SOLUTIONS TO PROBLEMS 1 (ODD NUMBERS)
1.1
(a) To two decimal places p = 124.68 (rounded up to the next even multiple);
(b) To two significant figures q = 400 ;
(c) To one significant figure r = 0.04 (the leading zeros are ignored).
1.3
In scientific notation,
(a) p = 1.246×103 to four significant figures;
(b) (−1.2×10−2 )×(1.31×105 ) = (−1.2×1.31)×105−2 , which to two decimal places is
1.57 ×103 ;
(c) 5×10−4 + 4 ×10−3 = 0.5×10−3 + 4 ×10−3 = 4.5×10−3.
1.5
(a)
123/49−1/361/2 = (2×2×3)3/4 (3×3)−1/3(3×2)1/2
= 23/4+3/4+1/233/4−1/3+1/2 = 25/2311/12,
(b) 31/252 25
−1/5
32/3 = 31/2−2/352+2/5 = 3−1/6512/5 ,
(c) 493/27−1/2 141/2 7−1/3 = 7 37−1/2 71/221/27−1/3 = 77/32−1/2 .
*1.7
(a) We have,
10
n=3
n
2
R
=
=
8
2
n =1
2n
R
=
=
2
0
so 10 = 10102 in base-2. Similarly,
0.31
n
n = −2 2
R
= 0.25
= 0.06
n = −4 2n
R
≈ 0.06
≈
0
so 0.31 ≈ 01012 in base-2, and finally 10.31 = 1010.01012 … in base-2.
(b) We have,
11012 = 23 + 22 + 20 = 13 in base-10
and
0.012 = 2−2 = 0.25 in base-10.
So 1101.012 = 13.25 in decimals.
S1.1
1.9
(a)
(b)
(c)
1.11
x x3
1
= x 1/2x 3/2x −5 = 3 ,
5
x
x
1+ 3 x
3+ x
=
(1 + 3 x )(3 − x )
(3 + x )(3 − x )
=
3 + 8 x − 3x
,
9−x
3a(3a)−5/3a 3/2
= 31/23−5/332 a 1/2a 3/2a −5/3a −2 = 35/6 a −5/3 .
(a 3)2
(
)(
)
(a) For x > 1.5 , the condition requires 2x − 3 < 5 , i.e. x < 4 . For x < 1.5 , it requires
3 − 2x < 5 , i.e. x > −1 . So −1 < x < 4 .
(b) For x 2 > 10 , we must have x 2 −10 < 6 , i.e. x 2 < 16. For x 2 < 10 , we must have
10 − x 2 < 6 , i.e. x 2 > 4 . So −4 < x < −2 or 2 < x < 4.
(c) We cannot multiply both sides by (x + 2) because this may be negative. But we
can multiply by (x + 2)2 > 0 . Then
x(x + 2) < 2(x + 2)2 ⇒ x 2 + 6x + 8 = (x + 2)(x + 4) > 0
⇒ x > −2 or x < −4.
1.13
The binomial expansion is
14 ⎛
⎞
(3a −b)14 = ∑ ⎜⎜⎜ 14 ⎟⎟⎟(3a)14−n (−b)n .
⎝ n ⎟⎠
n=0 ⎜
The term in a 3 therefore has 14 − n = 3 , i.e. n = 11 , and so
⎛ 14 ⎞⎟
14! 3 3 11
3
11
3 11
⎜⎜
⎟
⎜⎜⎝ 11 ⎟⎟⎠(3a) (−b) = − 11!3! 3 a b = −9828a b .
1.15
50 + 7 = 12 , so 5n + 7 is divisible by 4 for n = 0 . To generalise to all n ≥ 0 , we just
have to show that if 5n + 7 is divisible by 4 for any given n, then 5m + 7 is divisible
by 4, where m = n + 1 . To see this, we write
5m + 7 = 5×5n + 7 = 4 ×5n + (5n + 7)
which is divisible by 4 since both terms are divisible by 4.
1.17
(a) Squaring both sides and cross multiplying gives
y 2(x 2 − 2) = x 2 − 6 and hence x =
S1.2
2(y − 3)
,
y 2 −1
(b) y =
(x −1)(x − 2) (x −1)
2y + 1
=
, so after cross multiplying this gives x =
.
(x + 2)(x − 2) (x + 2)
1 −y
1/3
⎛(x + 2)(x − 2)⎞⎟
⎟⎟
(c) y = ⎜⎜⎜
⎟⎠
⎜⎝
x +2
2
= (x − 2)1/3 , so cubing both sides and cross multiplying
gives x = y 6 + 2 .
1.19
(a) Setting y = f (x) = (x − 2) (x + 2) , gives x = 2(y + 1) (1 −y) , and hence the inverse
function is
f −1(x) =
2(x + 1)
,
(1 − x)
(x ≠ 1) .
(b) Setting y = f (x) = 13 (x + 4)1/3 , gives x = 27y 3 − 4 and hence f −1(x) = 27x 3 − 4 .
1.21
The line y = 3x + 6 has slope 6 and so the slope of the perpendicular line to this has
slope −1 3 . Since it passes through the point (2, 2) , its equation is 3y + x = 8 and
this meets the line y = 3x + 6 at the point (−1,−3) . Finally, the distance between
(2, 2) and (−1,−3) is
1.23
(2 + 1)2 + (2 + 3)2 = 34 .
The equation of a circle with centre (x,y) = (a,b) and radius r is
(x −a)2 + (y −b)2 = r 2 .
(a) The equation is (x −1)2 + (y − 3)2 = 4
(b) The radius is now the distance from (1, 3) to (−2,5) , i.e.
r 2 = (1 + 2)2 + (3 − 5)2 = 13 ,
so the equation is (x −1)2 + (y − 3)2 = 13 .
1.25
The functions y = f (x) = 2.5 − 0.5x and y = f (x) = x 2 − 5x + 6
figure below. The hatched area corresponds to the region where
(2.5 − 0.5x) ≥ y(x) ≥ x 2 − 5x + 6 .
S1.3
are shown in the
S1.4