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.

1.    (20 Marks)    Find the derivative of the following function

 f(x) = æ è x2+1 x4+1 ö ø [1/3]

Solution. Taking ln, we obtain

 lnf(x)
 =
 ln æ è x2+1 x4+1 ö ø [1/3]
 =
 1 3 ln æ è x2+1 x4+1 ö ø
 =
 1 3 (ln(x2+1)-ln(x4+1)).

Taking the derivative, we have

 f¢(x)/f(x) = 1 3 é ë 2x x2+1 - 4x3 x4+1 ù û .

This implies f¢(x) = [1/3]([(x2+1)/(x4+1)])[1/3][[2x/(x2+1)]-[(4x3)/(x4+1)]].

2.    (20 Marks)    Evaluate the following limit

 lim x® 0+ æ è sinx x ö ø [1/(x2)]

Solution: Let f(x) = ([sinx/x])[1/(x2)]. Then we have lnf(x) = [(ln([sinx/x]))/(x2)] = [(lnsinx-lnx)/(x2)].

 lim x® 0+ lnf(x)
 =
 lim x® 0+ lnsinx-lnx x2 = lim x® 0+ (lnsinx-lnx)¢ 2x
 =
 lim x® 0+ (xcosx-sinx 2x2sinx = lim x® 0+ cosx-xsinx -cosx 2(2xsinx+x2cosx)
 =
 lim x® 0+ -sinx  2(2sinx+xcosx) = lim x® 0+ -cosx  2(3cosx-xsinx)
 =
 -1/6.

Hence limx® 0+lnf(x) = -1/6. Since f(x) = elnf(x) and g(x) = lnx is continuous, we have

 lim x® 0+ f(x) = lim x® 0+ elnf(x) = elimx®0+lnf(x) = e-1/6.

3.    (60 Marks)    Evaluate each of the following indefinite integrals.

(a) ò[(x+1)/((3x+1)1/3)] dx                   (b) òsin3x cos4x dx

(c) ò ln dx                                           (d) òexcos2x dx

Solution: (a) Let u = (3x+1)1/3. Then u3 = 3x+1 and 3u2du = 3dx. This implies

 dx = u2du   and x = (u3-1)/3.

 ó õ x+1 (3x+1)1/3 dx
 =
 ó õ (u3-1)/3+1 u .u2 du = 1 3 é ë ó õ (u3-1)u+3 u du ù û
 =
 1 3 æ è ó õ (u4+2u) du ö ø = 1 3 æ è 1 5 u5+u2 ö ø +C
 =
 1 15 u5+ 1 3 u2+C
 =
 1 15 (3x+1)5/3+ 1 3 (3x+1)2/3+C

(b)

 ó õ sin3x cos4x dx
 =
 ó õ sin2xcos4x sinx dx = - ó õ sin2xcos4x dcosx
 =
 - ó õ (1-cos2x)cos4x dcosx = ó õ (cos6x-cos4x) dcosx
 =
 1 7 cos7x- 1 5 cos5x+C.

(c) ò lnx dx = x ln xx.[1/x] dx = x ln xdx = x ln x-x+C.

(d)

 ó õ excos2x dx
 =
 ó õ cos2x d ex = excos2x- ó õ ex dcos2x
 =
 excos2x-2 ó õ ex(-sin2x) dx
 =
 excos2x+2 ó õ sin2x  d ex
 =
 excos2x+2ex sin2x-2 ó õ ex d sin2x
 =
 excos2x+2ex sin2x-2 ó õ ex (2cos2x) dx
 =
 excos2x+2ex sin2x-4 ó õ ex cos2x dx

This implies 5òexcos2x dx = excos2x+2ex sin2x+C1 and

 ó õ excos2x dx = 1 5 æ è excos2x+2ex sin2x+C1 ö ø = 1 5 æ è excos2x+2ex sin2x ö ø +C

File translated from TEX by TTH, version 2.79.
On 2 Apr 2001, 15:15.