424 STEP 5. Build Your Test-Taking Confidence
- 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.
- The correct answer is (C).
V=π
∫ 1
0
(x−x^2 )^2 dx
=π
∫ 1
0
(x^4 − 2 x^3 +x^2 )dx
=π
[
x^5
5
−
x^4
2
+
x^3
3
] 1
0
=
π
30
Section I Part B
- 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.
- 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.