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2011a_x1a_sols

# 2011a_x1a_sols - Spring 2011 Math 151 Exam I Version A...

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Spring 2011 Math 151 Exam I Version A 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 , 0 ) . 2. D 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 0 as x → ∞ , by the Squeeze Theorem, the limit is 0 . 4. A As we approach 4 from the right along the curve, the y -value approaches 4 . 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. E a = (6 i +3 j ) ( 3 i + j ) = 9 i +2 j . To form a unit vector ˆ a , multiply by the reciprocal of the magnitude: ˆ a = 1 9 2 + 2 2 (9 i + 2 j ) = 9 85 i + 2 85 j . 7. B lim x 5 - f ( x ) = 6 5 = 1 . lim x 5 + f ( x ) = 8 + 2(5) = 2 , and f (5) = 1 . Therefore, since lim x 5 - f ( x ) = f (5) negationslash = lim x 5 + f ( x ) , f is contin- uous only from the left.

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