CMI-CE6S - CC: AD [ 2011-12 ] 242: College Mathematics I...

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Unformatted text preview: CC: AD [ 2011-12 ] 242: College Mathematics I — Class Exercises [ Sol ] — Chapter 6: Differentiation and its Applications ª 6.1 Functions ª 6.2 Limit, Continuity and Differentiability ª 6.3 Differentiation ª 6.4 Applications of Differentiation Ch6: Differentiation and its Applications [ CE6S-1 ] q 6.1 Functions Class Ex. 1 (a) Starting from the definition of cosh x in terms of e x , find, in terms of natural logarithms, the values of x for which 3cosh x = 5. [ 3 ] (b) Show that, if x = sinh t , then t = ln ( x + p x 2 + 1 ) . [ 2 ] Solution (a) 3 ( e x + e − x 2 ) = 5 3 ( e x + 1 e x ) = 10 3 ( e x ) 2 − 10 e x + 3 = ( 3 e x − 1 )( e x − 3 ) = e x = 1 3 or 3 ∴ x = ln 1 3 = − ln3 or ln3 (b) 2 x = e t − e − t ( e t ) 2 − 2 xe t − 1 = e t = 2 x + p 4 x 2 + 4 2 ( e t > ) t = ln ( x + p x 2 + 1 ) AD: College Mathematics I [ 2011-12 ] Louis Lau 1 6.1 Functions [ CE6S-1 ] Class Ex. 2 Solve (a) sinh 2 x + 5 = 4cosh x [ 5 ] (b) 5sinh 2 x − 4sinh x cosh x + 1 = [ 5 ] giving your answer in terms of natural logarithms. Solution (a) sinh 2 x + 5 = 4cosh x cosh 2 x − 1 + 5 − 4cosh x = cosh 2 x − 4cosh x + 4 = ( cosh x − 2 ) 2 = cosh x = 2 e x + e − x 2 = 2 e 2 x − 4 e x + 1 = e x = 4 ± p 16 − 4 2 = 2 ± p 3 ∴ x = ln ( 2 + p 3 ) or ln ( 2 − p 3 ) = − ln ( 2 + p 3 ) [ ALT ] By formula, x = ± cosh − 1 ( 2 ) = ± ln ( 2 + p 2 2 − 1 ) . (b) 5sinh 2 x − 4sinh x cosh x + 1 = 5sinh 2 x − 4sinh x cosh x + cosh 2 x − sinh 2 x = 4sinh 2 x − 4sinh x cosh x + cosh 2 x = 4tanh 2 x − 4tanh x + 1 = ( 2tanh x − 1 ) 2 = tanh x = 1 2 ∴ x = tanh − 1 ( 1 2 ) = 1 2 ln ( 1 + 1 2 1 − 1 2 ) = 1 2 ln3 AD: College Mathematics I [ 2011-12 ] Louis Lau 2 6.1 Functions [ CE6S-1 ] Class Ex. 3 (a) Solve the equation sinh 2 x − cosh x = 1, giving your answer in terms of natural logarithm. [ 5 ] (b) Starting from the definition of sinh x , find the value of x , in terms of natural logarithm, for which sinh x = 2. [ 4 ] (c) Solve the equation 10cosh x + 2sinh x = 11. Give each answer in the form ln a where a is a rational number. [ 4 ] Solution (a) sinh 2 x − cosh x = 1 cosh 2 x − 1 − cosh x − 1 = ( cosh x − 2 )( cosh x + 1 ) = ∴ cosh x = 2 or − 1 (rejected) x = ± cosh − 1 2 = ± ln ( 2 + p 3 ) [ ln ( 2 ± p 3 )] (b) sinh x = e x − e − x 2 = 2 ( e x ) 2 − 4 ( e x ) − 1 = ∴ e x = 4 + p 16 + 4 2 = 2 + p 5 (as e x > 0) x = ln ( 2 + p 5 ) (c) 10 e x + e − x 2 + 2 e x − e − x 2 = 11 6 e 2 x − 11 e x + 4 = ( 2 e x − 1 )( 3 e x − 4 ) = e x = 1 2 or 4 3 ∴ x = ln 1 2 or ln 4 3 AD: College Mathematics I [ 2011-12 ] Louis Lau 3 Ch6: Differentiation and its Applications [ CE6S-2 ] q 6.2 Limit, Continuity and Differentiability Class Ex. 1 Let f be a real-valued function such that f ′ ( ) = 1 and f ( x + y ) = f ( x ) + f ( y ) + 2 xy for all x , y ∈ R Find lim h → f ( h ) h and f ′ ( x ) . [ 5 ] Solution lim h → f ( h ) h = lim h → f ( + h ) − f ( ) h = f ′ ( ) = 1....
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CMI-CE6S - CC: AD [ 2011-12 ] 242: College Mathematics I...

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