wtw 114 2020 final exam memo v1+v2 PDF

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Section A: Multiple Choice Questions [14 Marks available — 2 marks each]Question A1 the functionf: [− 3 ,5]→Rgraphed below: 4 - 3 - 2 - 1 1 2 3 4 5 6 x 2 1 12yf(x)Determine the Right Riemann SumR 4 offover [− 3 ,5] using 4 subintervals of equal length. (A)R 4 = 2 (B) R 4 = 4 (C) R 4 = 6 (D)R 4 = 8 (...

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Section A: Multiple Choice Questions [14 Marks available — 2 marks each]

Question A1. Consider the function f : [−3, 5] → R graphed below:

Determine the Right Riemann Sum R4 of f over [−3, 5] using 4 subintervals of equal length. (A) R4 = 2 (B) R4 = 4 (C) R4 = 6 (D) R4 = 8 (E) R4 = 0 Soln.

C

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Question A2. Let 0 < a < b and let f (x) = Consider the following statements:

x2 − a2 . bx(x − a)

(I) f has a vertical asymptote at exactly one value for x. (II) f has a horizontal asymptote as x → ∞. (III) f has no symmetry (that is, f is not even, odd, or periodic). Which of these statements are true? (A) Only II (B) Only III (C) Only I and II (D) Only I (E) All of the statements are true. Soln.

C

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Question A3. A function f : R → R is graphed below.

Given that

Z

d

f (x)dx = 0 and a

Z

c

f (x)dx = 4, determine b

Z

d a

|f (x)|dx.

(A) None of these. Z d (B) |f (x)|dx = 16 a

(C) (D)

Z

d

|f (x)|dx = 10

a

Z

d

Z

d

a

(E)

Soln.

a

|f (x)|dx = 12 |f (x)|dx = 6 A

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Question A4. Suppose −1, 0 and 1 are critical numbers of a function f and f ′′ (x) = 4x3 − 2x. Use the second derivative test to determine the critical numbers at which f has a local maximum. The set of such numbers is: (A) {−1} (B) {1} (C) {−1, 1} (D) {0} (E) None of these. Soln.

A

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Question A5. A function f : R → R is an odd function. It is also integrable on any closed interval. Z c Z b Z c f (x)dx. Suppose that 0 < a < b < c and that f (x)dx = 5 and f (x)dx = 5. Determine a

(A) (B) (C) (D)

Z Z Z

−a

−b

c −b

f (x)dx = −5

c

f (x)dx = 15 −b c

f (x)dx = 5 −b

Z

c

Z

c

f (x)dx = 10

−b

(E)

Soln.

f (x)dx = 0 −b

E

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Question A6. A function f : [a, b] → R is graphed below.

Consider the following statements about the function f : (I) f ′′ (x) < 0 for all x where f ′′ (x) exists. (II) Since f (a) = f (b), by Rolle’s Theorem there exists some c ∈ (a, b) such that f ′ (c) = 0. (III) The function f does not have a maximum on [a, b]. Which of the statements are true? (A) Only II (B) All of the statements are true. (C) Only I and II (D) Only I (E) Only III Soln.

E

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Question A7. Suppose f : R → R is a differentiable function and g : R → (0, ∞) is a differentiable function implicitly defined by f (sinh x) = ln(g(x) + 1). If f (0) = ln(5) and f ′ (0) = 5, determine g ′ (0). (A) g ′ (0) = 25 (B) g ′ (0) = 5 (C) g ′ (0) = 30 (D) g ′ (0) = 5 ln(5) (E) g ′ (0) = Soln.

1 5

A

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Section B: 28 Marks Available Show all steps and calculations and explain your answers. Question B1. [5 marks] Consider the function f : R → R, f (x) = 1 + 4xe4x . A rough sketch of this function is given below.

1.1. Prove that the function g(x) : [0, ∞) → [1, ∞), g(x) = f (x) is invertible. 1.2. Hence, explain why the inverse of g is differentiable. 1.3. Let c = g(11). Determine, with full reasons,

d dx

 g −1 (x)

[2.5] [1]

x=c

.

[1.5]

Soln. 1 We have that g ′ (x) = 4e4x (1 + 4x) > 0 for x ∈ (− 14 , ∞). Thus, g is strictly increasing on [0, ∞) and hence g is injective. We have that g(0) = 1 and limx→∞ 4xe4x = ∞. Thus, since g is increasing and continuous (it is the product of continuous functions), it follows that the range of g is [1, ∞). Therefore g is surjective. Accordingly, g is bijective and hence invertible. Soln. 2 Since g is a bijective and differentiable function such that g ′ (x) 6= 0 for any x ∈ [0, ∞), it follows by our theorem for the differentiability of the inverse of a function that g −1 is differentiable at every x ∈ [1, ∞) Soln. 3 We have that g(g −1 (x)) = x. Hence, by the chain rule, 1 d −1 g (x) = ′ −1 dx g (g (x))

for x ∈ [1, ∞).

Now g −1 (c) = 11 (since c = g(11)). Also g ′ (11) = 4e(4)(11) (1 + (4)(11)) = 180e44 . Thus 1 (g −1 )′ (c) = 180e 44

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Question B2. [6 marks] Suppose that f : R → R is a twice differentiable function. A sketch of the derivative f ′ of f is given below.

Sketch of f ′ 2.1. Use this sketch to draw sign lines for f ′ and f ′′ . Hence determine where f is increasing/decreasing and concave up/concave down. [2] 2.2. Determine if f has any local extrema and where they occur.

[1]

2.3. Determine if f has any points of inflection and where they occur.

[1]

2.4. Given that f (0) = 0, draw a rough sketch of f using all the information above. Clearly indicate any maxima, minima and points of inflection. [2] Soln. 1

Increasing on (−∞, A) and (0, D ) Decreasing on (A, 0) and (D, ∞)

Concave up on (B, C ) Concave down on (−∞, B) and (C, ∞)

Soln. 2 Local maxima at x = A and x = D. Local minima at x = 0. Soln. 3 Point of inflection at x = B and at x = C Soln. 4

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Question B3. Calculate lim (sin x)tan x . Show all your steps, and justify each step in full.

[5 marks]

x→0+

Soln.

Quiz 5

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Question B4. [5 marks] Find a value a > 0 such that the maximum vertical distance between y = ax2 +5 and y = x − 15 x2 on the interval [−1, 1] is exactly 113 . 15 Soln. Rough sketch with a = 1. (You can also sketch one or two variations, with smaller and larger a, to form a picture of what is happening.) 6 y y = x2 + 5 y = x − 15 x2 4

2

0 x −1

−0.5

0

0.5

1

Now the vertical distance d : [−1, 1] → R between the two curves is given by d(x) = ax2 + 5 − (x −

1 2 x ) 5

1 = (a + )x2 − x + 5 5 Thus d ′ (x) = 2(a + 15 )x − 1 and d ′ (x) = 0 exactly when x = Now we notice that d ′ (x) < 0 for x < have a local minimum at x =

1 . 2(a+ 51 )

1 2(a+ 51 )

1 . 2(a+ 15 )

and d ′ (x) > 0 for x >

1 . 2(a+ 15 )

Thus we

But we want to find the absolute maximum.

Since d is continuous on [−1, 2] we can apply the closed interval method to find the 1 maximum. Since we have a local minimum at x = we only need to consider the 2(a+ 15 ) endpoints. We have: 1 31 d(−1) = (a + ) + 1 + 5 = a + 5 5 21 1 d(1) = (a + ) − 1 + 5 = a + 5 5 So the absolute maximum is d(−1) = a+ 315 . Now we require that the absolute maximum is 113 . Thus when a = 34 then the absolute maximum between = 113 . Therefore a + 31 15 5 15 the two given curves is exactly 113 . 15

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Question B5.

[4 marks]

5.1. If f is continuous and

Z

8

f (x)dx = 4, find 0

Z

2

z 2 f (z 3 )dz.

[2]

0

5.2. Suppose g : R → R is a differentiable function. Determine

Z

g ′ (x) dx. 7g(x)2 + 17

[2]

Soln. 1 Variation of Q8 of Tut 13 Soln. 2 We have that

Thus

1 1 g ′ (x) g ′ (x) g ′ (x) = = 2 q 7 2 2 17 7g(x) + 17 17 17 g(x) + 1 7 g(x) +1 17 Z

g ′ (x) 1 q 2 17 7 g(x) +1 17 ! r 1 1 7 q arctan g(x) + c = 17 17 7

g ′ (x) dx = 7g(x)2 + 17

Z

17

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Question B6. [3 marks] The graphs of y = sin x and y = cos x are sketched below. Find the area of the shaded region.

Soln. Firstly we need to find the relevant points of intersection. That is we need to find x ∈ [0, 3π ] such that cos(x) = sin(x). Since this does not happen when sin(x) = 0, 2 x sin x = tan x = 1. Now tan x = 1, we can equivalently write this as cos = 1 or as cos x sin x 3π ] has solutions x = π and x = π + π x ∈ [0, 2 . 4 4 Accordingly, we need to integrate the vertical distance between the two function (top function minus bottom function) over the interval [ π4 , π + π4 ]. Thus Area =

Z

5π 4 π 4

(sin x − cos x)dx

 5π = (− cos x − sin x) π4 4

1 1 1 1 = ( √ + √ ) − (− √ − √ ) 2 2 2 2 √ 4 = √ = 2 2. 2

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Please note that this mock final assessment was set quite quickly and did not go through the usual moderation process — so there may be some misprints.

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Section A: Multiple Choice Questions [12 Marks available — 2 marks each]

Question A1. Consider the function f : [−6, 3] → R graphed below:

Determine the Right Riemann Sum R3 of f over [−6, 3] using 3 subintervals of equal length. (A) R3 = −12 (B) R3 = −9 (C) R3 = −3 (D) R3 = 3 (E) R3 = −6 Soln.

E

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Question A2. What definite integral is the following limit equal to lim n→∞

(A)

Z

(B)

Z

(C)

Z

(D)

X n 1 i=1

 ! 2i  cos2 1 + . n n

2

2 cos2 (1 + 2x) dx 0 2 2 x

1

cos2 (1 + x) dx

2

sin(1 + 2x) dx 0

Z

1

Z

1

cos2 (1 + 2x) dx

0

(E)

Soln.

2 sin(1 + x) dx 0

D

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Question A3. Suppose f : R → R is a continuous function. Suppose f has critical numbers x = a, x = b, x = c, and x = d. The second derivative of f exists everywhere except at x = d. The second derivative f ′′ is sketched below.

Consider the following statements (I) By the second derivative test, f has a local maximum at x = a. (II) By the second derivative test, f has a local maximum at x = c. (III) By the second derivative test, f has a local maximum at x = d . Which of these statements are true? (A) Only II (B) Only II and III (C) I, II, and III (D) None of the statements are true. (E) Only I Soln.

A

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Question A4. Let f : R → R be an invertible function. Indicate which of the following statements are true: (I) If f (0) = 0, then f −1 (0) = 0. (II) If f (0) = 0, then f −1 (0) is not defined. (III) If f −1 (1) = 2, then f (1) =

1 . 2

(A) Only I and III (B) Only III (C) Only II (D) Only II and III (E) Only I Soln.

E

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Question A5. The figure below shows the graph of a twice differentiable function g. Which one of the following is correct? y

y = g(x)

a •

• b

• c

• d

x

(A) g ′′ (c) < g(b) < g ′ (d) < g ′ (a) (B) g ′′ (c) < g ′ (a) < g ′ (d) < g(b) (C) g(b) < g ′′ (c) < g ′ (d) < g ′ (a) (D) g(b) < g ′ (d) < g ′ (a) < g ′′ (c) (E) g ′′ (c) < g ′ (d) < g ′ (a) < g(b) Soln.

A

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Question A6. The figure below shows the graph of a continuous function f : [0, 3] → R. Z e3 f (ln x) Evaluate the integral dx. 2x 1 y 2• y = f (x)

• 1

(A)

Z

(B)

Z

(C)

Z

(D)

Z

e3

f (ln x) dx = 1 2x

e3

3 f (ln x) dx = 2x 2

1

1 e3 1

1

e3

• 2

• 3

x

f (ln x) dx = 3 2x f (ln x) dx = 6 2x

(E) Insufficient information to compute the integral. Soln.

B

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Section B: 31 Marks Available Show all steps and calculations and explain your answers. Question B1. [3 marks] Let f : R → R be a differentiable function such that f ′ (1) = 4. Suppose the equation f (y + sin x) = 2f (y) − 2 sin x defines y implicitly as a differentiable function of x. Determine

dy dx

at the point (0, 1).

Soln. We start by differentiating the defining equation f (y + sin x) = 2f (y) − 2 sin x implicitly. By applying the chain and sum rules, we get f ′ (y + sin x)(y′ + cos x) = 2f ′ (y)y′ − 2 cos x. Thus (f ′ (y + sin x) − 2f ′ (y))y′ = −2 cos x − f ′ (y + sin x) cos x y′ =

−2 cos x − f ′ (y + sin x) cos x f ′ (y + sin x) − 2f ′ (y)

Accordingly, since we assume y(0) = 1, we get   y′ 

−2(1) − f ′ (1 + 0)(1) f ′ (1 + 0) − 2f ′ (1) x=0 −2 − 4 = 4 − 2(4) 3 −6 = . = −4 2 =

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Question B2. [3 marks] Use Rolle’s Theorem to show that the equation 2 sinh x = x has at most one solution. Provide a full explanation, justifying each step in your argument. Soln. Define function f : R → R by f (x) = 2 sinh x − x. Suppose that a and b with a < b are distinct solutions of the given equation. Therefore f (a) = 2 sinh a − a = 0 = 2 sinh b − b = f (b), whence • f (a) = f (b). Since f is differentiable on R, it follows that: • f is differentiable on (a, b) & • f is continuous on [a, b]. By Rolle’s Theorem, there exists c ∈ (a, b) such that f ′ (c) = 0. f ′ (c) = 0 ⇔ 2 cosh c − 1 = 0 ⇔ cosh c =

1 . 2

No such c can exist, however, because cosh x ≥ 1 for all x ∈ R. We thus have a contradiction. The given equation cannot therefore possess more than one solution, so that x = 0 is indeed its only solution.

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Question B3. Evaluate lim (1 + 2x )3x . Show all your steps and justify each step in full.

[4 marks]

x→∞

Soln. Let y = (1 + 2x )3x . Then ln y = 3x ln(1 + 2x ); we note that for x > 0 this is will defined. Then lim ln y = lim 3x ln(1 + x→∞

x→∞

= lim

lim ln y = lim

x→∞

d dx

1 3x

(“∞ · 0” indeterminate)

ln(1 + x2 )

(“ 00 ” indeterminate)

1 3x

x→∞

Now since ln(1 + 2x ) and apply L’Hospital’s rule:

2 ) x

are differentiable and

1 3x

6= 0 for all x > 0, we can safely

ln(1 + x2 ) d 1 dx 3x

x→∞

d 2 dx x

=

1+ x2 lim x→∞ − 1 2 3x

(chain rule)

= lim (−3x2 ) x→∞

= lim (−3x2 ) x→∞

6 x→∞ 1 +

= lim

d 2 dx x

1 + x2 − x22 1 + x2

2 x

=6

(since lim

x→∞

2 = 0 and using appropriate limit laws) x

Hence, since ex is a continuous function, it follows that lim y = lim eln y = elimx→∞ ln y = e6 .

x→∞

x→∞

Thus lim (1 + 2x )3x = e6 . x→∞

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Question B4. [4 marks] Consider the triangle in the sketch below. Find the maximum possible area of the triangle with vertices O, P and Q. (You may assume a > 0.)

y •

P y=

O

•

• a

16 +4

x2

• Q(2a, 0)

x

Soln.

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Question B5. [6 marks] Suppose that f : R → R is a twice differentiable function. A sketch of the derivative f ′ of f is given below.

Sketch of f ′ 5.1. Use this sketch to draw sign lines for f ′ and f ′′ . Hence determine where f is increasing/decreasing and concave up/concave down. [2] 5.2. Determine if f has any local extrema and where they occur.

[1]

5.3. Determine if f has any points of inflection and where they occur.

[1]

5.4. Given that f (0) = 0, draw a rough sketch of f using all the information above. Clearly indicate any maxima, minima and points of inflection. [2] Soln. 1

Increasing on (A, C) and (C, ∞) Decreasing on (−∞, A)

Concave up on (−∞, B) and (C, ∞) Concave down on (B, C ) Soln. 2 Local minima at x = A. Soln. 3 Point of inflection at x = B and at x = C Soln. 4

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Question B6. [3 marks] The figure below shows the graph of an integrable function f : [0, c] → R. It is given that Z c Z c |f (x)|dx = 5 and f (x)dx = −3. 0

Determine

Z

0

b

f (x)dx. Carefully explain your reasoning. a

y

• a

• b

x

• c

Soln. Let A1 , A2 , and A3 be the three areas bounded by f , x = 0, x = c, and the x-axis, as shown below. y

A3

A1 • a

• b

• c

x

A2

Then

Z

c

f (x)dx = A1 −A2 +A3 (since we get the negative of the area in an integral when Z b Z c |f (x)|dx = A1 + A2 + A3 and f (x) < 0 on an interval). Likewise f (x)dx = −A2 . 0

0

Now −A2 =

1 ((A1 − A2 + A3 ) − (A1 + A2 + A3 )). In other words 2 Z c  Z b Z c 1 f (x)dx = |f (x)|dx f (x)dx − 2 a 0 0 1 = (−3 − 5) 2 = −4

a

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Question B7. Let f : R → R be defined by

[4 marks] 

f (x) = Determine

Z

−5 if x ≤ 0 6 if x > 0

3π/2

sin x f (sin x)dx. 0

Soln. First, observe that

f (sin x) = So, for 0 < x <

3π , 2

( −5

if sin x ≤ 0

6 if sin x > 0.

we have  −5 if π ≤ x < 3π 2 f (sin x) =  6 if 0 < x < π.

Therefore Z

0

3π/2

sin x f (sin x)dx =

Z

π

6 sin xdx +

0

Z

3π/2

−5 sin xdx π

3π/2 π   = −6 cos x + 5 cos x 0

π

= −6(−1 − 1) + 5(0 + 1)

= 17

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Question B8. [4 marks] Suppose f : R → R is a differentiable function. Further suppose f (0) = 1 and f (1) = 2. Determine the following integral: Z 1 f ′ (x)(3f (x) + 1)17 dx. 0

Soln. Let u = 3f (x) + 1. Then du = 3f ′ (x)dx or equivalently 13 du = f ′ (x)dx. When x = 0, then u = 3f (0) + 1 = 4. When x = 1, then u = 3f (1) + 1 = 7. Accordingly, we can apply the substitution Z

1

f ′ (x)(3f (x) + 1)17 dx 0 Z 7 1 17 u du = 4 3 Z 1 7 17 u du = 3 4 1 1 18  7 = u  du 3 18 4 1 1 18 = (7 − 418 ) 3 18 718 − 418 = . 54

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END OF THE PAPER

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