94ç,ç‚Bø ‚25

94ç,ç‚Bø ‚25

(20%) 1. Evaluate the limits.

(a) lim

x→∞x(√

x²− x − x) sin 1 x . Solution:

x→∞lim x√

x²− x − x sin 1

x = lim

x→∞

√

x²− x − x√

x²− x + x

√x²− x + x· lim

x→∞

sin ¹_x

1 x

= −1

2· 1 = −1 2 (b) lim

a→0⁺

√1 a



x→alim 1 x− a

Z sin x sin a

√4

t²+ a² dt

 . Solution:

a→0lim⁺

√1 a



x→alim 1 x− a

Z sin x sin a

√4

t²+ a² dt



= lim

a→0⁺

√1 a

 d dx

Z sin x

·

√4

t² + a² dt    x=a



= lim

a→0⁺

√1a·p⁴

sin²a+ a²· cos a

= lim

a→0⁺

4

s

 sin a a

2

+ 1 · lim

a→0⁺cos a

= √⁴

2 · 1 = √⁴ 2 (c) lim

n→∞

π n

 sinπ

ncos cos π

n

+ sin2π n cos

 cos2π

n



+ · · · + sin π cos (cos π)

 Solution:

n→∞lim π n

 sinπ

n cos cosπ

n

+ sin2π n cos

 cos 2π

n



+ · · · + sin π cos (cos π)



= Z π

0

sin x cos (cos x) dx

= − sin (cos x)|^π0 = 2 sin 1 (20%) 2. (a) Find f (9) if

i.

Z f (x) 0

t² dt= x cos πx.

ii.

Z x² 0

f(t) dt = x sin πx.

Solution:

i. x cos πx = 1 3t³

f (x) 0

= 1 3f(x)³, 9 · cos 9π = 1

3f(9)³, f(9)³ = −27, f(9) = −3.

ii. d dx

Z x² 0

f(t) dt = d

dxxsin πx, 2x · f(x²) = sin πx + πx cos πx,

x= 3 ⇒ 2 · 3 · f(9) = sin 3π + 3π · cos 3π = −3π ⇒ f(9) = −π 2.



x= −3 ⇒ 2 · (−3) · f(9) = sin (−3π) − 3π cos (−3π) = 3π ⇒ f(9) = −π 2



(b) Evaluate

Z π²/4 π²/9

cos√ t ptsin√

t dt. Solution:

Let u = sin√

t, then du = cos√ t· 1

2√

t dt = cos√ t dt 2√

t . t= π²

4 ⇒ u = sinπ 2 = 1, t= π²

9 ⇒ u = sinπ 3 =

√3 2 . Hence

Z π²/4 π²/9

cos√ t ptsin√

t dt=

Z 1

√3 2

√1

u·2du = 2 Z 1

√3 2

u⁻¹²du= 4√ u

1

√3 2

= 4 1 −

√4

√3 2

! .

(10%) 3. Define f (x) = x³, x > 0, 0, x ≤ 0.

(a) Show that f is differentiable at x = 0.

(b) For a < 0 < b the mean value theorem states that there exists θ ∈ (0, 1) such that

f(b) − f(a) = f⁰(a + θ · (b − a)) · (b − a).

Find an expression of θ in terms of a and b.

Solution:

(a) lim

∆x→0⁺

f(∆x) − f(0)

∆x = lim

∆x→0⁺(∆x)³ = 0,

∆x→0lim⁻

f(∆x) − f(0)

∆x = lim

∆x→0⁻

0 − 0

∆x = 0

f Ê x= 0 õí¬ûbk˝ûb,] f Ê x = 0 õª}.

(b) ÄÑ a < 0, b > 0, FJ f (a) = 0, f (b) = b³,ÌMì܉A

b³ = f⁰(a + θ · (b − a)) · (b − a) (1) DÄ

f⁰(x) = 3x² ç x >0 0 ç x≤ 0

]J a + θ · (b − a) ≤ 0, † f⁰(a + θ · (b − a)) = 0,¥ (1) J¾.Ĥ

a+ θ · (b − a) > 0,â (1) )ƒ

b³ = 3(a + θ · (b − a))²· (b − a)

j)

θ = −a b− a +

s b³

3(b − a)³ C 1 b− a

s b³

3(b − a) − a

! .

(10%) 4. For each a > 0, b > 0 consider a solid whose base is an elliptic region given by

x²

a²+^y_b2² ≤ 1 and whose cross-sections perpendicular to the x-axis are squares. Let P be the inscribed polygon as shown in figures below.

Assume that the inscribed polygon has fixed perimeter 4. Show that the maxi- mum of the volume of the solid occurs when a = q

1

3 and b =q

2

3. Also find the maximum volume.

Solution:

A(x) = 2

ra²b²− b²x² a²

!2

= 4b²

 1 − x²

a²

 . V =

Z a

−a

4b²

 1 −x²

a²



dx= 8b²



x− x³ 3a²

a 0

= 16

3 ab² C 16

3 a− a³
a²+ b² = 1 ⇒ V = 16

3 a(1 − a²) = 16

3 (a − a³) (Note 0 < a < 1)

⇒ dV da = 16

3 (1 − 3a²) = 0 when a =r 1

3 thus b =r 2 3

!

The maximum of V is 16 3

r 1 3· 2

3 = 32 9√

3 C 32 27

√3

(10%) 5. Using a linear approximation to estimate the value cot 46^◦. Is your approximation accurate to within 0.1?

Solution:

L(46^◦) = fπ 4



+ f⁰π 4

 

46^◦− π 4



= 1 − 2 46π 180 − π

4



= 0.966.

By f⁰⁰(x) > 0, x ∈hπ 4,46^◦i

, L(46^◦) < f (46^◦) ,

By f⁰(x) < 0, x ∈ hπ 4,46^◦i

, f(46^◦) − 0.1 < fπ

4

− 0.1 = 0.9 < 0.966 = L (46^◦) .

(10%) 6. A cone of radius r centimeters and height h centimeters is lowered point first at a rate of 1 cm/s into a tall cylinder of radius R > r centimeters that is partially filled with water. How fast is the water level rising at the instant the cone is completely submerged?

Solution:

cone under water V = π

3

r h

2

x³, relation

πR²y= water + π 3

r h

2

x³, chain rule

dy dt =r

R

2 dx dt, relation

dy

dt + 1 = dx dt. Thus answer is

dy

dt = r² R²− r². (20%) 7. åÇæ} A^B s¶M

A Ñ/kæ, |(øüæÑåÇ

B Ñ˛ëÜ, X°çp“wl¬˙5à
~klF)®áTJ(™p1Åpw

Fú@5/k5U{
³lp“C„J(™p1‹Å/kæU6, vüæ.8

l}

A. Study the function y = f (x) = x^2/3(x − 2)² and answer the following questions.

i. The domain of y = f (x) is a)

ii. f⁰(x) = b) ,

y= f (x) has critical points at x = c) iii. f⁰⁰(x) = d)

iv. y = f (x) is increasing on intervals e) y= f (x) is decreasing on intervals f ) v. y = f (x) is concave upward on intervals g)

y= f (x) is concave downward on intervals h)

vi. y = f (x) has local maximum at (x, y) = i) y= f (x) has local minimum at (x, y) = j) y= f (x) has inflection points at x = k) vii. Does y = f (x) have any asymptote? l) viii. Sketch the graph of y = f (x) m) : Solution:

a)R b)4

3x⁻¹³(x − 2)(2x − 1) c)0,1

2,2 d)8

9x⁻⁴³ 5x²− 5x − 1
e)(0,1

2), (2, ∞) f)(−∞, 0), (1

2,2) g)(−∞,5 − 3√

5

10 ), (5 + 3√ 5 10 ,∞) h)(5 − 3√

5

10 ,0), (0,5 + 3√ 5 10 ) i)(1

2,9 4

1

√3

4) j)(0, 0), (2, 0) k)5 ± 3√

5 10 l) No m)

0.5 1 1.5 2 2.5 3

0.5 1 1.5 2