102-2007&2008-2-M10-March2008_2

102-2007&2008-2-M10-March2008_2 - x √ x 2-1-1 2-1...

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Kuwait University Math 102 March 27, 2008 First Exam Time: 90 minutes Calculators and Mobile Phones are not allowed. 1. Let f ( x ) = e x + e tan 1 x , −∞ < x < . i) Show that f 1 exists and fnd its domain. ii) Show that P (2 , 0) is on the graph oF f 1 and fnd the slope oF the tangent line to the graph oF f 1 at P (2 , 0). (5 Points) 2. Use logarithmic di±erentiation to fnd dy dx iF y = (cosh x ) sec 1 x r e x ln | x | . (4 Points) 3. a) Show that: d dx (sin 1 x ) = 1 1 x 2 , 1 < x < 1. b) Prove the identity: cos(2 sin 1 x ) = 1 2 x 2 . (2 Points Each) 4. Evaluate the Following integrals: i) i 1 x (9 + (ln x ) 2 ) dx. ii) i 6 x 2 x dx. iii) i tanh x r 4 cosh 2 x 1 dx. iv) i sec x sin x + 4 cos x dx. (3 Points Each)
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First Midterm Math102 , March 27, 2008 1. (a) f 0 ( x ) = e x + e tan - 1 x 1+ x 2 > 0 for all x , so f is increasing. Therefore f - 1 exists. lim x →-∞ ( e x + e tan - 1 x ) = 0 + e - π/ 2 = e - π/ 2 and lim x →∞ ( e x + e tan - 1 x ) = + e π/ 2 = , so D f - 1 = R f = ( e - π/ 2 , ). (b) f (0) = 2 f - 1 (2) = 0, so P (2 , 0) is on the graph of f - 1 . 2. ln y = sec - 1 lncosh x - 1 2 x - 1 2 lnln | x | , so 1 y dy dx = n sec - 1 x tanh x + lncosh x x x 2 - 1 - 1 2 - 1 2 x ln | x | o and dy dx = n sec - 1 x tanh x + lncosh x
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Unformatted text preview: x √ x 2-1-1 2-1 2 x ln | x | o (cosh x ) sec-1 x √ e x ln | x | 3. (a) y = sin-1 x ⇔ sin y = x ⇒ cos y · dy dx = 1 ⇒ dy dx = 1 cos y = 1 √ 1-sin 2 y = 1 √ 1-x 2 . (b) Let θ = sin-1 x ⇒ sin θ = x and so cos(2sin-1 x ) = cos(2 θ ) = cos 2 θ-sin 2 θ = 1-2sin 2 θ = 1-2 x 2 . 4. (a) Let u = ln x , then R dx x (9+(ln x ) 2 ) = R du 9+ u 2 = 1 3 tan-1 ( u 3 ) + C = 1 3 tan-1 ( ln x 3 ) + C (b) R 6 x 2-x dx = R 3 x 2 x 2-x dx = R 3 x dx = 3 x ln3 + C (c) Let u = 2cosh x , then R tanh x √ 4cosh 2 x-1 dx = R 2sinh x 2cosh x √ 4cosh 2 x-1 dx = R du u √ u 2-1 = sec-1 u + C = sec-1 (2cosh x ) + C (d) R sec x sin x +4cos x dx = R sec 2 x sec x (sin x +4cos x ) dx = R sec 2 x tan x +4 dx = ln | tan x + 4 | + C 1...
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102-2007&amp;amp;2008-2-M10-March2008_2 - x √ x 2-1-1 2-1...

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