For all you Calculus hot shots!

Test Preparation
1

<p>Test your Calculus skills!</p>

<p>If you get 5 of 10 you will get a “3” on the real AP test.
If you get 7 of 10 you will get a “4” on the real AP test.
If you get 9 of 10 you will get a “5” on the real AP test.</p>

<li> The graph of y = 3x² - x³ has a relative maximum at:</li>
</ol>

<p>a. (0,0) only b. (1,2) only c. (2,4) only d. (4,-16) only e. (0,0) and (2,4)</p>

<li> If f(x) = e^[sin(x)], how many zeros does f’(x) have on the closed interval [0,2pi]?</li>
</ol>

<p>a. 1 b. 2 c. 3 d. 4 e. 5</p>

<li> If f(x) = (4sinx + 2)^(.5), then f’(0) =</li>
</ol>

<p>a. -2 b. 0 c. 1 d. (sqrt2)/2 e. sqrt2</p>

<li> If “t” is measured in hours and f’(t) is measured in knots, then the integral of f’(t)dt from 0 to 2 = (note: 1 knot = 1 nautical mile/hour)</li>
</ol>

<p>a. f(2) knots b. f(2) - f(0) knots c. f(2) nautical miles d. f(2) - f(0) nautical miles e. f(2) - f(0) knots/hour</p>

<li> The equation of the tangent line to the curve of x² + y² = 169 at the point (5, -12) is </li>
</ol>

<p>a. 5y - 12x = -120
b. 5y - 12y = 119
c. 5x - 12y = 169
d. 12x + 5y = 0
e. 12x + 5y = 169</p>

<li> If the graph of f(x) = 2x² + k/x has a point of inflection at x = -1, then the value of k is</li>
</ol>

<p>a. -2
b. -1
c. 0
d. 1
e. 2</p>

2

<ol>
<li> What are all values of x for which the graph of y = (2)/(4-x) is concave down?</li>
</ol>

<p>a. no values of x
b. x<4
c. x>-4
d. x<-4
e. x>4</p>

<ol>
<li> A particle moves along the x-axis in such a way that its position at time t is given by x(t) = (1 - t)/(1+t). What is the acceleration of the particle at time t = 0?</li>
</ol>

<p>a. -4
b. -2
c. -3/5
d. 2
e. 4
9. If, for all real number x, f(x) = g(x) + 5 , then on any interval [a,b] the area of the region between the graphs of f and g is</p>

<p>a. 5
b. 5a+5b
c. 5b - 5a
d. 5a - 5b
e. 5ab</p>

<ol>
<li> If the integral from 0 to 6 of (x² - 2x + 2)dx is approximated by three inscribed rectangles of equal width on the x-axis, then the approximation is</li>
</ol>

<p>a. 24
b. 26
c. 28
d. 48
e. 76</p>

3

<p>The curve is better than 90% = 5...</p>

<p>Did you make that curve up??</p>

4

<p>this is a stupid question but hey AP's are not offered anywhere near me. AB or BC?</p>

5

<h2>My answers (don't look further down if you haven't done them yet)</h2>

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<p>c,b,e,d,c,e,e,e,e,d</p>

6

<p>I have different answers for 3, 9, and 10
I won't go into detail on 3, especially since my answer does not appear in the answers :)
for 9, i think that the equation is set up int(a,b) of f(x)-g(x), which can be rewritten g(x)+5-g(x). this gives you int(a,b) of 5 and an answer of 5b-5a, or c
for 10, i have never been good with approximations and knowing which rule to use when. Since it said that the rectangles were inscribed, I thought that the vertical side would be the leftmost side, so I got an answer of 28 (c)
Please correct my (definitely) flawed logic. Thank you!</p>

7

<p>looks pretty ABish since it doesnt have Series on it..</p>

8

<p>well for #10 if you use LRAM you get 28, but if you use RRAM you get 76, so the real answer must be somewhere in between?</p>

<p>oh shoot, i believe you're right on #9, it is
5(b-a), i meant to put that down but for some reason i chose the other one...?</p>

9

<h2>Here are my answers. Don't look down!!!</h2>

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<p>1.c 2.b 3.e 4.d 5.c 6.e 7.e 8.e 9.c 10. e/b</p>

<p>my questions regarding #10</p>

<p>you didn't say rhs or lhs or midpnt.</p>

<p>if you just take the integral, you get 'e'.
if you do the way where every single triangle must be inscibed, do this:</p>

<p>make table
x - 0 1 2 3 4 5 6
f(x) - 2 1 2 5 10 17 26</p>

<p>now confirm the minima of the parabola
minima = (x=1)</p>

<p>now you do:</p>

<p>2<em>1 + 2</em>2 + 2*10 = 26
and you get 'b'</p>

<p>on the AP Exam, I seriously doubt they would give questions as easy as this.
maybe perfect indicates a 4 or higher?</p>

10

<h2>Here are my answers. Don't look down!!!</h2>

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<p>1.c 2.b 3.e 4.d 5.c 6.e 7.e 8.e 9.c 10. e/b</p>

<p>my questions regarding #10</p>

<p>you didn't say rhs or lhs or midpnt.</p>

<p>if you just take the integral, you get 'e'.
if you do the way where every single triangle must be inscibed, do this:</p>

<p>make table
x - 0 1 2 3 4 5 6
f(x) - 2 1 2 5 10 17 26</p>

<p>now confirm the minima of the parabola
minima = (x=1)</p>

<p>now you do:</p>

<p>2<em>1 + 2</em>2 + 2*10 = 26
and you get 'b'</p>

<p>on the AP Exam, I seriously doubt they would give questions as easy as this.
maybe perfect indicates a 4 or higher?</p>

<p>Vintor: if you do LRAM, than 2nd two rectangle aren't entirely contained under the graph.</p>

11

<p>I'm a little confused about #5. I got 5/12 for the slope of the equation at that line; therefore: y-12 = 5/12(x-5) ? ? ? Thanks.</p>