math midterm 10

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Unformatted text preview: Saturday, June 12, 2010 Page 1 Mathematics 1225A (Online) Midterm Examination Students Name ( Print ) Students Number 1. Evaluate parenleftbigg 9 1 8 2 / 3 3 2 parenrightbigg 1 2 . A : 2 3 B : 1 C : 2 D : 2 E : 6 Solution: parenleftbigg 9 1 8 2 / 3 3 2 parenrightbigg 1 2 = parenleftbigg (3 2 ) 1 (2 3 ) 2 / 3 3 2 parenrightbigg 1 2 = parenleftbigg 3 2( 1) 2 3(2 / 3) 3 2 parenrightbigg 1 2 = parenleftBig 3 2 ( 2) 2 2 parenrightBig 1 2 = ( 3 2 2 ) 1 2 = ( 1(2 2 ) ) 1 2 = 2 2(1 / 2) = 2 Another Approach: 9 1 = 1 9 and since 8 1 / 3 = radicalbig [3]8 = 2 then 8 2 / 3 = 2 2 = 4. And of course 3 2 = 1 3 2 = 1 9 , so we have: parenleftbigg 9 1 8 2 / 3 3 2 parenrightbigg 1 / 2 = radicalBigg 1 9 4 1 9 = 4 = 2 2. Express log 2 9 2 log 2 6 as simply as possible. A : 2 B : 0 C : 1 8 D : 1 4 E : log 2 9 Solution: log 2 9 2 log 2 6 = log 2 9 log 2 6 2 = log 2 9 log 2 36 = log 2 parenleftbigg 9 36 parenrightbigg = log 2 parenleftbigg 1 4 parenrightbigg = log 2 2 2 = 2 Another Approach: log 2 9 2 log 2 6 = log 2 3 2 2 log 2 (3 2) = 2 log 2 3 2(log 2 3+log 2 2) = 2 log 2 3 2 log 2 3 2 log 2 2 = 0 2(1) = 2 3. If f ( x ) = 1 + ( 2 3 x 1 ) 4 and f ( a ) = 9 8 , what is the value of a ? A : 1 4 B : 1 12 C : 15 4 D : 97 32 E : ln 9 ln 8 4 + 3 Solution: We have f ( a ) = 1 + ( 2 3 a 1 ) 4 which must equal 9 8 . We see that: 1 + ( 2 3 a 1 ) 4 = 9 8 ( 2 3 a 1 ) 4 = 9 8 1 = 1 8 2 4(3 a 1) = 2 3 4(3 a 1) = 3 12 a 4 = 3 12 a = 1 a = 1 12 Mathematics 1225A (Online) Midterm Examination Saturday, June 12, 2010 Page 2 4. Which of the following most closely resembles the graph of y = log 2 x ? A : B : C : D : E : Solution: We know that for any b > 1, the graph of y = log 2 x is increasing everywhere, is asymptotic to the y-axis and approaches as x + , and that it passes through the point (1 , 0). That is, it looks like the graph in answer C. But if you couldnt remember that, we know that for f ( x ) = log 2 x , f is defined only for positive values of x , and has derivative f ( x ) = 1 x ln 2 . Since x > 0 everywhere in the domain of f (and hence also everywhere in the domain of f ), and ln 2 > 0, then f ( x ) > 0 throughout its domain, so f is increasing everywhere. Also, f (1) = log 2 1 = 0, so the graph of y = f ( x ) passes through (1 , 0). (And thats enough to eliminate the other graphs.) 5. Find f ( x ) if f ( x ) = x 2 ln | x | . A : 1 + 2 ln | x | B : 3 + 2 ln | x | C : x + 2 x ln | x | D : 2 x 2 E : 0 Solution: We know that d dx [ln | x | ] = 1 x . So using the product rule, we get: f ( x ) = x 2 parenleftbigg 1 x parenrightbigg + (2 x ) ln | x | = x + 2 x ln | x | Using the product rule again for the second term, we get: f ( x ) = 1 + 2 x parenleftbigg 1 x parenrightbigg + 2 ln | x | = 1 + 2 + 2 ln | x | = 3 + 2 ln |...
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