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Unformatted text preview: mao (tm23477) – Hw 14 – gualdani – (57180) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the value of f (1) when f ( x ) = 5 sin − 1 x + tan − 1 x . 1. f (1) = 11 4 π correct 2. f (1) = 9 4 π 3. f (1) = 7 4 π 4. f (1) = 5 4 π 5. f (1) = 3 4 π Explanation: Since sin − 1 (1) = π 2 , tan − 1 (1) = π 4 , we see that f (1) = parenleftBig 5 2 + 1 4 parenrightBig π = 11 4 π . 002 10.0 points Find the exact value of sin − 1 parenleftBig √ 3 2 parenrightBig in the interval parenleftBig , π 2 parenrightBig . 1. π 6 2. π 5 3. π 7 4. π 4 5. π 3 correct Explanation: We have to find x so that sin x = √ 3 2 , < x < π 2 . Known trig values thus ensure that x = π 3 . 003 10.0 points Simplify the expression y = sin parenleftbigg tan − 1 x √ 7 parenrightbigg by writing it in algebraic form. 1. y = x x 2 + 7 2. y = √ 7 √ x 2 + 7 3. y = x √ x 2 7 4. y = x √ x 2 + 7 correct 5. y = √ x 2 + 7 √ 7 Explanation: The given expression has the form y = sin θ where tan θ = x √ 7 , π 2 < θ < π 2 . To determine the value of sin θ given the value of tan θ , we can apply Pythagoras’ theorem to the right triangle √ 7 x θ radicalbig x 2 + 7 mao (tm23477) – Hw 14 – gualdani – (57180) 2 From this it follows that y = sin θ = x √ x 2 + 7 . Alternatively, we can use the trig identity csc 2 θ = 1 + cot 2 θ to determine sin θ . 004 10.0 points Simplify the expression f ( x ) = sin ( tan − 1 x ) . 1. f ( x ) = radicalbig 1 + x 2 2. f ( x ) = 1 √ 1 x 2 3. f ( x ) = radicalbig 1 x 2 4. f ( x ) = x √ 1 + x 2 correct 5. f ( x ) = x √ 1 x 2 6. f ( x ) = 1 √ 1 + x 2 Explanation: By definition f ( x ) = sin θ where x = tan θ . But then by Pythagoras, sin θ = x √ 1 + x 2 . Consequently, f ( x ) = x √ 1 + x 2 . 005 10.0 points Determine if lim x →∞ sin − 1 parenleftbigg 3 + 2 x 4 + 2 x parenrightbigg exists, and if it does, find its value. 1. limit does not exist 2. limit = 0 3. limit = π 3 4. limit = π 2 correct 5. limit = π 6 6. limit = π 4 Explanation: Since lim x →∞ 3 + 2 x 4 + 2 x = 1 , we see that lim x →∞ sin − 1 parenleftbigg 3 + 2 x 4 + 2 x parenrightbigg exists, and that the...
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This note was uploaded on 05/12/2010 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz

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