Addl Practice

8/7/2019 Addl Practice

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Hints/Solutions for Additional Practice Problems

1.(a) The statement is false, since both (xn) = (1, 0, 2, 0, 3, 0,...) and(yn) = (0, 1, 0, 2, 0, 3,...) are unbounded sequences in R but the sequence(xnyn) = (0, 0, 0,...) is convergent.

1.(b) The statement is true. If xn → 0, then there exists n0 ∈ N such that |xn| < 12

for all n ≥ n0 and so 0 ≤ |xnn| < (1

2)n for all n ≥ n0. Since ( 1

2)n → 0, by the sandwich

theorem, we get xnn → 0.

1.(c) The statement is true. If (xn) is convergent, then by limit rules for algebraicoperations (product rule), (x2

n) = (xnxn) is also convergent. Conversely, let (x2n) be

convergent. Then (x2n) is bounded, i.e. there exists M > 0 such that |x2

n| ≤ M for all

n ∈ N. This gives |xn| ≤√

M for all n ∈ N. So (xn) is bounded. Since it is given that(xn) is monotonic, we can conclude that (xn) is convergent.

1.(d) The statement is true. Let (xnk) be a convergent subsequence of (xn). Then(xnk) is bounded above, i.e. there exists M > 0 such that xnk ≤ M for all k ∈ N. Foreach k ∈ N, k ≤ nk and since (xn) is increasing, we get xk ≤ xnk ≤ M . Thus (xn) isbounded above and consequently (xn) is convergent.

1.(e) The statement is true. We have x2n = (1 − 12n

)sin nπ = 0 and

x4n+1 = (1 − 14n+1

) sin(2nπ + π2

) = 1 − 14n+1

for all n ∈ N. Hence x2n → 0 and x4n+1 → 1.Thus (xn) has two convergent subsequences (x2n) and (x4n+1) with different limits andtherefore (xn) is not convergent.

1.(f) The statement is true. Let xn =√

n2 + 1−

n for all n

∈N. Then xn > 0 for all n

∈N

and xn = 1√n2+1+n

=1n

1+ 1

n2+1

→ 0. Also, since

(n + 1)2 + 1 + (n + 1) > √n2 + 1 + n

for all n ∈ N, we get xn+1 < xn for all n ∈ N, i.e. the sequence (xn) is decreasing.

Therefore by Leibniz’s test,∞n=1

(−1)n+1xn is convergent and hence the given series is

convergent.

Again, if yn = 1n

for all n ∈ N, then limn→∞

xnyn

= limn→∞

1 1+ 1

n2+1

= 12

= 0. Since∞n=1

yn is

divergent, by the limit comparison test,∞n=1

xn is divergent, i.e.∞n=1

|(−1)n(√

n2 + 1 −n)|is divergent. Thus the given series is conditionally convergent.

1.(g) The statement is false. If f (x) = |x| for all x ∈ R and if g(x) =

1 if x ≥ 0,

−1 if x < 0,

then f : R → R is not differentiable at 0 and g : R → R is not differentiable at f (0) = 0,but (g ◦ f )(x) = 1 for all x ∈ R, so that g ◦ f is differentiable at 0.

1.(h) Let f (x) = x3 for all x ∈ R, so that f : R → R is differentiable. If possible,let there exist a, b ∈ R with a < 0 < b such that f (b) − f (a) = (b − a)f (0). Thenb3 − a3 = (b − a) · 0 = 0 ⇒ b3 = a3, which is not true, since a < 0 and b > 0. Thereforethe given statement is (in general) false.

1.(i) The statement is true. If possible, let f (x) = 0 for all x ∈ (0, ∞). Then bythe intermediate value property of the derivatives, either f (x) > 0 for all x ∈ (0, ∞) orf (x) < 0 for all x ∈ (0, ∞). We assume that f (x) > 0 for all x ∈ (0, ∞). (The other caseis almost similar.) Then f is strictly increasing on [0, ∞) and so f (x) > f (1) > f (0) = 0





6. Since∞n=1

xn is convergent, xn → 0, and so there exists n0 ∈ N such that |xn| < 1

for all n ≥ n0. Hence x2n ≤ |xn| for all n ≥ n0. Since

∞n=1

x2n is divergent, by the com-

parison test,∞n=1

|xn| must be divergent. Consequently∞n=1

xn is conditionally convergent.

7.(a) For all n ∈ N, 0 ≤ (√xn − 1n)2 = xn − 2

√xn

n + 1n2 . Hence

√xn

n ≤ 12(xn + 1n2 )

for all n ∈ N. Since both∞n=1

xn and∞n=1

1n2

converge,∞n=1

12

(xn + 1n2

) also converges.

Therefore by the comparison test,∞n=1

√xn

nconverges.

7.(b) Let an = xn+2n

xn+3nand bn = (2

3)n for all n ∈ N. Since

∞n=1

xn converges, xn → 0,

and so limn→∞

anbn

= limn→∞

1

2nxn+1

1

3nxn+1

= 1. Since∞n=1

bn converges, by the limit comparison test,

∞

n=1

an also converges.

(Alternative method: Since∞n=1

xn converges, xn → 0, and so there exists n0 ∈ N such

that |xn| < 1 for all n ≥ n0. Hence for all n ≥ n0, xn+2n

xn+3n< xn+2n

3n< (1

3)n + ( 2

3)n. Since

both∞n=1

(13

)n and∞n=1

(23

)n converge,∞n=1

[(13

)n + (23

)n] converges. Hence by the comparison

test,∞n=1

xn+2n

xn+3nconverges.)

8. If xn = 1n2

for all n ∈ N, then∞n=1

xn is convergent and∞n=1

xnn

=∞n=1

1n3/2

is

also convergent. On the other hand, if x1 = 0 and xn =

1

n(logn)2 for all n ≥ 2, then byCauchy’s condensation test,

∞n=1

xn =∞n=2

1n(logn)2

is convergent but∞n=1

xnn

=∞n=2

1n logn

is divergent.

9. Let xn = 2(4n+1)π

for all n ∈ N. The sequence (xn) converges to 0, but the se-

quence (f (xn)) = (2nπ + π2

) cannot converge because it is not bounded. Therefore f isnot continuous at 0.

10. Let f : R → R be defined by f (x) =

x if x ∈ Q,

x + 1 if x

∈R

\Q.

If y ∈ Q, then f (y) = y and if y ∈ R \Q, then y − 1 ∈ R \Q and f (y − 1) = y. Hence f is onto.Let x ∈ R. Then there exist sequences (rn) in Q and (tn) in R \Q such that rn → x andtn → x. Now f (rn) = rn → x and f (tn) = tn + 1 → x + 1. Since x = x + 1, it followsthat f cannot be continuous at x. Since x ∈ R was arbitrary, f is not continuous at anypoint of R.

11. Let x ∈ R \Q. Then there exists a sequence (rn) in Q such that rn → x. Since f iscontinuous at x, f (rn) → f (x). If f (x) is not an integer, then f (x) − [f (x)] > 0 and sothere exists n0 ∈ N such that |f (rn0) − f (x)| < 1

2(f (x) − [f (x)]), which is not possible,

because f (rn0) is an integer (by hypothesis). Therefore f (x) is an integer. Thus f (x) isan integer for each x ∈ R and by the intermediate value theorem, f : R → R must be a



constant function. Consequently f (13

) = f (12

) = 2.

(Alternative method for showing that f (x) is an integer : The sequence (f (rn)), being con-vergent, is a Cauchy sequence. Hence there exists n0 ∈ N such that |f (rn) − f (rn0)| < 1

2for all n ≥ n0. Since f (rn) is an integer for each n ∈ N (by hypothesis), we must havef (rn) = f (rn0) for all n ∈ N. Consequently f (rn) → f (rn0) and therefore f (x) = f (rn0),which is an integer.)

12. Let g(x) = f (x) − f (0) for all x ∈ R. The given condition gives 12

(f (x) + f (y)) =

f (12

(x + y)) = f (12

(x + y + 0)) = 12

(f (x + y) + f (0)) for all x, y ∈ R. So g(x + y) =f (x + y) − f (0) = f (x) + f (y) − 2f (0) = g(x) + g(y) for all x, y ∈ R.If n ∈ N, then g(n) = g(1 + · · · + 1) = g(1) + · · · + g(1) = ng(1). Also g(0) =g(0 + 0) = g(0) + g(0) ⇒ g(0) = 0. If m = −n, where n ∈ N, then 0 = g(0) =g(m + n) = g(m) + g(n) ⇒ g(m) = −g(n) = −ng(1) = mg(1). If r ∈ Q, then r = m

n

for some m ∈ Z, n ∈ N. So mg(1) = g(m) = g(mn

+ · · · + mn

) = g(mn

) + · · · + g(mn

) =ng(m

n) ⇒ g(m

n) = m

ng(1), i.e. g(r) = rg(1). Let x ∈ R. Then there exists a se-

quence (rn) in Q such that rn → x. Since f is continuous at x, g is also contin-uous at x. Hence g(x) = lim

n→∞g(rn) = lim

n→∞rng(1) = xg(1). Thus for all x

∈R,

f (x) − f (0) = x(f (1) − f (0)). Taking a = f (1) − f (0) ∈ R and b = f (0) ∈ R, we getf (x) = ax + b for all x ∈ R.

13. Since f : [0, 1] → R and g : [0, 1] → R are continuous, there exist x1, x2 ∈ [0, 1]such that f (x1) = sup{f (x) : x ∈ [0, 1]} and g(x2) = sup{g(x) : x ∈ [0, 1]}. Sincef (x1) = g(x2) (by hypothesis), we get f (x1) ≥ g(x1) and f (x2) ≤ g(x2). If f (x1) = g(x1)or f (x2) = g(x2), then the result follows immediately. So we may now assume thatf (x1) > g(x1) and f (x2) < g(x2). Let ϕ(x) = f (x) − g(x) for all x ∈ [0, 1]. Since f andg are continuous, ϕ : [0, 1] → R is continuous. Also ϕ(x1) > 0 and ϕ(x2) < 0. Hence bythe intermediate value theorem, there exists ξ between x1 and x2 such that ϕ(ξ) = 0,

i.e. f (ξ) = g(ξ).

14. For all x ∈ (0, 1], we have f (x) ≤ 1 + (1 − x) < 2. Hence 2 is an upper bound of {f (x) : x ∈ (0, 1]}. Therefore there exists u ∈ R such that u = sup{f (x) : x ∈ (0, 1]} ≤ 2.

Now 2(4n−1)π ∈ (0, 1] for all n ∈ N ⇒ u ≥ f

2

(4n−1)π

= 2− 2

(4n−1)π for all n ∈ N ⇒ u ≥ 2

(since limn→∞

2(4n−1)π = 0). Thus u = 2 and so (as seen at the beginning) f (x) < u for all

x ∈ (0, 1], i.e. there cannot exist any x0 ∈ (0, 1] such that f (x0) = sup{f (x) : x ∈ (0, 1]}.

15. Let α =n

i=1

αi. Then α = 0 and αiα

> 0 for i = 1,...,n. Since f : [a, b] → R is

continuous, there exist y, z ∈ [a, b] such that f (y) ≤ f (x) ≤ f (z) for all x ∈ [a, b]. In

particular, f (y) ≤ f (xi) ≤ f (z) for i = 1,...,n and soni=1

(αiα

)f (y) ≤ni=1

(αiα

)f (xi) ≤ni=1

(αiα

)f (z) ⇒ f (y) ≤ 1α

ni=1

αif (xi) ≤ f (z). By the intermediate value theorem, there

exists c between y and z (both inclusive) and so c ∈ [a, b] such that f (c) = 1α

ni=1

αif (xi),

i.e. f (c)α =ni=1

αif (xi).

(If we take α1 =

· · ·= αn = 1

n, then

n

i=1

αi = 1 and so applying the above result, we get

the required conclusion.)

16. Let p(x) = a0xn + a1xn−1 + · · · + an−1x + an for all x ∈ R, where ai ∈ R fori = 0, 1,...,n, n ∈ N is even and a0 = 0. Then p is infinitely differentiable (and so also



continuous) and p(n)(0) = n!a0. Since p(0) · p(n)(0) < 0, we have a0an < 0, i.e. a0 and an

are of different signs. Let us assume that a0 > 0, so that an < 0. (The case a0 < 0 andso an > 0 is almost similar.) Since p(x) = a0xn(1 + a1

a0· 1x

+ · · · + an−1

a0· 1xn−1 + an

a0· 1xn

)

for all x(= 0) ∈ R, we get limx→∞

p(x) = ∞ and limx→−∞

p(x) = ∞. So there exist x1 > 0

and x2 < 0 such that p(x1) > 0 and p(x2) > 0. Since p(0) = an < 0, by the intermediatevalue theorem, there exist ξ1

∈(x2, 0) and ξ2

∈(0, x1) such that p(ξ1) = 0 and p(ξ2) = 0.

20. Let f : R → R be defined by f (x) =

(x − 2)2 if x ∈ Q,

0 if x ∈ R \Q.

We have limx→2

f (x)−f (2)x−2 = lim

x→2

f (x)x−2 = 0, since

f (x)x−2

≤ |x − 2| for all x(= 2) ∈ R. Hence f

is differentiable at 2.Again, let x(= 2) ∈ R. Then there exist sequences (rn) in Q and (tn) in R \ Q suchthat rn → x and tn → x. Now f (rn) = (rn − 2)2 → (x − 2)2 and f (tn) → 0 (sincef (tn) = 0 for all n ∈ N). Since (x − 2)2 = 0, it follows that f cannot be continuous atx and consequently f cannot be differentiable at x. Therefore f is differentiable only at 2.

25. By the mean value theorem, there exist ξ1 ∈ (0,12) and ξ2 ∈ (

12 , 1) such that

f (12

) − f (0) = 12

f (ξ1) and f (1) − f (12

) = 12

f (ξ2). Hence ξ1, ξ2 ∈ [0, 1] with ξ1 = ξ2 suchthat f (ξ1) + f (ξ2) = 2[f (1) − f (0)] = 2.

27. Let a ∈ (0, 1), b ∈ R and let f (x) = x − a sin x − b for all x ∈ R. Thenf : R → R is differentiable (and also continuous) and f (x) = 1 − a cos x for allx ∈ R. Since a ∈ (0, 1), a cos x ≤ a < 1 for all x ∈ R and so f (x) = 0 for allx ∈ R. As a consequence of Rolle’s theorem, the equation f (x) = 0 has at most oneroot in R. Again, f (b + 1) = 1 − a sin(b + 1) > 0 (since a sin(b + 1) ≤ a < 1) andf (b − 1) = −1 − a sin(b − 1) < 0 (since a sin(b − 1) ≥ −a > −1). Hence by the interme-diate value theorem, the equation f (x) = 0 has at least one root in (b

−1, b + 1). Thus

the equation f (x) = 0, i.e. the equation a sin x + b = x has a unique root in R.

28. Let f (x) = ax4 + bx3 + cx2 − (a + b + c)x for all x ∈ R. Then f : R → R isdifferentiable and f (0) = 0 = f (1). Hence by Rolle’s theorem, the equation f (x) = 0,i.e. 4ax3 + 3bx2 + 2cx = a + b + c has at least one root in (0, 1).

30. Let f (x) = sin x − e−x for all x ∈ R. Then f : R → R is differentiable (alsocontinuous). Let a, b ∈ R with a < b be such that ea sin a = 1 = eb sin b. Thenf (a) = 0 = f (b). By Rolle’s theorem, there exists ξ ∈ (a, b) such that f (ξ) = 0, i.e.cos ξ + e−ξ = 0 ⇒ eξ(cos ξ + e−ξ) = 0 ⇒ eξ cos ξ + 1 = 0. Thus ξ ∈ (a, b) is a root of the

equation e

x

cos x + 1 = 0.

32. Let n ∈ N and let f n(x) = xn + x − 1 for all x ∈ [0, 1]. Then f n : [0, 1] → R

is differentiable and f n(x) = nxn−1 + 1 > 0 for all x ∈ [0, 1]. This shows that f n is astrictly increasing function on [0, 1] and so the equation f n(x) = 0 can have at most oneroot in [0, 1]. Again, since f n(0) = −1 < 0 and f n(1) = 1 > 0, by the intermediatevalue theorem, the equation f n(x) = 0 has at least one root in (0, 1). Thus the equationf n(x) = 0 has a unique root in [0, 1], which is denoted by xn.For each n ∈ N, 0 < xn < 1 ⇒ f n+1(xn) = xn+1

n +xn−1 < xnn+xn−1 = 0 = f n+1(xn+1) ⇒

xn < xn+1, since as shown above, f n+1 is strictly increasing on [0, 1]. Also xn ∈ (0, 1) forall n

∈N. Thus the sequence (xn) is increasing and bounded and consequently (xn) is

convergent. If = limn→∞xn, then 0 ≤ ≤ 1 (since 0 < xn < 1 for all n ∈ N). If possible,let < 1. Then there exists n0 ∈ N such that |xn − | < 1

2(1 − ) for all n ≥ n0. This

gives 0 < xnn < (1+

2)n for all n ≥ n0. Since 0 < 1+

2< 1, (1+

2)n → 0 and so xn

n → 0.



Now xnn + xn − 1 = 0 for all n ∈ N ⇒ lim

n→∞(xn

n + xn − 1) = 0 ⇒ − 1 = 0 ⇒ = 1, which

is a contradiction. Hence = 1.

33. By the mean value theorem, there exist x1 ∈ (a, c) and x2 ∈ (c, b) such that

f (x1) = f (c)−f (a)c−a = f (c)

c−a and f (x2) = f (b)−f (c)b−c = −f (c)

b−c . Again, by the mean value

theorem, there exists ξ

∈(x1, x2) (and so ξ

∈(a, b)) such that f (ξ) = f (x2)−f (x1)

x2−x1

=

− (b−a)f (c)(x2−x1)(b−c)(c−a) < 0, since f (c) > 0.

39. Let P = {x0, x1,...,xn}, where a = x0 < x1 < · · · < xn = b. Since L(f, P ) = U (f, P ),

we getni=1

(M i − mi)(xi − xi−1) = 0, where M i = sup{f (x) : x ∈ [xi−1, xi]} and

mi = inf {f (x) : x ∈ [xi−1, xi]} for i = 1, 2,...,n. Since M i ≥ mi and xi − xi−1 > 0 fori = 1, 2,...,n, it follows that M i − mi = 0, i.e. M i = mi for i = 1, 2,...,n. This impliesthat f is constant on [xi−1, xi] for each i ∈ {1, 2,...,n}. Hence f (x) = f (xi−1) = f (xi) forall x ∈ [xi−1, xi] (i = 1, 2,...,n). Consequently f (x) = f (a) for all x ∈ [a, b]. Therefore f

is a constant function.

42. Since f is Riemann integrable on [0, 1], f is bounded on [0, 1]. So there exists

M > 0 such that |f (x)| ≤ M for all x ∈ [0, 1]. Now |1 0

xnf (x) dx| ≤1 0

|xnf (x)| dx ≤

M 1 0

xn dx = M n+1

→ 0 as n → ∞. Hence it follows that limn→∞

1 0

xnf (x) dx = 0.

44. We have g(x) = xx 0

f (t) dt −x 0

tf (t) dt for all x ∈ R. Since f is continuous, by

the first fundamental theorem of calculus, g : R → R is differentiable and g(x) =x

0

f (t) dt + xf (x)

−xf (x) =

x

0

f (t) dt for all x

∈R. Again, since f is continuous, by the

first fundamental theorem of calculus, g : R → R is differentiable and g(x) = f (x) forall x ∈ R.

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