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Unformatted text preview: MATH 408N MIDTERM 1 Name: 9/22/2011 Bormashenko TA session: Show your work for all the problems. Good luck! (1) Let θ be an angle such that tan( θ ) = 1 2 and cos( θ ) < 0. (a) [5 pts] Calculate sec( θ ). (Please put this in the simplest form you can.) Solution: Recall that 1 + tan 2 ( θ ) = sec 2 ( θ ). Therefore, sec 2 ( θ ) = 1 + 1 2 2 = 1 + 1 4 = 5 4 Thus, sec( θ ) = ± r 5 4 = ± √ 5 2 Now we just need to figure out whether sec( θ ) is √ 5 / 2 or √ 5 / 2. Since cos( θ ) < 0, and sec( θ ) = 1 / cos( θ ), it must be true that sec( θ ) < 0. Therefore, sec( θ ) = √ 5 2 (b) [5 pts] Calculate sin( θ ). (This should also be put in simplest form.) Solution: This one can be done using part (a). Since tan( θ ) = sin( θ ) cos( θ ) = sin( θ ) · sec( θ ) by moving things to the other side, we get sin( θ ) = tan( θ ) sec( θ ) = 1 / 2 √ 5 / 2 = 1 √ 5 2 MATH 408N MIDTERM 1 (2) (a) [5 pts] Solve for x if 2 x +3 = 4 3 x 1 Solution: Writing everything as a power of 2, 2 x +3 = (2 2 ) 3 x 1 = 2 2(3 x 1) = 2 6 x 2 using exponent rules and expanding things out. To have powers of the same base be equal, the exponents have to be the same. Therefore, x + 3 = 6 x 2 ⇒ 5 = 5 x ⇒ x = 1 Thus, the answer is x = 1 . (b) [5 pts] Write the following quantity as a single logarithm: ln( x ) + ln( y 2 x ) 2ln( z ) Solution: Simplifying using logarithm rules, ln( x ) + ln( y 2 x ) 2ln( z ) = ln( x ) + ln( y 2 x ) ln( z 2 ) = ln x ( y 2 x ) z 2 MATH 408N MIDTERM 1 3 (3) The following is a graph of f ( x ): (a) [3 pts] Explain why f ( x ) is onetoone. Solution: f ( x ) is onetoone because it passes the horizontal line test. (b) [3 pts] Find the value of f 1 (3), explaining how you got it. Solution: f 1 (3) is the xvalue corresponds to y = 3. Therefore, f 1 (3) = 2 (c) [4 pts] Sketch the graph of f 1 ( x ) on these axes: Solution: Reflecting along the line y = x , we get: 4 MATH 408N MIDTERM 1 (4) Consider the f ( x ) in the following graph: (a) [5 pts] State whether lim x → 1 f ( x ) exists, and calculate it if it does....
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 Spring '11
 
 Math, lim, Continuous function, Limit of a function, lim g

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