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Quiz 1. Solution
1. (sol)
Let y = ex−e2−x then e2x− 2yex− 1 = 0
⇒ ex= y ±py2+ 1
Choose positive since ex>0
⇒ x = ln y +py2+ 1
⇒ f−1= ln x +√ x2+ 1
2. (sol) Since
x→−6lim+
2x + 12
x+ 6 = 2, lim
x→−6−
2x + 12
−x − 6 = −2 i.e the limit doesn’t exist.
3. (sol) (a) 2 (b) 0
(c) the limit doesn’t exist (d) 1
1
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