5 Steps to a 5 AP Calculus BC 2019

(Marvins-Underground-K-12) #1
424 STEP 5. Build Your Test-Taking Confidence


  1. The correct answer is (B).


1
n

[(
1
n

) 2
+

(
2
n

) 2
+···

(
n− 1
n

) 2 ]

represents the sum of the areas ofnrectangles
using LRAM each of width

1


n

. The heights of
the rectangles are the squares of the division
points, 0,


1


n

,


2


n

,


3


n

,...,


n− 1
n
, all of which are

between 0 and 1. Note that the first point from
the left isx=0. Thus,

nlim→∞

1


n

[(
1
n

) 2
+

(
2
n

) 2
+···

(
n− 1
n

) 2 ]

represents the area under they=x^2 from
0to1,or

∫ 1

0

x^2 dx.


  1. The correct answer is (C).


V=π

∫ 1

0

(x−x^2 )^2 dx


∫ 1

0

(x^4 − 2 x^3 +x^2 )dx


[
x^5
5


x^4
2

+


x^3
3

] 1

0
=
π
30

Section I Part B


  1. The correct answer is (C).


To assure thatf(x)=

{
ln(3−x)ifx< 2
a−bx ifx≥ 2
is differentiable atx=2, we must first be
certain that the function is continuous. As
x→2,
ln(3−x)→0, so we wanta− 2 b= 0
⇒a= 2 b. Continuity does not guarantee
differentiability, however; we must assure that
limh→ 0
f(2+h)−f(2)
h

exists. We must be certain

that limh→ 0 −
ln(3−(2+h))−ln(3−2)
h

is equal to limh→ 0 +
(a−b(x+h))−(a−bx)
h

.


hlim→ 0 −

ln(3−(2+h))−ln(3−2)
h
=hlim→ 0 −

ln(1−h)
h

=


0


0


. Thus, limh→ 0 −


(
1
1 −h

)
(−1)

=−1. limh→ 0 +
(a−b(2+h))−(a− 2 b)
h

=


−b⇒−b=− 1 ⇒b= 1 ⇒a=2.


  1. The correct answer is (B).
    Sinceh(x)=f(g(x)),h′(x)= f′(g(x))g′(x)
    andh′(2)=f′(g(2))g′(2)=f′(3)(1)=−1.

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