0 is a constant

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1091!D01-05í ®M1 ãTU

1. (18%) (a) (4%) Compute lim

x→∞

[x]

x , where [⋅] is the greatest integer function.

(b) (4%) Compute lim

x→0⁻

x

√

1 − cos kx, where k > 0 is a constant.

(c) (4%) Compute lim

x→0(csc x − 1 e^x−1).

(d) (6%) Find constants a, b ∈ R such that a ≠ 0, b > 0 and lim

x→0⁺(cos x)^a/x^b =3.

Solution:

(a) Since x − 1 < [x] ≤ x

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(3%)

, by Squeeze Theorem, the limit equals to 1

®

(1%)

.

(b)

x→0lim⁻

x

√

1 − cos kx

= lim

x→0⁻

x

√

1 + cos(kx)

∣sin(kx)∣

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(2%)

=

√ 2

°

(1%)

⋅ lim

x→0⁻

x

∣sin(kx)∣ =

√ 2 ⋅ lim

x→0⁻

x

−sin kx = −

√ 2 k

²

(1%)

Remark. -1% for omitting the absolute value.

(c)

limx→0(csc x − 1

e^x−1) =lim

x→0

e^x−1 − sin x (sin x)(e^x−1)

[⁰₀],L’H

= lim

x→0

e^x−cos x

(e^x−1) cos x + e^xsin x (2%)

[⁰₀],L’H

= lim

x→0

e^x+sin x

−(e^x−1) sin x + 2e^xcos x + e^xsin x

=1

2 (2%) (d) We need to solve lim

x→0⁺

a ln(cos x)

x^b =ln 3. (2%)

It is a ⁰₀ form since b > 0. Apply L’Hospital’s Rule to study

x→0lim⁺ a b

−sin x (cos x)x^b−1 = −

a b lim

x→0⁺

sin x

x^b−1 =ln 3 (2%) Clearly b = 2

±

(1%)

, a = −b ln 3 = −2 ln 3

´¹¹¹¹¹¸¹¹¹¹¹¶

(1%)

.

(2)

2. (8%) Compute the following derivatives.

(a) (4%) d

dx(2²^x+x^x²).

(b) (4%) d

dx(tan⁻¹( x a) +ln

√x + a

x − a), where a ≠ 0 is a constant.

Solution:

(a) By the chain rule

d

dx(2²^x) = (ln 2)²2^x⋅2²^x (2 pts) Let f (x) = x^x². ln ∣f (x)∣ = x²⋅ln ∣x∣...(*)

d

dx(*) ⇒ ^f_{f (x)}^′^(x)=2x ln ∣x∣ + x

(1 pt for trying to do logarithmic differentiation. e.g. compute ln ∣f (x)∣, and know that

d

dxln ∣f (x)∣ = ^f_{f (x)}^′^(x).)

⇒ f^′(x) = x^x²(2x ln ∣x∣ + x) (1 pt)

(It is O.K. to write ln x instead of ln ∣x∣. Because the domain of f (x) is {x∣x > 0}.) (b) _dx^d (tan⁻¹(^x_a)) = ¹

1+(^x_a)²

×¹_a = _x₂^a_+a₂ (2 pts.

Wrong ans: _dx^d (tan⁻¹(^x_a)) = ¹

1+(^x_a)²

= ^a

2

x²+a² ⇒ 1 pt) Compute _dx^d ln√

x+a x−a. sol 1: ln√

x+a

x−a is defined for ∣x∣ > ∣a∣.

For ∣x∣ > ∣a∣, ln√

x+a

x−a = ¹₂(ln ∣x + a∣ − ln ∣x − a∣)

d dx(ln√

x+a

x−a) = ¹₂_dx^d(ln ∣x + a∣ − ln ∣x − a∣) = ¹₂( ¹

x+a − ¹

x−a) = _x2^−a−a² (2 pts) sol 2:

d dx(ln√

x+a

x−a) = ^√¹_x+a

x−a

d dx(

√x+a

x−a) = ¹₂^x−a_x+a_(x−a)^−2a2 = _x₂^−a_−a₂ (2 pts) Hence _dx^d (tan⁻¹(^x

a) +ln√

x+a

x−a) =a [_x2+a¹ ² − ¹

x²−a²] = ^−2a

3

x⁴−a⁴

Page 2 of 9

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3. (10%) (a) (6%) Suppose that f (x) ≤ g(x) ≤ h(x) and f (x), h(x) are differentiable at a with f (a) = h(a), f^′(a) = h^′(a). Show that g(x) is differentiable at a and find g^′(a).

(b) (4%) Give an example of functions f (x), g(x), and h(x) such that f (x) ≤ g(x) ≤ h(x), f^′(a) = h^′(a) but g(x) is not differentiable at a.

Solution:

(a) (+1) First we show that g(a) = f (a) = h(a). Since f (x) ≤ g(x) ≤ h(x) with f (a) = h(a), we have g(a) = f (a).

(Use squeeze lemma, but do not carefully distinguish the sign: (+3)) We compute lim_x→a⁺ ^g(x)−g(a)_x−a . For x > a, we have

f (x) − f (a)

x − a ≤ g(x) − g(a)

x − a ≤ h(x) − h(a) x − a

followed by f (x) ≤ g(x) ≤ h(x), f (a) = g(a) = h(a) and x − a > 0. Since the lim_x→a⁺ of the left and right terms in the above inequalities exist and equal f^′(a) = h^′(a), this forces that lim_x→a⁺ ^g(x)−g(a)_x−a exists and equals f^′(a) = h^′(a).

For lim_x→a⁻ ^g(x)−g(a)_x−a , one repeats the argument with the reversed inequalities f (x) − f (a)

x − a ≥

g(x) − g(a) x − a ≥

h(x) − h(a) x − a . We conclude that g(x) is differentiable at a and g^′(a) = f^′(a).

(b) For example,

f (x) = x − 1, g(x) = [x], h(x) = x

and a is any integer. Then f^′(x) = 1 = h^′(x) but g(x) is not differentiable at a since it is not continuous at a.

(Sketch a graph without explanation: (+2); sketch a graph with correct explanations:

(+4).)

(4)

4. (12%) An observer stands at point P which is one meter from a straight path. Let O be the point on the path that is closest to P , and S be the point on the path that is one meter to the right of O. Two runners A and B start at S and run away from O along the path. Let θ be the observer’s angle of sight between the runners.

(a) (6%) Suppose that when A and B start at S, ^dθ_dt = ¹

4 rad/sec. Find the relative velocity between A and B at S.

(b) (6%) Suppose that A runs twice as fast as B. Find the maximum value of θ.

x y

P

S B A O

θ

1 m 1 m

Solution:

(a) Suppose that at time t(sec), A is A(t) meters to the right of O and B is B(t) meters to the right of O. Then A(0) = B(0) = 1.

θ(t) = tan⁻¹A(t) − tan⁻¹B(t).

(3 pts for assigning notations and the correct equation.)

dθ dt = ^A

′(t)

1+(A(t))² − ^B

′(t) 1+(B(t))²

(2 pts for differentiation) At t = 0, ^dθ_dt =¹₄ = ^A

′(0)

1+(A(0))² − ^B

′(0)

1+(B(0))² = ¹₂(A^′(0) − B^′(0)) i.e. A^′(0) = B^′(0) = ¹₂ m/s. (1 pt for plugging in t = 0)

Ans: The relative velocity between A and B at point S is ¹₂ m/s.

(b) Sol 1:

When B is x meters to the right of S, A is 2x meters to the right of S.

θ(x) = tan⁻¹(2x + 1) − tan⁻¹(x + 1) for x > 0.

(2 pts for assigning notations and deriving the correct equation with correct domain.)

dθ

dx = _1+(2x+1)² ₂ =_1+(x+1)¹ ₂ = ^−2x

2+2 (4x²+4x+2)(x²+2x+2),

dθ

dx =0 ⇒ x = ±1

(2 pts for computing ^dθ_dx. 1 pt for solving ^dθ_dx=0.)

dθ

dx >0 for 0 < x < 1 and _dx^dθ <0 for x > 1.

Hence θ obtains the absolute maximum when x = 1 i.e. when B is 1 meter and A is 2 meters to the right of S.

(1 pt for explaining that the critical number is the absolute maximum.) Sol 2:

Suppose that the velocity of B is v m/s and the velocity of A is 2v m/s.

Then offer t seconds B is 1 + vt meters to the right of O and A is 1 + 2vt meters to the right of O.

θ(t) = tan⁻¹(1 + 2vt) − tan⁻¹(1 + vt) for t > 0.

(2 pts for assigning notations and deriving the correct equation with correct domain.)

dθ

dt = _1+(1+2vt)^2v ₂ −_1+(1+vt)^v ₂ =v [(1+(1+2vt))^−2v²²^t(1+(1+vt))²⁺² ²] (2 pts)

dθ

dt =0 ⇒ vt = ±1 (1 pt)

dθ

dt >0 for 0 < vt < 1, ^dθ_dt <0 for vt > 1.

Hence θ obtains the absolute maximum when vt = 1 i.e. when B is 2 meters to the right of O and A is 3 meters to the right of O. (1 pt)

Page 4 of 9

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5. (14%) Consider the equation y⁵+1.009y³+y = 3.

(a) (6%) Show that the equation has exactly one real solution.

(b) (4%) Given y⁵+xy³+y = 3, find _dx^dy at (1, 1).

(c) (4%) Use a linear approximation to estimate the real root of y⁵+1.009y³+y = 3.

Solution:

(a) Let g (y) = y⁵+1.009y³+y − 3. Since

y→∞limg (y) = ∞, lim

y→−∞g (y) = −∞,

the Intermediate Value Theorem implies that g has real roots. (3 points) Moreover, since

g^′(y) = 5y⁴+3.027y²+1 > 0,

g is strictly increasing. In particular, g has at most one real root. (3 points) (b) By the argument in (a), y is implicitly defined as a function of x via the equation

y⁵+xy³+y = 3

near (x, y) = (1, 1). Differentiating both sides of the above equation with respect to x gives (5y⁴+3xy²+1)dy

dx+y³ =0 (3 points) . Substituting (x, y) = (1, 1) into the above equation gives

dy dx∣

(x,y)=(1,1)

= −1

9. (1 point)

(c) Let’s denote y = f (x). Note that f (1) = 1 and f^′(1) = −¹₉. (2 points) Then f (1.009) ≈ f (1) + f^′(1) ∗ 0.009 = 1 − 1

9∗0.009 = 0.999 (2 points) .

(6)

6. (18%) Suppose that f is differentiable and one-to-one on (−1, 1), f^′(x) = 1 + f²(x), and lim

x→0 f (x)

x

exists.

(a) (4%) Find f (0) and lim

x→0 f (x)

x .

(b) (4%) Show that f (x) is increasing on (−1, 1) and determine the concavity of y = f (x) on (−1, 1).

(c) (6%) Prove that f (x) ≥ x for x ∈ (0, 1). Then prove that f (x) ≥ x +^x₃³ for x ∈ (0, 1).

(d) (2%) Find _dx^d(f⁻¹(x)).

(e) (2%) Find f⁻¹(x) and f (x).

Solution:

(a) Let L = lim

x→0 f (x)

x . Since f is differentiable, it is continuous. Therefore, f (0) = lim

x→0f (x) = lim

x→0

f (x)

x ⋅x = L ⋅ 0 = 0, and hence

limx→0

f (x) x =lim

x→0

f (x) − f (0)

x − 0 =f^′(0) = 1 + f (0)²=1.

- Búf(0) = 0ï1

- Ð0f#'nlim

x→0f (x)ï1

- Búlim

x→0 f (x)

x =1ï1

- ëú0lim

x→0 f (x)

(b) Since f^′(x) = 1 + f (x)²⩾1 > 0, f is (strictly) increasing. To determine the concavity of f , we compute the 2nd derivative of f :

f^′′= (f^′)^′= (1 + f²)^′=2f f^′.

f^′′(x) and f (x) have the same sign for every x ∈ (−1, 1) since f^′(x) > 0. We have obtained in (a) that f (0) = 0. Therefore f (x) > 0 resp. < 0 if x ∈ (0, 1) resp. (−1, 0), as shows that f is concave upward resp. downward on (0, 1) resp. (−1, 0).

- úf^′>0&1dúF ↗↗ï2

- )(f^′=1 + f²f^′′&åd$·ùï1

- f^′′cº$·ù1¦cºï1

(c) Consider the function h(x) = f (x) − x. By (a) and (b) we have

h(0) = f (0) = 0 = 0 and h^′(x) = f^′(x) − 1 = f (x)²>0,

and hence h is increasing on [0, 1) and f (x) − x = h(x) > h(0) = 0 for x ∈ (0, 1). Now consider g(x) = f (x) − x −^x₃³. We have

g(0) = f (0) − 0 = 0 and g^′(x) = f^′(x) − 1 − x²=f (x)²−x²>0 (x ∈ (0, 1)),

where the last inequality holds by the first part of (c), as we just obtained. Therefore, f (x) − x −^x₃³ =g(x) > g(0) = 0 for every x ∈ (0, 1).

Page 6 of 9

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- G, èT3

(d) We have

(f⁻¹)^′(x) = 1

f^′(f⁻¹(x)) =

1 1 + f (f⁻¹(x))²

= 1 1 + x². - ú(f⁻¹)^′(x) = ¹

f^′(f⁻¹(x))ï1

- cº1

(e) From the result of (d) we see that f⁻¹ is a function whose derivative is _1+x¹2. We know one such function, namely, tan⁻¹x, and hence the derivative of f⁻¹(x) − tan⁻¹x is 0. A function on an interval with derivative 0 everywhere is a constant function by the mean value theorem. Therefore, f⁻¹(x) = tan x + C for some constant C. Finally, C = f⁻¹(0) − tan⁻¹0 = 0, and hence f⁻¹(x) = tan⁻¹x and f (x) = tan x.

- Ð0f⁻¹(x) = tan xï1

- n8xCUzï1

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7. (20%) Let f (x) = x(ln ∣x∣)².

(a) (2%) Find the domain of f (x). Is f an odd function or even function?

(b) (2%) Compute lim

x→0f (x).

(c) (4%) Compute f^′(x). Find the interval(s) of increase and interval(s) of decrease of f (x).

(d) (2%) Find local maximum and local minimum values of f (x).

(e) (4%) Compute f^′′(x). Find the interval(s) on which f (x) is concave upward. Find the interval(s) on which f (x) is concave downward.

(f) (2%) Find the point(s) of inflection of y = f (x).

(g) (1%) Find the asymptote(s) (vertical, horizontal, or slant) of y = f (x).

(h) (3%) Sketch the graph of f (x).

Solution:

(a) f is defined on R ∖ {0}. (1 point)

Since f (−x) = −f (x), f is an odd function. (1 point) (b)

x→0+lim f (x) = lim

x→0+

⎛

⎝ ln x

√1 x

⎞

⎠

2

= lim

x→0+4⎛

⎝ ln^√¹_x

√1 x

⎞

⎠

2

= lim

y→∞ 4 (ln y y )

2

=0. (2 points) Since f is odd, we have

x→0−lim f (x) = − lim

x→0+f (x) = 0.

Thus, lim_x→0 f (x) = 0.

(c)

f^′(x) = (ln ∣x∣)²+2 ln ∣x∣ = ln ∣x∣ (ln ∣x∣ + 2) . (2 points)

So f is increasing on (−∞, −1) ∪ (−e⁻², 0) ∪ (0, e⁻²) ∪ (1, ∞) and decreasing on (−1, −e⁻²) ∪ (e⁻², 1). (2 points)

(d) By the First Derivative Test, f has local maxima

f (−1) = 0, f (e⁻²) =4e⁻² (1 point) and local minima

f (−e⁻²) = −4e⁻², f (1) = 0. (1 point) (e)

f^′′(x) = 2

xln ∣x∣ + 2 x = 2

x(ln ∣x∣ + 1) . (2 points)

So f is concave upward on (−e⁻¹, 0) ∪ (e⁻¹, ∞) and concave downward on (−∞, −e⁻¹) ∪ (0, e⁻¹). (2 points)

(f) The inflection points of the graph of f are (−e⁻¹, −e⁻¹), (0, 0), and (e⁻¹, e⁻¹). (2 points) (g) Obviously there is no vertical asymptotes. Since

x→±∞lim f (x)

x = lim

x→±∞ (ln ∣x∣)² = ∞, f has no horizontal or slant asymptotes. (1 point)

Page 8 of 9

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(h) (3 points)