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Solutions HW 14

# Solutions HW 14 - mao(tm23477 Hw 14 gualdani(57180 This...

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mao (tm23477) – Hw 14 – gualdani – (57180) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind 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) = p 5 2 + 1 4 P π = 11 4 π . 002 10.0 points ±ind the exact value oF sin 1 p 3 2 P in the interval p 0 , π 2 P . 1. π 6 2. π 5 3. π 7 4. π 4 5. π 3 correct Explanation: We have to fnd x so that sin x = 3 2 , 0 < x < π 2 . Known trig values thus ensure that x = π 3 . 003 10.0 points SimpliFy the expression y = sin ± tan 1 x 7 ² 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 r x 2 + 7

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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 ) = r 1 + x 2 2. f ( x ) = 1 1 - x 2 3. f ( x ) = r 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 de±nition 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 p 3 + 2 x 4 + 2 x P exists, and if it does, ±nd 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 p 3 + 2 x 4 + 2 x P exists, and that the limit = sin 1 1 = π 2 .
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Solutions HW 14 - mao(tm23477 Hw 14 gualdani(57180 This...

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