2011a_x1b_sols

# 2011a_x1b_sols - Spring 2011 Math 151 Exam I Version B...

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Unformatted text preview: Spring 2011 Math 151 Exam I Version B Solutions 1. C Set x and y components equal and solve for t . t = 4 at (2 , − 3) , t = 0 at (0 , 1) , t = 1 at (1 , 0) , but − 1 = √ t has no solution, so the point not on the curve is ( − 1 , ) . 2. A cos x 1 − sin x · (1 + sin x ) (1 + sin x ) = cos x (1 + sin x ) 1 − sin 2 x = 1 + sin x cos x = sec x + tan x . 3. A Since − 1 ≤ cos x ≤ 1 , − 1 x ≤ cos x x ≤ 1 x Since ± 1 x → as x → ∞ , by the Squeeze Theorem, the limit is . 4. B As we approach 4 from the left along the curve, the y-value approaches 2 . 5. B Let f ( x ) = x 3 − x 2 + x . f is continuous since it is a polynomial, and f (2) = 6 , f (3) = 21 , so f (2) < 10 < f (3) . Therefore, by the Intermediate Value Theorem, there is a solution to f ( x ) = 10 on [ 2 , 3 ] . 6. C a = (6 i + 3 j ) − (3 i − j ) = 3 i + 4 j . To form a unit vector ˆ a , multiply by the reciprocal of the magnitude: ˆ a = 1 √ 3 2 + 4 2 (3 i + 4 j ) = 3 5 i + 4 5 j ....
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## This note was uploaded on 02/13/2012 for the course MATH 151 taught by Professor Artbelmonte during the Spring '06 term at Texas A&M.

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2011a_x1b_sols - Spring 2011 Math 151 Exam I Version B...

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