FA07Ma1aTest2Sol

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

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Unformatted text preview: Math 1a Test 2 Solutions Fall 2007 Problem 1. Let F ( x ) = Z cos x e- 2 y 3 dy. Find F (0). Solution 1. Let G ( u ) = Z u 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 . 3 (- sin(0)) = 0 . Problem 2. Show that there is some positive real x so that sin x x = x . Hint : Use Bolzanos Theorem. Solution 2. Consider the continuous function f : [ 4 , ] R defined by f ( x ) = sin x x- x . We have f ( 4 ) = 4 2- 4 > 0 because 2 2 < 16.(Recall that 2 < 10 and 2 < 3 2 ). On the other hand f ( ) =- < 0. So by Bolzanos theorem, x [ 4 , ] such that f ( x ) = 0. In other words sin x x = x . 1 2 Problem 3. Find lim x 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 LHopitals rule to obtain lim x ln(cos 5 x ) ln(cos 3 x ) = lim x...
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This note was uploaded on 02/22/2010 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.

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

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