Engg-Maths-MCQ-301 Laplace Transform PDF

Id Question The Laplace transform of 𝐹(𝑡) = 𝑒 𝑏𝑡 𝑠𝑖𝑛𝑎𝑡 is equal to 𝑎 A (𝑠 − 𝑏)2 + 𝑎2 𝑎 B (𝑠 + 𝑏)2 + 𝑎2 C 𝑏 (𝑠 − 𝑏)2 − 𝑎2 D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) =...

Id Question The Laplace transform of 𝐹(𝑡) = 𝑒 𝑏𝑡 𝑠𝑖𝑛𝑎𝑡 is equal to 𝑎 A (𝑠 − 𝑏)2 + 𝑎2 𝑎 B (𝑠 + 𝑏)2 + 𝑎2 C 𝑏 (𝑠 − 𝑏)2 − 𝑎2 D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = 𝒆−𝒃𝒕 𝒄𝒐𝒔𝒉𝒂𝒕 is equal to A 𝑠+𝑏 (𝑠 + 𝑏)2 + 𝑎2 B 𝑠+𝑏 (𝑠 + 𝑏)2 − 𝑎2 C 𝑠−𝑏 (𝑠 − 𝑏)2 + 𝑎2 D None. Answer Marks 1.5 Unit I Id Question If 𝐿{𝑓(𝑡)} = 𝑓(𝑠), then 𝐿{𝑒 −𝑎𝑡 𝑓(𝑡)} is equal to A 𝑓(𝑠) B 𝑓(𝑠 − 𝑎) C 𝑓(𝑠 + 𝑎) D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = 𝒆2𝒕 𝒕3 is equal to A 12 (𝑠 − 4)2 B 12 (𝑠 + 4)2 C 12 (𝑠 + 4)2 D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = 𝒔𝒊𝒏𝒉3𝒕 is equal to A 3 𝑠2 −9 B 3 𝑠2 + 9 C 9 2 𝑠 −9 D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = 𝒕 𝒄𝒐𝒔𝒂𝒕 is equal to A 𝑠 2 − 𝑎2 (𝑠 2 − 𝑎2 )2 B 𝑠 2 − 𝑎2 (𝑠 2 + 𝑎2 )2 C 𝑠 2 + 𝑎2 (𝑠 2 + 𝑎2 )2 D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = 𝒆𝒓𝒇(√𝒕) is equal to A 1 𝑠 √𝑠 2 + 1 B 1 𝑠 √𝑠 − 1 C 1 𝑠 √𝑠 + 1 D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = 𝒆3𝒕 𝒆𝒓𝒇(√𝒕) is equal to A 1 (𝑠 − 3) √𝑠 + 2 B 1 (𝑠 + 3) √𝑠 − 2 C 1 (𝑠 + 3) √𝑠 + 2 D 1 (𝑠 − 3) √𝑠 − 2 Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = 𝒆−𝒕 𝒆𝒓𝒇(√𝒕) is equal to A 1 (𝑠 + 1) √𝑠 B 1 (𝑠 − 1) √𝑠 C 1 (𝑠 + 1) √𝑠 + 2 D None. Answer Marks 1.5 Unit I Id 𝑓(𝑡) Question If 𝐿{𝑓(𝑡)} = 𝑓(𝑠), then 𝐿 ( ) is equal to 𝑡 𝑠 A 𝑓(𝑠) ∫ 𝑑𝑠 𝑠 ∞ ∞ B 𝑓(𝑠) ∫ 𝑑𝑠 𝑠 0 C ∞ 𝑓(𝑠) ∫𝑠 𝑠 𝑑𝑠 D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = (𝒂 + 𝒃𝒕)2 , where a & b are constants, is given by A (𝑎 + 𝑏𝑠)2 B 1 (𝑎 + 𝑏𝑠)2 C 𝑎2 2𝑎𝑏 2𝑏 2 + 2 + 3 𝑠 𝑠 𝑠 D 𝑎2 2𝑎𝑏 𝑏 2 + 2 + 3 𝑠 𝑠 𝑠 Answer Marks 1.5 Unit I Id Question If 𝐿{𝑓(𝑡)} = 𝑓(𝑠), then 𝐿{𝑒 𝑎𝑡 𝑓(𝑡)} is equal to A 𝑓 ̅ (𝑠 + 𝑎) B 𝑓 ̅ (𝑠 − 𝑎) C 𝑒 −𝑠𝑡 𝑓(𝑠) D None. Answer Marks 1.5 Unit I Id −1 Question The Laplace transform of 𝒇(𝒕) = 𝒕 2 is equal to A √𝜋 √𝑠 B √𝑠 √𝜋 C 0 D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = 𝒆∝𝒕 𝐜𝐨𝐬 𝒂𝒕 is equal to A 𝑠−∝ (𝑠−∝)2 + 𝑎2 B 𝑠+∝ (𝑠−∝)2 + 𝑎2 C 1 (𝑠−∝)2 D None. Answer Marks 1.5 Unit I Id 𝒕 𝒔𝒊𝒏𝒕 Question The Laplace transform of 𝑭(𝒕) = ∫0 𝒕 𝒅𝒕 is equal to cot−1 𝑠 A 𝑠 B tan−1 𝑠 𝑠 𝜋 C − tan−1 𝑠 2 D None. Answer Marks 1.5 Unit I Id Question 𝐿{ 𝑐𝑜𝑠ℎ𝑎𝑡 − 𝑐𝑜𝑠ℎ𝑏𝑡 } is equal to 𝑠 𝑠 A − 2 𝑠2 +𝑎 2 𝑠 + 𝑏2 𝑠 𝑠 B − 𝑠 2 − 𝑎2 𝑠 2 − 𝑏 2 C 𝑎 𝑏 − 𝑠 2 − 𝑎2 𝑠 2 − 𝑏 2 D None. Answer Marks 1.5 Unit I Id Question If 𝐿{𝑓(𝑡)} = 𝑓 (𝑠) , then 𝐿{ 𝑡 𝑓(𝑡)} is equal to A 𝑑𝑓 (𝑠) − 𝑑𝑠 ∞ B ∫𝑠 𝑓 (𝑠) 𝑑𝑠 C 𝑠 𝑓 (𝑠) − 𝑓(0) D None. Answer Marks 1.5 Unit I Id 𝑓(𝑡) Question If 𝐿{𝑓(𝑡)} = 𝑓(𝑠), then 𝐿 { 𝑡 } is equal to A 𝑑𝑓 (𝑠) − 𝑑𝑠 ∞ B ∫ 𝑓 (𝑠) 𝑑𝑠 𝑠 C 1 𝑓 (𝑠) 𝑠 D None. Answer Marks 1.5 Unit I Id Question If 𝐿{𝑓(𝑡)} = 𝑓 (𝑠), then 𝐿{ 𝑓(𝑎𝑡)} is equal to A 𝑒 −𝑎𝑠 𝑓 (𝑠) B 𝑓(𝑠 + 𝑎) C 1 𝑠 𝑓̅ ( ) 𝑎 𝑎 D None. Answer Marks 1.5 Unit I Id Question 𝐿{𝑒 −2𝑡 𝑠𝑖𝑛𝑡} is equal to A 1 𝑠2 + 1 B 𝑠+2 (𝑠 + 2)2 + 1 C 1 (𝑠 + 2)2 + 1 D None. Answer Marks 1.5 Unit I Id Question 𝐿{𝑒 −3𝑡 cos 2𝑡} is equal to A 𝑠+3 (𝑠 + 3)2 + 4 B 1 (𝑠 + 3)2 + 4 C 3 (𝑠 + 3)2 + 4 D None. Answer Marks 1.5 Unit I Id 𝑑𝑓 Question If 𝐿{𝑓(𝑡)} = 𝑓(𝑠), then 𝐿 { 𝑑𝑡 } is equal to A 𝑒 −𝑎𝑠 𝑓 (𝑠) B 𝑠 𝑓 (𝑠) − 𝑓(0) C 𝑠 𝑓 (𝑠) + 𝑓(0) D None. Answer Marks 1.5 Unit I Id Question 𝐿{cosh 𝑎𝑡} is equal to A 1 𝑠2 − 𝑎2 𝑎 B 𝑠 2 − 𝑎2 𝑠 C 𝑠 − 𝑎2 2 D None. Answer Marks 1.5 Unit I Id Question The Laplace transform of 𝑭(𝒕) = 𝒆−3𝒕 𝒔𝒊𝒏2𝒕 is equal to A 2 (𝑠 + 3)2 − 4 B 2 (𝑠 + 3)2 + 4 C 2 (𝑠 − 3)2 − 4 D 2 (𝑠 + 3)2 + 4 Answer Marks 1.5 Unit I Id sin 𝑡 𝑑 sin 𝑡 Question If 𝐿 { 𝑡 } = cot −1 𝑠, then 𝐿 {𝑑𝑡 ( 𝑡 )} is equal to A 𝑠 cot −1 𝑠 − 1 B s cot −1 𝑠 C 𝑠 cot −1 𝑠 + 1 D None. Answer Marks 1.5 Unit I Id 1 Question 𝑳−1 { } is equal to √𝒔+3 A 𝑒 −3𝑡 √𝜋𝑡 B 𝑒 3𝑡 √𝜋𝑡 C 𝑒𝑡 √𝜋𝑡 D None. Answer Marks 1.5 Unit II Id 1 Question The inverse Laplace transform of 𝒇(𝒔) = 𝒔2 +2𝒔 is given by A 1 − 𝑒 2𝑡 B 1 + 𝑒 2𝑡 C 1 − 𝑒 2𝑡 2 D 1 − 𝑒 −2𝑡 2 Answer Marks 1.5 Unit II Id 𝒔+1 Question The inverse Laplace transform of 𝒇(𝒔) = 𝐥𝐨𝐠 (𝒔−1) is given by A 2𝑐𝑜𝑠ℎ𝑡 𝑡 B 2𝑡𝑐𝑜𝑠𝑡 C 2𝑠𝑖𝑛ℎ𝑡 𝑡 D None. Answer Marks 1.5 Unit II Id 1 Question The inverse Laplace transform of 𝒇(𝒔) = (𝒔+3)5 is equal to A 𝑒 −3𝑡 𝑡 4 24 3𝑡 4 B 𝑒 𝑡 24 C 𝑒 −3𝑡 𝑡 4 D None. Answer Marks 1.5 Unit II Id 1 Question The inverse Laplace transform of 𝒇(𝒔) = 𝒔2 +4𝒔+13 is equal to A 1 −2𝑡 𝑒 sin 3𝑡 3 B 1 2𝑡 𝑒 sin 3𝑡 3 C 𝑒 −2𝑡 sin 3𝑡 D None. Answer Marks 1.5 Unit II Id 1 5 Question The inverse Laplace transform of 𝒇(𝒔) = (𝒔−4)5 + (𝒔−2)2 +52 is equal to A 𝑡4 𝑒 4𝑡 + 𝑒 2𝑡 𝑠𝑖𝑛5𝑡 24 𝑡4 B 𝑒 4𝑡 24 − 𝑒 2𝑡 𝑠𝑖𝑛5𝑡 C 𝑡3 𝑒 4𝑡 − 𝑒 2𝑡 𝑠𝑖𝑛5𝑡 24 D None. Answer Marks 1.5 Unit II Id 3𝒔+4 Question The inverse Laplace transform of 𝒇(𝒔) = 𝒔2 +9 is equal to A 4 3𝑐𝑜𝑠3𝑡 − 𝑠𝑖𝑛3𝑡 3 B 4 3𝑐𝑜𝑠3𝑡 + 𝑠𝑖𝑛3𝑡 3 C 3𝑐𝑜𝑠3𝑡 + 𝑠𝑖𝑛3𝑡 D None. Answer Marks 1.5 Unit II Id 𝒔2 −3𝒔+4 Question The inverse Laplace transform of 𝒇(𝒔) = is equal to 𝒔3 A 1 − 3𝑡 − 2𝑡 2 B 1 + 3𝑡 + 2𝑡 2 C 1 − 3𝑡 + 2𝑡 2 D None. Answer Marks 1.5 Unit II Id 𝒔+𝒂 Question The inverse Laplace transform of 𝒇(𝒔) = 𝐥𝐨𝐠 𝒔+𝒃 is equal to A 1 𝑎𝑡 (𝑒 − 𝑒 𝑏𝑡 ) 𝑡 B 1 −𝑎𝑡 (𝑒 + 𝑒 −𝑏𝑡 ) 𝑡 C 1 −𝑎𝑡 (𝑒 − 𝑒 −𝑏𝑡 ) 𝑡 D 1 − (𝑒 −𝑎𝑡 − 𝑒 −𝑏𝑡 ) 𝑡 Answer Marks 1.5 Unit II Id Question The inverse Laplace transform of 𝑓(𝒂𝒔) is equal to 1 𝑡 A 𝑎 𝑓 (𝑎 ) 1 𝑎 B 𝑎 𝑓 (𝑡 ) C 𝑡 𝑓( ) 𝑎 D None. Answer Marks 1.5 Unit II Id Question The inverse Laplace transform of 𝒇(𝒔) = 𝐜𝐨𝐭 −1 (𝒔) is equal to A 𝑠𝑖𝑛𝑡 𝑡 B 𝑐𝑜𝑠𝑡 𝑡 C 𝑠𝑖𝑛𝑡 D None. Answer Marks 1.5 Unit II Id 2 Question The inverse Laplace transform of 𝒇(𝒔) = 𝐭𝐚𝐧−1 (𝒔 ) is equal to A −1 𝑠𝑖𝑛2𝑡 𝑡 B 𝑠𝑖𝑛2𝑡 C 1 𝑠𝑖𝑛2𝑡 𝑡 D None. Answer Marks 1.5 Unit II Id 𝒔+3 Question The inverse Laplace transform of 𝒇(𝒔) = (𝒔+3)2 +4 is equal to A 𝑒 −3𝑡 𝑠𝑖𝑛2𝑡 B 𝑒 3𝑡 𝑠𝑖𝑛2𝑡 C 𝑒 −3𝑡 𝑐𝑜𝑠2𝑡 D None. Answer Marks 1.5 Unit II Id ̅̅̅̅̅ 3 Question |( ) 2 The inverse Laplace transform of 𝒇(𝒔) = 3 is equal to 𝒔2 3 A 𝑡2 −3 B 𝑡 2 1 C 𝑡2 D None. Answer Marks 1.5 Unit II Id 1 Question The inverse Laplace transform of 𝑓(𝑠) = (𝑠+3)2 is equal to A 𝑡 𝑒 −3𝑡 B 𝑒 −3𝑡 C 𝑡 −3𝑡 𝑒 2 D None. Answer Marks 1.5 Unit II Id 4𝑠 Question The inverse Laplace transform of 𝑓(𝑠) = 𝑠2 +16 is equal to A cos 4𝑡 B 4 cos 4𝑡 C 4 𝑠𝑖𝑛 4𝑡 D None. Answer Marks 1.5 Unit II Id 3 Question The inverse Laplace transform of 𝑓(𝑠) = 𝑠2 +25 is equal to A 3 𝑠𝑖𝑛 5𝑡 B 3 𝑠𝑖𝑛 5𝑡 5 C 3 cos 5𝑡 5 D None. Answer Marks 1.5 Unit II Id 1 Question The inverse Laplace transform of 𝑓(𝑠) = 3𝑠−4 is equal to A 1 −4𝑡 𝑒 3 3 4 B 𝑒 −3𝑡 C 1 4𝑡 𝑒3 3 D None. Answer Marks 1.5 Unit II Id 2 Question The inverse Laplace transform of 𝑓(𝑠) = 𝑠+2 is equal to A 2 𝑒 2𝑡 B 𝑒 −2𝑡 C 2 𝑒 −2𝑡 D None. Answer Marks 1.5 Unit II Id Question 𝑓(𝑠) If 𝐿−1 {𝑓(𝑠)} = 𝑓(𝑡), then 𝐿−1 { } is equal to 𝑠 𝑡 A ∫ 𝑓(𝑡) 𝑑𝑡 0 B −𝑡 𝑓(𝑡) C 1 𝑓(𝑡) 𝑡 D None. Answer Marks 1.5 Unit II Id 𝑠+𝑎 Question The inverse Laplace transform of 𝑓(𝑠) = log (𝑠+𝑏) is equal to A 𝑒 −𝑎𝑡 − 𝑒 −𝑏𝑡 𝑡 B 𝑒 −𝑏𝑡 − 𝑒 −𝑎𝑡 𝑡 C 𝑒 −𝑏𝑡 + 𝑒 −𝑎𝑡 𝑡 D None. Answer Marks 1.5 Unit II Id 3 Question The inverse Laplace transform of 𝑓(𝑠) = 𝑠4 is equal to A 𝑡3 B 𝑡3 3 C 𝑡3 2 D None. Answer Marks 1.5 Unit II Id 2𝑠+1 Question The inverse Laplace transform of 𝑓(𝑠) = 𝑠3 is equal to A 𝑡2 2𝑡 − 2 B 𝑡2 2𝑡 + 3 C 𝑡2 2𝑡 + 2 D None. Answer Marks 1.5 Unit II Id 1 Question The inverse Laplace transform of 𝑓(𝑠) = 3𝑠2 +27 is equal to A 1 𝑠𝑖𝑛 3𝑡 3 B 1 cos 3𝑡 3 C 1 𝑠𝑖𝑛 3𝑡 9 D None. Answer Marks 1.5 Unit II Id 1 Question The inverse Laplace transform of 𝒇(𝒔) = (𝒔+2)(𝒔−1) is equal to A 1 𝑡 (𝑒 − 𝑒 −2𝑡 ) 3 B 1 𝑡 (𝑒 + 𝑒 −2𝑡 ) 3 C 1 𝑡 (𝑒 − 𝑒 −2𝑡 ) 3 D None. Answer Marks 1.5 Unit II Id Question The Fourier cosine transform of 𝒇(𝒙) = 2𝒆−5𝒙 + 5𝒆−2𝒙 is A 10 10 + 2 𝑠2 + 25 𝑠 + 4 B 10 10 2 − 2 𝑠 + 25 𝑠 + 4 C 10 10 2 − 2 𝑠 − 25 𝑠 − 4 D None. Answer Marks 1.5 Unit III Id Question If 𝐹(𝑠) is the Fourier transform of 𝑓(𝑥) , then the Fourier transform of 𝑓(𝑎𝑥) is A 1 𝑠 𝐹( ) 𝑎 𝑎 𝑠 B 𝐹( ) 𝑎 C 1 𝐹(𝑠) 𝑎 D None. Answer Marks 1.5 Unit III Id Question The Fourier cosine transform of the function 𝑓(𝑡) is ∞ A 𝐹𝑐 (𝑠) = ∫ 𝑓(𝑡) cos 𝑠𝑡 0 ∞ B 𝐹𝑐 (𝑠) = ∫ 𝑓(𝑡) cos 𝑡 𝑑𝑡 0 ∞ C 𝐹𝑐 (𝑠) = ∫ 𝑓(𝑠𝑡) cos 𝑡 𝑑𝑡 0 D None. Answer Marks 1.5 Unit III Id Question Which of the following is correct representation of Fourier transform ∞ A F(s) = ∫ f(x)eisx dx −∞ ∞ B 1 F(s) = ∫ f(s)eisx ds 2π −∞ ∞ C 1 F(s) = ∫ f(s)eisx ds 2π 0 D None. Answer Marks 1.5 Unit III Id Question The Fourier sine transform is represented by ∞ A 𝐹𝑠 (𝑠) = ∫ 𝑓(𝑡) cos(𝑠𝑡) 𝑑𝑡 −∞ ∞ B 𝐹𝑠 (𝑠) = ∫ 𝑓(𝑡) sin(𝑠𝑡) 𝑑𝑡 0 ∞ C 𝐹𝑠 (𝑠) = ∫ 𝑓(𝑡) 𝑠𝑖𝑛(𝑠𝑡) 𝑑𝑡 −∞ D None. Answer Marks 1.5 Unit III Id Question If 𝐹{𝑓(𝑥)} = 𝐹(𝑠) , then 𝐹{𝑓(𝑥 − 𝑎)} is equal to A 𝑒 𝑖𝑠𝑎 B 𝑒 𝑖𝑠𝑎 𝐹(𝑠) C Both (𝑎) & (𝑏) D None. Answer Marks 1.5 Unit III Id Question The Fourier cosine transform of 𝒆−𝒙 is 𝑠 A 𝑠2 +1 𝑠 B 𝑠2 − 1 C 1 𝑠2 + 1 D None. Answer Marks 1.5 Unit III Id Question If 𝐹{𝑓(𝑥)} = 𝐹(𝑠) and 𝐹{𝑔(𝑥)} = 𝐺(𝑠) , then by parseval's identity 1 ∞ ̅̅̅̅̅̅ 𝑑𝑠 is equal to ∫ 𝐹(𝑠)𝐺(𝑠) 2𝜋 −∞ ∞ A ̅̅̅̅̅̅ 𝑑𝑥 ∫ 𝑓(𝑥)𝑔(𝑥) 0 1 ∞ B 2𝜋 −∞ ∫ 𝑓(𝑥) 𝑔(𝑥) 𝑑𝑥 ∞ C ̅̅̅̅̅̅ 𝑑𝑥 ∫ 𝑓(𝑥)𝑔(𝑥) −∞ D None. Answer Marks 1.5 Unit III Id 1 ∞ Question If 𝐹{𝑓(𝑥)} = 𝐹(𝑠) , then by parseval's identity ∫ [𝐹(𝑠)]2 2𝜋 −∞ 𝑑𝑠 is equal to ∞ A 1 ∫ [𝑓(𝑥)]2 𝑑𝑥 𝜋 0 ∞ B ∫ [𝑓(𝑥)]2 𝑑𝑥 −∞ ∞ C 2 2π ∫ (f(x)) dx 0 D None. Answer Marks 1.5 Unit III Id Question The Parseval's identities for Fourier cosine transform is ∞ ∞ A 2 ∫ 𝐹𝑐 (𝑠)𝐺𝑐 (𝑠)𝑑𝑠 = ∫ 𝑓(𝑥) 𝑔(𝑥)𝑑𝑥 𝜋 0 0 ∞ ∞ B ∫ 𝐹𝑐 (𝑠)𝐺𝑐 (𝑠)𝑑𝑠 = ∫ 𝑓(𝑥) 𝑔(𝑥)𝑑𝑥 0 0 ∞ ∞ C 2 ∫ 𝐹𝑐 (𝑠)𝐺𝑐 (𝑠)𝑑𝑠 = ∫ 𝑓(𝑥) 𝑔(𝑥)𝑑𝑥 𝜋 −∞ −∞ D None. Answer Marks 1.5 Unit III Id Question The Parseval's identity for Fourier sine transform is ∞ ∞ A 2 ∫ {𝐹𝑠 (𝑠)}2 𝑑𝑠 = ∫ {𝑓(𝑥)}2 𝑑𝑥 𝜋 0 0 ∞ ∞ B 2 ∫ {𝐹𝑠 (𝑠)}2 𝑑𝑠 = ∫ {𝑓(𝑥)}2 𝑑𝑥 𝜋 −∞ 0 ∞ ∞ C ∫ {𝐹𝑠 (𝑠)}2 𝑑𝑠 = ∫ {𝑓(𝑥)}2 𝑑𝑥 0 0 D None. Answer Marks 1.5 Unit III Id Question The inverse Fourier sine transform is given by ∞ A 1 𝑓(𝑥) = ∫ 𝐹𝑠 (𝑠) sin(𝑠𝑥)𝑑𝑠 𝜋 0 ∞ B 2 𝑓(𝑥) = ∫ 𝐹𝑠 (𝑠) sin(𝑠𝑥)𝑑𝑠 𝜋 0 ∞ C 2 𝑓(𝑥) = ∫ 𝐹𝑠 (𝑠) cos(𝑠𝑥)𝑑𝑠 𝜋 0 ∞ D 𝑓(𝑥) = ∫ 𝐹𝑠 (𝑠) sin(𝑠𝑥)𝑑𝑠 0 Answer Marks 1.5 Unit III Id Question The inverse Fourier cosine transform is ∞ A 𝑓(𝑥) = ∫ 𝐹𝑐 (𝑠) sin(𝑠𝑥)𝑑𝑠 0 ∞ B 2 𝑓(𝑥) = ∫ 𝐹𝑐 (𝑠) cos(𝑠𝑥)𝑑𝑠 𝜋 −∞ ∞ C 2 𝑓(𝑥) = ∫ 𝐹𝑐 (𝑠) cos(𝑠𝑥)𝑑𝑠 𝜋 0 D None. Answer Marks 1.5 Unit III Id 𝑠 Question If 𝐹𝑐 {𝑓(𝑎𝑥)} = 𝑘𝐹𝑐 (𝑎) , then k is equal to A 2 𝑎 B 𝑎 C 1 𝑎 D None. Answer Marks 1.5 Unit III Id Question In the Fourier integral representation of the function ∞ 𝑓(𝑥) = ∫0 [𝐴(𝜆) cos 𝜆𝑥 + 𝐵(𝜆) sin 𝜆𝑥] 𝑑𝜆 , 𝐴(𝜆) is given by ∞ A ∫ 𝑓(𝑡) cos 𝜆𝑡 𝑑𝑡 −∞ B 1 ∞ ∫ 𝑓(𝑡) cos 𝜆𝑡 𝑑𝑡 𝜋 −∞ ∞ C ∫ 𝑓(𝑡) sin 𝜆𝑡 𝑑𝑡 −∞ D None. Answer Marks 1.5 Unit III Id Question In the Fourier integral representation of the function ∞ 𝑓(𝑥) = ∫0 [𝐴(𝜆) cos 𝜆𝑥 + 𝐵(𝜆) sin 𝜆𝑥] 𝑑𝜆 , 𝐵(𝜆) is given by A 1 ∞ ∫ 𝑓(𝑡) sin 𝜆𝑡 𝑑𝑡 𝜋 −∞ B 1 ∞ ∫ 𝑓(𝑡) cos 𝜆𝑡 𝑑𝑡 𝜋 −∞ ∞ C ∫ 𝑓(𝑡) sin 𝜆𝑡 𝑑𝑡 −∞ D None. Answer Marks 1.5 Unit III Id Question In the Fourier cosine integral representation of the function ∞ 𝑓(𝑥) = ∫0 𝐴(𝜆) cos 𝜆𝑥 𝑑𝜆 , 𝐴(𝜆) is given by A 2 ∞ ∫ 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥 𝜋 0 B 1 ∞ ∫ 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥 𝜋 0 ∞ C ∫ 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥 0 D None. Answer Marks 1.5 Unit III Id Question In the Fourier sine integral representation of the function 𝑓(𝑥) = ∞ ∫0 𝐵(𝜆) sin 𝜆𝑥 𝑑𝜆 , 𝐵(𝜆) is given by ∞ A ∫0 𝑓(𝑥) sin 𝜆𝑥 𝑑𝑥 B 1 ∞ ∫ 𝑓(𝑥) sin 𝜆𝑥 𝑑𝑥 𝜋 0 C 2 ∞ ∫ 𝑓(𝑥) sin 𝜆𝑥 𝑑𝑥 𝜋 0 D None. Answer Marks 1.5 Unit III Id Question The Fourier integral theorem is given by A 1 ∞ ∞ 𝑓(𝑥) = ∫ ∫ 𝑓(𝑡) cos[𝜆(𝑡 − 𝑥)] 𝑑𝑡 𝑑𝜆 𝜋 0 −∞ B 1 ∞ ∞ 𝑓(𝑥) = ∫ ∫ 𝑓(𝑡) cos[𝜆(𝑡 − 𝑥)] 𝑑𝑡 𝑑𝜆 𝜋 −∞ −∞ C 1 ∞ ∞ 𝑓(𝑥) = ∫ ∫ 𝑓(𝑡) cos[𝜆(𝑡 − 𝑥)] 𝑑𝑡 𝑑𝜆 2𝜋 −∞ −∞ D None. Answer Marks 1.5 Unit III Id Question If the Fourier transform of 𝑓(𝑥) is 𝐹(𝑠), then 𝐹(𝑠) is equal to ∞ A 𝐹(𝑠) = ∫ 𝑓(𝑡) 𝑒 −𝑖𝑠𝑡 𝑑𝑡 −∞ ∞ B 𝐹(𝑠) = ∫ 𝑓(𝑡) 𝑒 𝑖𝑠𝑡 𝑑𝑡 −∞ C 1 ∞ 𝐹(𝑠) = ∫ 𝑓(𝑡) 𝑒 𝑖𝑠𝑡 𝑑𝑡 𝜋 −∞ D None. Answer Marks 1.5 Unit III Id Question If the Fourier transform of 𝑓(𝑥) is 𝐹(𝑠), then 𝑓(𝑥) is equal to A 1 ∞ 𝑓(𝑥) = ∫ 𝐹(𝑠) 𝑒 −𝑖𝑠𝑥 𝑑𝑠 2𝜋 −∞ B 1 ∞ 𝑓(𝑥) = ∫ 𝐹(𝑠) 𝑒 −𝑖𝑠𝑥 𝑑𝑠 𝜋 0 ∞ C 𝑓(𝑥) = ∫ 𝐹(𝑠) 𝑒 𝑖𝑠𝑥 𝑑𝑠 −∞ D None. Answer Marks 1.5 Unit III Id 1 Question The Fourier cosine transform of 𝑓(𝑥) = 1+𝑥 2 is equal to 𝜋 𝑠 A 𝑒 4 𝜋 𝑠 B 𝑒 2 𝜋 −𝑠 C 𝑒 2 D None. Answer Marks 1.5 Unit III Id ∞ sin 𝑡 Question The value of ∫0 𝑡 𝑑𝑡 is equal to 𝜋 A 4 𝜋 B 2 C 0 D None. Answer Marks 1.5 Unit III Id Question If the Fourier cosine transform of 𝑓(𝑥) is 𝐹𝑐 (𝑠), then A 1 𝑠 𝐹𝑐 {𝑓(𝑎𝑥)} = 𝐹𝑠 ( ) 𝑎 𝑎 B 1 𝑠 𝐹𝑐 {𝑓(𝑎𝑥)} = 𝐹𝑐 ( ) 𝑎 𝑎 𝑠 C 𝐹𝑐 {𝑓(𝑎𝑥)} = 𝐹𝑐 ( ) 𝑎 D None. Answer Marks 1.5 Unit III Id Question The Fourier cosine transform of 𝒆−𝒙 is 𝑠 A 𝑠2 +1 B 1 𝑠2 + 1 𝑠 C 𝑠2 − 1 D None. Answer Marks 1.5 Unit III Id 2 2 Question The order of the partial differential equation 𝜕z + 𝜕 2z + 𝜕 z = 1 is 𝜕x 𝜕x 𝜕x 𝜕y A 1 B 2 C 3 D None. Answer Marks 1.5 Unit IV Id 2 Question The degree of the partial differential equation 𝜕z + 𝜕 z2 = 1 is. 𝜕x 𝜕y A 2 B 0 C 1 D None. Answer Marks 1.5 Unit IV Id Question The degree of the partial differential equation 2 𝜕2 z 𝜕z 𝜕z a 2 [𝜕x2 + 𝜕y] + 𝜕y = sin(x + y) is A 1 B 2 C 3 D None. Answer Marks 1.5 Unit IV Id Question The order of the partial differential equation 𝜕2 z 𝜕z 2 𝜕x2 + (𝜕y) = 1 is A 2 B 0 C 1 D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating a& b from z = ax + (1 − a)y + b is 𝜕z 𝜕z A 𝜕x + 𝜕y = 1 𝜕z 𝜕z B 𝜕x − 𝜕y = 1 𝜕z 𝜕z C 𝜕x + 𝜕y = 0 D None Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating a& b from z = ax + by + ab is A z = xp + yq − pq B z = xp + yq + pq C z = xp − yq − pq D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating a and b from z = (x2 + a2 )(y2 + b2 ) is A 2xyz = pq B xyz = pq C 4xyz = pq D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating a and b from z = ax3 + by3 is A z = xp + yq B z = xp + yq + pq C 3z = xp + yq D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating the arbitrary function f from z = f(y2 − x2 ) is A yp + xq = 0 B yp − xq = 0 C xp + yq = 0 D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating the arbitrary function f from z = x + y + f(xy) is A px − qy = x − y B px + qy = x + y C py − qx = x + y D None. Answer Marks 1.5 Unit IV Id Question The general solution of 3p + 4q = 7 is given by A ɸ(4x − 3y ,7x − 3z) = 0 B ɸ(4x + 3y ,7x + 3z) = 0 C ɸ(4x − 3y ,7x + 3z) = 0 D None Answer Marks 1.5 Unit IV Id Question The general solution of xp + yq = z is given by x y A ɸ (y , z) = 0 B ɸ(xy , z ) = 0 C ɸ(xy , yz ) = 0 D None Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating arbitrary function from z = f(x + it) + g(x − it) is 𝜕2 z 𝜕2 z A 2 + 2 = 0 𝜕x 𝜕t 𝜕2 z 𝜕2 z B + 𝜕y2 = 0 𝜕x2 𝜕2 z 𝜕2 z C − 𝜕t2 = 0 𝜕x2 D None Answer Marks 1.5 Unit IV Id Question The partial differential equation for one dimensional heat equation is 𝜕2 u 𝜕u A 2 = 𝜕t 𝜕x 𝜕u 2 B 2𝜕 u 𝜕t =c 𝜕x2 𝜕2 u 𝜕2 u C = c 2 𝜕x2 𝜕t2 D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating the function from z = f(x2 − y2 ) A yp + xq = 0 B xp − yq = 0 C xp + yq = 0 D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating the function from z = eny ∅(x − y) A p − q = nz B p+q=n C p + q = nz D None. Answer Marks 1.5 Unit IV Id Question The general solution of 2p + 3q = a is given by A ɸ(3x − 2y , ay − 3z) = 0 B ɸ(3x + 2y , ay − 3z) = 0 C ɸ(3x − 2y , ay + 3z) = 0 D None. Answer Marks 1.5 Unit IV Id Question The general solution of zp = −x is given by A ɸ(x 2 + z 2 , y ) = 0 B ɸ(x 2 − z 2 , y ) = 0 C ɸ(x 2 + z 2 , 2 y ) = 0 D None. Answer Marks 1.5 Unit IV Id Question Temperature distribution of the plate in unsteady state is given by the equation 𝜕u 𝜕2 u 𝜕2 u A = c2 ( 2 + 2 ) 𝜕t 𝜕x 𝜕y 𝜕2 u 𝜕2 u 𝜕2 u B = c 2 ( 𝜕x2 + 𝜕y2 ) 𝜕t2 𝜕2 u 2 𝜕 u 2 𝜕2 u 𝜕2 u =c ( + + ) 𝜕t2 𝜕x 2 𝜕y 2 𝜕z2 C D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation for one dimensional wave equation is 𝜕2 y 𝜕y A 2 = 𝜕t 𝜕x 𝜕y 𝜕2 y B = c 2 𝜕x2 𝜕t 𝜕2 y 𝜕2 y C = c 2 𝜕x2 𝜕t2 D None. Answer Marks 1.5 Unit IV Id Question The Laplace equation in two dimension is 𝜕2 u 𝜕2 u A 2 + 2 = 0 𝜕x 𝜕y 𝜕2 u 𝜕2 u B − 𝜕y2 = 0 𝜕x2 𝜕2 u 𝜕2 u C = 𝜕x2 𝜕y2 D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation obtained by eliminating the constants a and b from z = (x2 − a)(y2 − b) is 𝜕z 𝜕z A 4xyz = (𝜕x) (𝜕y) 𝜕z 𝜕z B 4 = (𝜕x) (𝜕y) 𝜕z 𝜕z C 4xy = (𝜕x) (𝜕y) D None. Answer Marks 1.5 Unit IV Id Question The partial differential equation formed by eliminating the function y f from z = f (x) is 𝜕z 𝜕z A y (𝜕x) + x (𝜕y) = 0 𝜕z 𝜕z B (𝜕x) + (𝜕y) = 0 𝜕z 𝜕z C x (𝜕x) + y (𝜕y) = 0 D None. Answer Marks 1.5 Unit IV Id Question The general solution of the one dimensional heat flow equation 𝜕u 2 2𝜕 u 𝜕t = C 𝜕x2 is 2 2 A u = (c1 emx + c2 e−mx )c3 em c t B u = c1 (c2 x + c3 ) 2 c2 C u = (c1 cosmx + c2 sinmx)c3 e−m t D None. Answer Marks 1.5 Unit IV Id Question If u = c1 , v = c2 are the two solutions of Pp + Qq = R, then its general solution will be A ∅(u, v) = 1 B ∅(u, v) = −1 C ∅(u, v) = 0 D None. Answer Marks 1.5 Unit IV Id 𝒅2 𝒚 𝒅𝒚 Question The differential equation 𝒙2 𝒅𝒙2 + 𝒙 (𝒅𝒙) + (𝒙2 − 25)𝒚 = 0 is called A Bessel's differential equation of order 5 B Bessel's differential equation of order 4 C Bessel's differential equation of order 2 D None. Answer Marks 1.5 Unit V Id Question 𝑱−1 (𝒙) is equal to 2 A 2 √( ) 𝑐𝑜𝑠𝑥 𝜋𝑥 B 2 √( ) 𝑠𝑖𝑛𝑥 𝜋𝑥 C 𝜋𝑥 √( ) 𝑐𝑜𝑠𝑥 2 D None. Answer Marks 1.5 Unit V Id Question 𝑱1 (𝒙) is equal to 2 A 2 √( ) 𝑐𝑜𝑠𝑥 𝜋𝑥 B 2 √( ) 𝑠𝑖𝑛𝑥 𝜋𝑥 C 𝜋𝑥 √( ) 𝑐𝑜𝑠𝑥 2 D None. Answer Marks 1.5 Unit V Id 2 2 Question [ 𝑱1 (𝒙)] + [𝑱−1 (𝒙)] is equal to 2 2 2 A 𝜋𝑥 𝜋𝑥 B 2 C 1 𝜋𝑥 D None. Answer Marks 1.5 Unit V Id 𝒅 Question 𝒅𝒙 { 𝒙𝒏 𝑱𝒏 (𝒙)} is equal to A 𝑥 𝑛 𝐽𝑛 (𝑥) B 𝑥 𝑛 𝐽𝑛−1 (𝑥) C 𝑥 𝑛 𝐽𝑛+1 (𝑥) D None. Answer Marks 1.5 Unit V Id 𝒅 Question 𝒅𝒙 { 𝒙−𝒏 𝑱𝒏 (𝒙)} is equal to A −𝑥 −𝑛 𝐽𝑛 (𝑥) B −𝑥 𝑛 𝐽𝑛+1 (𝑥) C −𝑥 −𝑛 𝐽𝑛+1 (𝑥) D None. Answer Marks 1.5 Unit V Id Question The value of 𝑱−𝒏 (𝒙) is A (−1)𝑛 𝐽𝑛 (𝑥) B −1)𝑛−1 𝐽𝑛 (𝑥) C (−1)𝑛 𝐽𝑛+1 (𝑥) D None. Answer Marks 1.5 Unit V Id Question Which recurrence relation is true A 2𝑛 𝐽𝑛+1 (𝑥) = 𝐽 (𝑥) − 𝐽𝑛−1 (𝑥) 𝑥 𝑛 𝑛 B 𝐽𝑛+1 (𝑥) = 𝐽𝑛 (𝑥) − 𝐽𝑛−1 (𝑥) 𝑥 C 2𝑛 𝐽𝑛+1 (𝑥) = 𝐽 (𝑥) + 𝐽𝑛−1 (𝑥) 𝑥 𝑛 D None. Answer Marks 1.5 Unit V Id Question The Bessel equation of order zero is A 𝑑2 𝑦 𝑑𝑦 𝑥2 2 −𝑥 + (𝑥 2 − 𝑛2 )𝑦 = 0 𝑑𝑥 𝑑𝑥 B 𝑑2 𝑦 𝑑𝑦 𝑥 2+ − 𝑥𝑦 = 0 𝑑𝑥 𝑑𝑥 C 𝑑2 𝑦 𝑑𝑦 𝑥 2+ + 𝑥𝑦 = 0 𝑑𝑥 𝑑𝑥 D None. Answer Marks 1.5 Unit V Id Question The value of 𝑱0 (0) is A 0 B -1 C 1 D None. Answer Marks 1.5 Unit V Id Question Which recurrence relation is false 𝑛 A 𝐽𝑛 ′ (𝑥) + 𝐽𝑛 (𝑥) = 𝐽𝑛−1 (𝑥) 𝑥 𝑛 B ′ 𝐽𝑛 (𝑥) − 𝐽𝑛 (𝑥) = −𝐽𝑛+1 (𝑥) 𝑥 C 2𝐽𝑛 ′ (𝑥) = 𝐽𝑛−1 (𝑥) − 𝐽𝑛+1 (𝑥) D None. Answer Marks 1.5 Unit V Id Question 𝑱−𝒏 (𝒙) is equal to ∞ A (−1)𝑟 (𝑥)𝑛+2𝑟 ∑ (2)𝑛+2𝑟 𝛤𝑛 + 𝑟 + 1 𝑟=0 ∞ B (−1)𝑟 (𝑥)−𝑛+2𝑟 ∑ (2)−𝑛+2𝑟 𝛤 − 𝑛 + 𝑟 + 1 𝑟=0 ∞ C (−1)𝑟 (𝑥)2𝑟 ∑ (2)2𝑟 𝛤𝑛 + 𝑟 + 1 𝑟=0 D None. Answer Marks 1.5 Unit V Id Question If ∝ 𝑎𝑛𝑑 𝛽 are the roots of the equation 𝐽𝑛 (𝑥) = 0 , then the value of integral 1 ∫0 𝒙 𝑱𝒏 (∝ 𝒙)𝑱𝒏 (𝜷𝒙) 𝒅𝒙 if ∝ ≠ 𝜷 is A 0 B 1 1 C [𝐽𝑛+1 (∝)]2 2 D None. Answer Marks 1.5 Unit V Id Question If ∝ 𝑎𝑛𝑑 𝛽 are the roots of the equation 𝐽𝑛 (𝑥) = 0 , then the value of integral 1 ∫0 𝒙 𝑱𝒏 (∝ 𝒙)𝑱𝒏 (𝜷𝒙) 𝒅𝒙 if ∝ = 𝜷 is 1 A [𝐽𝑛+1 (∝)]2 2 B [𝐽𝑛+1 (∝)]2 1 C [𝐽 (∝)]2 2 𝑛−1 D None. Answer Marks 1.5 Unit V Id Question The value of 𝑱1 (𝒙) is 2 A 𝐽−1 (𝑥) 𝑡𝑎𝑛𝑥 2 B 𝐽−1 (𝑥) 𝑠𝑖𝑛𝑥 2 C 𝐽−1 (𝑥) 𝑐𝑜𝑡𝑥 2 D None. Answer Marks 1.5 Unit V Id Question 𝑱−5 (𝒙) is equal to 2 A 2 √( ) {( 3−𝑥 2 3 ) 𝑐𝑜𝑠𝑥 + 𝑥 𝑠𝑖𝑛𝑥} 𝜋𝑥 𝑥2 B 2 √( ) { ( 3+𝑥 2 3 ) 𝑐𝑜𝑠𝑥 + 𝑥 𝑠𝑖𝑛𝑥} 𝜋𝑥 𝑥2 C 2 √( ) {( 3−𝑥 2 3 ) 𝑠𝑖𝑛𝑥 − 𝑥 𝑐𝑜𝑠𝑥} 𝜋𝑥 𝑥2 D None. Answer Marks 1.5 Unit V Id 𝒅 Question 𝑱 (𝒙) 𝒅𝒙 0 is equal to A 𝐽1 (𝑥) B −𝐽1 (𝑥) C 𝐽0 (𝑥) D None. Answer Marks 1.5 Unit V Id Question 𝑱5 (𝒙) is equal to 2 A 2 √( ) {( 3−𝑥 2 3 ) 𝑠𝑖𝑛𝑥 − 𝑥 𝑐𝑜𝑠𝑥} 𝜋𝑥 𝑥2 B 2 √( ) {( 3−𝑥 2 3 ) 𝑠𝑖𝑛𝑥 + 𝑥 𝑐𝑜𝑠𝑥} 𝜋𝑥 𝑥2 C 2 √( ) {( 3−𝑥 2 1 ) 𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥} 𝜋𝑥 𝑥2 𝑥 D None. Answer Marks 1.5 Unit V Id Question 𝑱3 (𝒙) is equal to 2 A 2 √( ) {( 𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥)} 𝜋𝑥 𝑥 B 2 √( ) {( 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)} 𝜋𝑥 𝑥 C 2 √( ) {( 𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥)} 𝜋𝑥 𝑥 D None. Answer Marks 1.5 Unit V Id Question 𝑱−3 (𝒙) is equal to 2 A 2 √( ) {(− 𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥)} 𝜋𝑥 𝑥 B 2 √( ) {( 𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥)} 𝜋𝑥 𝑥 C 2 √( ) {( 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)} 𝜋𝑥 𝑥 D None. Answer Marks 1.5 Unit V Id Question 𝑱4 (𝒙) is equal to A 48 8 24 ( 3 − ) 𝐽1 (𝑥) − ( 2 − 1) 𝐽0 (𝑥) 𝑥 𝑥 𝑥 48 8 24 B (𝑥 3 − 𝑥) 𝐽1 (𝑥) + (𝑥 2 − 1) 𝐽0 (𝑥) C 48 8 24 ( 3 − ) 𝐽0 (𝑥) − ( 2 − 1) 𝐽1 (𝑥) 𝑥 𝑥 𝑥 D None. Answer Marks 1.5 Unit V Id Question 4 𝑱𝒏 ʺ (𝒙) is equal to A 𝐽𝑛−2 (𝑥) − 2𝐽𝑛 (𝑥) + 𝐽𝑛+2 (𝑥) B 𝐽𝑛−2 (𝑥) + 2𝐽𝑛 (𝑥) + 𝐽𝑛+2 (𝑥) C 𝐽𝑛−2 (𝑥) + 2𝐽𝑛 (𝑥) − 𝐽𝑛+2 (𝑥) D None. Answer Marks 1.5 Unit V Id 𝑑 Question 𝑑𝑥 [𝑥 𝐽1 (𝑥)] is equal to A 𝑥 𝐽0 (𝑥) B 𝐽0 (𝑥) C 𝐽1 (𝑥) D None. Answer Marks 1.5 Unit V Id 𝒅 Question 𝒅𝒙 [𝒙𝒏 𝑱𝒏 (𝒂𝒙)] is equal to A 𝑎 𝑥 𝑛 𝐽𝑛−1 (𝑎𝑥) B 𝑎 𝑥 −𝑛 𝐽𝑛−1 (𝑎𝑥) C 𝑎 𝑥 𝑛 𝐽𝑛+1 (𝑥) D None. Answer Marks 1.5 Unit V Id Question 𝑱1 ʺ (𝒙) is equal to A 1 𝐽1 (𝑥) − 𝐽2 (𝑥) 𝑥 B 1 𝐽1 (𝑥) + 𝐽2 (𝑥) 𝑥 C 𝐽1 (𝑥) − 𝐽2 (𝑥) D None. Answer Marks 1.5 Unit V Id Question Which of the following functions is an analytic function A f(z) = z̅ B f(z) = sinz C f(z) = Im(z) D None. Answer Marks 1.5 Unit VI Id Question The function f(z) = |z|2 is analytic at A everywhere B no where C origin D None Answer Marks 1.5 Unit VI Id Question If f(z)= u + iv is an analytic function , then f ′(z) is equal to 𝜕u 𝜕v A 𝜕x − i 𝜕x B 𝜕u 𝜕v +i 𝜕x 𝜕x 𝜕u 𝜕v C 𝜕x − i 𝜕y D None Answer Marks 1.5 Unit VI Id Question If the function u = ax3 + bx2 y + cxy 2 + dy3 is to be harmonic, if A c = 3d and b = 3a B c = −3a and b = −3d C c = 3a and b = 3d D None Answer Marks 1.5 Unit VI Id Question If the function 2x + x2 +∝ y2 is to be harmonic , then the value of ∝ will be A -1 B 1 C 2 D None Answer Marks 1.5 Unit VI Id Question The transformation w = az+b cz+d , where ad − bc ≠0 represents a transformation called A Magnification and rotation B Bilinear C Inversion D None Answer Marks 1.5 Unit VI Id Question The transformation w = cz represents a transformation called A Magnification and rotation B Translation C Inversion D None Answer Marks 1.5 Unit VI Id Question The analytic function f(z) = zz−1 2 +1 has singularities at A 1& − 1 B i & − i. C 1& − i D None Answer Marks 1.5 Unit VI Id Question The value of m for the function u = 2x − x2 + my2 to be harmonic is A 0 B 1 C 2 D None Answer Marks 1.5 Unit VI Id Question A function u( x, y) is said to be harmonic if 𝜕2 u 𝜕2 u A + 𝜕y2 = 0 𝜕x2 𝜕u 𝜕u B 𝜕x + 𝜕y = 0 𝜕2 u 𝜕2 u C − 𝜕x2 = 0 𝜕x2 D None Answer Marks 1.5 Unit VI Id Question Which of the following is a bilinear transformation 2z+1 A w = 4z+2 2z+1 B w = 4z−2 C w=z D None Answer Marks 1.5 Unit VI Id Question The transformation w = z + α is known as A Magnification and rotation B Translation C Inversion D None Answer Marks 1.5 Unit VI Id Question If real part of function f(z) constant, then f(z) is A Analytic function B Nowhere analytic function C Entire function D None Answer Marks 1.5 Unit VI Id Question The Cauchy - Riemann equations for f(z) = u(x, y) + iv(x, y) to be analytic are 𝜕2 u 𝜕2 u 𝜕2 v 𝜕2 v A 2 + 2 = 0; 2 + 2 = 0 𝜕x 𝜕y 𝜕x 𝜕y B 𝜕u 𝜕v 𝜕u 𝜕v =− ; =− 𝜕x 𝜕y 𝜕y 𝜕x 𝜕u 𝜕v 𝜕u 𝜕v C 𝜕x = 𝜕y ; 𝜕y = − 𝜕x D None Answer Marks 1.5 Unit VI Id Question If f(z) = u + iv is analyticin polar form, then 𝜕u 𝜕r is 𝜕v A 𝜕Ѳ B 𝜕v r 𝜕Ѳ C 1 𝜕v r 𝜕Ѳ D None Answer Marks 1.5 Unit VI Id Question Which of the following is true : A Re(z1 − z2 ) = Re(z1 ) − Re(z2 ) B Re(z1 z2 ) = Re(z1 )Re(z2 ) C |z1 − z2 | = |z1 | − |z2 | D None Answer Marks 1.5 Unit VI Id Question f(z) = z̅ is differentiable A Nowhere B only at z = 0 C Everywhere D None. Answer Marks 1.5 Unit VI Id Question The polar form of Cauchy - Riemann equations are 𝜕u 1 𝜕v 𝜕u 𝜕v A 𝜕Ѳ = ; r 𝜕r 𝜕r = r 𝜕Ѳ 𝜕u 𝜕v 𝜕u 1 𝜕v B 𝜕Ѳ = r 𝜕Ѳ ; 𝜕r = r 𝜕Ѳ 𝜕u 1 𝜕v 𝜕u 𝜕v C 𝜕r = ; r 𝜕Ѳ 𝜕Ѳ = −r 𝜕r D None. Answer Marks 1.5 Unit VI Id Question f(z) = ex (cosy − isiny)is A analytic B Not analytic Analytic when z = 0 C D None. Answer Marks 1.5 Unit VI Id Question The harmonic conjugate of u(x, y) = ey cosx is A −ey cosy + c B −ey sinx + c C ey sinx + c D None. Answer Marks 1.5 Unit VI Id Question Function u is said to be harmonic if and only if A uxx + uyy = 0 B uxx − uyy = 0 C ux + uy = 0 D None. Answer Marks 1.5 Unit VI Id Question If u and v are harmonic functions then f(z)= u+iv is A Analytic function B Need not be analytic function C Analytic function only at z=0 D None. Answer Marks 1.5 Unit VI Id Question If eax cosy is harmonic ,then a = A i B 0 C -1 D None. Answer Marks 1.5 Unit VI Id Question The function f(z) = |z| is a nonconstant A Nowhere analytic function B analytic function only at z = 0 C Everywhere analytic function D None. Answer Marks 1.5 Unit VI Id Question f(z) = |z̅|2 is differentiable A nowhere B only at z = 0 C everywhere D None. Answer Marks 1.5 Unit VI