Exam 1 - Part I – 28 questions – No Calculator Allowed

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Exam 1 - Part I – 28 questions – No Calculator Allowed - Solutions

1. Find

�

limx→0

6x5 − 8x3

9x3 − 6x5

A.

�

23

B.

�

−89

C.

�

43

D.

�

−83

E. nonexistent

�

B. limx→0

6x5 − 8x3

9x3 − 6x5 = limx→0

2x3 3x2 − 4( )3x3 3− 2x2( ) =

−89

2. Let f be a function such that

�

limx→4

f x( ) − f 4( )x − 4

= 2 . Which of the following must be true?

I. f is continuous at x = 4. II. f is differentiable at x = 4. III. The derivative of

�

′ f is continuous at x = 4.

A. I only B. II only C. I and II only D. I and III only E. I, II and III

�

C. Students should recognize that the statement tells them that the derivative of f exists at x = 4 and thus it must be true that f is continuous at x = 4 as well. The question is, what about ′ ′ f ? It is easy to construct a piecewise function for ′ f whose left and right limits are 2 at x = 4.

′ f x( ) =2x − 6,x ≤ 42,x > 4⎧ ⎨ ⎩

and thus ′ ′ f x( ) =2,x ≤ 40,x > 4⎧ ⎨ ⎩

so ′ ′ f is not continuous at x = 4.

And it is easy to generate constants so that f is continuous at x = 4 : f x( ) =x2 − 6x,x ≤ 42x −16,x > 4⎧ ⎨ ⎩

3. If

�

f x( ) = 2x + 1( ) x 2 − 3( )4, then ′ f x( ) =

A.

�

2 x 2 − 3( )3 x 2 + 4x −1( ) B.

�

4 2x +1( ) x 2 − 3( )3 C.

�

8x 2x +1( ) x 2 − 3( )3 D.

�

2 x 2 − 3( )3 3x 2 + x − 3( ) E.

�

2 x 2 − 3( )3 9x 2 + 4x − 3( )

�

E. Product/chain rule : ′ f x( ) = 2x + 1( )4 x 2 − 3( )32x( ) + 2 x 2 − 3( )4

= 2 x 2 − 3( )34x 2x + 1( ) + x 2 − 3[ ]

′ f x( ) = 2 x 2 − 3( )39x 2 + 4x − 3( )


4.

�

x 4 − x( )2∫ dx =

A.

�

x 9

9+ x 3

3+ C B.

�

x 9

9− x

6

6+ x 3

3+ C C.

�

x 9

9− x

6

3+ x 3

3+ C

D.

�

x 4 − x 2( )33

+ C E.

�

x 4 − x 2( )33 4x 3 − 2x( ) + C

�

C. x 8 − 2x 5 + x 2( )∫ dx = x 9

9− x

6

3+ x 3

3+ C

5.

�

ddx

tan t4 −1( ) dt1

x 3

∫ =

A.

�

sec2 x12 −1( ) B.

�

tan x 4 −1( ) C.

�

tan x12 −1( )

D.

�

3x 2 tan x12 −1( ) E.

�

12x11 tan x12 −1( )

�

D. ddx

tan t4 −1( ) dt1

x 3

∫ = tan x3( )4−1⎡

⎣ ⎢ ⎤ ⎦ ⎥ 3x2( ) = 3x2 tan x12 −1( )

6. Newton the cat begins to walk along a ledge at time t = 0. His velocity at

time t, 0 ≤ t ≤ 8, is given by the function whose graph is given in the figure to the right. What is Newton’s average speed from t = 0 to t = 8?

A. 0 B. 2 C. 3

D.

�

94

E. 5

�

C. Average Speed =v t( ) dt

0

8

∫8 − 0

=v t( ) dt + v t( ) dt + v t( ) dt − v t( ) dt

6

8

∫4

6

∫2

4

∫0

2

∫8 − 0

Average Speed =6 + 10 + 5 + 3

8=

248

= 3


7. The table below gives selected values of

�

v t( ) , of a particle moving along the x-axis. At time t = 0, the particle is at the origin. Which of the following could be the graph of the position,

�

x t( ) , of the particle for 0 ≤ t ≤ 4 ?

t 0 1 2 3 4

�

v t( ) 3 0 -1 1 3

A. B. C.

D. E.

D. Velocity is zero at t =1 and at 2 < t < 3 so the graph of x t( ) must have horizontal tangents at t =1 and at 2 < t < 3.

8. The graph of a twice-differentiable function f is shown in the figure

to the right. Which of the following is true? A.

�

f −2( ) < ′ f −2( ) < ′ ′ f −2( ) B.

�

f −2( ) < ′ ′ f −2( ) < ′ f −2( ) C.

�

′ ′ f −2( ) < ′ f −2( ) < f −2( ) D.

�

′ f −2( ) < f −2( ) < ′ ′ f −2( ) E.

�

′ ′ f −2( ) < f −2( ) < ′ f −2( )

�

D. Since f is decreasing, ′ f −2( ) < 0 and since f is concave up, ′ ′ f −2( ) > 0. Since f −2( ) = 0, the largest is ′ ′ f −2( ) and the smallest is ′ f −2( ).


9. What is the slope of the line tangent to the graph of

�

y = 1e2x x − 2( )

at x = 1 ?

A.

�

−2e

B.

�

−1e2

C. 0 D.

�

1e2

E.

�

−2e2

�

D. y = 1e2x x − 2( )

= e−2x

x − 2

′ y =x − 2( ) e−2x( ) −2( ) − e−2x

x − 2( )2 ⇒ ′ y 1( ) =2 e−2( ) − e−2

12 = 1e2

10. What are all the horizontal asymptotes of

�

f x( ) = 6 + 3ex

3− ex in the xy-plane?

A.

�

y = 3 only B.

�

y = −3 only C.

�

y = 2 only D.

�

y = −3 and y = 0 E.

�

y = −3 and y = 2

E. limx→∞

6 + 3ex

3− ex= −3 lim

x→−∞

6 + 03− 0

= 2

11. If f is continuous for all real numbers x and

�

f x( )1

4

∫ dx = 10, then f x − 2( ) + 2x[ ]3

6

∫ dx =

A. 37 B. 39 C. 35 D. 57 E. 25

�

A. f x − 2( ) + 2[ ]3

6

∫ dx = f x − 2( ) dx3

6

∫ + 2 dx3

6

∫ u = x − 2,du = dx x = 3,u = 1, x = 6,u = 4

f x − 2( ) dx3

6

∫ = f x( ) dx1

4

∫ = 10 2x dx3

6

∫ = x23

6= 36 − 9 = 27 ⇒10 + 27 = 37

12. If

�

f x( ) = x 3 + x 2 + x + 1 and g x( ) = f −1 x( ), what is the value of

�

′ g 4( ) ?

A.

�

185

B. 1 C.

�

157

D.

�

16

E. 24

�

D : Inverse : x = y 3 + y 2 + y + 1 = 4 ⇒ y = 1.

1 = 3y 2 + 2y + 1( ) dydx ⇒ dydx

= 13y 2 + 2y + 1

dydx y=1( )

= 13y 2 + 2y + 1

= 16


13. Find the y-intercept of the tangent line to

�

4 x + 2 y = x + y + 3 at the point (9, 4).

A. -2 B. 10 C.

�

52

D. 15 E.

�

49

�

B. 2x

+ 1ydydx

= 1+ dydx

⇒ dydx

1y−1

⎛

⎝ ⎜

⎞

⎠ ⎟ = 1− 2

x

dydx

=1− 2

x1y−1

⇒ dydx 9,4( )

=1− 2

312−1

= −23

Tangent line : y − 4 = −23

x − 9( )

y − intercept : y − 4 = −23

−9( ) ⇒ y = 10

14. The function f is continuous and non-linear for

�

−3 ≤ x ≤ 7 and

�

f −3( ) = 5 and f 7( ) = −5. If there is no value c, where

�

−3 < c < 7 , for which

�

′ f c( ) = −1, which of the following statements must be true?

A. For all k, where − 3< k < 7, ′f k( ) < −1. B. For all k, where − 3< k < 7, ′f k( ) > −1. C.

�

For some k, where − 3 < k < 7, ′ f k( ) = 0. D.

�

For − 3 < k < 7, ′ f k( ) exists. E.

�

For some k, where − 3 < k < 7, ′ f k( ) does not exist.

�

E. This is the Mean - Value Theorem which states that there must be some value k.

− 3 < k < 7 such that f 7( ) − f −3( )

7 + 3= −5 − 5

10= −1. Since there is no such value k,

then there must be a value k on − 3 < k < 7 where f is not differentiable.

15. If

�

x 2 + 2y 2 = 34 , find the behavior of the curve at (-4, 3). A. Increasing, concave up B. Increasing, concave down C. Decreasing, concave up D. Decreasing, concave down E. Decreasing, inflection point

�

B. 2x + 4y dydx

= 0 ⇒ dydx

= −2x4y

= −x2y

dydx −4,3( )

= 46

> 0 Curve is increasing.

d2ydx 2 =

−2y + x dydx( )

4y 2 d2ydx 2

−4,3( )=−6 − 4 2

3( )36

< 0 Curve is concave down.


16. A large block of ice in the shape of a cube is melting. All sides of the cube melt at the same rate. At the time that the block is s feet on each side, its surface area is decreasing at the rate of

�

24 ft2 hr . At what rate is the volume of the block decreasing at that time?

A.

�

12s ft 3 hr B.

�

6s ft 3 hr C.

�

4s ft 3 hr D.

�

2s ft 3 hr E.

�

s ft 3 hr

�

B. A = 6s2 ⇒ dAdt

= 12s dsdt

⇒−24 = 12s dsdt

⇒ dsdt

= −2s

ft hr

V = s3 ⇒ dVdt

= 3s2 dsdt

= 3s2 −2s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = −6s ft 3 hr

17. A squirrel climbs a telephone pole and starts to walk along the

telephone wire. Its velocity v of the squirrel at time t, 0 ≤ t ≤ 6 is given by the function whose graph is to the right. At what value of t does the squirrel change direction?

A. 1 and 5 only B. 2 only C. 2 and 4 only D. 1, 3, and 5 only E. 1, 3, 4 and 5 only

�

B. Direction changes ocurs when velocity switches from positive to negative or from negative to positive. This occurs at time t = 2 only.

18. The graph of

�

′ f x( ), the derivative of f , is shown to the right. Which of the following statements is not true?

A. f is increasing on 2 ≤ x ≤ 3. B. f has a local minimum at x = 1. C. f has a local maximum at x = 0. D. f is differentiable at x = 3. E. f is concave down on -2 ≤ x ≤ 1.

�

B. Local minima occur when ′ f x( ) switches from negative to positive. This occurs at x = 2. A. is true as f is increasing when ′ f x( ) > 0 C. is true as local maxima occur when ′ f x( ) switches from postitive to negative. D. is true as ′ f 3( ) exists. ′ ′ f 3( ) does not exist. E. is true as f is concave down when ′ f x( )is decreasing.


19. The difference in maximum acceleration and minimum acceleration attained on the interval 0 ≤ t ≤ 3 by the particle whose velocity is given by

�

v t( ) = 2t 3 −12t 2 +18t −1 is A. 4 B. 6 C. 18 D. 21 E. 24

�

E. a t( ) = 6t 2 − 24t + 18 ⇒ ′ a t( ) = 12t − 24 = 0 ⇒ t = 2

a 0( ) = 18, a 2( ) = −6, a 3( ) = 0 Maximum acceleration = 18 Minimum acceleration = −6 Range = 24

20. The function f is continuous on the closed interval [0, 8] and has the values given in the table below.

The trapezoidal approximation for

�

f x( ) dx0

8

∫ found with 3 subintervals is 20k. What is the value of k?

�

x 0 3 5 8f x( ) 5 k 2 7 10

A. 4 B.

�

±4 C. 8 D. -8 E. No values of k

A. 32

5+ k2( )+ 22k2 + 7( )+ 3

27+10( ) = 20k

15+ 3k2 + 2k2 +14 + 51= 40k⇒ 5k2 − 40k + 80 = 0

5 k2 −8k +16( ) = 0 ⇒ k − 4( )2 = 0 ⇒ k = 4

21.

�

sin2x1− sin2 2x

∫ dx =

A.

�

2ln cos2x( ) + C B.

�

12ln cos2x( ) + C C.

�

12sec2x tan2x + C

D.

�

12sec2x + C E.

�

2tan2x + C

�

D. sin2x1− sin2 2x

∫ dx = sin2xcos2 2x

∫ dx = tan2x sec2x dx =∫ 12

sec2x + C

22. The line

�

x + y = k , where k is a constant, is tangent to the graph of

�

y = 2x 3 − 9x 2 − x +1. What are the only possible values of k?

A. 1 only B. 0 and - 29 C. 1 and -29 D. 0 and 3 E. 1 and -26

�

E. y = −x + k ⇒ ′ y = −1 y = 2x 3 − 9x 2 − x + 1⇒ ′ y = 6x 2 −18x −1 = −1⇒ 6x x − 3( ) = 0 x = 0,y = 1 so x + y = 1 x = 3,y = −29 so x + y = −26


23. The graph of

�

y = ′ f x( ) , the derivative of f , is shown in the figure to the right. If

�

f 0( ) = −1, then f 1( ) =

A. 1 B. 2 C. 3 D. 4 E. 5

�

B. Method 1: ′ f x( )0

1

∫ dx = f 1( ) − f 0( ) ⇒ f 1( ) = f 0( ) + ′ f x( )0

1

∫ dx

′ f x( )0

1

∫ dx = area of trapezoid =12

5 + 1( ) = 3⇒ f 1( ) = −1 + 3 = 2

Method 2 : ′ f x( ) = −4x + 5 ⇒ f x( ) = −2x2 + 5x + C

f 0( ) = 0 + C = −1⇒ f x( ) = −2x2 + 5x −1⇒ f x( ) = −2 + 5 −1 = 2

24. The average value of

�

sin2 x cos x on the interval

�

π2, 3π2

⎡ ⎣ ⎢

⎤ ⎦ ⎥ is

A.

�

−23π

B.

�

23π

C. 0 D. -1 E. 1

�

A. favg =

sin2 x cos xπ 2

3π 2

∫ dx

3π2

−π2

u = sin x,du = cos x dx x =π2

,u = 1 x =3π2

,u = −1

favg =u2

1

-1

∫ du

π=u3

31

−1

=−23π

25. The functions f and g are differentiable and

�

f g x( )( ) = x 2 for all x. If

�

f 4( ) = 8, g 4( ) = 8, ′ f 8( ) = −2, what is the value of ′ g 4( )?

A.

�

−18

B.

�

−12

C.

�

−2 D.

�

−4 E. Insufficient data

�

D. ′ f g x( )( ) ′ g x( ) = 2x ⇒ ′ g x( ) = 2x′ f g x( )( )

′ g x( ) = 8′ f 8( )

= 8−2

= −4


26. Let f be a twice-differentiable function such that

�

f a( ) = b and f b( ) = a for two unknown constants a and b, a < b. Let

�

g x( ) = f f x( )( ). The Mean Value Theorem applied to

�

′ g on [a, b] guarantees a value k such that a < k < b such that

A.

�

′ g k( ) = 0 B.

�

′ ′ g k( ) = 0 C.

�

′ g k( ) =1 D.

�

′ ′ g k( ) = 1 E.

�

′ g k( ) = b − a

�

B. ′ g x( ) = ′ f f x( )( ) ⋅ ′ f x( ) ′ g a( ) = ′ f f a( )( ) ⋅ ′ f a( ) = ′ f b( ) ⋅ ′ f a( ) ′ g b( ) = ′ f f a( )( ) ⋅ ′ f b( ) = ′ f a( ) ⋅ ′ f b( ) so ′ g a( ) = ′ g b( ) Since f is twice differentiable, ′ g is differentiable so the MVT guarantees that

′ ′ g k( ) = ′ g b( ) − ′ g a( )b − a

= 0

27. The graph of

�

f x( ) = x 2 + 0.0001 − 0.01 is shown in the graph to the right. Which of the following statements are true?

�

I. limx→0

f x( ) = 0.

II. f is continuous at x = 0.III. f is differentiable at x = 0.

A. I only B. II only C. I and II only D. I, II, and III E. None are true

�

D. A trap problem. This looks like x which is not differentiable at x = 0.

But the function is given and ′ f x( ) =x

x2 + 0.0001 and ′ f 0( ) = 0.

28. What is the region enclosed by the graphs of

�

y = x − 4x 2 and y = −7x ?

A.

�

43

B.

�

163

C. 8 D.

�

683

E.

�

803

�

B. Even without a calculator, students should be able to sketch the line and parabola.

x − 4x2 = −7x⇒ 4x2 − 8x = 0 ⇒ 4x x − 2( ) = 0 ⇒ x = 0,x = 2

A = x − 4x2 + 7x( ) dx0

2

∫ = 8x − 4x2( ) dx0

2

∫ = 4x2 −4x3

3⎡

⎣ ⎢

⎤

⎦ ⎥

0

2

= 16 − 323

=163


Exam 1 - Part II – 17 questions – Calculators Allowed - Solutions

29. The slope field for the equation in the figure to the right could be

A.

�

dydx

= x + y 2 B.

�

dydx

= x − y 2 C.

�

dydx

= xy

D.

�

dydx

= x + y E.

�

dydx

= x 2 − y

�

E. Since the slope is 0 at 1,1( ), choices A and C are eliminated

Since the slope is 0 at −1,1( ), choice B is eliminated The slope field cannot be D because along the line y = −x, all slopes should be zero.

30. Let R be the region between the graphs of

�

y = 2sin x and y = cos x as shown in the figure to the right. The region R is the base of a solid with cross sections perpendicular to the x-axis as rectangles that are twice as high as wide. Find the volume of the solid.

A. 4.472 B. 7.854 C. 8.944 D. 15.708 E. 31.416

�

D. 2sin x = cos x⇒ x = 0.464,3.605

s = 2sin x − cos x. Area = 2s s( ) = 2s2. V = 2 2sin x − cos x( )2

0.464

3.605

∫ dx = 15.708

31. The first derivative of a function f is given by

�

′ f x( ) = x − 2( )esin 2x . How many points of inflection does the graph of f have on the interval 0 < x < 2π?

A. Two B. Three C. Four D. Five E. Six

�

B. ′ ′ f x( ) = 2cos2x x − 2( )esin 2x + esin 2x

As shown on the graph of ′ ′ f x( ), there are 3 values of x where it changes sign.


32. A conical tank is leaking water at the rate of

�

75in3 min. At the same time, water is being pumped into the tank at a constant rate. The tank’s height is 60 in while its top diameter is 20 inches. If the water level is rising at the rate of

�

5in min when the height of the water is 10 inches high, find the rate in which water is being pumped into the tank to the

nearest

�

in3 min . (The volume of a cone is given by

�

V = 13πr2h )

A.

�

44 in3 min B.

�

175 in3 min C.

�

250 in3 min D.

�

119 in3 min E.

�

31 in3 min

�

D. V = 13πr2h r

h= 10

60⇒ r = h

6

V = π3h2

36⎛

⎝ ⎜

⎞

⎠ ⎟ h = πh3

108

dVdt

− 75 = πh2

36dhdt

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 100π

365( ) = 43.633

dVdt

= 75 + 43.633 ≈119 in3 min.

33. At which points is the tangent line to the curve

�

8x 2 + 2y 2 = 6xy +14 vertical? I. (-2, -3) II (3, 8) III. (4, 6) A. I only B. II only C. III only D. I and II only E. I and III only

�

A. 8x 2 + 2y 2 = 6xy + 14

16x + 4y dydx

= 6x dydx

+ 6y⇒ dydx

= 6y −16x4y − 6x

= 3y − 8x2y − 3x

Tangent line vertical when dydx

is undefined so 2y = 3x.

−2,−3( ) is on the curve but 4,6( ) is not.

34. For the function f ,

�

′ f x( ) = 4 x − 3 and f 2( ) = 4. What is the approximation for

�

f 2.1( ) found by using the tangent line to the graph of f at x = 2.

A.

�

−2.6 B. 4.5 C. 4.8 D. 5.4 E. 9.4

�

B. Tangent line : ′ f 2( ) = 4 2( ) − 3 = 5 ⇒ y − 4 = 5 x − 2( ) ⇒ y 2.1( ) ≈ 5 2.1( ) − 6 = 4.5. Be careful of the trap answer (D). You aren't given the tangent line equation, just the formula for the slope of the tangent line.


35. Let

�

F x( ) be an antiderivative of

�

x 3 + x +1 . If

�

F 1( ) = −2.125, then F 4( ) =

A. -15.879 B. -11.629 C. 7.274 D. 15.879 E. 11.629

�

E. x3 + x + 1 dx1

4

∫ = F 4( ) − F 1( ) ⇒ F 4( ) = F 1( ) + x3 + x + 1 dx1

4

∫

F 4( ) = −2.125 + x3 + x + 1 dx1

4

∫ = −2.125 + 13.754 = 11.629

36. What is the area of the region in the first quadrant enclosed by the graph of

�

y = 2cos x,y = x, and the x-axis?

A. 0.816 B. 1.184 C. 1.529 D. 1.794 E. 1.999

�

A. A = x dx0

1.03

∫ + 2cos x dx1.03

π 2

∫ or A = cos−1 y2

⎛ ⎝

⎞ ⎠ − y

⎡ ⎣ ⎢

⎤ ⎦ ⎥ dx

0

1.03

∫ = 0.816

37. A particle moves along the x-axis so that at any time t > 0, its acceleration is given by

�

a t( ) = cos 1− 2t( ) . If the velocity of the particle is -2 at time t = 0, then the speed of the particle at t = 2 is

A. 0.613 B. 0.669 C. 1.331 D. 1.387 E. 1.796

�

D. v 2( ) = v 0( ) + a t( ) dt = −2 + 0.613 = −1.387 ⇒ Speed = 1.3870

2

∫

38. A right triangle has legs a and b and hypotenuse c. The lengths of leg a and leg b are changing but at a certain instant, the area of the triangle is not changing. Which statement must be true?

A. a = b B.

�

dadt

= dbdt

C.

�

dadt

= − dbdt

D.

�

a dadt

= −b dbdt

E.

�

a dbdt

= −b dadt

�

E. A = 12ab⇒ dA

dt= 1

2a dbdt

+ b dadt

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 0

a dbdt

= −b dadt


39. A particle moves along a straight line with velocity given by

�

v t( ) = t − 2.5cos2t . What is the acceleration of the particle at t = 2 ?

A. 0.168 B. 0.238 C. 0.451 D. 0.584 E. 1.450

�

B. a t( ) = 1− 2.5cos2t ln2.5( ) −sin2t( ) 2( ) ⇒ a t( ) = 1− 2.5cos4 ln2.5( ) −sin4( ) 2( ) = 0.238

40. If the graph of

�

′ f x( ) is shown to the right, which of the following could be the graph of

�

y = f x( )?

I. II. III. A. I only B. II only C. III only D. I and III only E. II and III only

�

A. Since ′ f x( ) > 0,x ≠ 0, f x( ) is increasing. But since ′ f 0( ) = 0

f x( ) has a horizontal tangent at x = 0. For III, ′ f 0( ) does not exist.

41. The rate at which the gasoline is changing in the tank of a hybrid car is modeled by

�

f t( ) = t + .5sin t − 2.5 gallons per hour, t hours after a 6-hour trip starts. At what time during the 6-hour trip was the gasoline in the tank going down most rapidly?

A. 0 B. 2.292 C. 3.228 D. 4.203 E. 6

�

A. ′ f t( ) = 12 t

+ .5cos t = 0. ⇒ t = 2.292 or t = 4.203.

Since the change is negative, we want the value of t when f t( ) is a minimum.

f 0( ) = −2.5 f 2.292( ) = −0.611 f 4.203( ) = −0.886 f 6( ) = −0.190


42. The expression

�

175

ln 7675

+ ln 7775

+ ln 7875

+ ...+ ln2⎛ ⎝ ⎜

⎞ ⎠ ⎟ is a Riemann sum approximation for

A.

�

ln x75

⎛ ⎝

⎞ ⎠ dx

1

2

∫ B.

�

ln x75

⎛ ⎝

⎞ ⎠ dx

76

150

∫ C.

�

175

ln x dx76

150

∫ D.

�

ln x dx1

2

∫ E.

�

175

ln x dx1

2

∫

�

D. This is a right Riemann sum. The base is 1

75 and the heights are ln 1 +

175

⎛ ⎝

⎞ ⎠ ,ln 1 +

275

⎛ ⎝

⎞ ⎠ ...ln 1 +

7575

⎛ ⎝

⎞ ⎠

So this is ln x dx1

2

∫ .

43. Let f be the function given by

�

f x( ) = cos t2 + t( ) dt0

x

∫ for − 2 ≤ x ≤ 2. Approximately, for what

percentage of values of x

�

for − 2 ≤ x ≤ 2 is

�

f x( ) decreasing?

A. 30% B. 26% C. 44% D. 50% E. 59%

�

B. ′ f x( ) =ddx

cos t2 + t( ) dt0

x

∫ = cos x2 + x( ) By the graphs below, ′ f x( ) < 0 for − 2 < x < 1.849 and 0.849 < x < 1.728 So ′ f x( ) < 0 for 1.849 + 2 + 1.728 − 0.849 = 1.030 values out of 4 = 26%


44. The rate of change of people waiting in line to buy tickets to a concert is given by

�

w t( ) = 100 t 3 − 4t 2 − t + 7( ) for 0 ≤ t ≤ 4. 800 people are already waiting in line when the box office opens at t = 0. Which of the following expressions give the change in people waiting in line when the line is getting shorter?

A.

�

′ w t( ) dt1.480

3.773

∫ B.

�

w t( ) dt1.480

3.773

∫ C.

�

800 − ′ w t( ) dt1.480

3.773

∫

D.

�

′ w t( ) dt0

2.786

∫ E.

�

w t( ) dt0

2.786

∫

�

B. Since w t( ) represents the rate of people waiting, w t( )t=a

t=b

∫ dt represents the change of

the number of people in line between 2 times. The line is getting shorter when its rate of change is negative. The 800 people already in line doesn't enter into it.

45. A particle moves along the x-axis so that its velocity

�

v t( ) =12te−2t − t +1. At t = 0, the particle is at position

�

x = 0.5. What is the total distance that the particle traveled from t = 0 to t = 3 ?

A. 1.448 B. 1.948 C. 2.911 D. 4.181 E. 4.681

�

D. Distance = v t( ) dt −0

1.69

∫ v t( ) dt or 1.69

3

∫ v t( ) dt = 4.181.0

3

∫ The starting position has no bearing on the answer.


Exam 1 – Answers and Description

# Ans. Topic # Ans. Topic1 B Limits/Horizontal Asymptotes 29 E Differential Equations2 C Definition of Derivative 30 D Area/Volume3 E Derivatives of Basic Functions 31 B Function Analysis4 C Integration Techniques 32 D Related Rates5 D Derivative of Accumulation Function 33 A Horizontal/Vertical Tangent Lines6 C Average Value of a Function 34 B Tangent Lines and Local Linear Approx.7 D Straight-Line Motion - Integrals 35 E Fundamental Thm of Calculus8 D Function Analysis 36 A Area/Volume9 D Tangent Lines and Local Linear Approx. 37 D Straight-Line Motion - Integrals

10 E Limits/Horizontal Asymptotes 38 E Related Rates11 A Fundamental Thm of Calculus 39 B Straight-Line Motion - Derivatives12 D Derivative of Inverse Function 40 A Function Analysis13 B Implicit Differentiation 41 A Absolute Extrema14 E Intermediate/Mean Value Theorem 42 D Riemann Sums15 B Implicit Differentiation 43 B Derivative of Accumulation Function16 B Related Rates 44 B Derivative as a Rate of Change17 B Straight-Line Motion - Derivatives 45 D Straight-Line Motion - Integrals18 B Function Analysis19 E Absolute Extrema20 A Riemann Sums21 D Integration Techniques22 E Tangent Lines and Local Linear Approx.23 B Derivative as a Rate of Change24 A Average Value of a Function25 D Derivatives of Basic Functions26 B Intermediate/Mean Value Theorem27 D Continuity/Differentiability28 B Area/Volume

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Exam 1 - Part I – 28 questions – No Calculator Allowed