Exam 1                              Friday, March 23, 2001

 
 

NAME:

Student Number:------            No.----           Marks ----
 
 


Instructions:

1. You have 50 minutes for this exam. You are not allowed to use any calculators.

2. This exam contains 3 questions and has a total of 100 marks.

3. Show all of your work. Your work must justify the answer that you give.
 

1.    (30 Marks)    Let f(x) = [(x+1)/(x-1)], x ¹ 1.

(i) Show that f is one to one on (, 1)È(1,¥).

(ii) Find the domain of the inverse function of f.

(iii) Find a formula for the inverse function.

(i) Proof. Assume that f(x) = f(x¢), x,x¢ Î (,1)È(1,¥). Then [(x+1)/(x-1)] = [(x¢+1)/(x¢-1)]. This implies 1+[2/(x-1)] = 1+[2/(x¢-1)] and x = x¢. Hence, f is one to one on (, 1)È(1,¥).

(ii) Let y = [(x+1)/(x-1)]. Then x = [(1+y)/(y-1)]. Hence, the domain of the inverse function is (,1)È(1,¥).

(iii) f-1(x) = [(1+x)/(x-1)], x ¹ 1.

2.    (50 Marks)    Evaluate the following limits.

   (a) limx® 3[(x2-5x+6)/(x2-8x+15)]                        (b) limx® ¥ [(5x-7)/(2x+Öx)] 

  (c) limx® 0 [(cos([(p)/4]+x)- cos[(p)/4])/x]              (d) limx® 0 [(cosx-cos3x)/(x2)]

Solution. (a) limx® 3[(x2-5x+6)/(x2-8x+15)] = limx® 3[((x-3)(x-2))/((x-3)(x-5))]
                                                                   = limx® 3[(x-2)/(x-5)] = -1/2.

(b) limx® ¥ [(5x-7)/(2x+Öx)] = limx® ¥ [(5-7/x)/(2+1/Öx)] = 5/2.

(c) limx® 0 [(cos([(p)/4]+x)-cos[(p)/4])/x]
     = limx® 0[(cos[(p)/4]cosx-sin[(p)/4]sinx-cos[(p)/4])/x]
     = (Ö2)/2) limx® 0[(cosx-1)/x]-(Ö2)/2) limx® 0[sinx/x]
     = -[(Ö2)/2].

(d) limx® 0 [(cosx-cos3x)/(x2)] = limx® 0 [2sin2x sinx/(x2)]
=  4limx® 0 [sin2x/2x]limx® 0 [sinx/x] = 4.

3    (20 Marks)    Determine whether the following function is continuous at x = 0.
 
 
f(x) =  ì
ï
ï
ï
í
ï
ï
ï
î
ê
ê
sinx
x
ê
ê
if x ¹ 0
1
 if x = 0.

Solution. We rewrite f as follows:
 
 
f(x) =  ì
ï
ï
ï
ï
ï
ï
ï
í
ï
ï
ï
ï
ï
ï
ï
î
|sinx|
x
if x > 0
- |sinx|
x
if x < 0
1
 if x = 0.

Then, we have
 
 

lim
x® 0+
f(x) = 
lim
x® 0+
|sinx|
x
= limx® 0+ sinx
x
= 1 and 
lim
x® 0-
f(x) = -
lim
x® 0-
|sinx|
x
= limx® 0- sinx
x
= 1.
.

This implies limx® 0f(x) = 1 = f(0). Hence, f is continuous at x = 0.
 
 


File translated from TEX by TTH, version 2.79.
On 3 Apr 2001, 10:05.