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FA07Ma1aTest2Sol

# FA07Ma1aTest2Sol - Math 1a Test 2 Solutions Fall 2007...

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Math 1a Test 2 Solutions Fall 2007 Problem 1. Let F ( x ) = cos x 0 e - 2 y 3 dy. Find F (0). Solution 1. Let G ( u ) = u 0 e - 2 y 3 dy By the fundamental theorem of calculus, G ( u ) = e - 2 u 3 By definition, if h ( x ) = cos x , F ( x ) = G ( h ( x )) so, by the chain rule, F (0) = G ( h (0)) h (0) = e - 2 . 0 3 ( - sin(0)) = 0 . Problem 2. Show that there is some positive real x 0 so that sin x 0 x 0 = x 0 . Hint : Use Bolzano’s Theorem. Solution 2. Consider the continuous function f : [ π 4 , π ] R defined by f ( x ) = sin x x - x . We have f ( π 4 ) = 4 π 2 - π 4 > 0 because π 2 2 < 16.(Recall that π 2 < 10 and 2 < 3 2 ). On the other hand f ( π ) = - π < 0. So by Bolzano’s theorem, x 0 [ π 4 , π ] such that f ( x 0 ) = 0. In other words sin x 0 x 0 = x 0 . 1

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2 Problem 3. Find lim x 0 ln(cos 5 x ) ln(cos 3 x ) . Solution 3. First way. When x = 0 the ratio is of the form 0 / 0, so we may apply L’Hopital’s rule to obtain lim x 0 ln(cos 5 x ) ln(cos 3 x ) = lim x 0 - 5 sin 5 x cos 5 x / - 3 sin 3 x cos 3 x = lim x 0 5 3 cos 3 x cos 5 x sin 5 x sin 3 x .
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