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GVR ActuarialExamPrep Online Exam P/1: Calculus

1

Calculus Refresher

Chapter 0 of the Study Guide contains practically all the Calculus you will need in this course. You should do all the problems at the end of Chapter 0 and the quiz at the end of this file. You may wish to review these topics several times as needed while we go through the course. The specific topics you should be very conversant with are: · The absolute value function · The exponential function · One-sided limits · Continuity · Techniques of differentiation, product, quotient and chain rules · Techniques of integration, especially substitution and parts · Double integrals (we will need this when we get to multivariate distributions)

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus · The absolute value function:

2

Without fail, every time you see the absolute value sign, write out the function explicitly without that sign. For instance if you see

|x - 5|, write it as |x - 5| = x - 5 if x - 5 0, i.e., x 5 5 - x if x - 5 < 0, i.e., x < 5

When you need to evaluate a definite integral involving the absolute value function, write the function in terms of an expression without the absolute value sign, split up the integral and evaluate it. This is illustrated in Example 0.4.1. · The exponential function: This is perhaps the most important function for this course. You should be thoroughly conversant with the properties E1-E6 on Page 1 of the Study Guide. · The logarithm function: This is the inverse of the exponential function. That means, y = ln x if and only if x = ey . exp{k ln x} = xk , x > 0. See properties L1-L5. · Get a lot of practice differentiating and integrating polynomials and the exponential function. Develop facility doing Problems such as 28 at the end of the chapter. Do not do these by rote. Get a grasp of what technique you are using. Frequently I will skip steps (especially in evaluating integrals). That is intentional. Filling in the steps is a good practice for the exam. You will retain the technical aspects better. · Geometric and exponential series: These two series are extremely important for this course. Especially observe the values over which the summation is performed. It is often a good idea to write out the first few terms of a given series. · Double integrals: It is always a good idea to draw a picture and identify the region over which the integral is defined. The method is described in Section 0.5.

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus

3

Quiz

1. Let g(x) = (x - 1)2 . Determine the set of all the values of x for which g(x) > 4. 2. Which of the following statements are true? I. If c > 0, then the graph of f (x + c) is obtained by shifting the graph of f (x) to the right by c units.

N N -k

II.

n=k b

f (n) =

n=0

f (n + k)

b/2

III.

a

f (x)dx =

a/2

f (2u)du

3. Let f (x) = exp{-c ln 2}.

Which of the following expressions is also equal to f (x)? I. (ln 2)c (ln 4)c

II. 2-c 3c III. 6c 4. You are given: f (x) = x+6

g(x) = |x - 20| Calculate |f (g(10)) - g(f (10))|.

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus 5. Solve for x: 2e2x + ex - 1 = 0. 6. The function F (x) is defined by 2 - x2 x 1 x + 3 x > 1.

4

F (x) =

Calculate

x1+

lim F (x) + lim F (x) + F (1). -

x1

7. If f (x) =

1 + x3 , calculate f (2).

8. If g(x) = |2x - 7|, calculate g (2). 9. Calculate

0 4

|x - 1|dx. 10. Evaluate

0

xe-x/2 dx

11. Let f (x) = You are also given that

0 x1

c 1<x<3 0 x3

f (x) dx = 1.

-

(a) Determine c. (b) Calculate M (t) =

-

f (x)etx dx.

(c) Determine whether or not M (t) is continuous at 0.

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus 12. Let

5

x2 x3 xn f (x) = x + + + ··· + + ··· 2! 3! n!

Calculate f (ln 2). 13. Which of the following shaded regions represents 1 < x + y < 2?

Figure 1: 14. Let D be the region 0 < x < y < 1. Evaluate x dA.

D

15. Let D be the region, 0 < y < z x < , z > 0. Let

F (z) =

D

e-x-2y dA.

Calculate F (1).

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus 16. If

0

6

f (x) = Evaluate

(x + 1) 1

2

- < x -1 -1 < x 5 x>5

6

f (x)dx.

-2

17. Let f (n) = c(0.8)n , n = 1, 2, 3, · · · You are given that

f (n) = 1.

n=1

Calculate

f (n).

n=3

18. Evaluate

2

dx . (x + 2)3

19. Let g(x) = Evaluate

0

0 0<x1 x - 1 x > 1.

e-x g(x) dx.

20. Let f (x) = and 0 elsewhere. Let

3x2 , if x > 0 (1 + x3 )2

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus

7

x

F (x) =

-

f (t)dt.

Calculate c such that F (c) = 0.9. 21. You are given f (m, n) = (0.4)m (0.5)n , m = 0, 1, 2, · · · ; n = 0, 1, 2 · · · . Calculate

f (m, n).

m=0 n=m

22. Let max(x, y) denote the larger of the numbers x, y. That is, max(x, y) = y if x y x if x > y.

Let D be the region 0 < x < 1, 0 < y < 2. Evaluate

max(x, y) dA.

D

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus

8

Solutions to Quiz

1. (x - 1)2 > 4 if and only if |x - 1| > 2. |x - 1| 2 in the interval -2 x-1 2, which is the same as -1 x 3. Therefore |x-1| > 2 if x < -1 or x > 3. 2. If g(x) = f (x + c) then g(-c) = f (0). Since c > 0, the value of g at c units to the left of the origin equals the value of f at the origin. So the graph of g(x) is obtained by shifting the graph of f to the left by c units. I is false.

N

f (n) = f (k) + f (k + 1) + · · · + f (N )

n=k

while

N -k

f (n + k) = f (k) + f (k + 1) + · · · + f (N ).

n=0

The two are the same. So II is true. Let x = 2u. Then dx = 2 du, u = a/2 when x = a, u = b/2 when x = b and b/2 b f (2u) du. f (x)dx = 2

a a/2

Therefore III is false (the factor of 2 is missing). 3. Since e-c ln 2 = eln 2

-c

= 2-c , all the expressions are equal.

4. f (g(10)) = g(10) + 6 = |10 - 20| + 6 = 4, and g(f (10)) = |f (10) - 20| = |4 - 20| = 16. So |f (g(10)) - g(f (10))| = |4 - 16| = 12. 5. 2e2x + ex - 1 = (2ex - 1)(ex + 1) = 0 implies that 2ex - 1 = 0. The other root is unacceptable because, the real exponential function cannot take negative values. Hence x = - ln 2. 6. By inspection, limx1- F (x) = 1, limx1+ F (x) = 2, F (1) = 1. Answer is 4. 7. f (x) = (1/2)(1 + x3 )-1/2 (3x2 ), f (2) = (1/2)(1/3)(12) = 2. 8. Near x = 2, 2x < 7. So g(x) = 7 - 2x and g (2) = -2.

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus 9. 10.

0 1 0 (1

9

- x)dx +

4 1 (x

- 1)dx = (1/2) + (9/2) = 5.

xe-x/2 dx =

0

x[-2e-x/2 ] dx

0

= x[-2e-x/2 ]| + 0

x [2e-x/2 ] dx

= 0+2

0

e-x/2 dx = 4.

11. (a)

3

c dx = 2c = 1

1

implies that c = 1/2. (b)

3

M (t) = (1/2)

1

etx dx =

e3t - et 2t

if t = 0. Note that this expression is undefined at t = 0. But that does not imply that M (t) is undefined at 0. In fact from the definition, as stated in Part (b) of the question,

M (0) =

-

f (x)e0 dx =

f (x) dx = 1.

-

because we are given in the problem that the integral of f over (-, ) is 1. (c) We need to check whether or not limt0 M (t) = M (0) = 1. To find the limit, use the Taylor series (Eq.(16)) for the exponential function near 0. e3t = 1 + 3t + (9/2)t2 + · · · et = 1 + t + (1/2)t2 + · · · e3t - et = 2t + 4t2 + · · · e3t - et M (t) = = 1 + 2t + · · · 2t lim M (t) = 1.

t0

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus Therefore M (t) is continuous at 0.1 12. f (x) = x + x2 x3 xn + + ··· + + · · · = ex - 1. 2! 3! n!

10

So f (ln 2) = eln 2 - 1 = 2 - 1 = 1. 13. By inspection the region is A. 14. The region of integration is shown shaded in the figure below: y goes

from 0 to 1 and for each fixed y, x goes from 0 to y. Therefore

1 y 1

x dA =

D 0 0

x dx dy =

0

y2 dy = 1/6. 2

15.

F (z) =

0

e-x

0

zx

e-2y dy dx = (1/2)

0

e-x (1-e-2zx ) dx = 1/2[1-(1+2z)-1 ].

Therefore F (z) = (1 + 2z)-2 , F (1) = 1/9.

1 Alternatively, one may us l'H^pital's rule. It says the following: If f (a) = 0 and o g(a) = 0 and limta f (t)/g (t) exists, then

ta

lim

f (t) f (t) = lim . g(t) ta g (t)

Observe that at t = 0, e3t - et = 0. Therefore e3t - et 3e3t - et = lim = 1. t0 t0 2t 2 lim

©2012 by G.V. Ramanathan

GVR ActuarialExamPrep Online Exam P/1: Calculus 16.

6 5

11

f (x)dx =

-2 -1

(x + 1)2 dx +

5

6

1 dx

= 17. 18.

2 n 1 (0.8)

(x + 1)3 5 |-1 + 1 = 73. 3

= (0.8)/0.2 = 4. So c = 0.25, 1 - f (1) - f (2) = 1 - (0.25)(0.8 + 0.64) = 0.64 dx = -(1/2)(x + 2)-2 | = 1/32. 2 3 (x + 2)

1

19. (x - 1)e-x dx = e-1 .

20.

c

F (c) =

0

3x2 1 dx = 1 - = 0.9, 3 )2 (1 + x 1 + c3

c3 = 9 and c = 2.08. 21. For fixed m sum over n first.

(0.4)m

n=m

(0.5)n = (0.4)m (0.5)m (2) = (2)(0.2)m .

Now sum over m. 2

(0.2)m = 2/0.8 = 2.5.

0

22. Split the integral along y = x and note that for y < x, max(x, y) = x and for y > x, max(x, y) = y. Therefore

1 x 2

max(x, y) dA =

D 0 1

[

0 x

max(x, y)dy +

x 2

max(x, y)dy] dx

=

0 1

[

0 0 1

xdy +

x

ydy] dx

= =

0

[x2 + 2 - x2 /2] dx [x2 /2 + 2]dx = 13/6.

©2012 by G.V. Ramanathan

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