AP Calculus AB Trivia Review Contest

<p>Because of severe lack of interest in a different competition thing I initiated early last week, I’ve decided to take an easier step in helping us start calculus review without taking great amounts of time while still keeping the competition element.</p>

<p>Every night leading up to the exam, meaning sometime between 8 PM - 10 PM EST, I will post 5-10 questions. Each question will be worth a varying number of points depending on the type. The rules are simple. If you know the answer to a question, then answer it as soon as possible. The first person to get it completely right will get the points. To ensure at least some fairness, you may only include the answer to one question per post. If you answer multiple questions per post, I’ll only check your first answer. In addition, I won’t tell you who’s right and who’s wrong until I post the next batch of questions, so just because someone answered a question doesn’t mean you shouldn’t take a stab even if you think it’s wrong (unless it’s a penalty type question). </p>

<p>Types of Questions
-True/False: *1 Point<a href=“Very%20straight%20forward.%20It’s%20either%20true%20or%20false.”>/i</a>
-Conceptual or Class Type Question: *2 Points<a href=“These%20questions%20aren’t%20up%20to%20par%20with%20the%20level%20of%20the%20AP%20exam,%20but%20they’re%20still%20good%20for%20review!”>/i</a>
-AP Multiple Choice: *3 Points, with a 1/2 point penalty for wrong answers<a href=“Just%20give%20the%20letter%20of%20the%20correct%20answer.%20If%20you%20don’t%20know%20the%20answer,%20don’t%20guess,%20you’ll%20just%20get%20penalized%20here.%20Only%20the%20first%20person%20that%20gets%20the%20answer%20right%20will%20get%20the%20points,%20but%20%5Bi%5Devery%5B/i%5D%20person%20that%20posts%20the%20wrong%20answer%20will%20get%20the%20.5%20point%20deduction.”>/i</a>
-AP Free Response: *Max 6 Points<a href=“If%20you%20see%20one%20of%20these,%20try%20to%20be%20as%20complete%20as%20you%20possibly%20can.%20%5Bi%5DEvery%5B/i%5D%20person%20that%20attempts%20these%20will%20be%20considered%20for%20points,%20so%20if%20you%20see%20one,%20take%20a%20crack%20at%20it!”>/i</a></p>

<p>What better way to review calculus than to get a little competitive spirit in here? None!</p>

<p>Try these; it’s the first day, so none of these are really mind-boggingly hard:</p>

<p>1) (T/F)- The derivitive of the basic natural exponential function (e^x) is the same as its antiderivitive.</p>

<p>2) (T/F)- The sum of two increasing functions is increasing.</p>

<p>3) (Conceptual)- Discuss the discontinuities of the function of f(x) = 1/(x^2 - 4).</p>

<p>4) (Conceptual)- Given that y=2(x^2 - 3x), what is dx/dt when x=1 given that dy/dt=5?</p>

<p>5) (Conceptual)- What is the limit of (sin x)/x as x approaches infinity?</p>

<p>6) (AP Multi-Choice)- What is the maximum value of the slope of f(x) = x^3 - 3x?
a) -1
b) 0
c) 1
d) 4
e) None of the above</p>

<p>7) (AP Multi-Choice)- Find the derivitive of the following: sin(6-2x)
a) 1-2cos^2(6-2x)
b) -2sin(6-2x)
c) cos(6-2x)
d) 2(2sin^2(3-x) - 1)
e) None of the above</p>

<p>ah wow, no players?</p>

<p>I would if I could put all of the answers in one post</p>

<h1>7, none of the above</h1>

<p>it would be -2cos(6-2x)</p>

<p>1 - true
2 - true (assuming u mean monotonically increasing, and not that one end is higher than another)
3 - 2 and -2 are discontinuous
4 - dx/dt=dy/dt/dy/dx so its 5/(-2)=-5/2
5 - 0 nice try with the trick, though
6 - It should be infinity so E
7 - Again E
These are prob mostly wrong i didnt check anything and just jotted down the answers</p>

<p>sorry didnt read about 1 post rule</p>

<p>Blah, I’ll extend this for another day, in case anyone else decides to enter.</p>

<ol>
<li>T</li>
</ol>

<p>This is same as asking if d(e^x) = int(e^x)? Answer is yes, it is.</p>

<ol>
<li>E</li>
</ol>

<p>sin(u) = cos(u)<em>du
So, using this, we can find sin(6-2x):
sin(6-2x) = cos(6-2x)</em>-2, which is E, or none of above.</p>

<ol>
<li>0</li>
</ol>

<p>To others who didn’t understand why this was a “trick” question, here’s why. There’s an identity for limit of (sin x)/x. Except, this identity includes x approaching 0, not infinity. Limit of (sin x)/x as x –> 0 is 1 (remember this). When x –> infinity, (sin x)/x will be 0 because even though x gets large, sin x only repeats within [-1,1]. So, this is why the answer is 0.</p>

<ol>
<li>E</li>
</ol>

<p>I haven’t seen any question like this. Questions usually ask people to find the maximum/minimum of f(x), not maximum/minimum of slope of f(x). What’s the difference? Slope of f(x) is same as first derivative of f(x). In this case, it is 3x^2-3. This is parabola that opens upward, which means that the maximum value is infinity.</p>

<ol>
<li>-5/2</li>
</ol>

<p>y = 2x^2-6x
dy/dt = 4x<em>dx/dt - 6</em>dx/dt
Substituting what problem gave
5 = 4(1)<em>dx/dt - 6</em>dx/dt
dx/dt = 5/-2</p>

<ol>
<li>+/- 2 (both of them)</li>
</ol>

<p>Discontinuities can occur in several occasions and one of the main ones is at asymptote. Vertical asymptotes occur in this function when denominator, x^2-4, is 0. Solving this equation, we get x = +/- 2.</p>

<ol>
<li>T</li>
</ol>

<p>I’m guessing it is but you might want to avoid questions like this. Unless I’m missing something critical, it’s pretty evident. Consider two functions like y = e^x and y = sqrt{x}. It’s clear that y = e^x + sqrt{x} increases.</p>

<p>I will say that all of the above are correct, except for one of them. Can anyone here figure out which one’s wrong and go for the steal?</p>

<p>Alright, time’s up for this set.</p>

<p>The answer to 7 is not E. It was designed to be very tricky, as it as actually also an identity in disguise.</p>

<p>f(x) = sin(6-2x)
f’(x) = -2cos(6-2x)
f’(x) = -2cos(2(3-x)) [Factor the 2 inside the cosine.]
f’(x) = -2 * (cos^2 (3-x) - sin^2 (3-x)) [cos(2x) = cos^2 (x) - sin^2 (x)]
f’(x) = -2 * ((1-sin^2 (3-x)) - sin^2 (3-x)) [cos^2 (x) = 1 - sin^2 (x)]
f’(x) = -2 * (1 - 2sin^2 (3-x))
f’(x) = 2(2sin^2 (3-x) -1) = Answer Choice D</p>

<p>Sorry about the trickiness of that question, but my calc teacher has told me that the AP people have used identities to disguise AP problems before.</p>

<p>Leaderboard and next set of questions will be up at 3 PM EST.</p>

<p>Leaderboard:

  1. mathwiz90 (10.5 pts)
  2. grayfalcon89 (10.5 pts)</p>

<p>It was the first round and the turnout was low, so I decided to let your “all answers in one post” slipup slide mathwiz, but from this point on, only one answer per post. Also, you don’t need to justify your responses unless the problem asks for it.</p>

<p>First person to get each question right gets the points, and everybody that attempts the free response will get the points that they earn.</p>

<p>1) (1 pt) T/F: If f is undefined at x=c, then the limit of f(x) as x approaches c doesn’t exist.</p>

<p>2) (1 pt) T/F: 4<em>(the integral on sinx</em>cosx*dx) = -cos2x + C</p>

<p>3) (2 pts): Find the slope of the tangent line to the graph of (x^2 + y^2)^2 = 4yx^2 at (1,1).</p>

<p>4) (2 pts): Find F(2) if F(x) = the integral of (t-5)dt given the lower bound 0 and the upper bound x.</p>

<p>5) (2 pts): Find the left hand limit of (absval(x))/x as x approaches 0.</p>

<p>6) (3 pts/ -.5 if wrong): Given that R is the region bounded by f(x) = 9 - x^2 and the x-axis, what line(s) will divide R into two regions of equal area?
I. x=0
II. y=4.8291
III. y=9/2
A) None of these
B) I Only
C) II Only
D) I and II Only
E) I and III Only</p>

<p>7) (FR: 6 Pts Max)
<a href=“http://xs514.xs.to/xs514/07145/7apr6.png[/url]”>http://xs514.xs.to/xs514/07145/7apr6.png&lt;/a&gt; <– This is the diagram you will refer to. Assme that everything in this graph is linear, and it is a graph of f '.
Additionally, f is a piecewise function that is not differentiable at (2,5).
a) Identify any points of inflection of f, if any. Justify your response.
b) What is the area of the region bounded by f, x=6, the x-axis, and the y-axis?</p>

<h1>4 is -8</h1>

<p>btw, i really like what your doing here; it should be some good practice for that exam</p>

<ol>
<li>F</li>
</ol>

<p>need 10 characters as minimum. :-D</p>

<p>i’'m kinda unsure about #3 (which is really bad considering how quickly the exam is approaching), but is it:
y-1=0(x-1)
OR more simply
y=1</p>

<ol>
<li>T</li>
</ol>

<p><em>cough cough cough</em> simple identity</p>