102-2007&amp;2008-2-M10-March2008_2

# 102-2007&amp;2008-2-M10-March2008_2 - x โ x 2-1-1 2-1...

This preview shows pages 1–2. Sign up to view the full content.

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)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x โ x 2-1-1 2-1 2 x ln | x | o (cosh x ) sec-1 x โ e x ln | x | 3. (a) y = sin-1 x โ sin y = x โ cos y ยท dy dx = 1 โ dy dx = 1 cos y = 1 โ 1-sin 2 y = 1 โ 1-x 2 . (b) Let ฮธ = sin-1 x โ sin ฮธ = x and so cos(2sin-1 x ) = cos(2 ฮธ ) = cos 2 ฮธ-sin 2 ฮธ = 1-2sin 2 ฮธ = 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 โ 4cosh 2 x-1 dx = R 2sinh x 2cosh x โ 4cosh 2 x-1 dx = R du u โ 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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

102-2007&amp;2008-2-M10-March2008_2 - x โ x 2-1-1 2-1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online