Mathematics 1G1. Autumn 2016
Theodore Voronov.
MATH19731. Autumn semester 2016-2017
Exercise sheet 1. Solutions.
1. The slope is
m=
5
7−2
= .
4−1
3
The equation of the line is then
5
y − 2 = (x − 1).
3
Rearranging, we get
1
5
y = x+ .
3
3
5
1
5
1
× 1 + and 7 = × 4 + .
3
3
3
3
5
1
(ii) The graph has gradient and crosses the y axis at (0, ).
3
3
5
1
(iii) When x = 3.111, y = × 3.111 + = 5.518.
3
3
5
1
3
1
(iv) When y = 0.425 we get 0.425 = x + . So x = (0.425 − ) = 0.055.
3
3
5
3
(i) 2 =
2. (i) We write parametric equation of a line through two points in the form (x, y) =
(1 − t)(x0 , y0 ) + t(x1 , y1 ) where (x0 , y0 ) is one point and (x1 , y1 ) is another point.
We can take (x0 , y0 ) = (4, −1) and (x1 , y1 ) = (−6, 5). Then the equation will be
(x, y) = (1 − t)(4, −1) + t(−6, 5), or: x = 4(1 − t) − 6t, y = −(1 − t) + 5t. After
simplification this gives x = 4 − 10t, y = −1 + 6t.
(ii) To eliminate t, multiply the first equation by 6 and the second, by 10, and add.
We obtain 6x + 10y = 14, or 3x + 5y = 7 (which is the answer).
(iii) By solving for y, we obtain y =
(iv) By solving for x, we obtain x =
7−3x
5
7−5y
3
=
=
7
5
7
3
− 53 x.
− 53 y.
2
4
3. (i) The line 2x + 3y = −4 has gradient − and crosses the y axis at y = − . The
3
3
3
11
line 3x − 4y = 11 has gradient and crosses the y axis at y = − . You can then
4
4
see that the lines cross at (about) (1, −2).
(ii) Multiply the first equation by 3 and the second equation by 2 to get
6x + 9y = −12,
6x − 8y = 22
Subtract one from the other to get 17y = −34. So y = −2 and then back substituting
we get x = 1. Therefore (1, −2) is the (exact) intersection point.
1
Mathematics 1G1. Autumn 2016
Theodore Voronov.
√
4. (i)( 5)3 = (51/2 )3 = 53/2 ≈ 11.180
(23 )5
215
215
(ii) √
=
=
= 227/2 ≈ 11585.238
4
1/2 )3
3/2
3
(2
2
( 4)
5. (graphs)
6. (i) It’s easy to use Pascal’s Triangle to get the coefficients:
(a + b)7 = a7 + 7a6 b + 21a5 b2 + 35a4 b3 + 35a3 b4 + 21a2 b5 + 7a6 b + b7
(ii) Taking a =
√
√
3 and b = − 2 we have:
√
√
√ 4
√ 3√
√ 2√ 2
√ √ 3 √ 4
( 3 − 2)4 = 3 − 4 3 2 + 6 3 2 − 4 3 2 + 2
Now
√
2
3 = 3,
√
√ √
√
2
2 = 2 and 3 2 = 6, so we can simplify
√
√
√
√
√
( 3 − 2)4 = 9 − 12 6 + 36 − 8 6 + 4 = 49 − 20 6
7. Using Pascal’s Triangle again:
(1 + x)6 = 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6
and letting x = 0.01 we have
(1.01)6 = 1 + 0.06 + 0.0015 + 0.000020 + ... ≈ 1.0615
Clearly all the terms from 0.000020 are insignificant to 4 decimal places.
8. (i) x = 2 is a root, hence 2x3 +x2 −13x+6 = (x−2)(2x2 +5x−3) = (x−2)(2x−1)(x+3)
(ii) x = 1 is a root, hence x3 −2x2 −5x+6 = (x−1)(x2 −x−6) = (x−1)(x+2)(x−3)
(iii) x = 2 is a root, hence
2x3 + x2 − 8x − 4 = (x − 2)(2x2 + 5x + 2) = (x − 2)(2x + 1)(x + 2)
(iv) x = 12 is a root, hence we can split off the factor 2x − 1, and 2x3 − x2 + 2x − 1 =
(2x − 1)(x2 + 1), where the quadratic here is irreducible.
9. (i) There are two distinct linear factors:
A
B
1
=
+
.
(x + 2)(x + 3)
x+2 x+3
Multiply both sides by the left hand denominator:
1=
A(x + 2)(x + 3) B(x + 2)(x + 3)
+
= A(x + 3) + B(x + 2)
(x + 2)
(x + 3)
2
Mathematics 1G1. Autumn 2016
Theodore Voronov.
Let x = −2, then A = 1.
Let x = −3 then, B = −1. Hence,
1
1
1
=
−
.
(x + 2)(x + 3)
x+2 x+3
2x − 3
+ 8x + 16
There is a repeated linear factor, as x2 + 8x + 16 = (x + 4)2 :
(ii)
x2
2x − 3
2x − 3
B
A
=
+
=
,
x2 + 8x + 16
(x + 4)2
x + 4 (x + 4)2
2x − 3 = A(x + 4) + B ,
giving A = 2 and B = −11. Thus
2
11
2x − 3
=
−
.
x2 + 8x + 16
x + 4 (x + 4)2
x2 + x − 1
x3 − x2
We factorize x3 − x2 = x2 (x − 1), with two repeated and one distinct factor:
(iii)
x2 + x − 1
A B
C
= + 2+
,
3
2
x −x
x x
x−1
x2 + x − 1 = Ax(x − 1) + B(x − 1) + Cx2 .
Now x = 0 gives −1 = −B, i.e., B = 1, and x = 1 gives C = 1. An examination of
the x2 –terms gives A = 0. Altogether:
1
1
x2 + x − 1
= 2+
.
3
2
x −x
x
x−1
Note that the substitution method will not yield all three constants.
2
(iv) 2
(x + 2)(x + 1)
The factor x2 + 2 is irreducible, so:
(x2
2=
2
Ax + B
C
= 2
+
,
+ 2)(x + 1)
x +2
x+1
(Ax + B)(x2 + 2)(x + 1) C(x2 + 2)(x + 1)
+
(x2 + 2)
(x + 1)
= (Ax + B)(x + 1) + C(x2 + 2) = Ax2 + Bx + Ax + B + Cx2 + 2C .
Now, for the x2 –term: 0 = A + C, the x–term: 0 = A + B and the constants:
2 = B + 2C. Thus, B = C and so B = C = 32 , with A = − 23 . Thus:
(x2
2
2 − 2x
2
=
+
.
2
+ 2)(x + 1)
3(x + 2) 3(x + 1)
3
Mathematics 1G1. Autumn 2016
Theodore Voronov.
10. The polynomial x2 + 2x − 3 factorizes to (x + 3)(x − 1), while x2 + 2x + 3 is irreducible.
Hence:
1
1
=
(x2 + 2x + 3)2 (x2 + 2x − 3)2
(x2 + 2x + 3)2 (x + 3)2 (x − 1)2 ,
and so the required decomposition is
Ax + B
Cx + D
F
H
E
G
+
+
+
+
+
.
x2 + 2x + 3 (x2 + 2x + 3)2 (x + 3) (x + 3)2 x − 1 (x − 1)2
4