5 Steps to a 5 AP Calculus BC 2019

28 STEP 2. Determine Your Test Readiness

3.4 Solutions to Diagnostic Test

Chapter 5

1. See Figure DS-1.
   Ifb=2, thenx=−1 would be a solution for
   f(x)=2.
   Ifb=0or−2,f(x)=2 would have two
   solutions.
   Thus,b=3, choice (A).

–2

–1

–2 0–1

1

4

2

3

(–2,4) (0,4)

y

x

y = 2

Figure DS-1

2. xlim→−∞

√

x^2 − 4

2 x

=xlim→−∞

√

x^2 − 4

/

(−

√

x^2 )

2 x

/

(−

√

x^2 )

(Note: asx→−∞,x=−

√

x^2 .)

=xlim→−∞

−

√

(x^2 −4)

/

x^2

2

=xlim→−∞

−

√

1 −(4/x^2 )

2

=−

√

1

2

=−

1

2

3. h(x)=

{√

x ifx> 4

x^2 −12 ifx≤ 4

xlim→ 4 +h(x)=xlim→ 4 +

√

x=

√

4 = 2

xlim→ 4 −h(x)=xlim→ 4 −(x^2 −12)=(4^2 −12)=^4

Since limx→ 4 +h(x)/=xlim→ 4 −h(x), thus limx→ 4 h(x)

does not exist.

4. f(x)=

∣∣

2 xex

∣∣

=

{

2 xex ifx≥ 0

− 2 xex ifx< 0

Ifx≥0,f′(x)= 2 ex+ex( 2 x)=

2 ex+ 2 xex

xlim→ 0 + f′(x)=xlim→ 0 +(^2 ex+^2 xex)=

2 e^0 + 0 = 2

Chapter 6

5. f(x)=−2 csc (5x)
   f′(x)=−2(−csc 5x)cot (5x)
   =10 csc (5x) cot (5x)
   f′

(

π

6

)

=10 csc

(

5 π

6

)

cot

(

5 π

6

)

=10(2)(−

√

3)=− 20

√

3

6. y=(x+1)(x−3)^2 ;
   dy
   dx
   =(1)(x−3)^2 +2(x−3)(x+1)
   =(x−3)^2 +2(x−3)(x+1)
   dy
   dx

∣∣

∣∣

x=− 1

=(− 1 −3)^2 +2(− 1 −3)(− 1 +1)

=(−4)^2 + 0 = 16

7. f′(x 1 )= lim
   Δx→ 0

f(x 1 +Δx)− f(x 1 )

Δx

Thus, lim

Δx→ 0

tan

(

π

4

+Δx

)

−tan

(

π

4

)

Δx

=

d

dx

(tanx)atx=

π

4

=sec^2

(

π

4

)

=(

√

2)^2 = 2

8. ByL’Hoˆpital’sRule, limx→π
   ex−eπ
   xe−πe
   =xlim→π
   ex
   exe−^1
   =xlim→π
   ex−^1
   xe−^1

=

eπ−^1

πe−^1

.

Chapter 7

9. See Figure DS-2 on the next page.


10. I. Since the graph ofgis decreasing and
    then increasing, it is not monotonic.
    II. Since the graph ofgis a smooth curve,
    g′is continuous.
    III. Since the graph ofgis concave upward,
    g′′>0.
    Thus, only statements II and III are true.