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(MATH023)9d63db - 023_final_2007.pdf
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Math 023 -Calculus I Final Examination, Fall Semester, 2007
Time Allowed: 2.5 Hours Total Marks: 100 dy
1. (12 Marks) Suppose that f is di.erentiable everywhere. Find of
dx
the following functions:
(a)
y = f(x2); (c) y = f (f(f(x))). f(x);

(b)
y = f(ex) e


2. (28 Marks) Evaluate the following limits:
(a) lim xa+ ln(x . a) ln(ex . ea); (c) lim n+ . cos n .n2 ;
(b) lim x0 2 . sin2 x . 2 cos x x4 ; (d) lim x0 . 1 ln(1 + tan x) . 1 x . .
3. (10 Marks) Derive the approximation formula
1 n an + x 1 a . x nan+1
for small x an . Use it to evaluate 1 10 1000 .
4. (6 Marks) Prove the following inequalities
1 2 x2 1 + x < x . ln(1 + x) < 1 2x 2 for x > 0.
These inequalities are reversed if .1 < x < 0.

5. (6 Marks) Find the global maximum and minimum of the function x . 1
in the interval [.1, 3].
x2 . x +1 3
x
6.
(14 Marks) Sketch the function y =.

(x . 1)2

7.
(6 Marks) Let y = a cos(ln x)+ b sin(ln x), with a and b being some constants. Show that


2
xy(n+2) + (2n + 1)xy(n+1) +(n 2 + 1)y(n) =0. 1
8.
(6 Marks) Show that an open cylinder (with the base surface but no the top surface) of given surface area has the greatest volume if its height is equal to the radius of its base.

9.
(6 Marks) Prove that the equation


2 n
x
1+ x + x+ + x= ae
2! n! has only one positive solution, where n 1 is an integer and 0 <a< 1.
10. (6 Marks) If f(x)g.(x) . f.(x)g(x) = 0 in an interval, then show that .between two consecutive solutions of f(x) = 0, there is exactly one solution of g(x) = 0.
Solutions
dy
1. (12 Marks) Suppose that f is di.erentiable everywhere. Find of
dx
the following functions:
(a)
y = f(x2); (c) y = f (f(f(x))). f(x);

(b)
y = f(ex) e


Solution
(a) dy = f.(x 2) 2x =2xf.(x 2)dx
(b)
dy f(x) ..
=[f(e x)]. e f(x) + f(e x) edx
xf(x)
= f.(e x) e e f(x) + f(e x) e f.(x)
(c) dy =[f(f(f(x)))].
dx = f.(f(f(x))) f.(f(x)) f.(x)
2. (28 Marks) Evaluate the following limits: ln(x . a) . .n2
(a) lim ; (c) lim cos ;
xa+ ln(ex . ea)n+ n
112 . sin2 x . 2 cos x
(b)
lim ; (d) lim . .

x0 x4 x0 ln(1 + tan x) x

(b)


Solution
(a)
lim xa+ ln(x . a) ln(ex . ea) = lim xa+ 1 x.a 1 ex.ea ex
= lim ex . ea lim 1
xa+ x . a xa+ ex
= e a 1 = 1.
ea

2 . sin2 x . 2 cos x .2 sin x cos x + 2 sin x
lim = lim
x0 x4 x0 4x3
sin x .2 cos x +2

= lim lim
x0 x x0 4x2 2 sin x 1
=1 lim = .
x0 8x 4
(c) We .rst consider the limit:
1 ln cos x 2
lim (cos x)x2 = lim e x
x0 x0
ln cos x
= exp lim
2
x0 x
1
(. sin x)
cos x
= exp lim
x0 2x
. 2 = e 2 .
1
A special case of the limit is to let x = and n +. Thus
n
..n2
1 . 2
2
lim cos = lim (cos x)x= e 2 .
n+ n x0
(d) Since, by lHospitals rule,
1
ln(