AP Calculus AB Trivia Review Contest

<p>I decided to start a little earlier since I think I’m heading away early. Hopefully that won’t pose an inconvenience. </p>

<p>Point Winners For Each Question:
1)grayfalcon89 (+1)
2)yillfog (+1)
3)grayfalcon89 (+2)
4)grayfalcon89 (+2)
5)grayfalcon89 (+2)
6)grayfalcon89 (+3)</p>

<p>Question Commentary
-T/F number 2 cannot be done in the calculator. Note that there is a discontinuity at x=0. If you divide it into two integrals, then the answer would come out as 2 anyway rather than -2, so the answer is false.
-Yes, the prime on MC #6 was a typo, whoops. I need to be more careful of that.</p>

<p>Overall Leaderboard

  1. grayfalcon89 (29.5 pts)
  2. mathwiz90 (10.5 pts)
  3. yillfog (6 pts)</p>

<p>Leaderboard Commentary
The lead is growing still :o</p>

<p> Question Set 4: Due Monday 9:30 PM EST </p>

<p>Remember, only answer one question per post.</p>

<p>New special! Now, anyone that correctly answers a question will get points regardless of if you’re right or not! Hopefully this will increase participation.</p>

<p>1) T/F: If the graph of a polynomial function has 3 x-intercepts, then it must have at least two points at which its tangent line is horizontal.</p>

<p>2) T/F: The integral of (2x+1)^2 dx is (1/3)(2x+1)^3 + C.</p>

<p>3) Concpt: Find the area of the region bounded by f(x) = x/(x^2 + 1), the x-axis, and x=3. (You must give me a precise answer, meaning your answer may not be purely a decimal.)</p>

<p>4) Concpt: Determine the open intervals on which h(x) = sqrt(x)*(x-3) is increasing.</p>

<p>5) Concpt: What is the limit as h approaches 0 of [(1-cos(h))^2] / h?</p>

<p>6) Concpt: Find a such that g is continuous on the entire real line.
g(x) = (x^2 – a^2)/(x-a) when x ≠ a, and g(x) = 8 when x = a.</p>

<p>7) Concpt: Find the equation of a line tangent to f(x) = x^3 + 2 and parallel to 3x-y-4=0.</p>

<p>8 ) Concpt: Evaluate the integral of (x^2)(sqrt(1-x))dx</p>

<p>9) FR: 7) FR: Let f(x) = c/x + x^2, where c is a nonzero constant.
a) Determine all values of the constant c such that f has a relative minimum, but no relative maximum.
b) Discuss the continuity of f.
[Somebody at least attempt that FR one]</p>

<h1>1 True</h1>

<p>i checked two different polynomial functions which had 3 x-intercepts, plotted their derivatives and saw that those deriv’s each had 2 x-intercepts</p>

<h1>2 False</h1>

<p>i got (1/6)((2x+1)^3) + C</p>

<h1>4 is (1, infinite)</h1>

<p>explanation: h’(x) is positive from (1, infinite)</p>

<h1>7 has two possible answers: (y-3)=3(x-1) and (y-1)=3(x+1)</h1>

<h1>2 False.</h1>

<p>I didn’t find integral but differentiated (1/3)(2x+1)^3 + C since I like differentiating better than integrating :-P.</p>

<p>3) ln(10)/2
5) 0</p>

<p>hey skp21, did you use some identity for #5?
can you elaborate on how you solved it so that a fool like me can understand:P</p>

<p>just use L’Hopital’ Rule. that is, take the derivative of the top and bottom then take the limit.</p>

<p>alrighty, thanks man;
we didnt learn that in my calc ab class this year but i picked it up from a math teacher pretty recently when he was helping me out with some calculus problems.
i wonder, is l’hospital’s rule AB material? my calc teacher has been prepping us for the exam since february but he said he still had one last thing to teach us. maybe l’hospital rule is that one more thing.
anyways, thx</p>

<p>1) true, if it’s a polynomial.
3 x -intercepts = at least 2 changes in direction</p>

<p>2) false, you have to use u-substition, and you pull out an extra 1/2 from the 2x b/c d(2x) = 2, and you need a 1/2 to equate it.</p>

<p>answer = 1/6 (2x+1) ^3 + C</p>

<p>3) area = int(x/(x^2+1)) from 0 to 3 (x = 0, y = 0, x=3 is a bound)</p>

<p>x / x^2 + 1 = d(arctanx), so int of that = arctan x.</p>

<p>area = arctanx |0–>3</p>

<p>so area = arctan 3 - arctan 0. idk if this is what you want as a precise answer, since it will end up going to a decimal, but here.</p>

<p>btw are we allowed to use graphers on every single one of these questions?</p>

<p>^yupporz,
unless stated otherwise me thinks</p>

<p>hope its not too late to bump</p>