇 𝝏𝒇 = 𝒙 𝝏𝒙 + 𝒚 𝝏𝒚 & 𝝏𝒖 𝝏𝒇 𝝏𝒇 = −𝒚 𝝏𝒙 + 𝒙 𝝏𝒚 𝝏𝒗 Solution: ➔ 𝜕𝑓 𝜕𝑢 Since ➔ 𝜕𝑓 𝜕𝑢 Then: ➔ 𝜕𝑓 𝜕𝑣 Since ➔ 𝜕𝑓 𝜕𝑣 Then: Page | 55 = 𝜕𝑓 𝜕𝑥 𝜕𝑥 𝜕𝑢 = 𝜕𝑥 𝜕𝑢 = 𝜕𝑥 + 𝜕𝑓 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑢 ∗ 𝑒 𝑢 𝑐𝑜𝑠 𝑣 + ∗ 𝜕𝑓 𝜕𝑥 𝜕𝑥 𝜕𝑣 +𝑦 + 𝜕𝑓 𝜕𝑦 𝜕𝑓 𝜕𝑦 ∗ 𝜕𝑓 𝜕𝑦 𝜕𝑦 𝜕𝑢 𝜕𝑓 𝜕𝑥 𝜕𝑓 is true 𝜕𝑦 𝜕𝑣 𝜕𝑣 ∗ (−𝑒 𝑢 𝑠𝑖𝑛 𝑣 ) + = −𝑦 𝜕𝑓 𝜕𝑥 +𝑥 = 𝑒 𝑢 𝑠𝑖𝑛 𝑣 = 𝑦 ∗ 𝑒 𝑢 𝑠𝑖𝑛 𝑣 = −𝑒 𝑢 𝑠𝑖𝑛 𝑣 = −𝑦 & 𝜕𝑣 = 𝜕𝑢 =𝑥 𝜕𝑓 𝜕𝑥 𝜕𝑥 = 𝑒 𝑢 𝑐𝑜𝑠 𝑣 = 𝑥 & 𝜕𝑓 𝜕𝑓 ∗ 𝜕𝑓 𝜕𝑦 𝜕𝑓 𝜕𝑦 𝜕𝑦 𝜕𝑣 = 𝑒 𝑢 𝑐𝑜𝑠 𝑣 = 𝑥 ∗ 𝑒 𝑢 𝑐𝑜𝑠 𝑣 is true 𝝏𝒛 𝝏𝒛 65] If 𝒛 = 𝒙𝟐 + 𝒚𝟐 when 𝒙 = 𝒓 𝐜𝐨𝐬 𝜽 , 𝒚 = 𝒓 𝐬𝐢𝐧 𝟐𝜽 , Find 𝝏𝒓 , 𝝏𝜽 respectively A- 𝟐𝒓(𝐜𝐨𝐬 𝜽)𝟐 + 𝟐𝒓(𝐬𝐢𝐧 𝟐𝜽)𝟐 , −𝒓𝟐 𝐬𝐢𝐧 𝟐𝜽 + 𝟐𝒓𝟐 𝐬𝐢𝐧 𝟒𝜽 B- 𝟐𝒓(𝐜𝐨𝐬 𝜽)𝟐 + 𝟐𝒓(𝐬𝐢𝐧 𝟐𝜽)𝟐 , 𝒓𝟐 𝐬𝐢𝐧 𝟐𝜽 + 𝟐𝒓𝟐 𝐬𝐢𝐧 𝟒𝜽 C- 𝟐(𝐜𝐨𝐬 𝜽)𝟐 + 𝟐𝒓(𝐬𝐢𝐧 𝟐𝜽)𝟐 , −𝒓𝟐 𝐬𝐢𝐧 𝟐𝜽 + 𝟐𝒓𝟐 𝐬𝐢𝐧 𝟒𝜽 D- None of them Solution: 𝜕𝑧 𝜕𝑟 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑟 = 𝜕𝑧 𝜕𝑥 ∗ = 2𝑥 , 𝜕𝑥 𝜕𝑟 𝜕𝑥 𝜕𝑟 + 𝜕𝑧 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑟 = cos 𝜃 , 𝜕𝑧 𝜕𝑦 = 2𝑦 , 𝜕𝑦 𝜕𝑟 = sin 2𝜃 = (2𝑥 ) ∗ (cos 𝜃 ) + (2𝑦) ∗ (sin 2𝜃 ) = 2𝑥 cos 𝜃 + 2𝑦 sin 2𝜃 By substitute ➔ 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 2𝜃 Then: 𝜕𝑧 𝜕𝑟 = 2(𝑟 cos 𝜃 ) ∗ cos 𝜃 + 2(𝑟 sin 2𝜃 ) ∗ sin 2𝜃 = 2𝑟(cos 𝜃 )2 + 2𝑟(sin 2𝜃 )2 𝜕𝑧 𝜕𝜃 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝜃 = 𝜕𝑧 𝜕𝑥 ∗ = 2𝑥 , 𝜕𝑥 𝜕𝜃 𝜕𝑥 𝜕𝜃 + 𝜕𝑧 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝜃 = −𝑟 sin 𝜃 , 𝜕𝑧 𝜕𝑦 = 2𝑦 , 𝜕𝑦 𝜕𝜃 = 2𝑟 cos 2𝜃 = (2𝑥 ) ∗ (−𝑟 sin 𝜃 ) + (2𝑦) ∗ (2𝑟 cos 2𝜃 ) = −2𝑟𝑥 sin 𝜃 + 4𝑟𝑦 cos 2𝜃 By substitute ➔ 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 2𝜃 Then: 𝜕𝑧 𝜕𝜃 = −2𝑟 ∗ (𝑟 cos 𝜃 ) ∗ sin 𝜃 + 4𝑟(𝑟 sin 2𝜃 ) ∗ cos 2𝜃 = −2𝑟 2 sin 𝜃 cos 𝜃 + 4𝑟 2 sin 2𝜃 cos 2𝜃 = −𝑟 2 sin 2𝜃 + 2𝑟 2 sin 4𝜃 Page | 56 𝝏𝒛 66] Find 𝝏𝒙 & 𝝏𝒛 𝝏𝒚 if 𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 + 𝟔𝒙𝒚𝒛 = 𝟏 respectively (𝒙𝟐 +𝟐𝒚𝒛) (𝒚𝟐 +𝟐𝒙𝒛) (𝒙𝟐 +𝟐𝒚𝒛) A- − (𝒛𝟐 +𝟐𝒙𝒚) , (𝒛𝟐 +𝟐𝒙𝒚) C- (𝒙𝟐 +𝟐𝒚𝒛) − (𝒛𝟐 +𝟐𝒙𝒚) , (𝒚𝟐 +𝟐𝒙𝒛) B- (𝒛𝟐 +𝟐𝒙𝒚) , − (𝒛𝟐+𝟐𝒙𝒚) (𝒚𝟐 +𝟐𝒙𝒛) − (𝒛𝟐 +𝟐𝒙𝒚) D- None of them Solution: ➔ 𝑥 3 + 𝑦 3 + 𝑧 3 + 6𝑥𝑦𝑧 − 1 = 0 𝜕𝑧 𝜕𝑥 =− Then: 𝜕𝑧 𝜕𝑦 =− 𝜕𝑓 𝜕𝑦 𝜕𝑓 𝜕𝑧 𝜕𝑦 𝑓𝑧 ➔ 𝑓𝑥 = 3𝑥 2 + 6𝑦𝑧 = 0 , 𝑓𝑧 = 3𝑧 2 + 6𝑥𝑦 = 0 = − (3𝑧 2 𝜕𝑥 𝜕𝑧 𝑓𝑥 (3𝑥 2 +6𝑦𝑧) 𝜕𝑧 =− Then: 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑧 =− 𝑓𝑦 𝑓𝑧 +6𝑥𝑦) (𝑥 2 +2𝑦𝑧) = − (𝑧 2 +2𝑥𝑦) ➔ 𝑓𝑦 = 3𝑦 2 + 6𝑥𝑧 = 0 , 𝑓𝑧 = 3𝑧 2 + 6𝑥𝑦 = 0 (3𝑦 2 +6𝑥𝑧) = − (3𝑧 2 +6𝑥𝑦) (𝑦 2 +2𝑥𝑧) = − (𝑧 2 +2𝑥𝑦) 67] If 𝒛 = 𝒇(𝒙, 𝒚) has continuous second-order partial derivatives and 𝒙 = 𝒓𝟐 + 𝒔𝟐 , 𝒚 = 𝟐𝒓𝒔 , Find : 𝝏𝒛 [1] 𝝏𝒓 𝝏𝒛 𝝏𝒛 𝝏𝒛 𝝏𝒛 𝝏𝒛 A- 𝝏𝒙 ∗ 𝟐𝒓 − 𝝏𝒚 ∗ 𝟐𝒔 C- 𝝏𝒙 ∗ 𝟐𝒓 + 𝝏𝒚 ∗ 𝟐𝒔 D- None of them Solution: 𝜕𝑧 𝜕𝑟 = 𝜕𝑧 𝜕𝑥 Page | 57 ∗ 𝜕𝑥 𝜕𝑟 + 𝜕𝑧 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑟 = 𝝏𝒛 B- 𝝏𝒙 ∗ 𝟐𝒓 + 𝝏𝒚 ∗ 𝟐𝒓 𝜕𝑧 𝜕𝑥 ∗ 2𝑟 + 𝜕𝑧 𝜕𝑦 ∗ 2𝑠 𝝏𝟐 𝒛 [2] 𝝏𝒓𝟐 𝝏𝒛 𝟐 𝟐𝝏 𝒛 𝟒𝒓 𝝏𝒙𝟐 𝝏𝒛 𝝏𝟐 𝒛 A- 𝟐 𝝏𝒙 + 𝝏𝟐 𝒛 𝟐 𝟐𝝏 𝒛 𝟒𝒔 𝝏𝒚𝟐 𝝏𝟐 𝒛 𝝏𝟐 𝒛 + 𝟖𝒓𝒔 𝝏𝒙𝝏𝒚 + 𝝏𝒛 B- 𝟐 𝝏𝒙 + C- 𝟐 𝝏𝒙 + 𝟒𝒓𝟐 𝝏𝒙𝟐 + 𝟖𝒓𝒔 𝝏𝒙𝝏𝒚 + 𝟒𝒔𝟐 𝝏𝒚𝟐 𝟐 𝟐𝝏 𝒛 𝟒𝒓 𝝏𝒙𝟐 D- None of them Solution: 𝜕2 𝑧 𝜕𝑟 2 𝜕 = 𝜕𝑧 𝜕 𝜕 = 𝜕 𝜕𝑟 𝜕 = Then: 𝜕𝑥 𝜕 = Then: 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕 + 2𝑟 ∗ 𝜕 𝜕 𝜕𝑦 𝜕 ∗ 𝜕𝑧 𝜕 𝜕𝑧 ( ) + 2𝑠 ∗ 𝜕𝑟 (𝜕𝑦 ) 𝜕𝑟 𝜕𝑥 𝜕𝑥 𝜕𝑧 𝜕 𝜕𝑦 𝜕𝑧 𝜕 𝜕𝑥 ∗ 𝜕𝑧 𝜕𝑥 ∗ 𝜕2 𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑟 2 ∗ 𝜕2 𝑧 𝜕𝑥 2 𝜕𝑥 𝜕𝑟 + ∗ 2𝑟 + + 𝜕 𝜕𝑦 𝜕2 𝑧 𝜕𝑥𝜕𝑦 ∗ 𝜕2 𝑧 𝜕𝑥𝜕𝑦 ∗ 𝜕𝑧 𝜕𝑥 ∗ 𝜕𝑦 𝜕𝑟 𝜕𝑦 𝜕𝑟 ∗ 2𝑠 𝜕𝑦 𝜕𝑟 𝜕𝑧 𝜕 𝜕𝑦 𝜕𝑧 𝜕 𝜕𝑥 𝜕𝑧 ( ) = 𝜕𝑦 ∗ 𝜕𝑟 ∗ (𝜕𝑦) + 𝜕𝑥 ∗ 𝜕𝑟 ∗ (𝜕𝑦) 𝜕𝑟 𝜕𝑦 = = Page | 58 𝜕𝑧 ( ) = 𝜕𝑥 ∗ 𝜕𝑟 ∗ (𝜕𝑥) + 𝜕𝑦 ∗ 𝜕𝑟 ∗ (𝜕𝑥) 𝜕𝑟 𝜕𝑥 = 𝜕𝑟 𝜕 𝜕𝑟 = 𝜕 𝜕𝑧 𝜕𝑥 ∗ = ➔ 𝜕𝑧 ( ∗ 2𝑟) + 𝜕𝑟 (𝜕𝑦 ∗ 2𝑠 ) 𝜕𝑟 𝜕𝑥 =2 ➔ 𝜕𝑧 ( ) = 𝜕𝑟 (𝜕𝑥 ∗ 2𝑟 + 𝜕𝑦 ∗ 2𝑠 ) 𝜕𝑟 𝜕𝑟 𝜕 𝜕𝑦 ∗ 𝜕𝑧 𝜕𝑦 ∗ 𝜕𝑦 𝜕𝑟 + 𝜕 𝜕𝑥 𝜕2 𝑧 𝜕2 𝑧 𝜕𝑦 𝜕𝑦𝜕𝑥 2 ∗ 2𝑠 + ∗ 𝜕𝑧 𝜕𝑦 ∗ 2𝑟 ∗ 𝜕𝑥 𝜕𝑟 = 𝜕2 𝑧 𝜕𝑦 𝜕𝑦 𝜕𝑟 2 ∗ + 𝜕2 𝑧 𝜕𝑦𝜕𝑥 ∗ 𝜕𝑥 𝜕𝑟 + 𝟐 𝟐𝝏 𝒛 𝟒𝒔 𝝏𝒚𝟐 ➔2 =2 =2 =2 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑥 Page | 59 + 2𝑟 ∗ 𝜕 𝜕𝑧 + 2𝑟 ∗ ( 𝜕2 𝑧 𝜕𝑥 2 + 4𝑟 2 + 𝜕 𝜕𝑧 ( ) + 2𝑠 ∗ 𝜕𝑟 (𝜕𝑦 ) 𝜕𝑟 𝜕𝑥 ∗ 2𝑟 + 𝜕2 𝑧 𝜕𝑥𝜕𝑦 𝜕2 𝑧 𝜕2 𝑧 𝜕𝑥 𝜕𝑥𝜕𝑦 2 + 4𝑟𝑠 2 2𝜕 𝑧 4𝑟 𝜕𝑥 2 + 8𝑟𝑠 𝜕2 𝑧 𝜕𝑥𝜕𝑦 ∗ 2𝑠) + 2𝑠 ∗ ( 𝜕𝑦 2 + 4𝑠 2 + 𝜕2 𝑧 ∗ 2𝑠 + 𝜕2 𝑧 𝜕2 𝑧 𝜕𝑦 𝜕𝑦𝜕𝑥 2 + 4𝑟𝑠 2 2𝜕 𝑧 4𝑠 𝜕𝑦 2 𝜕2 𝑧 𝜕𝑦𝜕𝑥 ∗ 2𝑟) 68] Find Fourier series for the function 𝒇(𝒙) , defined by −𝟏 𝒘𝒉𝒆𝒏 − 𝝅 ≤ 𝒙 ≤ 𝟎 𝒇(𝒙) = { 𝟏 𝒘𝒉𝒆𝒏 𝟎 ≤ 𝒙 ≤ 𝝅 𝟐 𝒏 A- ∑∞ 𝒏=𝟏 (𝒏𝝅 [(𝟏 + (−𝟏) )] 𝐬𝐢𝐧(𝒏𝒙)) 𝟐 𝒏 B- ∑∞ 𝒏=𝟏 (𝒏𝝅 [(𝟏 − (−𝟏) )] 𝐬𝐢𝐧(𝒏𝒙)) 𝟏 𝒏 C- ∑∞ 𝒏=𝟏 (𝒏𝝅 [(𝟏 − (−𝟏) )] 𝐬𝐢𝐧(𝒏𝒙)) D- None of them Solution: Even or Odd or Neither Even nor Odd? 𝑓(−𝑥 ) = { −1 1 , −𝜋 < −𝑥 < 0 , 0 < −𝑥 < 𝜋 Multiplying by −𝟏 −1 ,𝜋 > 𝑥 > 0 →{ 1 , 0 > 𝑥 > −𝜋 Then it is Odd ; Then 𝒂𝟎 , 𝒂𝒏 = 0 & 𝒃𝒏 has a value 1 𝑛𝜋𝑥 2 𝜆 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1 (𝑎𝑛 cos ( 1 2 𝑛𝜋𝑥 ) 𝜋 𝑓(𝑥) = 𝑎0 + ∑∞ 𝑛=1 (𝑎𝑛 cos ( 𝑛𝜋𝑥 ) + 𝑏𝑛 sin ( 𝑛𝜋𝑥 )) 𝜋 + 𝑏𝑛 sin ( 𝜆 )) ➔ 𝝀 = 𝝅 1 2 = 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 cos(𝑛𝑥 ) + 𝑏𝑛 sin(𝑛𝑥)) 1 = ∗ 0 + ∑∞ 𝑛=1(0 ∗ cos(𝑛𝑥) + 𝑏𝑛 sin(𝑛𝑥)) 2 = ∑∞ 𝑛=1(𝑏𝑛 sin(𝑛𝑥)) Page | 60 1 𝜋 0 1 𝜋 𝒃𝒏 = ∫−𝜋 𝑓(𝑥 ) sin(𝑛𝑥 ) 𝑑𝑥 = [∫−𝜋 𝑓(𝑥 ) sin(𝑛𝑥 ) + ∫0 𝑓(𝑥 ) sin(𝑛𝑥 )] 𝜋 𝜋 1 = [− [ − cos(𝑛𝑥) 0 𝜋 1 = [[ 𝜋 = = = = 1 𝑛𝜋 1 𝑛𝜋 1 𝑛𝜋 1 𝑛𝜋 ] 𝑛 −𝜋 cos(𝑛𝑥) 0 ] 𝑛 −𝜋 −[ +[ − cos(𝑛𝑥) 𝜋 𝑛 ] ] 0 cos(𝑛𝑥) 𝜋 ] ] 𝑛 0 [(cos 0 − cos 𝑛𝜋) − (cos(−𝑛𝜋) − cos 0)] [(1 − (−1)𝑛 ) − ((−1)𝑛 − 1)] [(1 − (−1)𝑛 ) − (−1)𝑛 + 1] [(2 − 2(−1)𝑛 )] = ∞ Then: ∑∞ 𝑛=1(𝑏𝑛 sin(𝑛𝑥 )) = ∑𝑛=1 ( 2 𝑛𝜋 𝟐 𝒏𝝅 [(𝟏 − (−𝟏)𝒏 )] [(1 − (−1)𝑛 )] sin(𝑛𝑥 )) Choice B 69] What are Fourier Coefficients? A- the terms that are presented in a Fourier Series B- the terms that are obtained through Fourier Series C- the terms which consist of the Fourier along with their Sine or Cosine values D- the terms which are of resemblance to Fourier transform in a Fourier Series are called Fourier Series Coefficients Solution: Choice C Page | 61 ‫ هو كده‬Fourier Series ‫ األصل بتاع‬، ‫بص يا سيدي للتوضيح بس‬ 1 𝑛𝜋𝑥 2 𝜆 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1 (𝑎𝑛 cos ( 1 0𝜋𝑥 2 𝜆 = 𝑎0 cos ( 𝑛𝜋𝑥 ) + 𝑏𝑛 sin ( 𝑛𝜋𝑥 ) + ∑∞ 𝑛=1 (𝑎𝑛 cos ( 1 𝑛𝜋𝑥 2 𝜆 1 𝑛𝜋𝑥 2 𝜆 = 𝑎0 ∗ 1 + ∑∞ 𝑛=1 (𝑎𝑛 cos ( = 𝑎0 ∗ 1 + ∑∞ 𝑛=1 (𝑎𝑛 cos ( 𝜆 𝜆 )) 𝑛𝜋𝑥 ) + 𝑏𝑛 sin ( 𝑛𝜋𝑥 ) + 𝑏𝑛 sin ( 𝜆 𝑛𝜋𝑥 ) + 𝑏𝑛 sin ( 𝜆 𝜆 )) )) )) Cosine terms ‫ و‬Sine ‫بالتالي األصل بتاع المتسلسلة دي ان فيها‬ 70] If the function 𝒇(𝒙) is even , then which of the following is zero? A- 𝒂𝒏 B- 𝒂𝟎 C- 𝒃𝒏 D- A & B Solution: Since 𝑓(𝑥 ) is even , then 𝒃𝒏 = 0 ➔ Choice C 71] The Fourier Series of an odd periodic function , contains only … A- Cosine terms C- Odd harmonics B- Sine terms D- Even harmonics Solution: Since 𝑓(𝑥 ) is odd , then 𝒂𝟎 , 𝒂𝒏 = 0 ➔ then 𝒃𝒏 has a value Then: The Fourier Series contains Sine terms Choice B Page | 62 72] If the function 𝒇(𝒙) is odd , then which of the following is zero? A- 𝒂𝒏 B- 𝒂𝟎 C- 𝒃𝒏 D- A & B Solution: Since 𝑓(𝑥 ) is odd , then 𝒂𝟎 , 𝒂𝒏 = 0 Choice D 73] The trigonometric Fourier Series of an even function does not have … A- Cosine terms C- Odd harmonics B- Sine terms D- Even harmonics Solution: Since 𝑓(𝑥 ) is even , then 𝒃𝒏 = 0 Then: The Fourier Series doesn’t contain Sine terms Choice B 74] Fourier Series representation can be used in case non-periodic signals too. True or False A-True E- False Solution: Choice E Page | 63 75] The Fourier Series of a real periodic function has only ( I ) Cosine terms if it is even ( II ) Sine terms if it is even ( III ) Cosine terms if it is odd ( IV ) Sine terms if it is odd Which of the above statement is correct ? A- I and III B- II and IV C- II and III D- I and IV Solution: If the function is even ➔ 𝒃𝒏 = 0 , 𝒂𝒏 , 𝒂𝟎 have values then: it consists of Cosine terms only If the function is odd ➔ 𝒂𝒏 , 𝒂𝟎 = 0 , 𝒃𝒏 has a value then: it consists of Sine terms only Choice D 76] Find fourier series for the function 𝒇(𝒙) , defined by 𝒇(𝒙) = when −𝝅 ≤ 𝒙 ≤ 𝝅 𝝅𝟐 𝟏 𝝅𝟐 𝟏 𝝅𝟐 𝟏 𝒏 A- 𝟏𝟐 + ∑∞ 𝒏=𝟏 ((−𝟏) 𝒏𝟐 𝐬𝐢𝐧(𝒏𝒙)) 𝒏 B- 𝟏𝟐 + ∑∞ 𝒏=𝟏 ((−𝟏) 𝒏 𝐜𝐨𝐬(𝒏𝒙)) 𝒏 C- 𝟏𝟐 + ∑∞ 𝒏=𝟏 ((−𝟏) 𝒏𝟐 𝐜𝐨𝐬(𝒏𝒙)) D- None of them Page | 64 𝒙𝟐 𝟒 , Solution: Even or Odd or Neither Even nor Odd? 𝑓(−𝑥 ) = (−𝒙)𝟐 𝟒 = 𝒙𝟐 𝟒 Then it is Even ; Then 𝒃𝒏 = 0 & 𝒂𝟎 , 𝒂𝒏 have values 1 𝑛𝜋𝑥 2 𝜆 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1 (𝑎𝑛 cos ( 𝑛𝜋𝑥 ) + 𝑏𝑛 sin ( 𝜆 )) ➔ 𝝀 = 𝝅 1 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 cos(𝑛𝑥 ) + 𝑏𝑛 sin(𝑛𝑥 )) 2 𝒂𝟎 = 𝜋 𝑓 (𝑥 ) 𝑑𝑥 ∫ 𝜋 −𝜋 1 = 𝜋 𝑥2 𝑑𝑥 ∫ 𝜋 −𝜋 4 1 1 𝑥3 𝜋 = [ 𝜋 12 𝒂𝒏 = = = Page | 65 + 𝜋3 12 −𝜋3 12 1 2𝜋3 ]= [ 𝜋 12 𝜋 𝑓 (𝑥 ) cos(𝑛𝑥 ) 𝑑𝑥 ∫ 𝜋 −𝜋 ]= 𝜋 𝑥2 ∫ 𝜋 −𝜋 4 1 = 𝑥 2 ∗ cos(𝑛𝑥 ) 𝑑𝑥 ∫ −𝜋 4𝜋 1 4𝜋 2 [(𝑥 ∗ 1 sin 𝑛𝑥 𝑛 − ((2𝜋 ∗ 4𝜋 1 4𝜋 [ 𝝅𝟐 𝟔 D ∗ cos(𝑛𝑥 ) 𝑑𝑥 𝜋 1 ] 6 1 [ = 𝜋 12 − 𝜋2 ((𝜋 2 ∗ = = [ 𝜋 12 −𝜋 1 𝜋3 Then: 𝒂𝟎 = 1 𝜋3 = [ ] 𝑛 − cos 𝑛𝑥 𝑛2 𝑛2 𝑛3 ) − (2 ∗ ((0) − (0)) − ((2𝜋 ∗ 𝑛2 𝑛 𝑛2 𝑛3 )) ) + (2𝜋 ∗ +((0) − (0)) )] cos 𝑛𝑥 + 𝜋 −𝜋 2𝑥 − )) − cos 𝑛∗−𝜋 − sin 𝑛∗−𝜋 −(−1)𝑛 𝑛3 sin 𝑛∗−𝜋 ) − (2 ∗ −𝜋 ∗ − sin 𝑛∗𝜋 − sin 𝑛𝑥 ) + (2 ∗ ) − ((−𝜋)2 ∗ − cos 𝑛𝜋 + ((2 ∗ 𝑥2 ) − (2𝑥 ∗ sin 𝑛𝜋 I 2 + )) 0 ] −(−1)𝑛 𝑛2 )) ] sin 𝑛𝑥 𝑛 − cos 𝑛𝑥 𝑛2 − sin 𝑛𝑥 𝑛3 = = = 1 4𝜋 1 4𝜋 1 4𝜋 [− (− (2𝜋 ∗ [((2𝜋 ∗ [ 4𝜋(−1)𝑛 𝑛2 (−1)𝑛 (−1)𝑛 𝑛2 ]= 𝑛2 ) − (2𝜋 ∗ ) + (2𝜋 ∗ (−1)𝑛 𝑛2 Then: 𝒂𝒏 = (−1)𝑛 (−1)𝑛 𝑛2 (−1)𝑛 𝑛2 = (−𝟏)𝒏 ))] 2𝜋(−1)𝑛 1 ))] = 4𝜋 [( 𝑛2 + 2𝜋(−1)𝑛 𝑛2 )] 𝟏 𝒏𝟐 1 𝑛2 1 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 cos(𝑛𝑥 ) + 𝑏𝑛 sin(𝑛𝑥 )) , 𝑏𝑛 = 0 2 1 1 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 cos(𝑛𝑥 ) + 0 ∗ sin(𝑛𝑥 )) = 𝑓 (𝑥 ) = 𝑎0 + 2 ∞ ∑𝑛=1(𝑎𝑛 cos(𝑛𝑥 )) 1 𝜋2 2 6 = ∗ 𝑛 + ∑∞ 𝑛=1 ((−1) 2 1 𝜋2 𝑛 12 ( ) 2 cos 𝑛𝑥 ) = 𝑛 + ∑∞ 𝑛=1 ((−1) 1 𝑛2 cos(𝑛𝑥 )) 1 Then: 𝑓(𝑥) = 2 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 cos(𝑛𝑥 ) + 𝑏𝑛 sin(𝑛𝑥 )) = 𝜋2 12 𝑛 + ∑∞ 𝑛=1 ((−1) 1 𝑛2 cos(𝑛𝑥 )) Choice C 77] Find Fourier series for the function 𝒇(𝒙) , defined by 𝒇(𝒙) = { 𝟏 A- 𝟒 + 𝟏 B- + 𝟐 𝟏 C- 𝟒 + 𝟎 , −𝟏 < 𝒙 < 𝟎 𝒙, 𝟎<𝒙<𝟏 (−𝟏)𝒏 −𝟏 ∞ ∑𝒏=𝟏 ( ( )𝟐 𝒄𝒐𝒔(𝒏𝝅𝒙) 𝒏𝝅 (−𝟏)𝒏 −𝟏 ∞ ∑𝒏=𝟏 ( 𝒄𝒐𝒔(𝒏𝝅𝒙) (𝒏𝝅)𝟐 (−𝟏)𝒏 −𝟏 ∞ ∑𝒏=𝟏 ( ( )𝟐 𝒄𝒐𝒔(𝒏𝝅𝒙) 𝒏𝝅 D- None of them Page | 66 − − + (−𝟏)𝒏 𝒏𝝅 (−𝟏)𝒏 𝒏𝝅 (−𝟏)𝒏 𝒏𝝅 𝒔𝒊𝒏(𝒏𝝅𝒙)) 𝒔𝒊𝒏(𝒏𝝅𝒙)) 𝒔𝒊𝒏(𝒏𝝅𝒙)) Solution: Even or Odd or Neither Even nor Odd? 𝑓(−𝑥 ) = { 0 −𝑥 , −1 < −𝑥 < 0 , 0 < −𝑥 < 1 Multiplying by −𝟏 0 →{ −𝑥 ,1 > 𝑥 > 0 , 0 > 𝑥 > −1 Then it is Neither even nor odd ; Then 𝑎0 , 𝑎𝑛 , 𝑏𝑛 have values 1 𝑛𝜋𝑥 2 𝜆 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1 (𝑎𝑛 𝑐𝑜𝑠 ( 𝑛𝜋𝑥 ) + 𝑏𝑛 𝑠𝑖𝑛 ( 𝜆 )) 2𝜆 = 2 , 𝜆 = 1 𝒂𝟎 = 𝜆 𝑓(𝑥 ) 𝑑𝑥 ∫ 𝜆 −𝜆 1 = 𝜆 1 𝒂𝒏 = ∫−𝜆 𝑓(𝑥 ) 𝑐𝑜𝑠 ( 𝜆 0 𝑓(𝑥 ) ( ∫ 1 −1 1 𝑛𝜋𝑥 𝜆 + 1 1 ∫0 𝑓(𝑥 )) 𝑑𝑥 =0+ 1 ∫0 𝑥 𝑑𝑥 𝑥2 1 =0+[ ] = 2 0 𝟏 𝟐 1 ) 𝑑𝑥 = 1 ∫−1 𝑓(𝑥 ) 𝑐𝑜𝑠(𝑛𝜋𝑥 ) 𝑑𝑥 0 1 = ∫−1 𝑓 (𝑥 ) 𝑐𝑜𝑠(𝑛𝜋𝑥 ) + ∫0 𝑓(𝑥 ) 𝑐𝑜𝑠(𝑛𝜋𝑥 ) 0 1 = ∫−1 0 ∗ 𝑐𝑜𝑠(𝑛𝜋𝑥 ) + ∫0 𝑥 ∗ 𝑐𝑜𝑠(𝑛𝜋𝑥 ) 1 ∫0 𝑥 ∗ 𝑐𝑜𝑠(𝑛𝜋𝑥 ) =𝑥∗ =[ 𝑥 𝑠𝑖𝑛(𝑛𝜋𝑥) =( 𝑛𝜋 𝑠𝑖𝑛(𝑛𝜋) 𝑛𝜋 + + 𝑠𝑖𝑛(𝑛𝜋𝑥 ) 𝑐𝑜𝑠(𝑛𝜋𝑥 ) 𝑥 𝑠𝑖𝑛(𝑛𝜋𝑥 ) 𝑐𝑜𝑠(𝑛𝜋𝑥 ) − (− = + ) (𝑛𝜋)2 (𝑛𝜋)2 𝑛𝜋 𝑛𝜋 𝑐𝑜𝑠(𝑛𝜋𝑥) 1 ] (𝑛𝜋)2 0 𝑐𝑜𝑠(𝑛𝜋) ) (𝑛𝜋)2 −( (−1)𝑛 D 0∗𝑠𝑖𝑛(𝑛𝜋∗0) = (0 + (𝑛𝜋)2 ) − (0 + 1) = = (−𝟏)𝒏 −𝟏 (𝒏𝝅)𝟐 Then: 𝒂𝒏 = 𝑛𝜋 (−1)𝑛 (𝑛𝜋)2 (−𝟏)𝒏 −𝟏 + 𝑐𝑜𝑠(𝑛𝜋∗0) ) (𝑛𝜋)2 + 1 − (𝒏𝝅)𝟐 0 Page | 67 𝑐𝑜𝑠(𝑛𝜋𝑥 ) 𝑥 1 − (𝑛𝜋)2 I 𝑠𝑖𝑛(𝑛𝜋𝑥 ) 𝑛𝜋 − 𝑐𝑜𝑠(𝑛𝜋𝑥 ) (𝑛𝜋)2 𝜆 1 𝒃𝒏 = ∫−𝜆 𝑓 (𝑥 ) 𝑠𝑖𝑛 ( 𝜆 𝑛𝜋𝑥 𝜆 1 1 ) 𝑑𝑥 = 1 ∫−1 𝑓(𝑥 ) 𝑠𝑖𝑛(𝑛𝜋𝑥 ) 𝑑𝑥 0 1 0 1 = ∫−1 𝑓(𝑥 ) 𝑠𝑖𝑛(𝑛𝜋𝑥 ) + ∫0 𝑓(𝑥 ) 𝑠𝑖𝑛(𝑛𝜋𝑥 ) = ∫−1 0 ∗ 𝑠𝑖𝑛(𝑛𝜋𝑥 ) + ∫0 𝑥 ∗ 𝑠𝑖𝑛(𝑛𝜋𝑥 ) 1 ∫0 𝑥 ∗ 𝑠𝑖𝑛(𝑛𝜋𝑥 ) = 𝑥 ∗ (− = [− = (− = 𝑐𝑜𝑠(𝑛𝜋𝑥) 𝑛𝜋 𝑥 𝑐𝑜𝑠(𝑛𝜋𝑥) 𝑛𝜋 𝑐𝑜𝑠(𝑛𝜋) 𝑛𝜋 −(−1)𝑛 𝑛𝜋 + + D (𝑛𝜋)2 ) − (− 𝑠𝑖𝑛(𝑛𝜋𝑥 ) 𝑥 𝑠𝑖𝑛(𝑛𝜋𝑥) 1 ] (𝑛𝜋)2 0 + 0 − (0) = Then: 𝒃𝒏 = 𝑠𝑖𝑛(𝑛𝜋𝑥) ) (𝑛𝜋)2 ) − (− 𝑠𝑖𝑛(𝑛𝜋) I + 0∗𝑐𝑜𝑠(𝑛𝜋∗0) 𝑛𝜋 + 𝑠𝑖𝑛(𝑛𝜋∗0) (𝑛𝜋)2 1 ) − −(−𝟏)𝒏 0 𝒏𝝅 −(−𝟏)𝒏 𝒏𝝅 1 Then: 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 𝑐𝑜𝑠 (𝑛𝜋𝑥 ) + 𝑏𝑛 𝑠𝑖𝑛 (𝑛𝜋𝑥 )) 2 1 1 (−1)𝑛 −1 2 2 (𝑛𝜋)2 = ∗ + ∑∞ 𝑛=1 ( 1 (−1)𝑛 −1 4 (𝑛𝜋)2 = + ∑∞ 𝑛=1 ( Choice A Page | 68 ∗ 𝑐𝑜𝑠(𝑛𝜋𝑥 ) + 𝑐𝑜𝑠(𝑛𝜋𝑥 ) − (−1)𝑛 𝑛𝜋 −(−1)𝑛 𝑛𝜋 ∗ 𝑠𝑖𝑛(𝑛𝜋𝑥 )) 𝑠𝑖𝑛(𝑛𝜋𝑥 )) − 𝑐𝑜𝑠(𝑛𝜋𝑥 ) 𝑛𝜋 − 𝑠𝑖𝑛(𝑛𝜋𝑥 ) (𝑛𝜋)2 78] Find Sin Series 𝒃𝒏 , 𝒇(𝒙) = 𝒆𝒌𝒙 , 𝟎 < 𝒙 < 𝝅 , 𝒌: constant 𝒏 𝒏 A- 𝒏𝟐 +𝒌 [−𝒆𝝅𝒌 (−𝟏)𝒏 + 𝟏] B- 𝒏𝟐 +𝒌𝟐 [−𝒆𝝅𝒌 (−𝟏)𝒏 − 𝟏] 𝒏 C- 𝒏𝟐 +𝒌𝟐 [−𝒆𝝅𝒌 (−𝟏)𝒏 + 𝟏] D- None of them Solution: 1 𝜋 𝜋 1 𝒃𝒏 = ∫0 𝑓 (𝑥 ) sin(𝑛𝑥 ) 𝑑𝑥 = ∫0 𝑒 𝑘𝑥 sin(𝑛𝑥 ) 𝑑𝑥 𝜋 𝜋 ∫ 𝑒 𝑘𝑥 sin(𝑛𝑥 ) = I 𝑢 = 𝑒 𝑘𝑥 , 𝑑𝑢 = 𝑘𝑒 𝑘𝑥 , 𝑑𝑣 = sin(𝑛𝑥 ) , 𝑣 = − 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 = − 𝑒 𝑘𝑥 cos(𝑛𝑥) − ∫− 𝑛 𝑘 cos(𝑛𝑥) 𝑛 cos(𝑛𝑥) 𝑛 ∗ 𝑘𝑒 𝑘𝑥 =− 𝑒 𝑘𝑥 cos(𝑛𝑥) 𝑛 + 𝒆𝒌𝒙 𝐜𝐨𝐬(𝒏𝒙) ∫ 𝑛 ∫ 𝒆𝒌𝒙 𝐜𝐨𝐬(𝒏𝒙) ➔ 𝑢 = 𝑒 𝑘𝑥 , 𝑑𝑢 = 𝑘𝑒 𝑘𝑥 , 𝑑𝑣 = cos(𝑛𝑥 ) , 𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 = 𝑒 𝑘𝑥 ∗ Then: − − ➔ − 𝑒 𝑘𝑥 cos(𝑛𝑥) 𝑛 𝑒 𝑘𝑥 cos(𝑛𝑥) 𝑛 − 𝑛 𝑛 𝑛 + [ 𝑛 + + 𝑘 𝑛2 𝑒 𝑘𝑥 I= 𝑛2 𝑘𝑒 𝑘𝑥 sin(𝑛𝑥)−𝑒 𝑘𝑥 𝑛 cos(𝑛𝑥) Page | 69 𝑛2 𝑛 ∗ = 𝑛 𝑘 − ∫ 𝑒 𝑘𝑥 sin(𝑛𝑥 ) 𝑛 𝑘 ∫ 𝑒 𝑘𝑥 sin(𝑛𝑥) = − 𝑛2 = I( sin(𝑛𝑥) 𝑛 − ∫ 𝑒 𝑘𝑥 sin(𝑛𝑥 )] = I 𝑛 = I (1 + 𝑛2 ∗ 𝑘𝑒 𝑘𝑥 = 𝑒 𝑘𝑥 ∗ 𝑘2 −𝑒 𝑘𝑥 𝑛 cos(𝑛𝑥)+𝑘𝑒 𝑘𝑥 sin(𝑛𝑥) 𝑛2 +𝑘 2 𝑛 sin(𝑛𝑥 ) = I + −𝑒 𝑘𝑥 𝑛 cos(𝑛𝑥)+𝑘𝑒 𝑘𝑥 sin(𝑛𝑥) I= 𝑛 𝑘𝑒 𝑘𝑥 sin(𝑛𝑥) 𝑛2 sin(𝑛𝑥) 𝑘 𝑒 𝑘𝑥 sin(𝑛𝑥) 𝑒 𝑘𝑥 sin(𝑛𝑥) − 2 𝑒 𝑘𝑥 cos(𝑛𝑥) −∫ 𝑛 𝑘 + 𝑒 𝑘𝑥 cos(𝑛𝑥) sin(𝑛𝑥) sin(𝑛𝑥) 2 + 𝑘2 𝑛2 𝑘2 𝑛2 2 ➔I = 1 𝑛2 +𝑘 2 𝑘 𝑘2 𝑛 𝑛2 𝑒 𝑘𝑥 sin(𝑛𝑥) − 2 ) 𝑛2 +𝑘 2 ) = I( 𝑛2 𝑛2 +𝑘 𝑛 + I 𝑘2 𝑛2 𝑒 𝑘𝑥 cos(𝑛𝑥) 𝑛2 ) 𝑘𝑒 𝑘𝑥 sin(𝑛𝑥)−𝑒 𝑘𝑥 𝑛 cos(𝑛𝑥) 𝑛2 +𝑘 2 ∗ (𝑘𝑒 𝑘𝑥 sin(𝑛𝑥 ) − 𝑒 𝑘𝑥 𝑛 cos(𝑛𝑥 )) I=I 1 ∗ [𝑘𝑒 𝑘𝑥 sin(𝑛𝑥 ) − 𝑒 𝑘𝑥 𝑛 cos(𝑛𝑥 )]𝜋0 𝑛2 +𝑘 2 = 1 𝑛2 +𝑘 2 ∗ [(𝑘𝑒 𝜋𝑘 sin(𝑛𝜋) − 𝑒 𝜋𝑘 𝑛 cos(𝑛𝜋)) − [𝑘𝑒 0∗𝑥 sin(𝑛 ∗ 0) − 𝑒 𝑘∗0 𝑛 cos(𝑛 ∗ 0)]] = 1 𝑛2 +𝑘 [(0 − 𝑒 𝜋𝑘 𝑛(−1)𝑛 ) − (0 − 𝑛)] = 2 Then: 𝒃𝒏 = = 𝟏 𝒏𝟐 +𝒌𝟐 𝒏 𝒏𝟐 +𝒌𝟐 𝟏 𝒏𝟐 +𝒌𝟐 [−𝒆𝝅𝒌 𝒏(−𝟏)𝒏 + 𝒏] [−𝒆𝝅𝒌 𝒏(−𝟏)𝒏 + 𝒏] [−𝒆𝝅𝒌 (−𝟏)𝒏 + 𝟏] Choice C 79] Find Fourier series for the function 𝒇(𝒙) , defined by 𝒇(𝒙) = { 𝟎 , −𝝅 < 𝒙 < 𝟎 𝟏, 𝟎<𝒙<𝝅 𝟏 𝟏 𝟏 𝟏 𝒏 A- 𝟐 + ∑∞ 𝒏=𝟏 ((− 𝝅𝒏 [(−𝟏) − 𝟏]) 𝐜𝐨𝐬(𝒏𝒙)) 𝒏 B- 𝟐 + ∑∞ 𝒏=𝟏 ((− 𝝅𝒏 [(−𝟏) − 𝟏]) 𝐬𝐢𝐧(𝒏𝒙)) 𝟏 𝒏 C- ∑∞ 𝒏=𝟏 ((− 𝝅𝒏 [(−𝟏) − 𝟏]) 𝐬𝐢𝐧(𝒏𝒙)) D- None of them Solution: Even or Odd or Neither Even nor Odd? 𝑓(−𝑥 ) = { 0 1 , −𝜋 < −𝑥 < 0 , 0 < −𝑥 < 𝜋 Multiplying by −𝟏 0 →{ 1 Page | 70 ,𝜋 > 𝑥 > 0 , 0 > 𝑥 > −𝜋 Then it is Neither ; Then 𝑎0 , 𝑎𝑛 , 𝑏𝑛 have values 1 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 cos(𝑛𝑥 ) + 𝑏𝑛 sin(𝑛𝑥 )) 2 1 𝜋 1 0 𝜋 0 1 𝜋 𝒂𝟎 = ∫−𝜋 𝑓 (𝑥 ) 𝑑𝑥 = [∫−𝜋 𝑓(𝑥 ) + ∫0 𝑓(𝑥 )] 𝑑𝑥 = [∫−𝜋 0 + ∫0 1] 𝑑𝑥 𝜋 𝜋 𝜋 1 1 𝜋 𝜋 = [0 + [𝑥 ]𝜋0 ] = [𝜋] = 𝟏 1 𝜋 0 1 𝜋 𝒂𝒏 = ∫−𝜋 𝑓(𝑥 ) cos(𝑛𝑥 ) 𝑑𝑥 = [∫−𝜋 𝑓(𝑥 ) cos(𝑛𝑥 ) + ∫0 𝑓(𝑥 ) cos(𝑛𝑥 )] 𝑑𝑥 𝜋 𝜋 0 1 𝜋 = [∫−𝜋 0 ∗ cos(𝑛𝑥 ) + ∫0 1 ∗ cos(𝑛𝑥 )] 𝜋 1 = [0 + 𝜋 = 1 𝜋𝑛 𝜋 ∫0 cos(𝑛𝑥 )] 1 sin(𝑛𝑥) 𝜋 = [ 𝜋 𝑛 ] = 0 1 𝜋𝑛 [sin(𝑛𝑥 )]𝜋0 [sin(𝑛𝜋) − sin(𝑛 ∗ 0)] = 𝟎 Then: 𝒂𝒏 = 𝟎 1 𝜋 0 1 𝜋 𝒃𝒏 = ∫−𝜋 𝑓(𝑥 ) sin(𝑛𝑥 ) 𝑑𝑥 = [∫−𝜋 𝑓(𝑥 ) sin(𝑛𝑥 ) + ∫0 𝑓(𝑥 ) sin(𝑛𝑥 )] 𝑑𝑥 𝜋 𝜋 0 1 𝜋 = [∫−𝜋 0 ∗ sin(𝑛𝑥 ) + ∫0 1 ∗ sin(𝑛𝑥 )] 𝜋 1 = [0 + 𝜋 =− =− Then: 𝒃𝒏 = − 𝟏 𝝅𝒏 1 𝜋𝑛 𝟏 𝝅𝒏 𝜋 ∫0 sin(𝑛𝑥 )] 1 − cos(𝑛𝑥) 𝜋 = [ 𝜋 [cos(𝑛𝜋) − cos(𝑛 ∗ 0)] [(−𝟏)𝒏 − 𝟏] [(−𝟏)𝒏 − 𝟏] 1 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 cos(𝑛𝑥 ) + 𝑏𝑛 sin(𝑛𝑥 )) 2 1 1 2 𝜋𝑛 = + ∑∞ 𝑛=1 ((− Choice B Page | 71 𝑛 [(−1)𝑛 − 1]) ∗ sin(𝑛𝑥 )) ] =− 0 1 𝜋𝑛 [cos(𝑛𝑥 )]𝜋0 80] Find Fourier series for the function 𝒇(𝒙) , defined by 𝒇(𝒙) = { 𝟎 , −𝝅 < 𝒙 < 𝟎 𝒔𝒊𝒏 𝒙 , 𝟎 < 𝒙 < 𝝅 A- 𝝅 + ∑∞ 𝒏=𝟏 (( 𝟏 𝟐𝝅 B- 𝝅 − ∑∞ 𝒏=𝟏 (( C- 𝝅 + ∑∞ 𝒏=𝟏 (( 𝟏 𝟐𝝅 𝟏 𝟐𝝅 [ −(−𝟏)𝒏 −𝟏 [ [ 𝟏+𝒏 −(−𝟏)𝒏 −𝟏 𝟏+𝒏 −(−𝟏)𝒏 −𝟏 𝟏+𝒏 − − + (−𝟏)𝒏 +𝟏 𝟏−𝒏 ]) 𝐜𝐨𝐬(𝒏𝒙)) (−𝟏)𝒏 +𝟏 𝟏−𝒏 (−𝟏)𝒏 +𝟏 𝟏−𝒏 ]) 𝐜𝐨𝐬(𝒏𝒙)) ]) 𝐜𝐨𝐬(𝒏𝒙)) D- None of them Solution: Even or Odd or Neither Even nor Odd? 𝑓(−𝑥 ) = { 0 sin 𝑥 , −𝜋 < −𝑥 < 0 , 0 < −𝑥 < 𝜋 Multiplying by −𝟏 0 →{ sin 𝑥 ,𝜋 > 𝑥 > 0 , 0 > 𝑥 > −𝜋 Then it is Neither ; Then 𝑎0 , 𝑎𝑛 , 𝑏𝑛 have values 1 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 cos(𝑛𝑥 ) + 𝑏𝑛 sin(𝑛𝑥 )) 2 1 𝜋 1 0 𝜋 𝜋 1 1 𝒂𝟎 = ∫−𝜋 𝑓 (𝑥 ) 𝑑𝑥 = [∫−𝜋 0 + ∫0 sin 𝑥 ] 𝑑𝑥 = [0 + ∫0 sin 𝑥 ] = [− cos 𝑥]𝜋0 𝜋 𝜋 𝜋 𝜋 1 1 𝟐 𝜋 𝜋 𝝅 = − [cos 𝜋 + cos 0] = − [−1 − 1] = 1 𝜋 1 0 𝜋 𝒂𝒏 = ∫−𝜋 𝑓(𝑥 ) cos(𝑛𝑥 ) 𝑑𝑥 = [∫−𝜋 𝑓(𝑥 ) cos(𝑛𝑥 ) + ∫0 𝑓(𝑥 ) cos(𝑛𝑥 )] 𝑑𝑥 𝜋 𝜋 1 0 𝜋 = [∫−𝜋 0 ∗ cos(𝑛𝑥 ) + ∫0 sin 𝑥 ∗ cos(𝑛𝑥 )] 𝑑𝑥 𝜋 1 𝜋 = [0 + ∫0 sin 𝑥 ∗ cos(𝑛𝑥 )] 𝜋 Page | 72 Since: 1] sin(A + B) = sin A cos B + cos A sin B 2] sin(A − B) = sin A cos B − cos A sin B By Adding 1 & 2 ➔ sin A cos B + cos A sin B + sin A cos B − cos A sin B = 2 sin A cos B = sin(A + B) + sin(A − B) Then: sin 𝑥 ∗ cos(𝑛𝑥 ) = sin A cos B ➔ 2 sin 𝑥 ∗ cos(𝑛𝑥 ) = 2 sin A cos B = sin(A + B) + sin(A − B) = sin(𝑥 + 𝑛𝑥 ) + sin(𝑥 − 𝑛𝑥 ) 𝜋 1 𝜋 1 ➔ 𝒂𝒏 = ∫−𝜋 𝑓(𝑥 ) cos(𝑛𝑥 ) 𝑑𝑥 = [∫0 (sin(𝑥 + 𝑛𝑥 ) + sin(𝑥 − 𝑛𝑥 ))] 𝜋 2𝜋 = = = 1 2𝜋 1 2𝜋 − cos(𝜋+𝑛𝜋) [( 1+𝑛 −(−1)𝑛 [( 1+𝑛 + + − cos(𝜋−𝑛𝜋) 1−𝑛 −(−1)𝑛 1−𝑛 1 2𝜋 [ − cos(𝑥+𝑛𝑥) 1+𝑛 − cos(0+𝑛∗0) )−( −1 1+𝑛 −1 1 𝟏 𝟐𝝅 [ −(−𝟏)𝒏 −𝟏 𝟏+𝒏 + − 𝟏 𝟐𝝅 [ 0 − cos(0−𝑛∗0) 1−𝑛 −(−1)𝑛 1 ] 1−𝑛 ) − (1+𝑛 + 1−𝑛)] = 2𝜋 [( = Then: 𝒂𝒏 = + − cos(𝑥−𝑛𝑥) 𝜋 1+𝑛 −(−𝟏)𝒏 −𝟏 𝟏+𝒏 − − )] (−1)𝑛 1−𝑛 −1 1 ) + (1+𝑛 − 1−𝑛)] (−𝟏)𝒏 +𝟏 𝟏−𝒏 ] (−𝟏)𝒏 +𝟏 𝟏−𝒏 𝜋 1 ] 0 𝜋 𝒃𝒏 = ∫−𝜋 𝑓(𝑥 ) sin(𝑛𝑥 ) 𝑑𝑥 = [∫−𝜋 𝑓(𝑥 ) sin(𝑛𝑥 ) + ∫0 𝑓(𝑥 ) sin(𝑛𝑥 )] 𝑑𝑥 𝜋 𝜋 1 0 𝜋 = [∫−𝜋 0 ∗ sin(𝑛𝑥 ) + ∫0 sin 𝑥 ∗ sin(𝑛𝑥 )] 𝜋 𝜋 1 = [0 + ∫0 sin 𝑥 ∗ sin(𝑛𝑥 )] 𝜋 Since: 1] cos(𝑥 − 𝑦) = cos 𝑥 cos 𝑦 + sin 𝑥 sin 𝑦 2] cos(𝑥 + 𝑦) = cos 𝑥 cos 𝑦 − sin 𝑥 sin 𝑦 By Subtracting 1 & 2 ➔ (cos 𝑥 cos 𝑦 + sin 𝑥 sin 𝑦) − (cos 𝑥 cos 𝑦 − sin 𝑥 sin 𝑦) = 2 sin 𝑥 sin 𝑦 = cos(𝑥 − 𝑦) − cos(𝑥 + 𝑦) Then: sin 𝑥 sin(𝑛𝑥 ) = Page | 73 cos(𝑥−𝑛𝑥)−cos(𝑥+𝑛𝑥) 2 1 𝜋 𝜋 𝜋 cos(𝑥−𝑛𝑥)−cos(𝑥+𝑛𝑥) 1 [∫0 sin 𝑥 ∗ sin(𝑛𝑥 )] = [∫0 𝜋 2 ]= 1 2𝜋 𝜋 [∫0 (cos(𝑥 − 𝑛𝑥 ) − cos(𝑥 + 𝑛𝑥 ))] = sin(𝑥−𝑛𝑥) sin(𝑥+𝑛𝑥) 𝜋 − ] 2𝜋 1−𝑛 1+𝑛 0 1 [ = 1 sin(𝜋−𝑛𝜋) sin(𝜋+𝑛𝜋) sin(0−𝑛∗0) sin(0+𝑛∗0) − )−( − )] 1−𝑛 1+𝑛 1−𝑛 1+𝑛 = 2𝜋 [( 1 2𝜋 [(0 − 0) − (0 − 0)] = 𝟎 Then: 𝒃𝒏 = 𝟎 1 𝑓(𝑥 ) = 𝑎0 + ∑∞ 𝑛=1(𝑎𝑛 cos(𝑛𝑥 ) + 𝑏𝑛 sin(𝑛𝑥 )) 2 1 2 1 2 𝜋 2𝜋 = ∗ + ∑∞ 𝑛=1 (( =𝜋+ [ −(−1)𝑛 −1 1+𝑛 − (−1)𝑛 +1 1−𝑛 ]) ∗ cos(𝑛𝑥 )) (−1)𝑛 +1 1 −(−1)𝑛 −1 ∞ ∑𝑛=1 (( [ − ]) cos(𝑛𝑥 )) 2𝜋 1+𝑛 1−𝑛 Choice A 81] What is the Fourier Series expansion of the function 𝒇(𝒙) in the interval (𝒄, 𝒄 + 𝟐𝝅) ? A- 𝒂𝟎 𝟐 + ∑∞ 𝒏=𝟏[𝒂𝒏 𝒄𝒐𝒔 𝒏𝒙 + 𝒃𝒏 𝒔𝒊𝒏 𝒏𝒙] B- 𝒂𝟎 + ∑∞ 𝒏=𝟏[𝒂𝒏 𝒄𝒐𝒔 𝒏𝒙 + 𝒃𝒏 𝒔𝒊𝒏 𝒏𝒙] C- 𝒂𝟎 𝟐 + ∑∞ 𝒏=𝟎[𝒂𝒏 𝒄𝒐𝒔 𝒏𝒙 + 𝒃𝒏 𝒔𝒊𝒏 𝒏𝒙] D- 𝒂𝟎 + ∑∞ 𝒏=𝟎[𝒂𝒏 𝒄𝒐𝒔 𝒏𝒙 + 𝒃𝒏 𝒔𝒊𝒏 𝒏𝒙] Solution: Choice A Page | 74 82] Study Stationary Points of 𝒇(𝒙, 𝒚) = 𝒙𝟐 + 𝒚𝟐 − 𝟐𝒙 − 𝟔𝒚 + 𝟏𝟒 Solution: 𝑓𝑥 = 2𝑥 − 2 = 0 , 𝑥 − 1 = 0 , then: 𝑥 = 1 𝑓𝑦 = 2𝑦 − 6 = 0 , 𝑦 − 3 = 0 , then: 𝑦 = 3 So, The Only Critical Point is (𝟏, 𝟑) 𝑓𝑥𝑥 = 2 , 𝑓𝑦𝑦 = 2 , 𝑓𝑥𝑦 = 0 2 D = 𝑓𝑥𝑥 (𝑎, 𝑏) ∗ 𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)] = 2 ∗ 2 − [0]2 = 4 At (𝟏, 𝟑) → D > 0 & 𝑓𝑥𝑥 (1,3) = 2 > 0 Then: (𝟏, 𝟑) is local minimum Page | 75 83] Study Stationary Points of 𝒇(𝒙, 𝒚) = 𝒙𝟒 + 𝒚𝟒 − 𝟒𝒙𝒚 + 𝟏 Solution: 𝑓𝑥 = 4𝑥 3 − 4𝑦 = 0 ➔ 𝑥 3 − 𝑦 = 0 ➔ 𝑥 3 = 𝑦 𝑓𝑦 = 4𝑦 3 − 4𝑥 = 0 ➔ 𝑦 3 − 𝑥 = 0 ➔ 𝑦 3 = 𝑥 By Solving those two equations: 𝑦 3 = 𝑥 ➔ (𝑥 3 )3 = 𝑥 ➔ 𝑥 9 = 𝑥 ➔ 𝑥 9 − 𝑥 = 0 ➔ 𝑥 (𝑥 8 − 1) = 0 𝑥 = 0 ; 𝑥 8 − 1 = 0 ➔ (𝑥 4 − 1)(𝑥 4 + 1) ➔ (𝑥 2 − 1)(𝑥 2 + 1)(𝑥 4 + 1) ➔ (𝑥 − 1)(𝑥 + 1)(𝑥 2 + 1)(𝑥 4 + 1) Then: 𝑥 = 1 , 𝑥 = −1 𝑥 = −1 , 0 , 1 𝑥 3 = 𝑦 ➔ (𝑦 3 )3 = 𝑦 ➔ 𝑦 9 = 𝑦 ➔ 𝑦 9 − 𝑦 = 0 ➔ 𝑦(𝑦 8 − 1) = 0 𝑦 = 0 ; 𝑦 8 − 1 = 0 ➔ (𝑦 4 − 1)(𝑦 4 + 1) ➔ (𝑦 2 − 1)(𝑦 2 + 1)(𝑦 4 + 1) ➔ (𝑦 − 1)(𝑦 + 1)(𝑦 2 + 1)(𝑦 4 + 1) Then: 𝑦 = 1 , 𝑦 = −1 𝑦 = −1 , 0 , 1 The three critical points are : (−𝟏, −𝟏) & (𝟎, 𝟎) & (𝟏, 𝟏) 𝑓𝑥𝑥 = 12𝑥 2 , 𝑓𝑦𝑦 = 12𝑦 2 , 𝑓𝑥𝑦 = −4 2 𝐷 = 𝑓𝑥𝑥 (𝑎, 𝑏) ∗ 𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)] = 12𝑥 2 ∗ 12𝑦 2 − [−4]2 = 0 144𝑥 2 𝑦 2 − 16 = 0 At (𝟎, 𝟎) Saddle Point ➔ 𝐷 = −16 At (𝟏, 𝟏) ➔ 𝐷 = 144 − 16 = 128 𝑓𝑥𝑥 at (1,1) = 12 > 0 ➔ then it is a local minimum value Page | 76 At (−𝟏, −𝟏) ➔ 𝐷 = 144 − 16 = 128 𝑓𝑥𝑥 at (−1, −1) = 12 > 0 ➔ then it is a local minimum value 84] Study Stationary Points of 𝒇(𝒙, 𝒚) = 𝒙𝟐 + 𝒙𝒚 + 𝒚𝟐 + 𝒚 Solution: 𝑓𝑥 = 2𝑥 + 𝑦 = 0 ➔ 𝑦 = −2𝑥 𝑓𝑦 = 𝑥 + 2𝑦 + 1 = 0 ➔ 𝑥 + 2 ∗ (−2𝑥 ) + 1 = 0 → 𝑥 − 4𝑥 + 1 = 0 → −3𝑥 + 1=0 Then: 1 = 3𝑥 ➔ 𝑥 = 1 3 1 −2 3 3 ➔ 𝑦 = −2𝑥 = −2 ∗ = 𝟏 −𝟐 The Critical Point is ( , 𝟑 𝟑 ) 𝑓𝑥𝑥 = 2 , 𝑓𝑦𝑦 = 2 , 𝑓𝑥𝑦 = 1 2 Then: 𝐷 = 𝑓𝑥𝑥 (𝑎, 𝑏) ∗ 𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)] = 2 ∗ 2 − [1]2 = 3 > 0 1 −2 𝑓𝑥𝑥 (𝑎, 𝑏) = 2 > 0 ➔ ( , 3 1 −2 𝑓(𝑎, 𝑏) = 𝑓 ( , 3 2 3 = 3 3 ) is a local minimum point. 1 2 1 −2 3 3 3 )=( ) + ∗ −1 3 Then: The minimum Value is Page | 77 −𝟏 𝟑 −2 2 −2 1 2 4 2 3 2 1 3 9 9 9 3 9 3 3 +( ) +( )= − + − = − = − 3 85] Study Stationary Points of 𝒇(𝒙, 𝒚) = 𝒙𝟑 + 𝟑𝒙𝟐 𝒚 − 𝟔𝒚𝟐 − 𝟔𝒙𝟐 + 𝟐 Solution: 𝑓𝑥 = 3𝑥 2 + 6𝑥𝑦 − 12𝑥 = 0 → 𝑓𝑥 = 𝑥 2 + 2𝑥𝑦 − 4𝑥 = 0 𝑓𝑦 = 3𝑥 2 − 12𝑦 ➔ 𝑥 2 − 4𝑦 = 0 , 𝑥 2 = 4𝑦 , 𝑦 = 2 𝑥 + 2𝑥 ∗ 𝑥2 4 𝑥2 4 − 4𝑥 = 0 multiply by 𝟐 2𝑥 2 + 𝑥 3 − 8𝑥 = 0 , 𝑥 (𝑥 2 + 2𝑥 − 8) = 0 Then: 𝑥 = 0 & 𝑥 2 + 2𝑥 − 8 = 0 𝑥 2 + 2𝑥 − 8 = (𝑥 − 2)(𝑥 + 4) = 0 ➔ 𝑥 = 2 , 𝑥 = −4 Then: 𝑥 = −4 , 0 , 2 𝑦= 𝑥2 4 ➔ at 𝑥 = −4 , 𝑦 = 4 At 𝒙 = 𝟎 , 𝒚 = 𝟎 At 𝒙 = 𝟐 , 𝒚 = 𝟏 Then the critical points are : (−4,4) , (0,0) , (2,1) 𝑓𝑥𝑥 = 6𝑥 + 6𝑦 − 12 , 𝑓𝑦𝑦 = −12 , 𝑓𝑥𝑦 = 6𝑥 2 𝐷 = 𝑓𝑥𝑥 ∗ 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) = (6𝑥 + 6𝑦 − 12 ) ∗ (−12) − (6𝑥)2 = −72𝑥 − 72𝑦 + 144 − 36𝑥 2 𝐷(−4,4) = −72 ∗ −4 − 72 ∗ 4 + 144 − 36 ∗ (−4)2 = 0 + 144 + 36 ∗ 16 = 𝑣𝑎𝑙𝑢𝑒 < 0 𝑓𝑥𝑥 (−4,4) = 6 ∗ −4 + 6 ∗ 4 − 12 = −12 < 0 Then: (−4,4) is a saddle point 𝐷(0,0) = −72 ∗ 0 − 72 ∗ 0 + 144 − 36 ∗ (0)2 = 144 > 0 𝑓𝑥𝑥 (0,0) = 6 ∗ 0 + 6 ∗ 0 − 12 = −12 < 0 Then: (0,0) is a local maximum point Page | 78 𝐷(2,1) = −72 ∗ 2 − 72 ∗ 1 + 144 − 36 ∗ (2)2 = 𝑣𝑎𝑙𝑢𝑒 < 0 Then: (2,1) is a Saddle point 86] Study Stationary Points of 𝒇(𝒙, 𝒚) = 𝒆𝒙 𝐜𝐨𝐬 𝒚 Solution: 𝑓𝑥 = 𝑒 𝑥 cos 𝑦 = 0 , 𝑦 = 90 , 270 && 𝑓𝑦 = −𝑒 𝑥 sin 𝑦 = 0 , 𝑦 = 0 , 180 There is no critical points 87] If 𝒇(𝒙, 𝒚) = 𝒙𝟐 + 𝒚𝟐 − 𝟐𝒙 − 𝟔𝒚 , then has extreme value at : A- (𝟎, 𝟎) B- (𝟏, 𝟏) C- (𝟑, 𝟏) D- (𝟏, 𝟑) Solution: 𝑓𝑥 = 2𝑥 − 2 = 0 ➔ 2(𝑥 − 1) = 0 ➔ 𝑥 = 1 𝑓𝑦 = 2𝑦 − 6 = 0 ➔ 2(𝑦 − 3) = 0 ➔ 𝑦 = 3 Then The critical point is : (1,3) 𝑓𝑥𝑥 = 2 , 𝑓𝑦𝑦 = 2 , 𝑓𝑥𝑦 = 0 2 𝐷 = 𝑓𝑥𝑥 (𝑎, 𝑏) ∗ 𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)] = 2 ∗ 2 − 02 = 4 > 0 Then at (1,3) it is local minimum point Choice D Page | 79 88] Stationary points of the function 𝒖(𝒙, 𝒚) are obtained by : A- 𝒖𝒙 = 𝟎 B- 𝒖𝒙 = 𝟎 and 𝒖𝒚 = 𝟎 C- 𝒖𝒚 = 𝟎 D- None of these Solution: Choice B 89] If 𝒇𝒙𝒙 (𝒂, 𝒃) = 𝒔 , 𝒇𝒚𝒚 (𝒂, 𝒃) = 𝒕 , 𝒇𝒙𝒚 (𝒂, 𝒃) = 𝒓 , and ∆= 𝒔𝒕 − 𝒓𝟐 then 𝒇(𝒂, 𝒃) is the maximum at A- ∆> 𝟎 and 𝒓 < 𝟎 B- ∆> 𝟎 and 𝒔 < 𝟎 C- ∆< 𝟎 and 𝒔 < 𝟎 D- None of these Solution: Suppose 𝐷 = ∆= 𝑓𝑥𝑥 (𝑎, 𝑏) ∗ 𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)] 2 = 𝑠𝑡 − 𝑟 2 It is maximum at 𝐷 > 0 , 𝑓𝑥𝑥 (𝑎, 𝑏) = 𝑠 < 0 Choice B 90] If 𝒇(𝒙, 𝒚) = 𝒙𝟐 + 𝒚𝟐 , then has extreme value at : A- (𝟎, 𝟎) B- (𝟏, 𝟏) C- (𝟑, 𝟏) D- (𝟏, 𝟑) Solution: 𝑓𝑥 = 2𝑥 = 0 ➔ 𝑥 = 0 𝑓𝑦 = 2𝑦 = 0 ➔ 𝑦 = 0 Then The critical point is : (0,0) Page | 80 Choice A 91] If 𝒇𝒙𝒙 (𝒂, 𝒃) = 𝒔 , 𝒇𝒚𝒚 (𝒂, 𝒃) = 𝒕 , 𝒇𝒙𝒚 (𝒂, 𝒃) = 𝒓 , and ∆= 𝒔𝒕 − 𝒓𝟐 then 𝒇(𝒂, 𝒃) is the minimum at A- ∆> 𝟎 and 𝒓 < 𝟎 B- ∆> 𝟎 and 𝒔 < 𝟎 C- ∆< 𝟎 and 𝒔 < 𝟎 D- None of these Solution: Suppose 𝐷 = ∆= 𝑓𝑥𝑥 (𝑎, 𝑏) ∗ 𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)] = 𝑠𝑡 − 𝑟 2 It is minimum at 𝐷 > 0 , 𝑓𝑥𝑥 (𝑎, 𝑏) = 𝑠 > 0 Choice D Page | 81 2 92] If 𝒇(𝒙, 𝒚) = 𝒆𝒙+𝒚 , then find Taylor’s expansion at the point (𝟎, 𝟎) for 𝒏 = 𝟑 𝟏 𝟏 𝟏 𝟏 A- 𝟏 + (𝒙 + 𝒚) + 𝟐! (𝒙𝟐 + 𝒚𝟐 + 𝟐𝒙𝒚) + 𝟑! [𝒙𝟑 + 𝟑𝒙𝟐 𝒚 + 𝟑𝒙𝒚𝟐 + 𝒚𝟑 ]𝒆𝜽(𝒙+𝒚) B- 𝟏 + (𝒙 + 𝒚) + 𝟐! (𝒙𝟐 + 𝒚𝟐 + 𝟐𝒙𝒚) + 𝟑! [𝒙𝟑 + 𝟑𝒙𝟐 𝒚 + 𝟑𝒙𝒚𝟐 + 𝒚𝟑 ] 𝟏 C- 𝟏 + (𝒙 + 𝒚) + 𝟐! (𝒙𝟐 + 𝒚𝟐 + 𝟐𝒙𝒚) + [𝒙𝟑 + 𝟑𝒙𝟐 𝒚 + 𝟑𝒙𝒚𝟐 + 𝒚𝟑 ]𝒆𝜽(𝒙+𝒚) D- None of these Solution: 𝑓(𝑥, 𝑦) = 𝑓(𝑎, 𝑏) + [(𝑥 − 𝑎) 𝑏) 𝜕 𝜕𝑥 + (𝑦 − 𝑏) 𝜕 1 𝜕𝑦 1 𝜕 ] 𝑓(𝑎, 𝑏) + 2! [(𝑥 − 𝑎) 𝜕𝑥 + (𝑦 − 𝜕 2 𝜕𝑦 ] 𝑓(𝑎, 𝑏) + 𝑅3 At (𝒂, 𝒃) = (0,0) 𝜕 𝜕 1 1 𝜕 𝜕 2 = 𝑓(0,0) + [(𝑥 − 𝑎) + (𝑦 − 𝑏) ] 𝑓(0,0) + [(𝑥 − 𝑎) + (𝑦 − 𝑏) ] 𝑓(0,0) + 𝑅3 𝜕𝑥 𝜕𝑦 2! 𝜕𝑥 𝜕𝑦 𝑓(0,0) = 𝑒 0+0 = 𝑒 0 = 1 𝜕 𝜕𝑥 = 𝜕 𝜕𝑦 = [(𝑥 − 𝑎) 𝜕2 𝜕𝑥 2 𝜕 𝜕𝑥 = 𝜕2 𝜕𝑦 2 = 𝑒 𝑥+𝑦 ➔ at (0,0) 𝑒 𝑥+𝑦 = 1 + (𝑦 − 𝑏) 𝜕 1 𝜕𝑦 ] 𝑓(0,0) = [(𝑥 − 0) 𝜕 𝜕 1 ( 𝑦 − 0) ] + 𝜕𝑥 𝜕𝑦 = ((𝑥 − 0) ∗ 1 + (𝑦 − 0) ∗ 1) = (𝒙 + 𝒚) 1 𝜕 2 𝜕 1 𝜕 2 𝜕 [(𝑥 − 𝑎) 𝜕𝑥 + (𝑦 − 𝑏) 𝜕𝑦] 𝑓(0,0) = 2! [(𝑥 − 0) 𝜕𝑥 + (𝑦 − 0) 𝜕𝑦] 2! = 𝑅3 = 1 3! [(𝑥 − 𝑎) At (𝒂, 𝒃) = (0,0) Page | 82 𝜕 𝜕𝑥 + (𝑦 − 𝑏) 1 2! 𝜕 3 𝜕𝑦 [(𝑥) ∗ 1 + (𝑦) ∗ 1]2 = 𝟏 𝟐! (𝒙𝟐 + 𝒚𝟐 + 𝟐𝒙𝒚) ] 𝑓 ((𝑎 + 𝜃(𝑥 − 𝑎)), (𝑏 + 𝜃(𝑦 − 𝑏))) 1 Then: 𝑅3 = 1 = 3! 1 = 3! 3! 𝜕 [(𝑥 − 0) [(𝑥) [(𝑥) 𝜕 𝜕𝑥 𝜕 𝜕𝑥 𝜕𝑥 + (𝑦) + (𝑦) + (𝑦 − 0) 𝜕𝑦 ] 𝑓 ((0 + 𝜃(𝑥 − 0)), (0 + 𝜃(𝑦 − 0))) 𝜕 3 𝜕𝑦 ] 𝑓(𝜃(𝑥), 𝜃(𝑦)) 𝜕 2 𝜕𝑦 ] [(𝑥) 𝜕 2 1 𝜕 3 𝜕 𝜕𝑥 + (𝑦) 𝜕 𝜕 𝜕𝑦 ] 𝜕 2 𝜕 𝜕 𝜕 ([(𝑥) 𝜕𝑥] + 2 [((𝑥) 𝜕𝑥) ((𝑦) 𝜕𝑦)] + [(𝑦) 𝜕𝑦] ) ∗ [(𝑥) 𝜕𝑥 + (𝑦) 𝜕𝑦] 3! = = 𝑥2 1 3! 𝜕2 𝜕 𝜕𝑥 𝜕𝑥 2 ∗𝑥 + 𝑥2 𝜕2 𝜕𝑥 2 ∗𝑦 𝜕 𝜕𝑦 + 2 [((𝑥) +𝑦 2 [ = 𝑥3 ➔ 𝜕3 𝜕𝑥 3 𝜕3 𝜕3 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 = 𝑓𝑥𝑥𝑥 = 𝑒 𝑥+𝑦 , 𝜕3 𝑒 𝑥+𝑦 + 𝑥 2𝑦 3 𝜕𝑦 2 𝜕𝑥 𝜕3 𝜕𝑥 2 𝜕𝑦 = 𝑓𝑦𝑦𝑥 = 𝑒 𝑥+𝑦 , At (𝟎, 𝟎) , then Then: 𝑥 3 + 2(𝑥 2 𝑦) 𝜕3 𝜕𝑥 3 = 𝜕3 𝜕𝑥 2 𝜕𝑦 𝜕3 𝜕3 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 + 𝑥 2𝑦 3 = 𝜕 𝜕𝑥 𝜕 𝜕2 𝜕𝑦 2 ∗𝑥 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 𝜕 𝜕𝑥 + 𝑦2 𝜕𝑦 3 𝜕2 𝜕𝑦 2 + 2(𝑥𝑦 2 ) 𝜕3 = 𝑓𝑥𝑥𝑦 = 𝑒 𝑥+𝑦 , 𝜕3 𝜕 𝜕 𝜕 𝜕𝑥𝜕𝑦 2 𝜕 ∗𝑦 𝜕3 𝜕𝑥𝜕𝑦 ] 𝜕𝑦 + 𝑦2𝑥 2 𝜕3 𝜕𝑦 2 𝜕𝑥 = 𝑓𝑥𝑦𝑦 = 𝑒 𝑥+𝑦 , + 𝑦3 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 𝜕3 𝜕𝑦 3 = 𝑓𝑥𝑦𝑥 = = 𝑓𝑦𝑦𝑦 = 𝑒 𝑥+𝑦 𝜕3 𝜕𝑥𝜕𝑦 2 + 2(𝑥 2 𝑦) = 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 = 𝜕3 𝜕𝑦 2 𝜕𝑥 + 2(𝑥𝑦 2 ) = 𝜕3 𝜕𝑦 3 𝜕3 𝜕𝑥𝜕𝑦 = 𝑒 0+0 = 1 + 𝑦2𝑥 2 𝜕3 𝜕𝑦 2 𝜕𝑥 + 𝑦3 𝜕3 𝜕𝑦 3 = 𝑥 3 (𝑒 𝑥+𝑦 ) + 𝑥 2 𝑦(𝑒 𝑥+𝑦 ) + 2(𝑥 2 𝑦)(𝑒 𝑥+𝑦 ) + 2(𝑥𝑦 2 )(𝑒 𝑥+𝑦 ) + 𝑦 2 𝑥(𝑒 𝑥+𝑦 ) + 𝑦 3 (𝑒 𝑥+𝑦 ) = 𝑥 3 (𝑒 𝑥+𝑦 ) + 3(𝑥 2 𝑦)(𝑒 𝑥+𝑦 ) + 3(𝑥𝑦 2 )(𝑒 𝑥+𝑦 ) + 𝑦 3 (𝑒 𝑥+𝑦 ) = 𝑒 𝑥+𝑦 (𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 ) = 1 ∗ (𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 ) Then: 𝑅3 = = 1 3! 1 3! [(𝑥) 𝜕 𝜕𝑥 + (𝑦) 𝜕 3 𝜕𝑦 ] 𝑓(𝜃(𝑥), 𝜃(𝑦)) [𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 ]𝑓 (𝜃(𝑥), 𝜃(𝑦)) ➔ 𝑓(𝜃(𝑥), 𝜃(𝑦)) = 𝑒 𝜃𝑥+𝜃𝑦 = 𝑒 𝜃(𝑥+𝑦) 1 3! [𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 ]𝑓(𝜃(𝑥), 𝜃(𝑦)) = Page | 83 𝜕 ) ((𝑦) 𝜕𝑦)] ∗ 𝑥 𝜕𝑥 + 2 [((𝑥) 𝜕𝑥) ((𝑦) 𝜕𝑦)] ∗ 𝑦 𝜕𝑦 𝟏 𝟑! [𝒙𝟑 + 𝟑𝒙𝟐 𝒚 + 𝟑𝒙𝒚𝟐 + 𝒚𝟑 ]𝒆𝜽(𝒙+𝒚) 𝑓(𝑥, 𝑦) = 𝑓(𝑎, 𝑏) + [(𝑥 − 𝑎) 𝑏) 𝜕 𝜕𝑥 + (𝑦 − 𝑏) 𝜕 1 𝜕𝑦 1 𝜕 ] 𝑓(𝑎, 𝑏) + 2! [(𝑥 − 𝑎) 𝜕𝑥 + (𝑦 − 𝜕 2 𝜕𝑦 ] 𝑓(𝑎, 𝑏) + 𝑅3 𝟏 = 𝟏 + (𝒙 + 𝒚) + 𝟐! (𝒙𝟐 + 𝒚𝟐 + 𝟐𝒙𝒚) + 𝟏 [𝒙𝟑 + 𝟑𝒙𝟐 𝒚 + 𝟑𝒙𝒚𝟐 + 𝒚𝟑 ]𝒆𝜽(𝒙+𝒚) 𝟑! Choice A 93] If 𝒇(𝒙, 𝒚) = 𝐜𝐨𝐬(𝒙 + 𝒚) , then find Taylor’s expansion at the point (𝟎, 𝟎) for 𝒏 = 𝟑 𝟏 𝟏 A- −𝟏 − 𝟐! (𝒙 + 𝒚)𝟐 + 𝟑! (𝒙 + 𝒚)𝟑 𝐬𝐢𝐧 𝜽(𝒙 + 𝒚) 𝟏 𝟏 B- 𝟏 − 𝟐! (𝒙 + 𝒚)𝟐 − 𝟑! (𝒙 + 𝒚)𝟑 𝐬𝐢𝐧 𝜽(𝒙 + 𝒚) C- 𝟏 − (𝒙+𝒚)𝟐 𝟐! + (𝒙+𝒚)𝟑 𝐬𝐢𝐧 𝜽(𝒙 + 𝒚) 𝟑! D- None of these Solution: 𝑓(𝑥, 𝑦) = 𝑓(𝑎, 𝑏) + [(𝑥 − 𝑎) 𝑏) 𝜕 𝜕𝑥 + (𝑦 − 𝑏) 𝜕 1 1 𝜕 ] 𝑓(𝑎, 𝑏) + 2! [(𝑥 − 𝑎) 𝜕𝑥 + (𝑦 − 𝜕𝑦 𝜕 2 𝜕𝑦 ] 𝑓(𝑎, 𝑏) + 𝑅3 At (𝒂, 𝒃) = (0,0) = 𝑓(0,0) + [(𝑥 − 𝑎) 𝜕 𝜕𝑥 + (𝑦 − 𝑏) 𝜕 1 1 𝜕 2 𝜕 ] 𝑓(0,0) + 2! [(𝑥 − 𝑎) 𝜕𝑥 + (𝑦 − 𝑏) 𝜕𝑦] 𝑓(0,0) + 𝑅3 𝜕𝑦 𝑓(0,0) = cos(0 + 0) = cos(0) = 1 𝜕 𝜕𝑥 = − sin(𝑥 + 𝑦) , 𝜕2 𝜕𝑥𝜕𝑦 𝜕 𝜕𝑦 = − sin(𝑥 + 𝑦) , 𝜕2 𝜕𝑥 2 = − cos(𝑥 + 𝑦) , 𝜕2 𝜕𝑦 2 = − cos(𝑥 + 𝑦) , = − cos(𝑥 + 𝑦) At (𝟎, 𝟎) ➔ [(𝑥 − 𝑎) Page | 84 𝜕 𝜕𝑥 𝜕 𝜕𝑥 = 𝜕 𝜕𝑦 =0 , + (𝑦 − 𝑏) 𝜕 1 𝜕𝑦 𝜕2 𝜕𝑥 2 = 𝜕2 𝜕𝑦 2 = 𝜕2 𝜕𝑥𝜕𝑦 = −1 ] 𝑓(0,0) = [(𝑥 − 0) ∗ 0 + (𝑦 − 0) ∗ 0]1 = 𝟎 1 2! [(𝑥 − 𝑎) [ (𝑦 − 0) 𝜕 𝜕𝑥 𝜕 2 + (𝑦 − 𝑏 ) 𝜕𝑦 𝜕𝑦 2 2 𝜕 (𝑥 2 2! 𝜕𝑥 1 1 𝜕 𝜕𝑥 𝜕𝑥 ] + 2 [((𝑥 − 0) ) (( 𝑦 − 0 ) 𝜕 𝜕𝑦 )] + 1 Then: 𝑅3 = 1 3! 1 3! 3! 𝜕 [(𝑥 − 0) 1 [(𝑥) [(𝑥) 𝜕 𝜕𝑥 𝜕 𝜕𝑥 + (𝑦) + (𝑦) 𝜕𝑥𝜕𝑦 𝜕2 ] + 𝑦2 𝜕𝑦2 ) 𝟏 2! 𝟐! 𝜕 3 𝜕𝑦 ] 𝑓 ((0 + 𝜃(𝑥 − 0)), (0 + 𝜃(𝑦 − 0))) 𝜕 3 𝜕𝑦 ] 𝑓(𝜃(𝑥), 𝜃(𝑦)) 𝜕 2 𝜕𝑦 ] [(𝑥) 𝜕 2 1 𝜕2 (−𝑥 2 − 2𝑥𝑦 − 𝑦 2 ) = − (𝒙𝟐 + 𝟐𝒙𝒚 + 𝒚𝟐 ) + (𝑦 − 0) 𝜕𝑥 + 2 [𝑥𝑦 (𝑥2 ∗ −1 + 2[𝑥𝑦 ∗ −1] + 𝑦2 ∗ −1) 2! = = 𝜕 2 ] ) = = 2! ([(𝑥 − 0) 𝜕 2 = = 1 ] 𝑓 (0,0) = 𝜕 𝜕𝑥 + (𝑦) 𝜕 𝜕 𝜕𝑦 ] 𝜕 2 𝜕 𝜕 𝜕 ([(𝑥) 𝜕𝑥] + 2 [((𝑥) 𝜕𝑥) ((𝑦) 𝜕𝑦)] + [(𝑦) 𝜕𝑦] ) ∗ [(𝑥) 𝜕𝑥 + (𝑦) 𝜕𝑦] 3! = 𝑥2 1 3! 𝜕2 𝜕𝑥 2 ∗𝑥 𝜕 𝜕𝑥 + 𝑥2 𝜕2 𝜕𝑥 2 ∗𝑦 𝜕 𝜕𝑦 +𝑦 2 [ = 𝑥3 ➔ + 2 [((𝑥) 𝜕3 𝜕𝑥 3 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 𝜕3 𝜕3 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 + 𝑥 2𝑦 3 + 2(𝑥 2 𝑦) = 𝑓𝑥𝑥𝑥 = sin(𝑥 + 𝑦) , = 𝑓𝑥𝑦𝑥 = sin(𝑥 + 𝑦) At ((𝜽(𝒙), 𝜽(𝒚))) , then 𝜕3 𝜕𝑥 3 𝜕3 𝜕𝑥 2 𝜕𝑦 𝜕3 𝜕𝑦 2 𝜕𝑥 = 𝜕 𝜕 𝜕 𝜕2 𝜕 𝜕𝑦 𝜕𝑥 2 ∗𝑥 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 + 𝑦2 𝜕2 𝜕𝑦 2 ∗𝑦 𝜕3 + 2(𝑥𝑦 2 ) 𝜕𝑥𝜕𝑦 = 𝑓𝑥𝑥𝑦 = sin(𝑥 + 𝑦) , = 𝑓𝑦𝑦𝑥 = sin(𝑥 + 𝑦) , 𝜕3 𝜕𝑥 2 𝜕𝑦 = 𝜕 𝜕 𝜕3 𝜕𝑥𝜕𝑦 2 = 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 = 𝜕 ] 𝜕𝑦 + 𝑦2𝑥 2 𝜕3 𝜕𝑥𝜕𝑦 2 𝜕3 𝜕𝑦 3 𝜕3 𝜕𝑦 2 𝜕𝑥 + 𝑦3 𝜕3 𝜕𝑦 3 = 𝑓𝑥𝑦𝑦 = sin(𝑥 + 𝑦) , = 𝑓𝑦𝑦𝑦 = sin(𝑥 + 𝑦) 𝜕3 𝜕𝑦 2 𝜕𝑥 = 𝜕3 𝜕𝑦 3 = sin(𝜃(𝑥) + 𝜃(𝑦)) = sin 𝜃(𝑥 + 𝑦) Then: 𝑥 3 𝜕3 𝜕3 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 + 𝑥 2𝑦 3 + 2(𝑥 2 𝑦) 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 + 2(𝑥𝑦 2 ) 𝜕3 𝜕𝑥𝜕𝑦 + 𝑦2𝑥 2 𝜕3 𝜕𝑦 2 𝜕𝑥 = 𝑥 3 ∗ sin 𝜃(𝑥 + 𝑦) + 𝑥 2 𝑦 ∗ sin 𝜃(𝑥 + 𝑦) + 2(𝑥 2 𝑦) ∗ sin 𝜃(𝑥 + 𝑦) +2(𝑥𝑦 2 ) ∗ sin 𝜃(𝑥 + 𝑦) + 𝑦 2 𝑥 ∗ sin 𝜃(𝑥 + 𝑦) + 𝑦 3 ∗ sin 𝜃(𝑥 + 𝑦) = sin 𝜃(𝑥 + 𝑦) [𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 ] Page | 85 𝜕 ) ((𝑦) 𝜕𝑦)] ∗ 𝑥 𝜕𝑥 + 2 [((𝑥) 𝜕𝑥) ((𝑦) 𝜕𝑦)] ∗ 𝑦 𝜕𝑦 𝜕𝑥 + 𝑦3 𝜕3 𝜕𝑦 3 Then: 𝑅3 = = 𝟏 𝟑! 1 3! [(𝑥) 𝜕 𝜕𝑥 + (𝑦) 𝜕 3 𝜕𝑦 [𝒙𝟑 + 𝟑𝒙𝟐 𝒚 + 𝟑𝒙𝒚𝟐 + 𝒚𝟑 ] 𝐬𝐢𝐧 𝜽(𝒙 + 𝒚) 𝑓(𝑥, 𝑦) = 𝑓(𝑎, 𝑏) + [(𝑥 − 𝑎) 𝑏) ] 𝑓(𝜃(𝑥), 𝜃(𝑦)) 𝜕 𝜕𝑥 + (𝑦 − 𝑏) 𝜕 1 𝜕𝑦 1 𝜕 ] 𝑓(𝑎, 𝑏) + 2! [(𝑥 − 𝑎) 𝜕𝑥 + (𝑦 − 𝜕 2 𝜕𝑦 ] 𝑓(𝑎, 𝑏) + 𝑅3 1 1 = 1 + 0 + (− (𝑥 2 + 2𝑥𝑦 + 𝑦 2 )) + [𝑥3 + 3𝑥2 𝑦 + 3𝑥𝑦2 + 𝑦3 ] sin 𝜃(𝑥 + 𝑦) 2! 3! 𝟏 = 𝟏 − (𝒙𝟐 + 𝟐𝒙𝒚 + 𝒚𝟐 ) + 𝟐! 𝟏 = 𝟏 − (𝒙 + 𝒚 )𝟐 + 𝟐! 𝟏 [𝒙𝟑 𝟑! + 𝟑𝒙𝟐 𝒚 + 𝟑𝒙𝒚𝟐 + 𝒚𝟑 ] 𝐬𝐢𝐧 𝜽(𝒙 + 𝒚) 𝟏 (𝒙 + 𝒚)𝟑 𝐬𝐢𝐧 𝜽(𝒙 + 𝒚) 𝟑! Choice C 94] Expand 𝒇(𝒙, 𝒚) = 𝒙𝟐 + 𝒙𝒚 + 𝒚𝟐 in powers of (𝒙 − 𝟐) , (𝒚 − 𝟑) A- 𝟕𝒙 + 𝟗𝒚 + (𝒙 − 𝟐)𝟐 + (𝒙 − 𝟐)(𝒚 − 𝟑) + (𝒚 − 𝟑)𝟐 B- −𝟏𝟗 + 𝟕𝒙 + 𝟖𝒚 + (𝒙 − 𝟐)𝟐 + (𝒙 − 𝟐)(𝒚 − 𝟑) + (𝒚 − 𝟑)𝟐 C- 𝟏𝟗 + 𝟕𝒙 + 𝟗𝒚 + (𝒙 − 𝟐)𝟐 + (𝒙 − 𝟐)(𝒚 − 𝟑) + (𝒚 − 𝟑)𝟐 D- None of these Solution: 𝑥−2=0➔𝑥 =2 𝑦−3=0➔𝑦 =3 𝑓(2,3) = 19 Then point is (𝟐, 𝟑) 𝑓𝑥 = 2𝑥 + 𝑦 , 𝑓𝑥 (2,3) = 7 𝑓𝑦 = 𝑥 + 2𝑦 , 𝑓𝑥 (2,3) = 8 𝑓𝑥𝑥 = 2 , 𝑓𝑥 (2,3) = 2 Page | 86 𝑓𝑦𝑦 = 2 , 𝑓𝑥 (2,3) = 2 𝑓𝑥𝑦 = 1 , 𝑓𝑥 (2,3) = 1 𝑓𝑦𝑥 = 1 , 𝑓𝑥 (2,3) = 1 𝑓𝑥𝑥𝑥 = 𝑓𝑦𝑦𝑦 = 𝑓𝑥𝑦𝑦 = 𝑓𝑦𝑦𝑥 = 0 All The third higher order partial derivatives is Zero Then 𝒏 = 𝟐 𝑓(𝑥, 𝑦) = 𝑓(2,3) + [(𝑥 − 𝑎) 𝜕 𝜕𝑥 + (𝑦 − 𝑏 ) 𝜕 1 𝜕𝑦 1 𝜕 2! 𝜕𝑥 ] 𝑓 (2,3) + [(𝑥 − 𝑎) + 𝜕 2 (𝑦 − 𝑏 ) 𝜕𝑦 ] 𝑓(2,3) 𝑓(2,3) = 19 𝜕 𝜕 = 2𝑥 + 𝑦 ➔ at (𝟐, 𝟑) , 2 ∗ 2 + 3 = 7 𝜕𝑥 𝜕2 𝜕𝑥 2 𝜕2 = 2 ➔ at (𝟐, 𝟑) , equals 2 𝜕2 𝜕𝑥𝜕𝑦 𝜕𝑦 𝜕𝑦 2 = 𝑥 + 2𝑦 ➔ at (𝟐, 𝟑) , 2 + 2 ∗ 3 = 8 = 2 ➔ at (𝟐, 𝟑) , equals 2 = 1 ➔ at (𝟐, 𝟑) , equals 1 [(𝑥 − 𝑎) 𝜕 𝜕𝑥 + (𝑦 − 𝑏) 𝜕 1 𝜕𝑦 ] 𝑓(2,3) = [(𝑥 − 2) ∗ 7 + (𝑦 − 3) ∗ 8]1 = 7𝑥 − 14 + 8𝑦 − 24 = 7𝑥 + 8𝑦 − 38 1 2! [(𝑥 − 𝑎) [ (𝑦 − 3) 𝜕 𝜕𝑥 + (𝑦 − 𝑏 ) 𝜕 2 𝜕𝑦 ] 𝑓(2,3) = 1 2! ([(𝑥 − 2) 𝜕 2 𝜕 𝜕𝑥 𝜕𝑥 ] + 2 [((𝑥 − 2) ) (( 𝑦 − 3 ) )] + 𝜕 2 𝜕𝑦 ] ) = = 1 2! 1 2! ((𝑥 − 2)2 𝜕2 𝜕𝑥2 + 2 [(𝑥 − 2)(𝑦 − 3) 𝜕2 ] + (𝑦 𝜕𝑥𝜕𝑦 − 3)2 𝜕2 ) 𝜕𝑦2 ((𝑥 − 2)2 ∗ 2 + 2[(𝑥 − 2)(𝑦 − 3) ∗ 1] + (𝑦 − 3)2 ∗ 2) = (𝑥 − 2)2 + (𝑥 − 2)(𝑦 − 3) + (𝑦 − 3)2 Page | 87 𝜕 𝜕𝑦 Then: 𝑓(2,3) + [(𝑥 − 𝑎) 𝜕 𝜕𝑥 + (𝑦 − 𝑏) 𝜕 1 𝜕𝑦 1 𝜕 2! 𝜕𝑥 ] 𝑓(2,3) + [(𝑥 − 𝑎) + (𝑦 − 𝑏) 𝜕 2 𝜕𝑦 ] 𝑓(2,3) = 19 + 7𝑥 + 8𝑦 − 38 + (𝑥 − 2)2 + (𝑥 − 2)(𝑦 − 3) + (𝑦 − 3)2 = −𝟏𝟗 + 𝟕𝒙 + 𝟖𝒚 + (𝒙 − 𝟐)𝟐 + (𝒙 − 𝟐)(𝒚 − 𝟑) + (𝒚 − 𝟑)𝟐 Choice B 95] Find the first and second degree Taylor polynomial of 𝒇(𝒙, 𝒚) = 𝟐 𝟐 𝒆−𝒙 −𝒚 at (𝟎, 𝟎) A- 𝟏 − 𝒙𝟐 − 𝒚𝟑 B- 𝟏 − 𝒙𝟐 − 𝒚𝟐 C- 𝟏𝟏 − 𝒙𝟐 + 𝒚𝟐 D- None of these Solution: 𝑓(𝑥, 𝑦) = 𝑓(0,0) + [(𝑥 − 𝑎) (𝑦 − 𝑏 ) 𝜕𝑥 𝜕 𝜕𝑦 𝜕𝑥 + (𝑦 − 𝑏 ) 𝜕𝑦 2 −02 ∗ −2𝑥 ➔ at (𝟎, 𝟎) , 𝑒 −0 = 𝑒 −𝑥 2 −𝑦 2 ∗ −2𝑦 ➔ at (𝟎, 𝟎) , 𝑒 −0 = (𝑒 −𝑥 2 −𝑦 2 𝜕𝑦 2 = (𝑒 −𝑥 2 −𝑦2 𝜕𝑥𝜕𝑦 = 𝑒 −𝑥 Page | 88 2! 𝜕𝑥 + 2 −𝑦2 2 −02 − 2𝑒 −0 2 −02 − 2𝑒 −0 ∗ −2 ∗ 0 = 0 2 −02 2 −𝑦 2 2 −02 ∗ −2𝑦 ∗ −2𝑦) + (𝑒 −𝑥 At (𝟎, 𝟎) , 4 ∗ 02 ∗ 𝑒 −0 𝜕2 2 −02 ∗ −2𝑥 ∗ −2𝑥) + (𝑒 −𝑥 At (𝟎, 𝟎) , 4 ∗ 02 ∗ 𝑒 −0 𝜕2 𝜕 =1 2 −𝑦 2 𝜕𝑥 2 𝜕𝑦 1 ] 𝑓 (0,0) + [(𝑥 − 𝑎) ] 𝑓(0,0) = 𝑒 −𝑥 𝜕2 𝜕 1 𝜕 2 𝑓(0,0) = 𝑒 −0 𝜕 𝜕 ∗ −2) = 4𝑥 2 𝑒 −𝑥 2 −𝑦2 − 2𝑒 −𝑥 2 −𝑦 2 2 −𝑦2 − 2𝑒 −𝑥 = 0 − 2 = −2 2 −𝑦 2 2 −02 ∗ −2 ∗ 0 = 0 ∗ −2) = 4𝑦 2 𝑒 −𝑥 = 0 − 2 = −2 ∗ −2𝑥 ∗ −2𝑦 ➔ at (0,0) , 𝑒 −𝑥 2 −𝑦 2 ∗ −2𝑥 ∗ −2𝑦 = 0 2 −𝑦 2 [ (𝑥 − 𝑎 ) 1 2! 𝜕 𝜕𝑥 [(𝑥 − 𝑎) [ (𝑦 − 0) 𝜕 𝜕𝑥 + (𝑦 − 𝑏 ) + (𝑦 − 𝑏 ) 𝜕 1 𝜕𝑦 ] 𝑓 (0,0) = [(𝑥 − 0) ∗ 0 + (𝑦 − 0) ∗ 0]1 = 0 𝜕 2 𝜕𝑦 ] 𝑓(0,0) = 2! ([(𝑥 − 0) 𝜕 2 𝜕 𝜕𝑥 𝜕𝑥 ] + 2 [((𝑥 − 0) ) (( 𝑦 − 0 ) 𝜕 𝜕𝑦 )] + 𝜕 2 𝜕𝑦 ] ) = = = Then: 𝑓(0,0) + [(𝑥 − 𝑎) 𝑏) 1 𝜕 𝜕𝑥 1 2! 1 2! 1 2! (𝑥 2 𝜕2 𝜕2 𝜕𝑥 𝜕𝑥𝜕𝑦 + 2 [𝑥𝑦 2 ] + 𝑦2 𝜕2 𝜕𝑦 2 ) (𝑥 2 ∗ −2 + 2[𝑥𝑦 ∗ 0] + 𝑦 2 ∗ −2) (−2𝑥 2 − 2𝑦 2 ) = −𝑥 2 − 𝑦 2 + (𝑦 − 𝑏 ) 𝜕 1 𝜕𝑦 1 𝜕 2! 𝜕𝑥 ] 𝑓 (0,0) + [(𝑥 − 𝑎) + (𝑦 − 𝜕 2 𝜕𝑦 ] 𝑓(0,0) = 1 + 0 + (−𝑥 2 − 𝑦 2 ) = 𝟏 − 𝒙𝟐 − 𝒚𝟐 Choice B 96] Find the first and second degree Taylor polynomial of 𝒇(𝒙, 𝒚) = 𝒙𝒆𝒚 at (𝟏, 𝟎) 𝟏 A- 𝒙 − 𝒚 + 𝟐! (𝟐𝒙𝒚 − 𝟐𝒚 + 𝒚𝟐 ) 𝟏 B- 𝒙 + 𝒚 + 𝟐 (𝟐𝒙𝒚 − 𝟐𝒚 + 𝒚𝟐 ) C- 𝒙 + 𝒚 + 𝟏 𝟑! (𝟐𝒙𝒚 − 𝟐𝒚 + 𝒚𝟐 ) D- None of these Solution: 𝑓(𝑥, 𝑦) = 𝑓(1,0) + [(𝑥 − 𝑎) (𝑦 − 𝑏 ) Page | 89 𝜕 2 𝜕𝑦 ] 𝑓(1,0) 𝜕 𝜕𝑥 + (𝑦 − 𝑏 ) 𝜕 1 𝜕𝑦 1 𝜕 2! 𝜕𝑥 ] 𝑓 (1,0) + [(𝑥 − 𝑎) + 𝑓(1,0) = 𝑥𝑒 𝑦 = 1 ∗ 𝑒 0 = 1 𝜕 𝜕𝑥 𝜕 𝜕𝑦 = 𝑒 𝑦 ➔ at (𝟏, 𝟎) , 𝑒 0 = 1 = 𝑥𝑒 𝑦 ➔ at (𝟏, 𝟎) , 1 ∗ 𝑒 0 = 1 𝜕2 𝜕𝑥 2 𝜕2 𝜕𝑦 2 =0 = 𝑥𝑒 𝑦 ➔ at (𝟏, 𝟎) , 1 ∗ 𝑒 0 = 1 𝜕2 𝜕𝑥𝜕𝑦 = 𝑒 𝑦 ➔ at (𝟏, 𝟎) , 𝑒 0 = 1 𝜕 [(𝑥 − 𝑎) 1 2! [(𝑥 − 𝑎) [ (𝑦 − 0) 𝜕𝑥 𝜕 𝜕𝑥 + (𝑦 − 𝑏) + (𝑦 − 𝑏 ) 𝜕 1 𝜕𝑦 ] 𝑓(1,0) = [(𝑥 − 1) ∗ 1 + (𝑦 − 0) ∗ 1]1 = 𝑥 − 1 + 𝑦 𝜕 2 𝜕𝑦 ] 𝑓(1,0) = 2! ([(𝑥 − 1) 𝜕𝑦 = = Then: 𝑓(1,0) + [(𝑥 − 𝑎) 𝜕 𝜕𝑥 1 2! 1 2! 1 2! 1 2! (( 𝑥 − 1 ) 2 𝜕2 𝜕𝑥 2 𝜕𝑥 ] + 2 [((𝑥 − 1) ) (( 𝑦 − 0 ) 𝜕 𝜕𝑦 + (𝑦 − 𝑏 ) 1 2! Choice B 𝟐! 𝜕𝑥𝜕𝑦 ] + 𝑦2 𝜕2 𝜕𝑦 2 ) (2𝑥𝑦 − 2𝑦 + 𝑦 2 ) 𝜕 1 𝜕𝑦 (𝟐𝒙𝒚 − 𝟐𝒚 + 𝒚𝟐 ) 1 𝜕 2! 𝜕𝑥 ] 𝑓 (1,0) + [(𝑥 − 𝑎) = 1 + 𝑥 − 1 + 𝑦 + (2𝑥𝑦 − 2𝑦 + 𝑦 2 ) 𝟏 𝜕2 (0 + 2[(𝑥 − 1)𝑦] + 𝑦 2 ) ] 𝑓(1,0) =𝒙+𝒚+ + 2 [ (𝑥 − 1)𝑦 ((𝑥 − 1)2 ∗ 0 + 2[(𝑥 − 1)𝑦 ∗ 1] + 𝑦 2 ∗ 1) 𝜕 2 Page | 90 𝜕 𝜕𝑥 ] ) = 𝜕𝑦 𝜕 2 𝜕 2 = 𝑏) 1 + (𝑦 − )] + 97] Obtain Taylor’s expansion of 𝒇(𝒙, 𝒚) = 𝒆𝒙 𝐜𝐨𝐬 𝒚 at point (𝟎, 𝟎) for 𝒏 = 𝟑 𝟏 𝟏 A- 𝒙𝟐 + 𝟐! (𝒙𝟐 − 𝒚𝟐 ) + 𝟑! [𝒆𝜽(𝒙) ((𝒄𝒐𝒔 𝜽(𝒚) (𝒙𝟑 − 𝟑𝒙𝒚𝟐 )) + (𝒔𝒊𝒏 𝜽(𝒚) (𝒚𝟑 − 𝟑𝒙𝟐 𝒚)))] 𝟏 B- 𝒙 + 𝟐! (𝒙𝟐 − 𝒚𝟐 ) + [𝒆𝜽(𝒙) ((𝒄𝒐𝒔 𝜽(𝒚) (𝒙𝟑 − 𝟑𝒙𝒚𝟐 )) + (𝒔𝒊𝒏 𝜽(𝒚) (𝒚𝟑 − 𝟑𝒙𝟐 𝒚)))] 𝟏 𝟏 C- 𝟏 + 𝟐! (𝒙𝟐 − 𝒚𝟐 ) + 𝟑! [𝒆𝜽(𝒙) ((𝒄𝒐𝒔 𝜽(𝒚) (𝒙𝟑 − 𝟑𝒙𝒚𝟐 )) + (𝒔𝒊𝒏 𝜽(𝒚) (𝒚𝟑 − 𝟑𝒙𝟐 𝒚)))] D- None of these Solution: 𝑓(𝑥, 𝑦) = 𝑓(0,0) + [(𝑥 − 𝑎) 𝜕 𝜕𝑥 𝑅3 𝒇(𝟎, 𝟎) = 𝑒 0 cos 0 = 1 𝜕 𝜕𝑥 = 𝑒 𝑥 cos 𝑦 At (𝟎, 𝟎) ➔ 𝑒 0 cos 0 = 1 𝜕 𝜕𝑦 = −𝑒 𝑥 sin 𝑦 At (𝟎, 𝟎) ➔ −𝑒 0 sin 0 = 0 𝜕2 𝜕𝑥 2 = 𝑒 𝑥 cos 𝑦 At (𝟎, 𝟎) ➔ 𝑒 0 cos 0 = 1 𝜕2 𝜕𝑦 2 = −𝑒 𝑥 cos 𝑦 Page | 91 + (𝑦 − 𝑏) 𝜕 1 1 𝜕 𝜕 2 ] 𝑓(0,0) + 2! [(𝑥 − 𝑎) 𝜕𝑥 + (𝑦 − 𝑏) 𝜕𝑦] 𝑓(0,0) + 𝜕𝑦 At (𝟎, 𝟎) ➔ −𝑒 0 cos 0 = −1 𝜕2 𝜕𝑥𝜕𝑦 = −𝑒 𝑥 sin 𝑦 At (𝟎, 𝟎) ➔ −𝑒 0 sin 0 = 0 𝜕 [(𝑥 − 𝑎) 1 2! [(𝑥 − 𝑎) [ (𝑦 − 0) 𝜕𝑥 𝜕 + (𝑦 − 𝑏) + (𝑦 − 𝑏 ) 𝜕𝑥 𝜕 1 𝜕𝑦 ] 𝑓(0,0) = [(𝑥 − 0) ∗ 1 + (𝑦 − 0) ∗ 0]1 = 𝒙 𝜕 2 𝜕𝑦 ] 𝑓(1,1) = 𝜕𝑦 = Then: 𝑹𝟑 = = 𝜕 2 𝜕 𝜕𝑥 𝜕𝑥 ] + 2 [((𝑥 − 0) ) (( 𝑦 − 0 ) 𝜕 𝜕𝑦 )] + ] ) = = 2! ([(𝑥 − 0) 𝜕 2 = = 1 1 3! 1 3! 1 3! [(𝑥) [(𝑥) [(𝑥 − 0) 𝜕 𝜕𝑥 𝜕 𝜕𝑥 + (𝑦) + (𝑦) 𝜕 𝜕𝑥 2! 1 2! 𝟏 𝟐! (( 𝑥 ) 2 𝜕2 𝜕2 𝜕𝑥 𝜕𝑥𝜕𝑦 + 2 [(𝑥)(𝑦) 2 ] + (𝑦 )2 𝜕2 𝜕𝑦 2 ) ((𝑥)2 ∗ 1 + 2[(𝑥)(𝑦) ∗ 0] + (𝑦)2 ∗ −1) (𝒙𝟐 − 𝒚𝟐 ) + (𝑦 − 0) 𝜕 3 𝜕𝑦 ] 𝑓 ((0 + 𝜃(𝑥 − 0)), (0 + 𝜃(𝑦 − 0))) 𝜕 3 𝜕𝑦 ] 𝑓(𝜃(𝑥), 𝜃(𝑦)) 𝜕 2 𝜕𝑦 ] [(𝑥) 𝜕 2 1 1 𝜕 𝜕𝑥 + (𝑦) 𝜕 𝜕 𝜕𝑦 ] 𝜕 2 𝜕 𝜕 𝜕 ([(𝑥) 𝜕𝑥] + 2 [((𝑥) 𝜕𝑥) ((𝑦) 𝜕𝑦)] + [(𝑦) 𝜕𝑦] ) ∗ [(𝑥) 𝜕𝑥 + (𝑦) 𝜕𝑦] 3! = 𝑥2 1 3! 𝜕2 𝜕𝑥 2 ∗𝑥 𝜕 𝜕𝑥 + 𝑥2 𝜕2 𝜕𝑥 2 ∗𝑦 𝜕 𝜕𝑦 + 2 [((𝑥) +𝑦 2 [ = 𝑥3 Page | 92 𝜕3 𝜕3 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 + 𝑥 2𝑦 3 + 2(𝑥 2 𝑦) 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 ) ((𝑦) 𝜕𝑦)] ∗ 𝑥 𝜕𝑥 + 2 [((𝑥) 𝜕𝑥) ((𝑦) 𝜕𝑦)] ∗ 𝑦 𝜕𝑦 𝜕𝑥 𝜕2 𝜕 𝜕𝑦 𝜕𝑥 2 ∗𝑥 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 + 𝑦2 𝜕2 𝜕𝑦 2 + 2(𝑥𝑦 2 ) ∗𝑦 𝜕3 𝜕𝑥𝜕𝑦 𝜕 ] 𝜕𝑦 + 𝑦2𝑥 2 𝜕3 𝜕𝑦 2 𝜕𝑥 + 𝑦3 𝜕3 𝜕𝑦 3 𝜕3 𝜕𝑥 3 = 𝑒 𝑥 cos 𝑦 , 𝜕3 𝜕𝑦 2 𝜕𝑥 3 3 𝜕 𝑥 𝜕𝑥 3 𝜕3 𝜕𝑥 2 𝜕𝑦 = −𝑒 𝑥 cos 𝑦 , 𝜕3 2 +𝑥 𝑦 𝜕𝑥 2 𝜕𝑦 = −𝑒 𝑥 sin 𝑦 , 𝜕3 𝜕𝑦 3 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 = −𝑒 𝑥 sin 𝑦 , 𝜕3 𝜕𝑥𝜕𝑦 2 = −𝑒 𝑥 cos 𝑦 = 𝑒 𝑥 sin 𝑦 + 2( 𝑥 2 𝑦 ) 𝜕3 𝜕𝑥𝜕𝑦𝜕𝑥 + 2(𝑥𝑦 2 ) 𝜕3 𝜕𝑥𝜕𝑦 2 𝜕3 2 +𝑦 𝑥 𝜕𝑦 2 𝜕𝑥 + 3 3 𝜕 𝑦 𝜕𝑦 3 = 𝑥 3 ∗ 𝑒 𝑥 cos 𝑦 + 𝑥 2 𝑦 ∗ (−𝑒 𝑥 sin 𝑦) + 2(𝑥 2 𝑦) ∗ (−𝑒 𝑥 sin 𝑦) +2(𝑥𝑦 2 ) ∗ (−𝑒 𝑥 cos 𝑦) + 𝑦 2 𝑥 ∗ (−𝑒 𝑥 cos 𝑦) + 𝑦 3 ∗ 𝑒 𝑥 sin 𝑦 = 𝑥 3 𝑒 𝑥 cos 𝑦 − 3𝑥 2 𝑦𝑒 𝑥 sin 𝑦 − 3𝑥𝑦 2 𝑒 𝑥 cos 𝑦 + 𝑦 3 𝑒 𝑥 sin 𝑦 = 𝑒 𝑥 cos 𝑦 (𝑥 3 − 3𝑥𝑦 2 ) + 𝑒 𝑥 sin 𝑦 (𝑦 3 − 3𝑥 2 𝑦) At ((𝟎 + 𝜽(𝒙 − 𝟎)), (𝟎 + 𝜽(𝒚 − 𝟎))) = (𝜽(𝒙), 𝜽(𝒚)) Substitute in 𝑒 𝑥 cos 𝑦 & 𝑒 𝑥 sin 𝑦 Only ➔ 𝑒 𝜃(𝑥) cos 𝜃 (𝑦) (𝑥 3 − 3𝑥𝑦 2 ) + 𝑒 𝜃(𝑥) sin 𝜃 (𝑦) (𝑦 3 − 3𝑥 2 𝑦) Then: 𝑅3 = 1 3! [𝑒 𝜃(𝑥) ((cos 𝜃 (𝑦) (𝑥 3 − 3𝑥𝑦 2 )) + (sin 𝜃 (𝑦) (𝑦 3 − 3𝑥 2 𝑦)))] 𝜕 𝜕 1 1 𝜕 Then: 𝑓(0,0) + [(𝑥 − 𝑎) 𝜕𝑥 + (𝑦 − 𝑏) 𝜕𝑦] 𝑓(0,0) + 2! [(𝑥 − 𝑎) 𝜕𝑥 + (𝑦 − 𝑏) 𝜕 2 ] 𝑓(0,0) + 𝜕𝑦 =𝟏+ 𝟏 𝟐! (𝒙𝟐 − 𝒚𝟐 ) + 𝟑𝒙𝟐 𝒚)))] Choice C Page | 93 𝑅3 𝟏 𝟑! [𝒆𝜽(𝒙) ((𝒄𝒐𝒔 𝜽(𝒚) (𝒙𝟑 − 𝟑𝒙𝒚𝟐 )) + (𝒔𝒊𝒏 𝜽(𝒚) (𝒚𝟑 − ∞ 98] Determine ∫𝒂 𝒆−𝒙 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at 𝒆−𝒂 B- Diverged at ∞ D- Converged at 𝒆𝒂 C- Diverged at −∞ Solution: ∞ 𝑡 ∫𝑎 𝑒 −𝑥 𝑑𝑥 = lim ∫𝑎 𝑒 −𝑥 𝑑𝑥 = lim [−𝑒 −𝑥 ]𝑡𝑎 = lim [(−𝑒 −𝑡 ) − (−𝑒 −𝑎 )] 𝑡→∞ 𝑡→∞ 𝑡→∞ = lim [(−𝑒 −𝑡 ) + 𝑒 −𝑎 ] = (−𝑒 −∞ ) + 𝑒 −𝑎 = − 𝑡→∞ 1 𝑒∞ + 𝑒 −𝑎 = 0 + 𝑒 −𝑎 = 𝑒 −𝑎 Since the result is a number not ∞ or −∞ , then it converged Then: this improper integral Converged ∞ 𝟏 99] Determine ∫𝟒 𝒙−𝟐 Choice A 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged B- Diverged at ∞ C- Diverged at −∞ D- None of these Solution: ∞ 1 ∫4 𝑥−2 𝑡 1 𝑑𝑥 = lim ∫4 𝑡→∞ 𝑥−2 𝑑𝑥 = lim [ln|𝑥 − 2|]𝑡4 = lim [(ln|𝑡 − 2|) − (ln|4 − 2|)] 𝑡→∞ 𝑡→∞ = lim [(ln|𝑡 − 2|) − (ln|2|)] 𝑡→∞ = [(ln|∞ − 2|) − (ln|2|)] = [(ln|∞|) − (ln|2|)] = [∞ − (ln|2|)] = ∞ Then: this improper integral Diverged Page | 94 Choice B ∞ 𝟏 100] Determine ∫−∞ 𝟏+𝒙𝟐 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at 𝝅 B- Diverged at ∞ C- Diverged at −∞ D- Converged at 𝟐 𝝅 Solution: ∞ 𝑐 1 −1 lim [tan 𝑡→∞ 𝑥 ]𝑡𝑐 𝑡 1 ∫−∞ 1+𝑥 2 𝑑𝑥 = lim ∫𝑡 1+𝑥 𝑡→−∞ 2 𝑑𝑥 + lim ∫𝑐 𝑡→∞ 1 1+𝑥 2 𝑑𝑥 = lim [tan−1 𝑥 ]𝑐𝑡 + 𝑡→−∞ = lim [tan−1 𝑐 − tan−1 𝑡] + lim [tan−1 𝑡 − tan−1 𝑐 ] 𝑡→−∞ 𝑡→∞ Let 𝑐 = 0 or any number in the Domain of tan ➔ lim [tan−1 0 − tan−1 𝑡] + lim [tan−1 𝑡 − tan−1 0] 𝑡→−∞ 𝑡→∞ = lim [0 − tan−1 𝑡] + lim [tan−1 𝑡 − 0] = − tan−1 −∞ + tan−1 ∞ 𝑡→−∞ 𝑡→∞ 𝜋 𝜋 = − (− ) + = 𝜋 2 2 Then: this improper integral Converged 𝟖 101] Determine ∫𝟎 𝟑𝒙 𝟑 √𝟔𝟒−𝒙𝟐 Choice A 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at 𝟑𝟔 B- Diverged at ∞ C- Diverged at −∞ D- Converged at −𝟑𝟔 Solution: 8 ∫0 3𝑥 3 √64−𝑥 𝑡 3 lim− ∫0 𝑡→8 Page | 95 𝑡 𝑑𝑥 = lim− ∫0 2 𝑥 𝑡→8 1 (64−𝑥 2 )3 𝑑𝑥 3𝑥 3 √64−𝑥 𝑡 𝑑𝑥 = 3 ∗ lim− ∫0 2 𝑡→8 𝑥 3 √64−𝑥 2 𝑑𝑥 = = 𝑡 3 lim− ∫0 𝑥 𝑡→8 ∗ (64 − 𝑥 1 2 )−3 𝑑𝑥 = 3 ∗ 1 −3 𝑡 3 = − lim− ∫0 −2𝑥 (64 − 𝑥 2 ) 2 𝑡→8 3 3 = − lim− [(64 − 𝑥 2 𝑡→8 2 9 = − lim− [(64 − 𝑡 4 𝑡→8 2 2 )3 2 2 )3 3 𝑑𝑥 = − lim− [ ∗ −2 ∗ (64 − 𝑥 1 − +1 2 (64−𝑥 ) 3 2 3 2 𝑡→8 𝑡 9 ] = − lim− [(64 − 𝑥 4 𝑡→8 0 − (64 − 0 2 9 𝑡 − lim− ∫0 𝑥 2 𝑡→8 1 2 2 2 )3 2 2 )3 1 2 )−3 𝑑𝑥 𝑡 ] 0 𝑡 ] 0 ] 2 9 3 = − [(64 − 82 )3 − (64)3 ] = − [(0)3 − √(64)2 ] 4 4 9 9 3 = − [− √4096 ] = − ∗ −16 = +36 4 4 Then: this improper integral Converged Choice A 𝟑 𝟏 102] Determine ∫−𝟏 𝒙𝟐 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged B- Diverged C- Diverged at −∞ D- None of these Solution: 3 1 0 1 3 1 ∫−1 𝑥 2 𝑑𝑥 = ∫−1 𝑥 2 𝑑𝑥 + ∫0 𝑥 𝑡 2 𝑑𝑥 = lim− ∫−1 𝑡→0 3 1 1 𝑥 2 𝑑𝑥 + lim+ ∫𝑡 𝑡→0 𝑡 3 𝑥2 𝑑𝑥 = lim− ∫−1(𝑥 )−2 𝑑𝑥 + lim+ ∫𝑡 (𝑥 )−2 𝑑𝑥 𝑡→0 𝑡→0 1 𝑡 1 3 = lim− [− ] 𝑡→0 1 1 𝑥 −1 1 + lim+ [− ] 𝑥 𝑡 𝑡→0 1 = lim− [(− ) − (− )] + lim+ [(− ) − (− )] 𝑡 −1 3 𝑡 𝑡→0 Page | 96 𝑡→0 1 1 1 = lim− [(− ) − 1] + lim+ [(− ) + ] 𝑡 3 𝑡 𝑡→0 𝑡→0 1 1 1 = lim− (− ) − lim−(1) + lim+ (− ) + lim+ ( ) 𝑡 3 𝑡 𝑡→0 𝑡→0 1 𝑡→0 1 𝑡→0 1 1 1 1 = lim− (− ) − 1 − + lim+ ( ) = lim− (− ) − 1 − + lim+ ( ) 𝑡 3 𝑡 0 3 0 𝑡→0 𝑡→0 𝑡→0 𝑡→0 1 = −∞ − 1 − + ∞ 3 Then: this improper integral Diverged 𝟏 103] Determine ∫𝟎 𝟏 √𝟏−𝒙𝟐 Choice B 𝒅𝒙 converges or diverges , if converges Find its value . 𝝅 A- Converged at 𝟐 B- Diverged C- Diverged at −∞ D- Converged at 𝝅 Solution: 1 ∫0 1 √1−𝑥 2 𝑡 𝑑𝑥 = lim− ∫0 𝑡→1 1 √1−𝑥 2 𝑑𝑥 = lim−[sin−1 𝑥 ]𝑡0 = lim−[sin−1 𝑡 − sin−1 0] 𝑡→1 𝑡→1 𝜋 𝜋 2 2 = [sin−1 1 − sin−1 0] = − 0 = Then: this improper integral Converged 𝟐 104] Determine ∫𝟏 𝟏 𝒙 𝐥𝐧 𝒙 Choice A 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged B- Diverged at ∞ C- Diverged at −∞ D- None of these Page | 97 Solution: 2 1 ∫1 𝑥 ln 𝑥 𝑑𝑥 = 2 1 lim+ ∫𝑡 𝑑𝑥 𝑥 ln 𝑥 𝑡→1 1 = 2 lim+ ∫𝑡 𝑥 ln 𝑥 𝑡→1 𝑑𝑥 = lim+[ln|ln|𝑥 ||]2𝑡 𝑡→1 = lim+[(ln|ln|2||) − (ln|ln|𝑡||)] 𝑡→1 = (ln|ln|2||) − (ln|ln|1||) = ln|ln|2|| − ln|0| = ln|ln|2|| − (−∞) = ∞ Then: this improper integral Diverged Choice B 𝟐𝟕 𝟏 105] Determine ∫𝟎 𝟑 √𝒙𝟐 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at −𝟗 B- Diverged at ∞ C- Diverged at −∞ D- Converged at 𝟗 Solution: 27 1 ∫0 3 √𝑥 27 1 𝑑𝑥 = ∫0 2 27 1 ∫0 2 𝑑𝑥 𝑥3 = 2 𝑑𝑥 𝑥3 27 1 𝑙𝑖𝑚+ ∫𝑡 2 𝑑𝑥 𝑡→0 𝑥3 1 = 27 −2 𝑙𝑖𝑚 ∫ 𝑥 3 𝑑𝑥 𝑡→0+ 𝑡 2 − +1 𝑥 3 = 𝑙𝑖𝑚+ [ 𝑡→0 2 −3+1 1 27 ] = 𝑙𝑖𝑚+ [3𝑥 ] 𝑡 1 = 𝑙𝑖𝑚+ [(3 ∗ 273 ) − (3 ∗ 𝑡 3 )] = 𝑙𝑖𝑚+ [(3 ∗ 3) − (3 ∗ 𝑡 3 )] 𝑡→0 𝑡→0 1 3 1 3 = 𝑙𝑖𝑚+ [9 − (3 ∗ 𝑡 )] = 9 − (3 ∗ 0 ) = 9 𝑡→0 Then: this improper integral Converged Page | 98 Choice D 1 3 𝑡→0 27 𝑡 𝟒 𝟏 106] Determine ∫𝟎 √𝒙 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at 𝟎 B- Diverged at ∞ C- Converged at 𝟒 D- Converged at −𝟒 Solution: 4 1 ∫0 𝑥 𝑑𝑥 √ = 4 1 𝑙𝑖𝑚+ ∫𝑡 𝑑𝑥 √𝑥 𝑡→0 4 1 = 𝑙𝑖𝑚+ [ 𝑡→0 − +1 𝑥 2 1 −2+1 = 4 1 𝑙𝑖𝑚+ ∫𝑡 1 𝑑𝑥 𝑡→0 𝑥2 1 ] = 𝑙𝑖𝑚+ [ 𝑡→0 𝑡 𝑥2 1 2 = 4 −1 𝑙𝑖𝑚 ∫ 𝑥 2 𝑑𝑥 𝑡→0+ 𝑡 1 − +1 𝑥 2 = 𝑙𝑖𝑚+ [ 𝑡→0 1 2 − +1 4 ] 𝑡 4 4 ] = 𝑙𝑖𝑚+[2√𝑥]𝑡 = 𝑙𝑖𝑚+[(2√4) − (2√𝑡)] 𝑡→0 𝑡 𝑡→0 = 𝑙𝑖𝑚+[4 − (2√𝑡)] = 4 − (2√0) = 4 𝑡→0 Then: this improper integral Converged Choice C 𝟗 107] Determine ∫𝟎 𝟏 √𝟗−𝒙 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at −𝟔 B- Diverged at ∞ C- Converged at 𝟔 D- Converged at 𝟒 Solution: 9 1 ∫0 √9−𝑥 𝑑𝑥 = 𝑡 1 𝑙𝑖𝑚− ∫0 𝑑𝑥 √9−𝑥 𝑡→9 1 − +1 (9−𝑥) 2 1 𝑡→9− (−2+1)∗(−1) = 𝑙𝑖𝑚 [ = 𝑡 𝑡 𝑙𝑖𝑚− ∫0 𝑡→9 1 1 1 (9−𝑥)2 1 −2 𝑡→9− ] = 𝑙𝑖𝑚 [ 0 1 2 𝑑𝑥 = (9−𝑥)2 1 2 𝑡 𝑙𝑖𝑚− ∫0 (9 𝑡→9 − 𝑥) 𝑡 Page | 99 𝑑𝑥 1 2 ] = −2 𝑙𝑖𝑚− [(9 − 𝑥 ) ] 0 𝑡→9 1 2 1 2 = −2 𝑙𝑖𝑚− [(9 − 𝑡) − (9 − 0) ] = −2 [(9 − 9) − (9) ] 𝑡→9 1 −2 𝑡 0 1 1 = −2 [(0)2 − (9)2 ] = −2[−√9] = −2[−3] = 6 Then: this improper integral Converged Choice C 𝟒 𝟒𝒙 108] Determine ∫𝟏 𝒙𝟐 −𝟏 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at −𝟔 B- Diverged at ∞ C- Converged at 𝟔 D- Converged at 𝟒 Solution: 4 4𝑥 ∫1 4 4𝑥 𝑥 2 −1 𝑑𝑥 = 𝑙𝑖𝑚+ ∫𝑡 𝑡→1 𝑥 2 −1 𝑑𝑥 Let 𝑥 2 − 1 = 𝑢 , 2𝑥 𝑑𝑥 = 𝑑𝑢 ➔ 𝑑𝑥 = 4 4𝑥 𝑙𝑖𝑚+ ∫𝑡 𝑡→1 𝑥 2 −1 4 4𝑥 𝑑𝑥 = 𝑙𝑖𝑚+ ∫𝑡 𝑡→1 𝑢 ∗ 𝑑𝑢 2𝑥 𝑑𝑢 2𝑥 42 = 𝑙𝑖𝑚+ ∫𝑡 𝑡→1 𝑢 𝑑𝑢 = 2 ∗ 𝑙𝑖𝑚+[ln|𝑢|]4𝑡 𝑡→1 = 2 𝑙𝑖𝑚+[ln|𝑥 2 − 1|]4𝑡 = 2 𝑙𝑖𝑚+[ln|42 − 1| − ln|𝑡 2 − 1|] 𝑡→1 𝑡→1 = 2 𝑙𝑖𝑚+[ln|15| − ln|𝑡 2 − 1|] = 2[ln|15| − ln|12 − 1|] 𝑡→1 = 2[ln|15| − ln|0|] = 2 ln|15| − 2 ln|0| = 2 ln|15| − (−∞) = ∞ Then: this improper integral Diverged Choice B ∞ 109] Determine ∫𝟎 𝒙 𝒆−𝒙 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at −𝟏 B- Diverged at −∞ C- Converged at 𝟏 D- Converged at 𝟎 Page | 100 Solution: ∞ 𝑡 ∫0 𝑥 𝑒 −𝑥 𝑑𝑥 = 𝑙𝑖𝑚− ∫0 𝑥 𝑒 −𝑥 𝑑𝑥 𝑡→∞ = 𝑙𝑖𝑚−[𝑥 ∗ (−𝑒 −𝑥 ) − 𝑒 −𝑥 ]𝑡0 𝑡→∞ 𝐈 𝑥 𝑒 −𝑥 1 = 𝑙𝑖𝑚−[−𝑥𝑒 −𝑥 − 𝑒 −𝑥 ]𝑡0 = 𝑙𝑖𝑚−[−𝑒 −𝑥 (𝑥 + 1)]𝑡0 𝑡→∞ 𝐃 𝑡→∞ + − −𝑒 −𝑥 𝑒 −𝑥 0 = 𝑙𝑖𝑚−[(−𝑒−𝑡 (𝑡 + 1)) − (−𝑒−0 (0 + 1))] 𝑡→∞ = 𝑙𝑖𝑚−[(−𝑒−𝑡 (𝑡 + 1)) − (−1)] = 𝑙𝑖𝑚−[(−𝑒−𝑡 (𝑡 + 1)) + 1] 𝑡→∞ 𝑡→∞ = 𝑙𝑖𝑚−(−𝑒−𝑡 (𝑡 + 1)) + 𝑙𝑖𝑚− 1 = 𝑙𝑖𝑚− (− 𝑡→∞ 𝑡→∞ 𝑡+1 = − 𝑙𝑖𝑚− ( 𝑡→∞ 𝑒𝑡 𝑡→∞ 𝑡+1 𝑒𝑡 )+1 )+1 ∞+1 By Substitution in limits ➔ ( 𝑒∞ ∞ )=∞ Then We will Use L’hopital Rule 𝑡+1 − 𝑙𝑖𝑚− ( 𝑡→∞ 1+0 ) + 1 = − 𝑙𝑖𝑚− ( 𝑒𝑡 𝑡→∞ 𝑒𝑡 1 ) + 1 = − 𝑙𝑖𝑚− (𝑒∞) + 1 = −0 + 1 = 1 𝑡→∞ Then: this improper integral Converged Choice C ∞ 110] Determine ∫𝟏 𝟏 (𝟒𝒙+𝟏)𝟑 𝒅𝒙 converges or diverges , if converges Find its value . 𝟏 A- Converged at 𝟐𝟎𝟎 B- Diverged at −∞ C- Converged at −𝟓 ∗ 𝟏𝟎−𝟑 D- Converged at 𝟐𝟎𝟎 Solution: ∞ ∫1 1 𝑑𝑥 (4𝑥+1)3 𝑡 𝑡 1 = 𝑙𝑖𝑚− ∫1 (4𝑥+1)3 𝑑𝑥 = 𝑙𝑖𝑚− ∫1 (4𝑥 + 1)−3 𝑑𝑥 𝑡→∞ (4𝑥+1)−3+1 𝑡→∞ 𝑡 = 𝑙𝑖𝑚− [ (−3+1)∗(4) ] = 𝑙𝑖𝑚− [ 𝑡→∞ Page | 101 1 𝑡→∞ (4𝑥+1)−2 −8 𝑡 ] = 𝑙𝑖𝑚− [− 1 𝑡→∞ 1 𝑡 ] 8(4𝑥+1)2 1 = 𝑙𝑖𝑚− [(− 𝑡→∞ 1 1 8(4𝑡+1)2 = 𝑙𝑖𝑚− [(− 𝑡→∞ 1 1 ) − (− 8(4∗1+1)2)] = 𝑙𝑖𝑚− [(− 8(4𝑡+1)2 ) − (− 8∗25)] 𝑡→∞ 1 1 1 1 ) + 200] = 𝑙𝑖𝑚− [(− 8(4∗∞+1)2 ) + 200] 8(4𝑡+1)2 𝑡→∞ 1 1 1 1 1 1 = 𝑙𝑖𝑚− [(− ) + ] = 𝑙𝑖𝑚− (− ) + 𝑙𝑖𝑚− =0+ = ∞ 200 ∞ 200 200 200 𝑡→∞ 𝑡→∞ 𝑡→∞ Then: this improper integral Converged Choice A 𝟐 111] Determine ∫−𝟐 𝟏 √𝟐+𝒙 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at 𝟎 B- Diverged at −∞ C- Converged at −𝟒 D- Converged at 𝟒 Solution: 2 1 ∫−2 √2+𝑥 𝑑𝑥 = 2 1 𝑙𝑖𝑚+ ∫𝑡 𝑑𝑥 √2+𝑥 𝑡→−2 1 (2+𝑥)−2+1 = 𝑙𝑖𝑚+ [ 𝑡→−2 1 −2+1 = 2 𝑙𝑖𝑚+ ∫𝑡 𝑡→−2 2 1 (2+𝑥)2 𝑡→−2 1 2 1 2 𝑑𝑥 = (2+𝑥)2 1 ] = 𝑙𝑖𝑚+ [ 𝑡 1 2 𝑙𝑖𝑚+ ∫𝑡 (2 𝑡→−2 2 + 𝑥) 1 2 ] = 2 𝑙𝑖𝑚+ [(2 + 𝑥 ) ] 𝑡→−2 𝑡 1 2 1 −2 𝑑𝑥 2 𝑡 = 2 𝑙𝑖𝑚+ [((2 + 2) ) − ((2 + 𝑡) )] = 2 𝑙𝑖𝑚+[(√4) − (√2 + 𝑡)] 𝑡→−2 𝑡→−2 = 2[(√4) − (√2 − 2)] = 2[(2) − (0)] = 4 Then: this improper integral Converged Choice D Page | 102 𝟎 112] Determine ∫−𝝅 𝐬𝐞𝐜 𝟐 𝒙 𝒅𝒙 converges or diverges , if converges 𝟐 Find its value . A- Converged at ∞ B- Diverged at −∞ 𝝅 C- Converged at − 𝟐 D- Diverged at ∞ Solution: 0 ∫−𝜋 sec 2 𝑥 𝑑𝑥 2 0 0 1 𝜋 1 1 1 Since ∫−𝜋 sec 2 𝑥 𝑑𝑥 = ∫−𝜋 2 𝑑𝑥 then: at 𝑥 = − ➔ 2 = 2 𝜋 = cos 𝑥 2 cos 𝑥 0 cos − 2 2 2 0 0 Then: ∫−𝜋 sec 2 𝑥 𝑑𝑥 = 𝑙𝑖𝑚+ ∫𝑡 sec 2 𝑥 𝑑𝑥 = 𝑙𝑖𝑚+[tan 𝑥 ]0𝑡 𝜋 2 𝜋 𝑡→− 2 𝑡→− 2 = 𝑙𝑖𝑚+[(tan 0) − (tan 𝑡)] = 𝑙𝑖𝑚+[0 − (tan 𝑡)] 𝜋 𝜋 𝑡→− 2 𝑡→− 2 = 𝑙𝑖𝑚+ 0 − 𝑙𝑖𝑚+ tan 𝑡 = 0 − 𝑙𝑖𝑚+ tan − 𝜋 𝑡→− 2 𝜋 1 2 0 𝜋 𝑡→− 2 Since tan − = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝜋 Then: 𝑙𝑖𝑚+ tan − = ∞ 𝜋 𝑡→− 2 2 𝜋 ➔ 0 − 𝑙𝑖𝑚+ tan − = 0 − ∞ = −∞ 𝜋 𝑡→− 2 2 Then: this improper integral Diverged Choice B Page | 103 𝜋 𝑡→− 2 𝜋 2 ∞ 𝟏 113] Determine ∫𝟎 𝟒+𝒙𝟐 𝒅𝒙 converges or diverges , if converges Find its value . 𝝅 A- Converged at 𝟐 C- Converged at B- Diverged at −∞ 𝟐𝝅 D- None of them 𝟒 Solution: ∞ ∫0 1 4+𝑥 2 ∞ ∫0 1 4+𝑥 2 𝑑𝑥 at 𝑥 = ∞ ➔ 𝑡 1 4+∞2 = 1 ∞ 1 𝑑𝑥 = 𝑙𝑖𝑚− ∫0 𝑑𝑥 4+𝑥 2 𝑡→∞ 1 𝑥 1 ∫ 4+𝑥 2 𝑑𝑥 ➔ Suppose 𝑢 = 2 ➔ 𝑥 = 2𝑢 , 𝑑𝑢 = 2 𝑑𝑥 ➔ 2𝑑𝑢 = 𝑑𝑥 1 1 2 1 1 ∫ 4+𝑥 2 𝑑𝑥 = ∫ 4+(2𝑢)2 ∗ 2𝑑𝑢 = ∫ 4+4𝑢2 𝑑𝑢 = 2 ∫ 4+4𝑢2 𝑑𝑢 = 2 ∫ 4(1+𝑢2 ) 𝑑𝑢 = 2∫ 1 2 1 1 1 1 −1 𝑑𝑢 = 𝑑𝑢 = 𝑑𝑢 = tan 𝑢 ∫ ∫ 2 2 2 4(1+𝑢 ) 4 1+𝑢 2 1+𝑢 2 1 𝑥 = tan−1 ( ) 2 2 𝑡 1 𝑙𝑖𝑚− ∫0 𝑑𝑥 4+𝑥 2 𝑡→∞ = = 1 𝑥 tan−1 (2) = 𝑙𝑖𝑚− [ 𝑡→∞ ] = 2 0 𝑡 𝑥 1 𝑥 𝑙𝑖𝑚− [tan−1 ( )] 2 2 𝑡→∞ 1 𝑡 0 𝑡→∞ ∞ 2 𝑡→∞ 1 𝑡→∞ 1 𝜋 𝜋 2 2 4 Then: this improper integral Converged Page | 104 0 𝑙𝑖𝑚− [(tan−1 ( )) − (0)] = 𝑙𝑖𝑚−(tan−1 (∞)) 2 2 = ∗ = Choice D 𝑡 0 𝑙𝑖𝑚− [tan−1 ( )] = 𝑙𝑖𝑚− [(tan−1 ( )) − (tan−1 ( ))] 2 2 2 2 2 𝑡→∞ 1 𝑡 ∞ 𝒍𝒏 𝒙 114] Determine ∫𝟏 𝒅𝒙 converges or diverges , if converges Find 𝒙𝟑 its value . 𝟏 A- Converged at 𝟐 B- Diverged at ∞ 𝟏 C- Converged at 𝟒 D- Converged at 𝟎 Solution: ∞ 𝑙𝑛 𝑥 ∫1 𝑥3 ∞ 𝑙𝑛 𝑥 ∫1 ∫ 𝑥3 𝑙𝑛 𝑥 𝑥3 𝑑𝑥 at 𝑥 = ∞ ➔ 𝑙𝑛 ∞ 𝑡 𝑙𝑛 𝑥 𝑑𝑥 = 𝑙𝑖𝑚− ∫1 = ∫ 𝑙𝑛 𝑥 ∗ 𝑥 𝟏 −3 ➔ 𝒖 = 𝒍𝒏 𝒙 , 𝒅𝒖 = , 𝒅𝒗 = 𝒙 2 −𝟑 𝒙 𝒖𝒗 − ∫ 𝒗 𝒅𝒖 = 𝑙𝑛 𝑥 ∗ (− 1 ∞ 𝑑𝑥 𝑥3 𝑡→∞ ∞ = ∞3 1 1 2𝑥 2 1 ,𝒗= 𝑙𝑛 𝑥 1 𝒙−𝟐 −𝟐 =− 1 𝟏 𝟐𝒙𝟐 𝑙𝑛 𝑥 ) − ∫ (− 2𝑥 2 ∗ 𝑥) = − 2𝑥 2 + 2 ∫ 𝑥 3 = − 2𝑥 2 + ∫ 𝑥 −3 =− Then: ∫ 𝑙𝑛 𝑥 𝑥3 𝑙𝑛 𝑥 2𝑥 =− 1 2 + ∗ 2 𝑥 −2 −2 𝑙𝑛 𝑥 1 2𝑥 4𝑥 2 𝑡 𝑙𝑛 𝑥 ➔ 𝑙𝑖𝑚− ∫1 3 𝑑𝑥 𝑥 𝑡→∞ 𝑙𝑛 1 1 2 − =− = 𝑙𝑖𝑚− [− 𝑡→∞ 𝑙𝑛 𝑥 2𝑥 2 − 𝑙𝑛 𝑥 1 2𝑥 4𝑥 2 − 1 4𝑥 2 𝑡 2 ] = 𝑙𝑖𝑚− [(− 𝑡→∞ 1 𝑙𝑛 𝑡 1 2𝑡 4𝑡 2 2 − )− (− 2∗12 − 4∗12 )] = 𝑙𝑖𝑚− [(− 𝑡→∞ = 𝑙𝑖𝑚− (− 𝑡→∞ = − 𝑙𝑖𝑚− 𝑡→∞ Since 𝑙𝑖𝑚− 𝑙𝑛 ∞ 𝑡→∞ 2∞2 𝑙𝑛 𝑡 2𝑡 2 ➔ Page | 105 1 𝑡 4𝑡 = 1 4𝑡 2 = ∞ ∞ 𝑙𝑛 𝑡 2𝑡 2 − 𝑙𝑛 𝑡 2𝑡 2 𝑙𝑛 𝑡 2𝑡 2 − 1 1 𝑙𝑛 𝑡 1 4𝑡 2 𝑡→∞ 1 ) + 𝑙𝑖𝑚− 4 − 𝑙𝑖𝑚− 𝑡→∞ 1 4𝑡 2 1 𝑡→∞ 𝑡→∞ 1 4𝑡 2 1 𝑙𝑛 ∞ 𝑡→∞ 4 2∞2 + 𝑙𝑖𝑚− = − 𝑙𝑖𝑚− , then Use L’hopital rule , then 𝑙𝑖𝑚− 1 ) − (0 − 4)] = 𝑙𝑖𝑚− [(− 2𝑡 2 − 4𝑡 2 ) + 4] 4𝑡 2 = 𝑙𝑖𝑚− 1 𝑡→∞ 4∗∞2 =0 𝑡→∞ − 𝑙𝑖𝑚− 𝑡→∞ 1 4∞2 + 1 4 1 𝑙𝑖𝑚− 𝑡→∞ 4∞2 =0 Then: − 𝑙𝑖𝑚− 𝑡→∞ 𝑙𝑛 ∞ 2∞2 − 𝑙𝑖𝑚− 𝑡→∞ 1 4∞2 1 1 1 4 4 4 + =0+0+ = Then: this improper integral Converged Choice C 𝟏 𝟏 115] Determine ∫−𝟑 (𝒙+𝟐)𝟐 𝒅𝒙 converges or diverges , if converges Find its value . 𝟏 A- Converged at 𝟐 B- Diverged 𝟏 C- Converged at 𝟒 D- Converged at 𝟎 Solution: 1 1 1 ∫−3 (𝑥+2)2 𝑑𝑥 , at 𝑥 = −2 ➔ 0 1 −2 1 ∫−3 (𝑥+2)2 𝑑𝑥 = ∫−3 1 1 (𝑥+2)2 −2 1 𝑑𝑥 + ∫−2 (𝑥+2)2 𝑑𝑥 = 𝑙𝑖𝑚− ∫−3 𝑡→−2 −2 1 1 (𝑥+2)2 𝑡→−2 1 = 𝑙𝑖𝑚− ∫−3 (𝑥 + 2)−2 𝑑𝑥 + 𝑙𝑖𝑚+ ∫−2(𝑥 + 2)−2 𝑑𝑥 𝑡→−2 𝑡→−2 = 𝑙𝑖𝑚− [ (𝑥+2)−2+1 𝑡→−2 −2+1 = 𝑙𝑖𝑚− [− 𝑡→−2 1 𝑡→−2 = 𝑙𝑖𝑚− [(− = 𝑙𝑖𝑚− (− 𝑡→−2 = 𝑙𝑖𝑚− (− 𝑡→−2 4 𝑡 𝑥+2 −3 = 𝑙𝑖𝑚− [(− 𝑡→−2 ] 𝑡 ] 1 −3 Page | 106 (𝑥+2)−2+1 −2+1 𝑡→−2 + 𝑙𝑖𝑚+ [− 𝑡→−2 1 ] 1 ] 𝑡 1 𝑥+2 𝑡 1 1 1 𝑡+2 ) − (− −3+2)] + 𝑙𝑖𝑚+ [(− 1+2) − (− 𝑡+2)] 1 1 𝑡→−2 1 ) − (1)] + 𝑙𝑖𝑚+ [(− 3) + 𝑡+2] 𝑡+2 𝑡→−2 1 1 1 ) − 𝑙𝑖𝑚−(1) + 𝑙𝑖𝑚+ (− 3) + 𝑙𝑖𝑚+ 𝑡+2 𝑡+2 𝑡→−2 1 𝑡→−2 1 𝑡→−2 1 1 4 1 ) − 1 − 3 + 𝑙𝑖𝑚+ −2+2 = 𝑙𝑖𝑚− (− 0) − 3 + 𝑙𝑖𝑚+ 0 −2+2 = −∞ − + ∞ ≠ 0 3 + 𝑙𝑖𝑚+ [ 𝑡→−2 𝑡→−2 1 𝑑𝑥 + 𝑙𝑖𝑚+ ∫−2 (𝑥+2)2 𝑑𝑥 𝑡→−2 Then: this improper integral Diverged Choice B 𝟏 116] Determine ∫𝟎 𝒙 𝒍𝒏 𝒙 𝒅𝒙 converges or diverges , if converges Find its value . 𝟕 A- Converged at − 𝟐𝟖 B- Diverged at ∞ 𝟏 𝟏 C- Converged at 𝟒 D- Converged at − 𝟐 Solution: 1 ∫0 𝑥 𝑙𝑛 𝑥 𝑑𝑥 at 𝑥 = 0 ➔ 0 𝑙𝑛 0 = 0 ∗ 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 1 1 ∫0 𝑥 𝑙𝑛 𝑥 𝑑𝑥 = 𝑙𝑖𝑚+ ∫𝑡 𝑥 𝑙𝑛 𝑥 𝑑𝑥 𝑡→0 𝒙𝟐 𝟏 ∫ 𝑥 𝑙𝑛 𝑥 ➔ 𝒖 = 𝒍𝒏 𝒙 , 𝒅𝒖 = 𝒙 , 𝒅𝒗 = 𝒙 , 𝒗 = 𝒖𝒗 − ∫ 𝒗 𝒅𝒖 = 𝑙𝑛 𝑥 ∗ 𝑥2 2 −∫ 𝑥2 2 1 𝑥 2 𝑙𝑛 𝑥 𝑥 2 ∗ = 𝟐 1 − ∫𝑥 = 2 = 1 𝑙𝑖𝑚+ ∫𝑡 𝑥 𝑙𝑛 𝑥 𝑑𝑥 𝑡→0 = 𝑙𝑖𝑚+ [ 𝑡→0 𝑥 2 𝑙𝑛 𝑥 2 − 𝑥2 1 2 2 𝑡→0 1 4 2 𝑡→0 4 𝑡→0 1 𝒕𝟐 𝒍𝒏 𝒕 = − − 𝒍𝒊𝒎+ 4 1 𝒕→𝟎 𝟐 Page | 107 2 𝑡2 + ] 4 𝑡2 + ] 4 𝑡 2 𝑙𝑛 𝑡 2 +0 = − − (−∞) = ∞ 4 𝑡 2 𝑙𝑛 𝑡 2 𝑡 2 𝑙𝑛 𝑡 = − 𝑙𝑖𝑚+ − 𝑙𝑖𝑚+ 2 4 𝑡 2 𝑙𝑛 𝑡 1 1 2 𝑥2 12 𝑡→0 𝑡→0 𝑥2 − )−( 4 = 𝑙𝑖𝑚+ [(0 − ) − 4 = 𝑙𝑖𝑚+ [− − − 1 − ∗ 2 𝑥 2 𝑙𝑛 𝑥 12 𝑙𝑛 1 ] = 𝑙𝑖𝑚+ [( 4 𝑡 𝑥 2 𝑙𝑛 𝑥 + 𝑙𝑖𝑚+ 𝑡→0 𝑡2 4 𝑡2 − )] 4 Then: this improper integral Diverged Choice B 𝝅 𝟐 117] Determine ∫𝟎 𝟏 𝟏−𝐬𝐢𝐧 𝒙 𝒅𝒙 converges or diverges , if converges Find its value . 𝟕 A- Converged at − 𝟐𝟖 B- Diverged at ∞ 𝟏 𝟏 C- Converged at 𝟒 D- Converged at − 𝟐 Solution: 𝜋 2 ∫0 ➔ 1 1−𝑠𝑖𝑛 𝑥 1 1−𝑠𝑖𝑛 𝑥 𝜋 1 2 𝜋 1−𝑠𝑖𝑛 2 𝑑𝑥 , at 𝑥 = ➔ ∗ 1+𝑠𝑖𝑛 𝑥 1+𝑠𝑖𝑛 𝑥 = 1+𝑠𝑖𝑛 𝑥 1−𝑠𝑖𝑛2 𝑥 = 1+𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 2 𝑥 = = 1 = 1−1 1 𝑐𝑜𝑠 2 𝑥 + 1 0 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 2 𝑥 = 𝑠𝑒𝑐 2 𝑥 + 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 ∗ 𝑡𝑎𝑛 𝑥 𝑠𝑒𝑐 𝑥 Then: ∫(𝑠𝑒𝑐2 𝑥 + 𝑡𝑎𝑛 𝑥 𝑠𝑒𝑐 𝑥) = 𝑡𝑎𝑛 𝑥 + 𝑠𝑒𝑐 𝑥 𝜋 2 ∫0 𝑡 1 1−𝑠𝑖𝑛 𝑥 𝑑𝑥 = 𝑙𝑖𝑚 ∫ 𝜋− 0 𝑡→ 2 1 1−𝑠𝑖𝑛 𝑥 [𝑡𝑎𝑛 𝑥 + 𝑠𝑒𝑐 𝑥 ]𝑡0 𝑑𝑥 = 𝑙𝑖𝑚 𝜋− 𝑡→ 2 [(𝑡𝑎𝑛 𝑡 + 𝑠𝑒𝑐 𝑡) − (𝑡𝑎𝑛 0 + 𝑠𝑒𝑐 0)] = 𝑙𝑖𝑚 𝜋− 𝑡→ 2 = 𝑙𝑖𝑚 𝑡𝑎𝑛 𝑡 + 𝑙𝑖𝑚 𝑠𝑒𝑐 𝑡 − 𝑙𝑖𝑚 𝑡𝑎𝑛 0 − 𝑙𝑖𝑚 𝑠𝑒𝑐 0 𝜋− 𝜋− 𝜋− 𝜋− 𝑡→ 2 𝑡→ 2 𝜋 𝑡→ 2 𝑡→ 2 𝜋 = 𝑙𝑖𝑚 𝑡𝑎𝑛 + 𝑙𝑖𝑚 𝑠𝑒𝑐 − 0 − 𝑙𝑖𝑚 𝜋− 𝜋− 𝜋− 𝑡→ 2 2 𝑡→ 2 2 𝜋 𝜋 𝑡→ 2 𝜋 𝑡→ 2 𝜋 𝑙𝑖𝑚 𝑡𝑎𝑛 = ∞ , 𝑙𝑖𝑚 𝑠𝑒𝑐 = ∞ 𝜋− 𝜋− 𝑡→ 2 2 𝑡→ 2 2 =∞+∞−1=∞ Page | 108 2 cos 0 2 = 𝑙𝑖𝑚 𝑡𝑎𝑛 + 𝑙𝑖𝑚 𝑠𝑒𝑐 − 0 − 𝑙𝑖𝑚 𝜋− 𝜋− 𝜋− 𝑡→ 2 1 1 𝑡→ 2 cos 0 1 𝑐𝑜𝑠 𝑥 = 𝑠𝑒𝑐 2 𝑥 + Then: this improper integral Diverged Choice B ∞ 118] Determine ∫𝟎 𝟏 (𝒙−𝟑)𝟐 𝒅𝒙 converges or diverges , if converges Find its value . 𝟕 A- Converged at − 𝟐𝟖 B- Diverged at ∞ 𝟏 𝟏 C- Converged at 𝟒 D- Converged at − 𝟐 Solution: ∞ ∫0 ∞ ∫0 1 𝑑𝑥 (𝑥−3)2 , at 𝑥 = 3 ➔ (3−3)2 = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 1 𝑑𝑥 (𝑥−3)2 = 𝑙𝑖𝑚− ∫0 (𝑥−3)2 𝑑𝑥 + 𝑙𝑖𝑚+ ∫𝑡 1 ∫ (𝑥−3)2 𝑑𝑥 𝑡 1 0 𝑡 ∞ 1 𝑡→3 𝑡→3 = ∫(𝑥 − 3)−2 𝑑𝑥 = ∞ 1 𝑙𝑖𝑚− ∫0 (𝑥−3)2 𝑑𝑥 + 𝑙𝑖𝑚+ ∫𝑡 𝑡→3 1 𝑡→3 𝑡→3 = 𝑙𝑖𝑚− [(− = 𝑙𝑖𝑚− (− 𝑡→3 = 𝑙𝑖𝑚− (− 𝑡→3 −2+1 1 𝑑𝑥 (𝑥−3)2 = 𝑙𝑖𝑚− [(− 𝑡→3 (𝑥−3)−2+1 𝑡→3 1 1 𝑥−3 𝑡 1 ] + 𝑙𝑖𝑚+ [− 𝑥−3 0 1 𝑡→3 1 1 ] ∞ 𝑥−3 𝑡 1 ) − (− 0−3)] + 𝑙𝑖𝑚+ [(− ∞−3) − (− 𝑡−3)] 𝑡−3 𝑡→3 1 1 1 1 ) − 3] + 𝑙𝑖𝑚+ [(− ∞−3) + 𝑡−3] 𝑡−3 𝑡→3 1 1 1 1 ) − 𝑙𝑖𝑚− 3 − 𝑙𝑖𝑚+ ∞−3 + 𝑙𝑖𝑚+ 𝑡−3 𝑡−3 𝑡→3 1 𝑡→3 1 𝑡→3 1 ) − 3 − 0 + 𝑙𝑖𝑚+ 3−3 3−3 1 3 Then: this improper integral Diverged Page | 109 = −(𝑥 − 3)−1 = − = 𝑙𝑖𝑚− [− =∞− −0+∞=∞ Choice B 1 𝑑𝑥 (𝑥−3)2 𝑡→3 𝟔 119] Determine ∫𝟎 𝒙 (𝒙𝟐 −𝟒)𝟐 𝒅𝒙 converges or diverges , if converges Find its value . 𝟕 A- Converged at − 𝟐𝟖 B- Diverged 𝟏 𝟏 C- Converged at 𝟒 D- Converged at − 𝟐 Solution: 6 ∫0 𝑥 (𝑥 2 −4) 2 𝑑𝑥 , at 𝑥 = 2 ➔ (22 2 2 −4)2 = = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 0 𝑥 𝑑𝑢 ∫ (𝑥 2 −4)2 𝑑𝑥 , 𝑥 2 − 4 = 𝑢 , 2𝑥 𝑑𝑥 = 𝑑𝑢 , 𝑑𝑥 = 𝑥 ∫ (𝑢)2 6 ∫0 ∗ 𝑑𝑢 2𝑥 𝑥 (𝑥 2 −4)2 =∫ 1 1 𝑑𝑢 = ∫ 𝑢 2𝑢2 2 𝑡 𝑑𝑥 = 𝑙𝑖𝑚− ∫0 (𝑥 2 𝑡→2 = 𝑙𝑖𝑚− [− 𝑡→2 = 𝑙𝑖𝑚− [(− = 𝑙𝑖𝑚− (− 𝑡→2 = 𝑙𝑖𝑚− (− 𝑡→2 𝑢−2+1 2 −2+1 𝑑𝑢 = ∗ 6 𝑥 −4)2 𝑑𝑥 + 𝑙𝑖𝑚+ ∫𝑡 𝑡→2 ] + 𝑙𝑖𝑚+ [− 2(𝑥 2 −4) 0 𝑡→2 1 𝑡 1 = 𝑙𝑖𝑚− [(− 𝑡→2 −2 2𝑥 𝑡→2 1 𝑥 (𝑥 2 −4)2 1 ] 2 2𝑢 =− 1 2(𝑥 2 −4) 𝑑𝑥 6 1 1 1 ) − (− 2(02 −4))] + 𝑙𝑖𝑚+ [(− 2(62 −4)) − (− 2(𝑡 2 −4))] 2(𝑡 2 −4) 𝑡→2 1 1 1 1 ) − 8] + 𝑙𝑖𝑚+ [(− 64) + 2(𝑡 2 −4)] 2(𝑡 2 −4) 𝑡→2 1 1 1 1 ) − 𝑙𝑖𝑚− 8 − 𝑙𝑖𝑚− 64 + 𝑙𝑖𝑚− 2(𝑡 2 −4) 2(𝑡 2 −4) 𝑡→2 1 1 𝑡→2 𝑡→2 1 1 ) − 8 − 64 + 𝑙𝑖𝑚− 2(22 −4) 2(22 −4) 𝑡→2 1 1 𝑡→2 1 1 𝑡→2 1 1 8 64 = −∞ − − +∞ Then: this improper integral Diverged Page | 110 1 2(𝑥 2 −4) 𝑡 = 𝑙𝑖𝑚− (− ) − 8 − 64 + 𝑙𝑖𝑚− 2∗(0) 2∗(0) Choice B 1 = − 𝑢−1 = − ∞ 𝟏 120] Determine ∫𝟏 𝒙𝟐 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at −𝟏 B- Diverged at −∞ 𝟒 C- Converged at 𝟒 D- Converged at 𝟎 Solution: ∞ 1 ∫1 𝑥2 𝑑𝑥 , at 𝑥 = ∞ ➔ ∞ 1 ∫1 𝑥 2 𝑑𝑥 = 1 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 ∞2 𝑡 1 𝑙𝑖𝑚− ∫1 2 𝑑𝑥 𝑥 𝑡→∞ = 𝑡 𝑙𝑖𝑚− ∫1 𝑥 −2 𝑑𝑥 𝑡→∞ 1 1 1 = 𝑙𝑖𝑚− [ 𝑡→∞ 𝑥 −2+1 𝑡 1 𝑡 ] = 𝑙𝑖𝑚− [− ] −2+1 1 𝑡→∞ 𝑥 1 = 𝑙𝑖𝑚− [(− ) − (− )] = 𝑙𝑖𝑚− [(− ) + 1] 𝑡 1 𝑡 𝑡→∞ 𝑡→∞ 1 1 = 𝑙𝑖𝑚− (− ) + 𝑙𝑖𝑚− 1 = − 𝑙𝑖𝑚− ( ) + 1 𝑡 ∞ 𝑡→∞ 𝑡→∞ 𝑡→∞ =0+1=1 Then: this improper integral Converged Choice C ∞ 𝟏 121] Determine ∫𝟏 𝟒 √𝒙 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at −𝟏 B- Diverged at −∞ C- Converged at 𝟒 D- Diverged at ∞ Solution: ∞ 1 ∫1 4 √𝑥 ∞ 1 ∫1 4 √𝑥 𝑑𝑥 , at 𝑥 = 1 ∞ 𝑡 1 𝑑𝑥 = 𝑙𝑖𝑚− ∫1 Page | 111 𝑡→∞ 4 √𝑥 𝑡 1 𝑑𝑥 = 𝑙𝑖𝑚− ∫1 𝑡→∞ 1 𝑥4 𝑡 𝑑𝑥 = 𝑙𝑖𝑚− ∫1 𝑥 𝑡→∞ 1 −4 1 − +1 𝑥 4 1 𝑡→∞− −4+1 𝑑𝑥 = 𝑙𝑖𝑚 [ 𝑡 ] 1 𝑡 3 = 𝑙𝑖𝑚− [ 𝑥4 𝑡→∞ 3 4 ] = 1 4 3 4 𝑡 𝑙𝑖𝑚 [𝑥 ] = 3 𝑡→∞− 1 4 3 4 3 4 𝑙𝑖𝑚 [(𝑡 ) − (1 )] = ∞ 3 𝑡→∞− Then: this improper integral Diverged Choice D ∞ 𝟏 122] Determine ∫𝟐𝟐 𝒆𝒙 +𝟕 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at −𝟏 B- Diverged at −∞ C- Converged at 𝟒 D- Diverged Solution: ∞ 1 1 1 ∫22 𝑒 𝑥+7 𝑑𝑥 , at 𝑥 = ∞ ➔ 𝑒 ∞ +7 = ∞ = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 1 ∫ 𝑒 𝑥 +7 𝑑𝑥 , 𝑢 = 𝑒 𝑥 + 7 , 𝑒 𝑥 = 𝑢 − 7 , 𝑑𝑢 = 𝑒 𝑥 𝑑𝑥 , 𝑑𝑥 = 1 1 𝑑𝑢 𝑑𝑢 𝑒𝑥 1 ∫ 𝑒 𝑥 +7 𝑑𝑥 = ∫ 𝑢 ∗ 𝑒 𝑥 = ∫ 𝑢(𝑢−7) 𝑑𝑢 1 = 𝑢(𝑢−7) A 𝑢 + B 𝑢−7 By multiplying in two sides by : 𝑢(𝑢 − 7) ➔ 1 = A(𝑢 − 7) + B𝑢 At 𝒖 = 𝟕 ➔ 1 = 0 + B ∗ 7 , then: B = 1 7 At 𝒖 = 𝟎 ➔1 = A(0 − 7) , then: A = − Then: 1 𝑢(𝑢−7) =− 1 1 + 7𝑢 1 1 7 1 7(𝑢−7) 1 1 1 ∫ 𝑢(𝑢−7) 𝑑𝑢 = ∫ (− 7𝑢 + 7(𝑢−7)) 𝑑𝑢 = − ∫ 7𝑢 𝑑𝑢 + ∫ 7(𝑢−7) 1 1 1 1 = − ∫ 𝑑𝑢 + ∫ (𝑢−7) 7 𝑢 7 Page | 112 1 1 1 1 7 7 7 7 = − ln|𝑢| + ln|𝑢 − 7| = − ln|𝑒 𝑥 + 7| + ln|𝑒 𝑥 + 7 − 7| 1 1 7 7 = − ln|𝑒 𝑥 + 7| + ln|𝑒 𝑥 | ∞ 1 ➔∫22 𝑥 𝑑𝑥 𝑒 +7 = 𝑡 1 𝑙𝑖𝑚− ∫22 𝑥 𝑑𝑥 𝑒 +7 𝑡→∞ 1 1 𝑥 𝑥| = 𝑙𝑖𝑚− [− ln|𝑒 + 7| + ln|𝑒 ] 𝑡→∞ 7 7 1 1 1 7 7 7 𝑡 22 = 𝑙𝑖𝑚− [(− ln|𝑒 ∞ + 7| + ln|𝑒 ∞ |) − (− ln|𝑒 22 + 7| + 𝑡→∞ 1 7 ln|𝑒 22 |)] 1 1 7 7 = 𝑙𝑖𝑚− [(−∞ + ∞) − ( ln|𝑒 22 + 7| − ln|𝑒 22 |)] 𝑡→∞ Then: this improper integral Diverged Choice D ∞ 𝟏 123] Determine ∫𝟐 𝒙−𝟏 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at −𝟏 B- Diverged at ∞ C- Converged at 𝟒 D- None of them Solution: ∞ 1 ∫2 𝑥−1 𝑑𝑥 , at 𝑥 = ∞ ➔ 1 ∞−1 = 1 ∞ = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 1 ∫ 𝑥−1 = ln|𝑥 − 1| ∞ 1 ∫2 𝑥−1 𝑡 1 𝑑𝑥 = 𝑙𝑖𝑚− ∫2 𝑡→∞ 𝑥−1 𝑑𝑥 = 𝑙𝑖𝑚−[ln|𝑥 − 1|]𝑡2 = 𝑙𝑖𝑚−[(ln|𝑡 − 1|) − 𝑡→∞ 𝑡→∞ (ln|2 − 1|)] = 𝑙𝑖𝑚−[(ln|𝑡 − 1|) − (ln|2 − 1|)] = ∞ − 0 = ∞ 𝑡→∞ Then: this improper integral Diverged Choice B Page | 113 𝟎 𝟏 124] Determine ∫−∞ 𝟓𝒙 𝒅𝒙 converges or diverges , if converges Find its value . A- Converged at −𝟏 B- Diverged at ∞ C- ∞ D- None of them Solution: 0 1 5−𝑥 0 ∫−∞ 5𝑥 𝑑𝑥 = ∫−∞ 5−𝑥 𝑑𝑥 = [−1∗ln 5] = [(− = [(− 0 −∞ = [− 1 ln 5∗5 𝑥] 0 −∞ 1 ) − (− ln 5∗5−∞)] ln 5∗50 1 1 1 5∞ 1 ) + ln 5∗5−∞] = − ln 5 + ln 5 = − ln 5 + ∞ = ∞ ln 5 Then: this improper integral Diverged Page | 114 1 Choice C 125] Write the equation 𝟕𝒙𝟐 + 𝟑𝒚𝟐 = 𝟔𝟑 in standard form and sketch a graph of the ellipse A- 𝒙𝟐 𝒚𝟐 + 𝟐𝟏 = 𝟏 B- C- 𝟕𝒙𝟐 + 𝟑𝒚𝟐 = 𝟏 D- 𝟗 𝒙𝟐 𝟗 𝒙𝟐 𝟗 𝒚𝟐 − 𝟐𝟏 = 𝟏 𝒚 + 𝟐𝟏 = 𝟏 Solution: 7𝑥 2 + 3𝑦 2 = 63 , By dividing 63 in two sides 7𝑥 2 +3𝑦2 =63 63 7𝑥 2 63 + 3𝑦 2 63 = = 𝑥2 9 7𝑥 2 63 + + 𝑦2 21 3𝑦 2 63 =1 = 1 ➔ its called Standard Form 𝑎2 = 21 ➔ 𝑎 = √21 𝑏 2 = 9 ➔ 𝑏 = √9 = 3 𝑐 = √𝑎2 − 𝑏 2 = √21 − 9 = √12 = 2√3 (𝟎, √𝟐𝟏) 𝒗𝟏 1] Center = (0,0) 2] Major axis ➔ 𝑦 − 𝑎𝑥𝑖𝑠 𝒇𝟏 3] 𝑎 = √21 ≈ 4.58 , 𝑏 = 3 4] 𝑐 = 2√3 ≈ 3.46 (𝟎, 𝟐√𝟑) (𝟎, 𝟎) (𝟑, 𝟎) (−𝟑, 𝟎) Choice A (𝟎, −𝟐√𝟑) 𝒇𝟐 (𝟎, −√𝟐𝟏) Page | 115 𝒗𝟐 126] If the length of the horizontal axis of an ellipse is 20 and the length of the vertical axis is 16 , Write the equation of the ellipse centered at the origin in standard form . 𝒙𝟐 𝒚𝟐 𝒙𝟐 𝟖𝒚𝟐 𝒙𝟐 A- 𝟏𝟎𝟎 − 𝟔𝟒 = 𝟏 C- 𝟏𝟎𝟎 + 𝟔𝟒 𝒚𝟐 B- 𝟏𝟎𝟎 + 𝟔𝟒 =𝟏 D- None of them Solution: The horizontal axis = 20 Then : 2𝑎 = 20 , 𝑎 = 10 The vertical axis = 16 Then : 2𝑏 = 16 , 𝑏 = 8 ➔ 𝑐 = √𝑎2 − 𝑏 2 = √102 − 82 = 6 Center = (0,0) Then: The Standard Form = Choice D Page | 116 𝑥2 100 + 𝑦2 64 = 1 , Because 𝑥 𝑎𝑥𝑖𝑠 is the major axis 127] Write the equation 𝟐𝒙𝟐 + 𝟖𝒚𝟐 − 𝟐𝟎𝒙 + 𝟒𝟖𝒚 = −𝟏𝟎𝟔 in Standard Form . AC- 𝒙𝟐 + 𝟖 (𝒚+𝟑)𝟐 (𝒙−𝟓)𝟐 𝟖 =𝟏 𝟐 + (𝒚+𝟑)𝟐 𝟐 B- =𝟏 (𝒙−𝟓)𝟐 𝟖 − (𝒚+𝟑)𝟐 𝟐 D- None of them Solution: → 2𝑥 2 + 8𝑦 2 − 20𝑥 + 48𝑦 = −106 → 2𝑥 2 − 20𝑥 + 8𝑦 2 + 48𝑦 = −106 ➔ 2(𝑥 2 − 10𝑥 ) + 8(𝑦 2 + 6𝑦) = −106 By Completing Squaring 2((𝑥 − 5)2 − 25) + 8((𝑦 + 3)2 − 9) 2(𝑥 − 5)2 − 50 + 8(𝑦 + 3)2 − 72 = −106 2(𝑥 − 5)2 + 8(𝑦 + 3)2 = −106 + 72 + 50 2(𝑥 − 5)2 + 8(𝑦 + 3)2 = 16 By Dividing 16 (𝑥−5)2 8 (𝑥−5)2 8 + + (𝑦+3)2 2 (𝑦+3)2 2 = 1 ➔ Standard Form of the ellipse = (𝑥−5)2 8 2 + (𝑦−(−3)) 2 = (𝑥−ℎ)2 𝑎2 + (𝑦−𝑘)2 𝑏2 Then: ℎ = 5 , 𝑘 = −3 , 𝑎2 = 8 ➔ 𝑎 = √8 = 2√2 , 𝑏 2 = 2 ➔ 𝑏 = √2 𝑐 = √𝑎2 − 𝑏 2 = √8 − 2 = √6 Then: Major axis is Parallel to X-axis Length of major axis = 2𝑎 = 4√2 Length of minor axis = 2𝑏 = 2√2 Page | 117 Choice C 128] Write the equation of the ellipse in Standard Form of the vertices of the ellipse (𝟐, 𝟏𝟗) , (𝟐, −𝟕) and Co vertices of the Ellipse (−𝟑, 𝟔) , (𝟕, 𝟔) (𝒙−𝟐)𝟐 AC- (𝟏𝟑)𝟐 (𝒙−𝟐)𝟐 (𝟓 )𝟐 + + (𝒚−𝟔)𝟐 (𝟓 )𝟐 (𝒚−𝟔)𝟐 (𝟏𝟑)𝟐 (𝒙−𝟐)𝟐 =𝟏 B- =𝟏 D- None of them (𝟓 )𝟐 + (𝒚−𝟔)𝟐 (𝟏𝟑)𝟐 Solution: Since the vertices is (2,19) , (2, −7) , Then the major axis is Parallel to Y-axis Co-vertices is (−3,6) , (7,6) , Then the minor axis is parallel to X-axis 𝑥1 +𝑥2 𝑦1 +𝑦2 Midpoint = ( 2 , 2 2+2 19−7 )=( , 2 2 4 12 ) = (2 , 2 ) = (2,6) OR 𝑥1 +𝑥2 𝑦1 +𝑦2 Midpoint = ( 2 , 2 −3+7 6+6 )=( 2 , 2 4 12 ) = (2 , 2 ) = (2,6) Then: Center = (2,6) ➔ ℎ = 2 , 𝑘 = 6 (2,19) Length of major axis = 2𝑎 = 19 + 7 = 26 ➔ 𝑎 = 13 Length of minor axis = 2𝑏 = 3 + 7 = 10 ➔ 𝑏 = 5 The Standard Form is ➔ (𝑥−ℎ)2 𝑏2 Choice C Page | 118 + (𝑦−𝑘)2 𝑎2 = (𝑥−ℎ)2 𝑏2 (𝑥−2)2 (5)2 + + (𝑦−𝑘)2 𝑎2 (𝑦−6)2 (13)2 =1 =1 (−3,6) (ℎ, 𝑘) (7,6) 129] If the length of the horizontal axis is 𝟖 and the length of the vertical axis is 𝟏𝟎 , what is the equation of the ellipse centered at the origin ? 𝒙𝟐 A- 𝟐𝟓 + 𝒙𝟐 𝒚𝟐 𝟒 =𝟏 𝒙𝟐 𝒚𝟐 B- 𝟐𝟓 − 𝟏𝟔 = 𝟏 𝒚𝟐 C- 𝟏𝟔 + 𝟐𝟓 = 𝟏 D- None of them Solution: The vertical axis = 10 , then: 2𝑎 = 10 ➔ 𝑎 = 5 The horizontal axis = 8 , then: 2𝑏 = 8 ➔ 𝑏 = 4 𝑐 = √𝑎2 − 𝑏 2 = √52 − 42 = 3 Center = (0,0) Then : The Standard equation = Choice C Page | 119 𝑥2 𝑦2 𝑏 𝑎2 2 + =1➔ 𝑥2 𝑦2 4 52 2 + =1 130] Write the equation of the ellipse with vertices (−𝟑, 𝟏) , (𝟏, 𝟏) and co-vertices (−𝟏, 𝟐) , (−𝟏, 𝟎) . AC- (𝒙+𝟏)𝟐 𝟒 (𝒙+𝟏)𝟐 𝟒 + (𝒚−𝟏)𝟐 𝟏 = 𝟐𝟎 B- (𝒙+𝟏)𝟐 𝟒 + (𝒚 − 𝟏)𝟐 = 𝟏 − (𝒚−𝟏)𝟐 𝟏 =𝟏 D- None of them Solution: Since The vertex is (−3,1) , (1,1) , then: the major axis is parallel to X-axis The co-vertex is (−1,2) , (−1,0) , then: the major axis is parallel to Y-axis Midpoint = ( 𝑥1 +𝑥2 𝑦1 +𝑦2 2 , 2 )=( −3+1 1+1 2 , 2 −1+(−1) 2+0 ) = (−1,1) OR ( 2 ➔ Center = (−1,1) , then: ℎ = −1 , 𝑘 = 1 The length of the major axis = 2𝑎 = 3 + 1 = 4 ➔ 𝑎 = 2 The length of the minor axis = 2𝑏 = 1 + 1 = 2 ➔ 𝑏 = 1 The Standard Form = ➔ (𝑥−ℎ)2 𝑎2 Choice C Page | 120 + (𝑦−𝑘)2 𝑏2 (𝑥−ℎ)2 𝑎2 + (𝑦−𝑘)2 2 = (𝑥−(−1)) 22 + 𝑏2 (𝑦−1)2 12 =1 = (𝑥+1)2 4 + (𝑦−1)2 1 =1 , 2 ) = (−1,1) 131] Write the equation 𝒙𝟐 + 𝟒𝒚𝟐 = 𝟖 in Standard Form . AC- 𝒙𝟐 𝟖 𝒙𝟐 𝟖 + − 𝒚𝟐 𝟐 𝒚𝟐 𝟐 𝒙𝟐 + 𝒚𝟑 =𝟏 B- =𝟏 D- None of them 𝟖 𝟐 =𝟏 Solution: ➔ 𝑥 2 +4𝑦 2 =8 8 = 𝑥2 8 + 4𝑦 2 8 8 𝑥2 8 8 = ➔ + 𝑦2 =1 2 Choice A 132] Write the equation 𝟐𝟓𝒙𝟐 + 𝟒𝒚𝟐 + 𝟓𝟎𝒙 − 𝟑𝟐𝒚 = 𝟏𝟏 in Standard Form . AC- (𝒙+𝟏)𝟐 𝟒 (𝒙+𝟏)𝟐 𝟒 + + (𝒚−𝟒)𝟐 𝟐𝟓 =𝟏 (𝒚−𝟒)𝟐 𝟒 B- (𝒙+𝟏)𝟐 𝟏𝟎𝟎 + (𝒚−𝟒)𝟐 𝟐𝟓 =𝟏 D- None of them Solution: 25𝑥 2 + 4𝑦 2 + 50𝑥 − 32𝑦 = 11 ➔ 25𝑥 2 + 50𝑥 + 4𝑦 2 − 32𝑦 = 11 → 25(𝑥 2 + 2𝑥 ) + 4(𝑦 2 − 8𝑦) = 11 → 25((𝑥 + 1)2 − 1 ) + 4((𝑦 − 4)2 − 16) = 11 → 25(𝑥 + 1)2 − 25 + 4(𝑦 − 4)2 − 4 ∗ 16 = 11 → 25(𝑥 + 1)2 − 25 + 4(𝑦 − 4)2 − 64 − 11 = 0 → 25(𝑥 + 1)2 + 4(𝑦 − 4)2 − 100 = 0 → 25(𝑥 + 1)2 + 4(𝑦 − 4)2 = 100 Page | 121 → 25(𝑥 + 1)2 + 4(𝑦 − 4)2 = 100 → → 25(𝑥+1)2 100 (𝑥+1)2 4 Choice A Page | 122 + + 4(𝑦−4)2 100 (𝑦−4)2 25 = =1 100 100 Dividing by 𝟏𝟎𝟎 133] Find the vertex, focus and directrix of the parabola 𝒚𝟐 = 𝟏𝟔𝒙 respectively A- (𝟎, 𝟎) , (−𝟒, 𝟎) , 𝟒 B- (𝟎, 𝟎) , (𝟒, 𝟎) , −𝟒 C- the origin point , (𝟒, 𝟎) , 𝟒 D- None of them Solution: Axis of Symmetry is on 𝑥 − 𝑎𝑥𝑖𝑠 ➔ (𝑦 − 𝑘)2 = ±4𝑝(𝑥 − ℎ) ➔ 𝑦 2 = 16𝑥 Then: (𝑦 − 𝑘)2 = 𝑦 2 ➔ 𝑘 = 0 Then: 4𝑝(𝑥 − ℎ) = 16𝑥 ➔ ℎ = 0 , 4𝑝 = 16 → 𝑝 = 4 Then: 𝒗𝒆𝒓𝒕𝒆𝒙 = (ℎ, 𝑘) = (0,0) Since 𝑝 = 4 & Sign (+) Then: 𝒇𝒐𝒄𝒖𝒔 = (ℎ + 𝑝, 𝑘) = (0 + 4,0) = (4,0) The 𝑫𝒊𝒓𝒆𝒄𝒕𝒓𝒊𝒙 ➔ 𝑥 = ℎ − 𝑝 → 𝑥 = 0 − 4 = −4 Choice B Page | 123 134] Find the vertex, focus and directrix of the parabola respectively A- (𝟎, 𝟎) , (−𝟒, 𝟖) , −𝟒 B- (𝟎, 𝟏) , (𝟓, 𝟎) , −𝟐 C- the origin point , (−𝟐, 𝟎) , 𝟐 D- None of them Solution: ➔ −𝑦 2 8 = 𝑥 → −𝑦 2 = 8𝑥 → 𝑦 2 = −8𝑥 Axis of Symmetry is on 𝒙 − 𝒂𝒙𝒊𝒔 ➔ (𝑦 − 𝑘)2 = ±4𝑝(𝑥 − ℎ) ➔ 𝑦 2 = −8𝑥 Then: (𝑦 − 𝑘)2 = 𝑦 2 ➔ 𝑘 = 0 Then: 4𝑝(𝑥 − ℎ) = −8𝑥 ➔ ℎ = 0 , 4𝑝 = 8 → 𝑝 = 2 Then: 𝒗𝒆𝒓𝒕𝒆𝒙 = (ℎ, 𝑘) = (0,0) Since 𝑝 = 2 & Sign (−) Then: 𝒇𝒐𝒄𝒖𝒔 = (ℎ − 𝑝, 𝑘) = (0 − 2,0) = (−2,0) The 𝑫𝒊𝒓𝒆𝒄𝒕𝒓𝒊𝒙 ➔ 𝑥 = ℎ + 𝑝 → 𝑥 = 0 + 2 = 2 Choice C Page | 124 −𝒚𝟐 𝟖 =𝒙 135] Find the vertex, focus and directrix of the equation 𝟏 𝒚 = (𝒙𝟐 − 𝟒𝒙 + 𝟐𝟐) respectively 𝟔 A- (𝟎, 𝟎) , (−𝟒, 𝟖) , −𝟒 𝟕 B- (𝟎, 𝟎) , (𝟐 , 𝟑) , 𝟎 𝟗 𝟑 C- (𝟐, 𝟑) , (𝟑, 𝟐) , 𝟐 D- None of them Solution: 1 𝑦 = (𝑥 2 − 4𝑥 + 22) → 6𝑦 = 𝑥 2 − 4𝑥 + 22 6 𝑥 2 − 4𝑥 + 22 = [(𝑥 − 2)2 − 4] + 22 ➔ 6𝑦 = 𝑥 2 − 4𝑥 + 22 By Completing Squaring = [(𝑥 − 2)2 − 4] + 22 = (𝑥 − 2)2 + 18 6 3 4 2 6𝑦 − 18 = (𝑥 − 2)2 → 6(𝑦 − 3) = (𝑥 − 2)2 ➔ 4𝑝 = 6 → 𝑝 = = Then: 𝒗𝒆𝒓𝒕𝒆𝒙 = (ℎ, 𝑘) = (2,3) 3 Since 𝑝 = & Sign (+) 2 3 9 Then: 𝒇𝒐𝒄𝒖𝒔 = (ℎ, 𝑘 + 𝑝) = (2,3 + ) = (2, ) 2 2 3 3 2 2 The 𝑫𝒊𝒓𝒆𝒄𝒕𝒓𝒊𝒙 ➔ 𝑦 = 𝑘 − 𝑝 → 𝑦 = 3 − = Choice D Page | 125 136] The point (𝟎, 𝟕) is the focus of a parabola that its vertex (𝟐, 𝟕). Find the equation of the parabola. A- (𝒚 − 𝟕)𝟐 = −𝟖(𝒙 − 𝟐) B- (𝒚 − 𝟕)𝟐 = 𝟖(𝒙 − 𝟐) C- (𝒚 − 𝟕)𝟐 = −𝟖(𝒙 + 𝟐) D- None of them Solution: 𝒗𝒆𝒓𝒕𝒆𝒙 = (ℎ, 𝑘) = (2,7) 𝒇𝒐𝒄𝒖𝒔 = (0,7) 1] Axis of symmetry is parallel to 𝑥 − 𝑎𝑥𝑖𝑠 2] 𝑣𝑒𝑟𝑡𝑒𝑥 = (2,7) 3] Since 𝑣𝑒𝑟𝑡𝑒𝑥 = (2,7) & 𝐹𝑜𝑐𝑢𝑠 = (0,7) , then 2 → 0 ➔ Sign (−) 4] 𝐹𝑜𝑐𝑢𝑠 = (2 − 𝑝, 7) → 𝑝 = 2 (𝑦 − 𝑘)2 = ±4𝑝(𝑥 − ℎ) (𝑦 − 𝑘)2 = ±4𝑝(𝑥 − ℎ) → (𝑦 − 7)2 = −4 ∗ 2(𝑥 − 2) → (𝒚 − 𝟕)𝟐 = −𝟖(𝒙 − 𝟐) Choice A Page | 126 137] The point (𝟔, 𝟐) is the vertex of a parabola that its Directrix equals 𝒙 = 𝟒 . Find the equation of the parabola. A- (𝒚 − 𝟐)𝟐 = 𝟖(𝒙 − 𝟔) B- (𝒚 − 𝟐)𝟐 = −𝟖(𝒙 − 𝟔) C- (𝒚 − 𝟐)𝟐 = 𝟐(𝒙 − 𝟔) D- None of them Solution: Since the Directrix ➔ 𝑥 = 4 Then: the equation ➔ (𝑦 − 𝑘)2 = ±4𝑝(𝑥 − ℎ) Since the 𝒗𝒆𝒓𝒕𝒆𝒙 = (6,2) ➔ ℎ = 6 , 𝑘 = 2 𝑫𝒊𝒓𝒆𝒄𝒕𝒓𝒊𝒙 ➔ 𝑥 = 4 From the sketch (𝟔, 𝟐) Then: 𝑝 → (+) 𝑠𝑖𝑔𝑛 𝑫𝒊𝒓𝒆𝒄𝒕𝒓𝒊𝒙 ➔ 𝑥 = 4 = ℎ − 𝑝 → 6 − 𝑝 = 4 Then: 𝑝 = 2 𝑻𝒉𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 ➔ (𝑦 − 2)2 = 4 ∗ 2(𝑥 − 6) → (𝑦 − 2)2 = 8(𝑥 − 6) Choice A Page | 127 Equation: 𝒙 = 𝟒 (𝟎, 𝟎) (𝟒, 𝟎) 138] Find the vertex, focus and directrix of the parabola 𝒚𝟐 = 𝟏𝟐𝒙 − 𝟖𝒚 + 𝟖 respectively . A- (−𝟐, −𝟒) , (−𝟏, 𝟒) , −𝟓 B- (−𝟐, −𝟒) , (𝟏, −𝟒) , 𝟓 C- (−𝟐, −𝟒) , (𝟏, −𝟒) , −𝟓 D- None of them Solution: 𝑦 2 = 12𝑥 − 8𝑦 + 8 ➔ 𝑦 2 + 8𝑦 = 12𝑥 + 8 𝑦 2 + 8𝑦 By Completing Squaring → (𝑦 + 4)2 − 16 Then: (𝑦 + 4)2 − 16 = 12𝑥 + 8 ➔ (𝑦 + 4)2 = 12𝑥 + 8 + 16 ➔ (𝑦 + 4)2 = 12𝑥 + 24 ➔ (𝑦 + 4)2 = 12(𝑥 + 2) Since the equation: (𝑦 − 𝑘)2 = ±4𝑝(𝑥 − ℎ) → (𝑦 + 4)2 = 12(𝑥 + 2) 𝑘 = −4 , ℎ = −2 , 𝑝 = 12 4 = 3 , 𝑠𝑖𝑔𝑛 = (+) Then: 𝒗𝒆𝒓𝒕𝒆𝒙 = (ℎ, 𝑘) = (−2, −4) Since 𝑝 = 3 & Sign (+) Then: 𝒇𝒐𝒄𝒖𝒔 = (ℎ + 𝑝, 𝑘) = (−2 + 3, −4) = (1, −4) The 𝑫𝒊𝒓𝒆𝒄𝒕𝒓𝒊𝒙 ➔ 𝑥 = ℎ − 𝑝 → 𝑥 = −2 − 3 = −5 Choice C Page | 128 139] Find the vertex, focus and directrix of the parabola 𝒚 = 𝒙𝟐 respectively . A- (𝟎, 𝟎) , (𝟏, 𝟎) , −𝟓 B- the origin point , (𝟎, 𝟏) , −𝟐 C- (𝟎, 𝟎) , (𝟏, 𝟎) , −𝟏 D- None of them Solution: 𝑦= 𝑥2 4 ➔ 𝑥 2 = 4𝑦 (𝑥 − ℎ)2 = ±4𝑝(𝑦 − 𝑘) → ℎ = 0 , 𝑘 = 0 , 4𝑝 = 4 ➔ 𝑝 = 1 & Sign (+) Then: 𝒗𝒆𝒓𝒕𝒆𝒙 = (ℎ, 𝑘) = (0,0) Since 𝑝 = 1 & Sign (+) Then: 𝒇𝒐𝒄𝒖𝒔 = (ℎ, 𝑘 + 𝑝) = (0,0 + 1) = (0,1) The 𝑫𝒊𝒓𝒆𝒄𝒕𝒓𝒊𝒙 ➔ 𝑦 = 𝑘 − 𝑝 → 𝑦 = 0 − 1 = −1 Choice D Page | 129 𝟒 140] Find the sum of the first n-terms of series : 𝟏 ∗ 𝟑 + 𝟐 ∗ 𝟒 + 𝟑 ∗ 𝟓 + ⋯ Solution: 𝑢𝑘 = 𝑘 (𝑘 + 2) ➔ 𝑢𝑘 = 𝑘 2 + 2𝑘 ∑𝑛𝑘=1 𝑢𝑘 = ∑𝑛𝑘=1 𝑘 2 + 2 ∑𝑛𝑘=1 𝑘 ∑𝑛𝑘=1 𝑢𝑘 = 𝑛(𝑛+1)(2𝑛+1) ∑𝑛𝑘=1 𝑢𝑘 = 𝑛(𝑛+1)(2𝑛+1) 6 6 +2∗ + 𝑛(𝑛+1) 2 6∗𝑛(𝑛+1) 6 = 𝑛(𝑛+1)(2𝑛+1) 6 + 𝑛(𝑛 + 1) 𝑛(𝑛+1)(2𝑛+1)+6𝑛(𝑛+1) = 6 = = (𝑛(𝑛+1))[(2𝑛+1)+6] 6 (𝑛(𝑛+1))(2𝑛+7) 6 141] Find the sum of the first n-terms of series : 𝟑 ∗ 𝟏𝟐 + 𝟒 ∗ 𝟐𝟐 + 𝟓 ∗ 𝟑𝟐 + ⋯ Solution: 𝑢𝑘 = (𝑘 + 2)𝑘 2 ➔ 𝑢𝑘 = 𝑘 3 + 2𝑘 2 ∑𝑛𝑘=1 𝑢𝑘 = ∑𝑛𝑘=1 𝑘 3 + 2 ∑𝑛𝑘=1 𝑘 2 ∑𝑛𝑘=1 𝑢𝑘 = = = Page | 130 𝑛2 (𝑛+1)2 4 +2∗ 𝑛(𝑛+1)(2𝑛+1) 6 = 3∗(𝑛2 (𝑛+1)2 )+4(𝑛(𝑛+1)(2𝑛+1)) 4∗3 𝑛(𝑛+1)[3𝑛(𝑛+1)+4(2𝑛+1)] 12 = 𝑛2 (𝑛+1)2 = 4 + 𝑛(𝑛+1)(2𝑛+1) 3 3𝑛2 (𝑛+1)2 +4𝑛(𝑛+1)(2𝑛+1) 12 𝑛(𝑛+1)[3𝑛2 +3𝑛+8𝑛+4] 12 = 𝑛(𝑛+1)[3𝑛2 +11𝑛+4] 12 𝟏 𝟏 𝟏 𝟏 𝟏 142] Test the convergence of the series : ∑∞ 𝒏=𝟏 𝟐𝒏−𝟏 = 𝟏 + 𝟐 + 𝟒 + 𝟖 + ⋯ + 𝟐𝒏−𝟏 Solution: ∑∞ 𝑛=1 1 2𝑛−1 1 1 1 1 2 4 8 2𝑛−1 = 1 + + + + ⋯+ 1 1 1 1 1 1 2 4 2 8 4 2 Then: 𝑎 = 1 , 𝑟 = ÷ 1 = ÷ = ÷ = 1 Since |𝑟| = < 1 ➔ then it is converge 2 S𝑛 = 𝑎 1−𝑟 = 1 1 1−2 = 1 1 2 =2 143] Discuss the nature of the series : ∑∞𝒏=𝟏(−𝟏)𝒏−𝟏 = 𝟏 − 𝟏 + 𝟏 − 𝟏 + ⋯ + (−𝟏)𝒏−𝟏 + ⋯ Solution: S1 = 1 0 𝑒𝑣𝑒𝑛 S2 = 1 − 1 = 0 S𝑛 = { S3 = 1 − 1 + 1 = 1 1 𝑜𝑑𝑑 S4 = 1 − 1 + 1 − 1 = 0 Then the given series is Oscillating −𝒏+𝟐 144] Discuss the nature of the series : ∑∞ ∗ 𝟒𝒏+𝟏 𝒏=𝟏 𝟗 Solution: −𝑛+2 −(𝑛−2) ∑∞ ∗ 4𝑛+1 = ∑∞ ∗ 4𝑛+1 = ∑∞ 𝑛=1 9 𝑛=1 9 𝑛=1 = ∑∞ 𝑛=1 Then: 𝑎 = 16 ∗ 9 , 𝑟 = Page | 131 4 9 4𝑛−1 ∗42 ∗91 9𝑛−1 4𝑛+1 9𝑛−2 = ∑∞ 𝑛=1 4𝑛−1 ∗42 9𝑛−1 ∗9−1 4 𝑛−1 = ∑∞ 𝑛=1 16 ∗ 9 ∗ (9) 4 𝑛−1 ∑∞ 𝑛=1 16 ∗ 9 ∗ (9) ➔ S𝑛 = 𝑎 1−𝑟 = 16∗9 4 1−9 𝑛−1 = ∑∞ [Geometric Series] 𝑛=1 𝑎 (𝑟) 16∗9 = 5 9 = 16∗92 5 = 1296 5 = 259.2 4 16∗92 9 5 Since |𝑟| = < 1 ➔ then it is converge and its value equals 145] Determine if series converge or diverge , if converge find its 𝟏 value : ∑∞ 𝒏=𝟎 𝟐 𝒏 +𝟑𝒏+𝟐 Solution: ∑∞ 𝑛=0 1 1 𝑛2 +3𝑛+2 1 (𝑛+1)(𝑛+2) = ∑∞ 𝑛=0 (𝑛+1)(𝑛+2) A B = (𝑛+1) + (𝑛+2) By multiplying in two sides by : (𝒏 + 𝟏)(𝒏 + 𝟐) 1 = A(𝑛 + 2) + B(𝑛 + 1) At 𝒏 = −𝟏 ➔1 = A(−1 + 2) + 0 → 1 = A(1) , then: A = 1 At 𝒏 = −𝟐 ➔1 = 0 + B(−2 + 1) → 1 = B(−1) , then: B = −1 1 1 1 Then: (𝑛+1)(𝑛+2) = (𝑛+1) − (𝑛+2) 1 1 − ) (𝑛+1) (𝑛+2) ∑∞ 𝑛=0 ( 1 (𝑛+1) ∑∞ 𝑛=0 ( 1 1 1 1 1 1 1 1 1 1 2 2 3 3 4 4 5 ➔ S𝑛 = 1 − 1 𝑛+2 get the limit where 𝒏 approaches to ∞ 1 lim S𝑛 = lim (1 − ) = 1 − lim 𝑛+2 𝑛→∞ Page | 132 1 1 − (𝑛+2)) = ( − ) + ( − ) + ( − ) + ( − ) + ⋯ + ((𝑛+1) − (𝑛+2)) 𝑛→∞ 1 𝑛→∞ 𝑛+2 = 1 − 0 = 1 [Converges] 146] Determine if series converge or diverge , if converge find its 𝟏 value : ∑∞ 𝒏=𝟏 𝟐 𝒏 +𝟒𝒏+𝟑 Solution: ∑∞ 𝑛=1 1 1 𝑛2 +4𝑛+3 1 (𝑛+1)(𝑛+3) = ∑∞ 𝑛=1 (𝑛+1)(𝑛+3) A B = (𝑛+1) + (𝑛+3) By multiplying in two sides by : (𝒏 + 𝟏)(𝒏 + 𝟑) 1 = A(𝑛 + 3) + B(𝑛 + 1) At 𝒏 = −𝟑 ➔1 = 0 + B(−3 + 1) → 1 = B(−2) , then: B = − At 𝒏 = −𝟏 ➔1 = A(−1 + 3) → 1 = A(2) , then: A = 1 Then: (𝑛+1)(𝑛+3) = 1 ∑∞ 𝑛=1 ( 2(𝑛+1) − 1 2(𝑛+1) 1 2(𝑛+3) − 1 2(6) ➔ S𝑛 = 1 2( 2 + ) 1 2( 3 1 1 + ) 2(𝑛+1) 1 1 − − − 1 )+( 2(4) 1 2(8) 1 2(3) 1 − )+( 2(5) Page | 133 5 12 +0−0= 1 2(4) − 1 2(𝑛+3) 12 )+( 2(6) 1 2(5) − 1 2(7) ) get the limit where 𝒏 approaches to ∞ 1 5 1 𝑛→∞ 2(𝑛+1) 5 1 1 1 − ) 2(𝑛+1) 2(𝑛+3) 𝑛→∞ = 2 ) + ⋯+ ( lim S𝑛 = lim ( + + − ) = 12 + lim 4 6 2(𝑛+1) 2(𝑛+3) 𝑛→∞ 1 2(𝑛+3) 2(2) +( 2 1 1 )=( 1 [Converges] − lim 1 𝑛→∞ 2(𝑛+3) 𝟖 𝟐𝟕 𝟔𝟒 147] Test Series 𝟏 − 𝟗 + 𝟐𝟖 − 𝟔𝟓 + ⋯ Solution: 𝑢𝑛 = 𝑛3 𝑛3 +1 ➔ lim 𝑢𝑛 = lim 𝑛→∞ 𝑛→∞ 3 𝑛 = lim 𝑛3 +1 𝑛→∞ 𝑛3 𝑛3 3 𝑛 1 + 𝑛3 𝑛3 = lim 𝑛→∞ 1 1 1+ 3 𝑛 1 = 1+0 = 1 ≠ 0 [Diverges] 𝟏 148] Test Convergence ∑∞ 𝒏=𝟏 𝒏 Solution: ∑∞ 𝑛=1 1 𝑛1 , Since 𝑛 > 0 && 𝑝 = 1 Then: it is divergence 149] Test Convergence ∑∞ 𝒏=𝟏 𝟏 √𝒏 Solution: ∑∞ 𝑛=1 1 √𝑛 = ∑∞ 𝑛=1 1 1 𝑛2 Then: it is divergence Page | 134 1 , Since 𝑛 > 0 && 𝑝 = < 1 2 𝒏 150] Test Convergence ∑∞ 𝒏=𝟏 𝒏𝟐 −𝐜𝐨𝐬𝟐 𝒏 Solution: ∑∞ 𝑛=1 𝑛 𝑛2 −cos2 𝑛 𝑛 𝑛2 −cos2 𝑛 1 𝑛 𝑛 > 𝑛2 = 1 𝑛 by using p-test Since 𝒑 = 𝟏 , then: it is divergence Then: ∑∞ 𝑛=1 𝑛 𝑛2 −cos2 𝑛 is divergence 151] Test Convergence ∑∞ 𝒏=𝟏 𝒏𝟐 +𝟐 𝒏𝟒 +𝟓 Solution: 𝑛2 +2 𝑛4 +5 1 𝑛2 + < 2 𝑛4 𝑛2 +2 𝑛4 𝑛2 =( 𝑛4 + 2 𝑛4 = 𝟏 𝒏𝟐 + 𝟐 𝒏𝟒 ) by using p-test Since 𝒑 = 𝟐 & 𝒑 = 𝟒 , then: convergence + convergence = convergence Then: 𝑛2 +2 𝑛4 +5 Page | 135 is convergence 152] Test Convergence ∑∞ 𝒏=𝟏 𝐬𝐢𝐧 𝒏 𝒏𝟑 Solution: ∑∞ 𝑛=1 sin 𝑛 𝑛3 Since : −1 < sin 𝑛 < 1 ➔ |sin 𝑛| = 1 ∑∞ 𝑛=1 |sin 𝑛| 𝑛3 = ∑∞ 𝑛=1 1 𝑛3 By using p test Since 𝒑 = 𝟑 > 𝟏 , then: it is convergence 𝟏 𝟏 𝟏 𝟏 153] Test Convergence 𝟐 + 𝟓 + 𝟏𝟎 + 𝟏𝟕 + ⋯ Solution: Then: 𝑎𝑛 = ∑∞ 𝑛=1 1 𝑛2 1 𝑛2 +1 1 𝑛2 +1 , by using comparison ➔ 1 𝑛2 +1 by using p test Since 𝒑 = 𝟐 > 𝟏 , then: it is convergence ➔ ∑∞ 𝑛=1 Page | 136 1 𝑛2 +1 is convergence < 1 𝑛2 𝟐𝟒 𝟏 𝟑𝟒 154] Test Convergence 𝒆 + 𝒆𝟒 + 𝒆𝟗 + ⋯ + 𝒏𝟒 𝒆𝒏 𝟐 Solution: 𝑎𝑛 = 𝑎𝑛+1 𝑎𝑛 𝑛4 𝑒𝑛 = 2 , 𝑎𝑛+1 = (𝑛+1)4 𝑒 (𝑛+1)2 ÷ (𝑛+1)4 𝑒 (𝑛+1) 𝑛4 𝑒 = 𝑛2 2 (𝑛+1)4 𝑒 (𝑛+1)2 ∗ 𝑒𝑛 2 𝑛4 =( 𝑛+1 4 𝑛 ) ∗ 𝑒𝑛 𝑒 2 (𝑛+1)2 𝑛+1 4 =( 𝑛+1 4 =( lim | 𝑛→∞ 𝑎𝑛+1 𝑎𝑛 𝑛+1 4 | = lim |( 𝑛+1 4 ) ∗ 𝑒 −2𝑛−1 = ( 1+ 1 ) ∗ 𝑒 2𝑛+1 | = lim |( 𝑛 𝑛→∞ 𝑛 𝑛→∞ 𝒙 𝒙𝟐 𝒙𝟑 1 1 𝑛 Solution: 𝑎𝑛 = 𝑎𝑛+1 𝑎𝑛 = lim | 𝑛→∞ 𝑛! , 𝑎𝑛+1 = 𝑥 𝑛+1 𝑥𝑛 ÷ (𝑛+1)! 𝑛! 𝑎𝑛+1 𝑎𝑛 Page | 137 𝑥 𝑛+1 (𝑛+1)! = | = lim | 𝑥 𝑛 ∗𝑥 (𝑛+1)𝑛! 𝑥 𝑛→∞ 𝑛+1 |=0 ∗ 𝑛! 𝑥𝑛 = 𝑥 𝑛+1 [converge] 𝑛 2 −(𝑛2 +2𝑛+1) 1 ) ∗ 𝑒 2𝑛+1 4 1 ) ∗ 𝑒 2𝑛+1 | = 0 155] Test Convergence 𝟏 + 𝟏! + 𝟐! + 𝟑! + ⋯ 𝑥𝑛 𝑛 ) ∗ 𝑒𝑛 [converge]

Math-2 Final Revision (1)

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