Download AP Calculus Multiple Choice: BC Edition – Solutions

AP Calculus Multiple Choice:
BC Edition – Solutions
J. Sloan
March 8, 2014
Z3
1)
dx
is
(1 − x)2
0
A) −
3
2
B) −
1
2
C)
1
2
D)
3
2
E) Divergent
This function inside the integral has vertical asymptotes at x = 1, and the integral bounds
contain this value. We cannot integrate through asymptotes like this and the the integral is
divergent, so the answer is E.
√
2) Which of the following integrals gives the length of the graph y = sin( x) between
x = a and x = b where 0 < a < b?
Zb q
√
A)
x + cos2 ( x) dx
Zb q
√
B)
1 + cos2 ( x) dx
a
Zb r
√
√
1
cos2 ( x) dx
C)
sin2 ( x) +
4x
a
Zb r
√
1
cos2 ( x) dx
D)
1+
4x
a
a
Zb r
E)
√
1 + cos2 ( x)
dx
4x
a
The arc length of a differentiable function f (x) over the interval (a, b) is defined as
s=
Zb p
1 + [f 0 (x)]2 dx.
a
0
We need f (x), which is given as
√
cos( x)
√
f (x) =
2 x
for our function f (x). Then piecing this together gives
s
√ 2
Zb
Zb r
√
cos( x)
1
√
x=
1+
dx =
1+
cos2 ( x) dx
4x
2 x
0
a
a
1
and thus the answer is D.
3) Which of the following integrals represents the area enclosed by the smaller loop of the
graph of r = 1 + 2 sin(θ)?
1
A)
2
11π/6
Z
2
(1 + 2 sin(θ)) dθ
1
B)
2
7π/6
1
D)
2
11π/6
Z
(1 + 2 sin(θ)) dθ
1
C)
2
2
(1 + 2 sin(θ)) dθ
1
E)
2
−π/6
(1 + 2 sin(θ))2 dθ
−π/6
7π/6
11π/6
Z
7π/6
Z
−π/6
Z
(1 + 2 sin(θ)) dθ
7π/6
Use your graphing calculator to graph this function in polar mode. You will see this.
y
r = 1 + 2 sin(θ)
x
What we want is the area of the tiny loop stemming off of the origin. We first need to find
the angles which bound that loop. These will be angles for which the raius r is 0, since this
occurs at the origin. Thus we solve
0 = 1 + 2 sin(θ)
−1 = 2 sin(θ)
1
sin(θ) = −
2
7π 11π
θ=
,
.
6 6
So these are our bounds. The area within a general polar curve r(θ) between two angle
bounds α and β is given as
Zβ
1
A=
r(θ)2 dθ.
2
α
If we plug everything in to the above, we get that the area enclosed is
1
2
11π/6
Z
(1 + 2 sin(θ))2 dθ,
7π/6
2
and thus the answer is A.
4) A curve is described by the parametric equations x = t2 + 2t and y = t3 + t2 . An
equation of the line tangent to the curve at the point determined by t = 1 is
A) 2x − 3y = 0
B) 4x − 5y = 2
C) 4x − y = 10
D) 5x − 4y = 7
E) 5x − y = 13
In order to write the equation of a line tangent to the parametric curve, we need to find the
slope, in other words the derivative dy/dx at t = 1. For parametric curves,
dy dx
dy
=
.
dx
dt dt
Easily enough,
dx
dy
dy
3t2 + 2t
= 2t + 2,
= 3t2 + 2t,
=
,
dt
dt
dx
2t + 2
dy 5
3(1)2 + 2(1)
= .
=
dx t=1
2(1) + 2
4
and
In order to use point slope, we need an actual point! Plugging t = 1 into x(t) and y(t)
respectively gives (x, y) = (3, 2). Putting this all together gives
5
y − 2 = (x − 3)
4
4y − 8 = 5x − 15
5x − 4y = 7,
and thus the answer is D.
5) The table gives selected values for the derivative of a function g on
the intervale −1 ≤ x ≤ 2. If g(−1) = −2 and Euler’s method with a step
size of 1.5 is used to approximate g(2), what is the resulting approximation?
A) −6.5
B) −1.5
C) 1.5
D) 2.5
E) 3
x g 0 (x)
−1.0
2
−0.5
4
0.0
3
0.5
1
1.0
0
1.5
−3
2.0
−6
We are asked to use Euler’s method on the interval [−1, 2] with step size
1.5. This means that we will need the derivative values at x = −1 and
x = 0.5, which are respectively 2 and 1 as per the table. Using Euler’s method gives
−2 + 1.5(2) + 1.5(1) = 2.5.
Thus the answer is D.
3
6) What are all values of x for which the series
∞
X
n3n
n=1
A) All x except x = 0
D) |x| > 3
B) |x| = 3
xn
converges?
C) −3 ≤ x ≤ 3
E) The series diverges for all x
Whenever we get power series like this and are asked to find out where it converges, the
thing to do is apply a ratio test. Taking the limit of the ratio of each term to the previous
gives
3
(n + 1)3n+1 xn
3(n + 1)
= .
lim
= lim
n+1
n
n→∞
n→∞
x
n3
xn
x
The condition for convergence is that the absolute value of the above limit is less than 1. So
3
< 1 =⇒ |x| > 3.
x
This makes sense, because plugging in a particular value for x makes the series semi geometric
(if you ignore the factor of n). As long as the base of the factor in the denominator is greater
than 3, the ratio of terms decreases almost geometrically, and is thus convergent. In light of
this, the answer is D.
7) Which of the following series converge to 2?
I.
∞
X
2n
n+3
n=1
∞
X
−8
II.
(−3)n
n=1
∞
X
1
III.
2n
n=0
A) I only
B) II only
C) II only
D) I and III only
I. This series doesn’t even converge, because lim
n→∞
at all, it certain doesn’t converge to 2. ×
E) II and III only
2n
= 2 6= 0. If it doesn’t converge
n+3
II. This is a geometric series with r = −1/3. Using the standard sum formula, and
remembering to account for the starting index of 1,
∞
X
−8
1
3 4
−8
=
−8
−
1
=
−8
−
=
= 2.
n
(−3)
1
−
(−1/3)
4
4
−4
n=1
4
III. This is a geometric series that is much simpler than the last:
∞
X
1
1
= 2.
=
n
2
1
−
1/2
n=0
Thus both series II and III converge to 2, and the answer is E.
8) The third degree Taylor polynomial about x = 0 of ln(1 − x) is
A) −x −
x2 x3
−
2
3
B) 1 − x +
x2
2
C) x −
x2 x3
+
2
3
D) −1 + x −
x2
2
E) −x +
x2 x3
−
2
3
It’s useful to know the Maclaurin expansion for ln(1 − x) off the top of your head, but in
case you don’t, it’s not difficult to construct, so we’ll do that here. For notational purposes
I will define f (x) = ln(1 − x). Finding derivative values at x = 0 is as follows:
f (x) = ln(1 − x)
1
f 0 (x) = −
1−x
1
f 00 (x) =
(1 − x)2
2
f 000 (x) = −
(1 − x)3
f (0) = ln(1) = 0
f 0 (0) = −1
f 00 (0) = 1
f 000 (0) = −2
Thus putting this together gives that
1
1
2
x + x2 − x3
1!
2!
3!
1 2 1 3
= −x + x − x
2
3
P3 (x) = 0 −
and thus the answer is E.
5
9) For a series S, let
1 1
1
1
1
1
1
1
1
+ −
+ −
+ −
+
−
+ · · · + an + · · · ,
9 2 25 4 49 8 81 16 121

1

if n is odd 
2(n−1)/2
Which of the following statements are true?
−1

if
n
is
even

(n + 1)2
S =1−
where an =





I. S converges because the terms of S alternate and lim an = 1.
n→∞
II. S diverges because it is not true that |an+1 | < |an | for all n.
III. S converges although it is not true that |an+1 | < |an | for all n.
A) None
B) I only
C) II only
D) III only
E) I and III only
Your first reaction to this problem is probably to want to choose answer “F) I have no idea”,
but nonetheless this is doable if you pick it apart. This series can really be thought of as
two series combined into one. One of the series is
1+
1 1 1
1
+ + +
+ ···
2 4 8 16
and the other is
1
1
1
1
1
−
−
−
− ···
− −
9 25 49 81 121
The first series is a geometric series with common ratio 1/2, and thus this converges. The
second series can be compared to a p-series with p = 2, and thus it too converges. The sum
of two convergent series is also convergent. This leaves choices I and III still valid, but we
need to examine them individually to double check their explanations. It turns out that I
makes no sense because in order for a series to converge as stated, it absolutely must be true
that
lim an = 0,
n→∞
whereas the answer says that the limit is 1. Throw option I out. III is indeed true because
we have shown that the series converges even though the absolute value of each term is not
necessarily smaller than that of the previous. III is the only correct statement so the answer
is D.
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