Solution Manual For Algebra and Trigonometry, 3rd Edition

Solutions for Focus on Problem Solving at www.stewartmath.com

ALGEBRA & TRIGONOMETRY

3

e

CONTENTS Focus on Problem Solving 1 Focus on Problem Solving 2 Focus on Problem Solving 3 Focus on Problem Solving 4 Focus on Problem Solving 5 Focus on Problem Solving 6 Focus on Problem Solving 7 Focus on Problem Solving 8 Focus on Problem Solving 9 Focus on Problem Solving 10 Focus on Problem Solving 11 Focus on Problem Solving 12 Focus on Problem Solving 13

1 5 9 12 14 19 22 26 32 35 38 42 45

Focus on Problem Solving 1 1.

Suppose that n3 − 1 is a prime number p. Then n3 − 1 = p ⇔ (n − 1)(n 2 + n + 1) = p .

Since p is a prime number one of the factors on the right-hand side of this equation must be 1. We consider two cases.

Case 1: n − 1 = 1 . So n = 2 . But then n 2 + n + 1 = 4 + 2 + 1 = 7 , which is a prime number.

Case 2: n 2 + n + 1 = 1 . Solving we get n = 0 . But 0 is not a natural number, so this case is impossible.

We conclude that the only natural number n for which n3 − 1 is a prime number is the natural number 2.

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1

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7.

(a)

(b)

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14.

(a) (b) (c)

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(d)

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Focus on Problem Solving 2 1.

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Focus on Problem Solving 3 1.

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Focus on Problem Solving 4 1.

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Focus on Problem Solving 5 1.

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100 = A + x = (a + b + c − ab − ac − bc + abc) + x ⇔ 100 = 68 + 52 + 52 − 33 − 35 − 37 + 30 + x = 97 + x ⇔ x = 100 − 97 = 3

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Focus on Problem Solving 6 1.

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Focus on Problem Solving 7

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Focus on Problem Solving 8 1.

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Focus on Problem Solving 9 1.

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7.

By Viete’s relations in Problem 6, we have x3 − (a + b + c) x 2 + (ab + ac + bc) x − abc = ( x − a)( x − b)( x − c)

The roots of the polynomial are a, b, and c. We see from Viete’s relations that the coefficient of x 2 is the sum of the roots and the constant term is the product of the roots.

So, if a, b, and c are the roots of P ( x) = x3 − 8 x 2 + 7 x − 6 then we must have − ( a + b + c ) = −8

and

− abc = −6

So the sum of the roots is 8 and the product of the roots is 6.

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Focus on Problem Solving 10 1.

2.

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35

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36

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Focus on Problem Solving 11 1.

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39

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40

8.

We verify the identity:

(a) (b) (c)

13 = 22 + 32 41 = 42 + 52

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Focus on Problem Solving 12 1. To maximize the number of pieces made by a cut, the cut must intersect each of the preceding cuts. (This is always possible because all the cuts have different slopes, so we cut along a line with a slope different than all the slopes of the preceding cuts.) If n cuts have already been made, then the next cut intersects the existing n cuts (lines) thereby creating n + 1 new pieces. Let P(n) = number of pieces after n cuts

The above observations show that

P(n + 1) = P(n) + n + 1 Since P (0) = 1 , we can make a table of the number of pieces after the first few cuts. n 0 1 2 3 4 5

P ( n) 1 1+1 1+1+ 2 1+1+ 2 + 3 1+1+ 2 + 3 + 4 1+1+ 2 + 3 + 4 + 5

We see a pattern in the table that leads us to make the following conjecture: P (n) = 1 + (1 + 2 + 3 + " + n) n(n + 1) P (n) = 1 + 2

(We have used the fact that the sum of the first n natural numbers is

n(n + 1) . See 2

Stewart, Redlin, Watson, Precalculus, 6e, p. 818.)

We now use mathematical induction to prove the conjecture that the number of pieces P(n) after n cuts is given by the above formula. For the induction step we use the fact that the number of pieces after n + 1 cuts is equal to the number of pieces after n cuts plus n + 1 . That is, P(n + 1) = P (n) + n + 1 , as observed above.

Step 1 P (1) is true. This is because P (1) = 2 , which is the correct number of pieces after 1 cut.

Step 2 Suppose that P (k ) is true. That is, suppose that after k cuts there are P (k ) pieces where k (k + 1) P(k ) = 1 + 2 By the above we know that

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P (k + 1) = P(k ) + k + 1

Property of cuts

k (k + 1) + (k + 1) Induction hypothesis 2 ⎛k ⎞ Factor k + 1 from the last two terms = 1 + (k + 1) ⎜ + 1⎟ ⎝2 ⎠ (k + 1)(k + 2) = 1+ Common denominator 2 = 1+

This shows that the formula holds for k + 1 . We conclude by mathematical induction that the formula holds for all natural numbers n.

2.

3. The formula for the sum of the first k natural numbers is

k (k + 1) 2 Since the square contains the numbers from 1 to n 2 , the sum of all the entries in the square is n 2 (n 2 + 1) 1 + 2 + 3 + " + n2 = 2 1+ 2 + 3 +" + k =

Now, since S is the sum of each column, and there are n columns in the square, the sum of all the entries is nS . By the above formula we have nS = 1 + 2 + 3 + " + n 2 =

n 2 (n 2 + 1) 2

Solving for S we get n(n 2 + 1) 2 For example, for a 3 × 3 square the sum S is S=

For a 4 × 4 the sum S is

3(32 + 1) = 15 S= 2 S=

4(42 + 1) = 34 2

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5.

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Focus on Problem Solving 13 1. Let L be the required limit. Then, using the property that H n = H n −1 + 2 H n − 2 we have ⎛ 2 H n−2 ⎞ ⎛ ⎞ Hn H + 2 H n−2 2 2 = lim n −1 = lim ⎜1 + = lim ⎜ 1 + = 1+ ⎟ ⎟ n →∞ H n →∞ n →∞ n →∞ H n −1 H n −1 ⎠ L n −1 ⎝ ⎝ H n −1 / H n − 2 ⎠

L = lim

2 ⇔ L2 = L + 2 ⇔ L2 − L − 2 = 0 . By the quadratic formula we get L 1± 1+ 8 L= = −1, 2. Since the sequence contains only positive terms, the limit is 2.

2

So L = 1 +

2. (a)

(b)

1⎛ N ⎞ Let L be the required limit. Then, using the property that rn = ⎜ rn −1 + ⎟ we 2⎝ rn −1 ⎠ have ⎛ 1⎛ N ⎞ 1 N ⎞ 1⎛ N⎞ L = lim rn = lim ⎜ rn −1 + = lim ⎜ rn −1 + = ⎜L+ ⎟ ⎟ ⎟ n →∞ n →∞ 2 rn −1 ⎠ 2 n→∞ ⎝ rn −1 ⎠ 2 ⎝ L⎠ ⎝

N⎞ N N 1⎛ So L = ⎜ L + ⎟ ⇔ 2 L = L + ⇔ L = ⇔ L2 = N ⇔ L = ± N . Since the L⎠ L L 2⎝ sequence contains only positive numbers, the limit is N .

3.

4.

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5. (a)

(b)

6. (a)

(b)

(c)

As n gets large, the areas of the polygons An get closer to the area of the circle, but never exceed it, so lim An = π . To prove that the limit is π we can use the fact that n →∞

sin x =1 x →0 x

lim

(See Stewart, Redlin, Watson, Precalculus, 6e, p. 395 for a graphical justification for this limit, or Stewart, Calculus, Early Transcendentals, 7e, for a proof.) Now, as 2π 2π n → ∞ we have → 0 . In the following calculations we let be represented n n by x, so as n → ∞ we have x → 0 , and we use the above displayed limit. sin 2π 2π sin x n = lim π 2π n = lim π = π ⋅1 = π lim An = lim sin 0 →∞ → n →∞ n →∞ 2 n x n x n This proves that lim An = π . n →∞

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