calculus

1. SMN6014 DIFFERENTIAL AND INTEGRAL CALCULUS
Assignment 1: Fundamentals (Solutions)
1. Find the values of x such that
x2
3x 2 < 10 2x
Solution:
x2
3x 2 < 10 2x
x2
x 12 < 0
(x 4) (x + 3) < 0
Case 1: x 4 > 0 and x + 3 < 0:
x 4 > 0 ) x > 4
x + 3 < 0 ) x < 3
Impossible.
Case 2: x 4 < 0 and x + 3 > 0:
x 4 < 0 ) x < 4
x + 3 > 0 ) x > 3
3 < x < 4
2. Prove that the square of any odd integer is odd.
Solution: Suppose a is and odd integer. Then, a = 2k + 1 for some k 2 Z:
a2
= (2k + 1)2
= 4k2
+ 4k + 1
= 2 k2
+ 2k + 1
= 2l + 1; l 2 Z
which is odd.
3. Prove that there is no rational number whose square is 2:
Solution: We’ll prove by contradiction. Suppose there is a rational number, say q;
whose the square is 2: Let q =
a
b
; where a; b 2 Z such that q is in the lowest
form (i.e. no common factors, that is gcd (a; b) = 1):
q2
=
a
b
2
= 2
a2
b2
= 2
a2
= 2b2
1

2. a2
is even, and so a is even. Let a = 2k: Then
a2
= 2b2
2 2k2
= 2b2
b2
= 2k2
Hence, b2
is also even, and b is even. So, a and b has common factors which is
a contradiction.
4. Let y = 4x2
8x+9: Determine at what x the graph of this equation has a minimum.
Solution: Use completing the square to get the quadratic equation in the form of
y = a (x h)2
+ k
4x2
8x + 9 = 4 x2
2x +
9
4
= 4 (x 1)2
+
9
4
4
4
= 4 (x 1)2
+
5
4
= 4 (x 1)2
+ 5
x = 1:
5. Prove that, for all n 2 N;
a + ar + ar2
+ ar3
+ + arn 1
=
a (rn
1)
r 1
; r 6= 1
Solution: For n = 1 :
ar1 1
= a
a (r1
1)
r 1
= a
Suppose
12
+ 22
+ 32
+ + ark 1
=
a rk
1
r 1
then,
a rk
1
r 1
+ ark
=
ark
a + ark
(r 1)
r 1
=
ark
(1 + r 1) a
r 1
=
ark+1
a
r 1
=
a rk+1
1
r 1
; r 6= 1
2