Answers to Problems
1
Plotting functions
1. Plot y = x2 over the range of x from −5 to +5.
The first step is to determine a sensible range for the y values. Do this by
making a table of the values of y for each x.
x
y
−5
25
−4
16
−3
9
−2
4
−1
1
0
0
1
1
2
4
3
9
4
16
5
25
25
20
15
10
5
0
-4
-2
0
2
4
What do you notice about the symmetry of the function about the y-axis?
It is symmetric.
1
2. Plot y = −3x2 + 6x over same range.
20
0
-20
-40
-60
-80
-100
-120
-4
-2
0
2
4
3. Plot y = x3 + 2 over same range.
150
100
50
0
-50
-100
-150
-4
-2
0
2
4
What do you notice abouth the symmetry of the function about the yaxis?
It is anti-symmetric.
2
4. Plot y = x3 + x2 + 2x over same range.
200
150
100
50
0
-50
-100
-150
-4
-2
0
2
4
5. Plot r = sin(θ) for θ = −180, −135, −90, −45 . . . + 135, +180.
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
θ
r
−180
0
-150
-100
−135
−0.7
−90
−1
-50
0
−45
−0.7
0
0
3
50
45
0.7
90
1
100
135
0.7
150
180
0
6. Plot r = sin(2θ).
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-150
-100
-50
0
50
100
150
-100
-50
0
50
100
150
7. Plot r = cos(θ).
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-150
4
8. Plot r = cos(θ + 30).
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-150
-100
-50
0
50
100
150
-100
-50
0
50
100
150
9. Plot r = sin(θ + 30).
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-150
5
10. Plot r = cos(θ + 90).
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-150
-100
-50
0
50
100
150
-100
-50
0
50
100
150
11. Plot r = sin(θ + 90).
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-150
6
12. Plot r = tan(θ).
80
60
40
20
0
-20
-40
-60
-80
-150
-100
-50
0
50
100
150
What do you notice abouth the symmetry of these functions about the
θ-axis? What do you notice about the plots with (θ + 90)?
The functions repeat periodically. The sin and cos functions vary
between −1 and 1. The tan function goes to infinity for certain
values of θ. Thinking about the functions in trigonometry should
make it clear why this is so.
13. Extend one or more of the above plots for some θ < −180 and θ > 180
The functions just repeat even further.
7
2
Numbers
1. Classify the following numbers as integer, rational or irrational:
π
1 √
1, 0, 0.5, 2, 2.0, π, , 0.3, 0.333, , 7
2
3
1, 0, 2 are integers; 0.5, 0.3, 0.333 and 2.0 are rational provided
√
we know their values exactly; π is irrational, 31 is rational, 7 is
irrational.
2. Write the following numbers to 3 significant figures (by rounding as appropriate):
0.25666, 0.5999, 0.1, 2, 121, 176.455
0.257, 0.600, 0.1, 2, 121, 176. One could argue for 0.100 and 2.00
also.
3. Write the same numbers as in the previous question to 2 decimal places.
0.26, 0.60, 0.10, 2.00, 121.00, 176.00. We may be claiming accuracy we don’t have in real life here, though.
4. (a) Round 7.455 to 3 significant figures.
7.46
(b) Round the result of above to 2 significant figures.
7.5
(c) Round the result of above to 1 significant figure.
8
(d) Round 7.455 to 1 significant figure.
7
What do you notice is different from rounding in several steps to rounding
in one step?
Repeated rounding accumulates small errors. We should round
once only: when we give our final answer.
5. Write the following numbers in scientific notation with 3 significant figures:
1, 100, 1024, 97.45679, 0.0045, 1.00004, 10.345
1.00 × 100 or 1.00, 1.00 × 102 , 1.02 × 103 , 9.75 × 101 , 4.5 × 10−3,
1.00 × 100 or 1.00, 1.03 × 101 .
6. Evaluate the following:
2.6 × 10−4 + 4.5 × 10−3
4.76 × 10−3
8
7. Evaluate the following to 2 decimal places:
1.45788 × 10−2 + 6.58945 × 102
6.59 × 102
8. Evaluate the following to 3 decimal places:
1.45788 × 10−2 × 6.58945 × 102
9.607 × 100
9. Evaluate the following to 2 decimal places:
6.6 × 10−3 − 8.5 × 10−1
−0.84
10. The mass of 1l of water is 1kg. What size cube is 1l of water? If the
mass per mole of water is 18g and there are 6.02 × 1023 molecules of water in one mole, how many water molecules are there in 1g? How many
water molecules are there in 1m3 ? What is the average volume of one
water molecule? If this volume where a cube, what length would each size
be? Sketch a small picture of a water molecule with bond angle 90◦ and
bond length 1Å. Is the size of the cube calculated consistent with the space
occupied by your sketch? (Obviously water molecules do not sit in individual little cubes in water, but the values you obtain should be comparable).
10cm cube.
18g≡ 6.02 × 1023 molecules.
1g is therefore ≡ 6.02 × 1023/18 = 4.63 × 1022
In 1m3 there are 4.63 × 1022 × 1003 = 4.63 × 1028
So average volume per molecule is
1
4.63×1028
= 2.15 × 10−29m3
A
cube with this volume would have edges of length
p
3
2.15 × 10−29 = 2.27 × 10−10
1Å≡ 1 × 10−10 m. Your picture can tell you only that a single
water molecule occupies roughly a cube of this order of magnitude (same power of 10). The size of the cube you calculated
should also be in agreement to an order of magnitude. It is
therefore a reasonable, if rough, approximation.
9
3
Algebra
Expand out the brackets in the following and simplify where possible
1. 2 × (3 + 4) (14)
2. 3 × (13 − 7) (18)
3. 5(3x + 5) (15x + 25)
4. 7x(6x + 4) (42x2 + 28x)
5. (x + 3)(x + 5) (x2 + 8x + 15)
6. (x + 4)(5 − x) (−x2 + x + 20)
7. (2x + 6)(6x + 3) (12x2 + 42x + 18)
8. (x − 1)2 (x2 − 2x + 1)
9. (x + 1)2 (x2 + 2x + 1)
10. (x + 1)(x − 3)2 (x2 − 2x − 3)
Find the roots of the following equations by using the formula x =
√
−b± b2 −4ac
.
2a
11. y = x2 − 5x + 6 (2 and 3)
12. y = x2 + 8x + 7 (−1 and −7)
13. y = 2x2 + 15x + 12 (−0.91 and −6.59)
14. y = 3x + 12x + 5 (− 13 )
Work out the following
15.
2
3
−
4
6
(0)
16.
2
3
×
4
6
( 49 )
17.
2
7
+
2
−5
18.
ab2
bc2
19.
1
a
20.
a
µ2
×
−
2
c
+
4
(− 35
)
ac
gb
2
( acg )
( c−2a
ac )
b
µ
)
( a+µb
µ2
21. Rearrange the following
to give an expression for y in terms of x.
q
1
1
x = 1−y2 (y = 1 − x )
Solve the following sets of simultaneous equations for both unknowns.
10
22. y = 2x + 2, y = 3x + 6 (x = −4 and y = −6)
23. y = −x + 2, y = 2x + 4 (x =
−2
3
and y = 83 )
24. y = x + 2, y = 2x + 4 (x = −2 and y = 0)
25. What goes wrong when you try to solve the equations y = 2x + 2, y =
2x + 6? Why?
You get an impossible situation where one number is equal to
another, eg. by rearranging first equation and substituting into
second, you get 0 = 4.
These equations are not solvable because the two lines described
by them are parallel.
25. solve the equations y = x + 1, y = x2 − 4. Sketch the graphs y = x + 1
and y = x2 − 4 to see if your answer is reasonable. (x = 2.79, −1.79 and
y = 3.79, −0.79)
10
5
0
-5
-10
-10
-5
0
Looks about right.
11
5
10
4
Exponentials and Logarithms
1. Express the following in their simplest form
(a)
x2
x3
(x−1 or
(b)
x
x3
(x−5 )
(c)
x2
x3
(d)
x 2 ×x 3
(e)
y 2 ×x 2
(f)
x0.5
x2.6
(g)
x
(h)
−19
x−1.5 ×x 3
(x 6 )
x3
2x
−3x
)
42x (2
−2
1
1
x
)
5
(x− 2 )
4
5
7
x2
9
1
(x 3 )
3
2
y3
(x
16
3
)
(x−2.1 )
−1.5
x3
(x−4.5)
4
(i)
4
44
3
(j) x 3 × x 5 ÷ x−1 (x 15 )
(k) x(y 3 ) × y 2 (x5 )3 (x15 y 5 )
2. Evaluate the following
(a)
(b)
22
53
(0.032)
2.2
2
35
2
(0.019)
(c)
2
44
(d)
1.52
4.24.4
(0.015625)
(0.004)
1
3
(e) (42 + 3 2 ) 5 (5.614)
(f) (x[(x + 2)(x + 3)(x − 5)]0.2 for x = 1.56 (-3.48)
4
(g) (x[(x − 2)1.2 (2x + 1)(x + 2) 5 ]2.3 for x = 4.0 (114672)
3. Write down the following as the simplest power possible
√
1
(a) 2 x (x 2 )
1
√
(b) 4 y (y 4 )
√
1
(c) 3 x + 1 (x 3 + 1)
√
1
(d) 3 x + 2 ((x + 2) 3 )
p
19
4
2
x5 × x 3 (x 6 )
(e)
√
ml
(f) n xm × xl (x n )
√ √
ml
q
(g) p xm xl (x pq )
12
(h)
(i)
l(m+1)
x2 √
√
(x2(1− p(q−1) ) )
p m+1 q−1
2l
x
x
√
√
5
p
12
−14m
y 4 x2m
√ 2 (x 15 y 5p )
√
3 4m 5
x
y
(j) (x3 )5 (x15 )
(k) (x3 )3 (x9 )
1
(l) (x3 ) 3 (x)
1
(m) (y 2 )2 (y)
x
1
(n) (y 2 )x (y 2 )
√
3
e
(o) x3 (x e )
√
y
(p) x xy (x x )
√
x
(q) 3 ex (e 3 )
√
x
(r) n xx (x n )
4. Evaluate the following
√
(a) 2 4 (2)
√
(b) 3 125 (5)
√
1
(c) 3 5 (125)
√
−1
(d) 4 6 (7.71 × 10−4 )
p
(e) 4 5(4 + 5)(4 + 2) (4.36)
√
(f) ( 5 4)3.5 (2.64)
√
5
(g) 43.5 (2.64)
√
√
1.2
5.64.2 × .5 4 + 0.5 (6649.83)
(h)
5. Simplify first then evaluate
√
(a) ( 5 4)3.5 (40.7 , 2.64)
√
(b) ( 2 x)3 for x = 1.2 (1.21.5, 1.31)
√
√
16
(c) ( 3 x)5 × 5 x)3 for x = 1.5 (1.5 15 , 1.54)
1
3
5
(d) x 2 × x 5 × x 2 for x = 2 (2
4
3
3
7
18
5
, 12.13)
2
3
1
for x = 2, y = 3, r = 1 (2
(e) x(x x x ) ÷ √
r
1.1
y
27
7
× 31.1, 36.05)
6. The half life of a reaction given by the rate equation [A] = [A0 ]e−kt is
defined as the time (t 12 ) taken for [A] to fall from [A0 ] to 21 [A0 ].
(a) Estimate (by examining the plots) the half life for the curves in figure 1 for k = 0.1, 0.2. (7s, 3.5s)
(b) Estimate the time taken for [A] to fall from 12 [A0 ] to 41 [A0 ] for these
cases also. (7s, 3.5s)
13
[A]
0
k=0.025
k=0.05
0.5[A]
0
k=0.1
k=0.2
0
0
2
4
6
8
10
12
14 t / s
Figure 1: Plot of [A] = [A0 ]e−kt for k = 0.025, 0.05, 0.1 and 0.2
(c) Estimate the time taken for [A] to fall from 14 [A0 ] to 81 [A0 ] for these
cases also. (7s, 3.5s)
(d) What do you notice about these answers? (All the same! It is the
time for the concentration to halve in all cases.)
7. Calculate the following (assume log ≡ log10 ).
(a) log(103 ) (3)
(b) log(103 × 104 ) (7)
(c) 3 log(10) (3)
(d) 4 log(100) (8)
(e) 2 log(7) + 3 log(4) (3.50)
(f) 2 log(7) (1.69)
(g) 102 log(7) (49)
(h) ln(5) + ln(4) (3.00)
(i) ln(20) (1.30)
(j) ln(−20) (Error! Not doable. No positive real number (in this
case e) raised to any power can give a negative number.)
8. Simplify the following where possible
14
(a) log(2) + log(3) + log(5 + 4) (log(54))
(b) ln(2) + log(4) + ln(6) (log(4) + ln(12))
(c) ln(2.4) + ln(5.4) + ln(8.8) (ln(114.048))
(d) ln(x) + ln(y) + ln(z) (ln(xyz))
x
))
(e) ln(x) − ln(y) − ln(z) (ln( yz
(f) log(x) − log(y) + log(z) (log( xz
))
y
(g) log(ab) − log(c) + log(d) (log( abd
))
c
ab
) − log(q) + log(p) (log( abp
))
(h) log( cd
cdq
9. Consider the equation [A] = [A0 ]e−kt .
(a) Manipulate [A] = [A0 ]e−kt to give an expression for t.
)
(t = ln([A0 ]−ln([A]))
k
(b) Give an expression for the half life in terms of rate constant k.
)
(t 12 = ln(2)
k
(c) What is the half life using your expression when k = 0.1, 0.2? (6.93s,
3.47s)
(d) How do your answers compare with estimate from quesion 6a? (They
should be in good agreement.)
10. pH is defined as − log10 [H+ ] where [H+ ] is the concentration of H+ ions.
Calculate the pH of the following acids:
(a) 10−3 M HCl (3)
(b) 10−4 M HCl (4)
(c) 10−4 M H2 SO4 (3.69)
(d) 2M H2 SO4 (-0.3)
11. What is [H+ ] in a solution with a pH of 4.2? (6.3 × 10−5 M)
12. In question 9a you took logs of [A] = [A0 ]e−kt to give an expression for t.
(a) Rearrange this (or go back some steps) to obtain an expression for
ln[A]. (ln([A]) = ln([A0 ]) − kt)
(b) For constant positive k, sketch the form of the graph given by this
expression. Plot ln[A] vertically and t horizontally. (A straight line
sloping downwards as t increases.)
(c) How would you determine the rate constant k? (The slope of the
graph.)
(d) How would you determine the initial concentration [A0 ]? (The point
where the line cuts the vertical (ln[A]) axis; the intercept of
the graph.)
15
5
Differentiation
Differentiate the following functions with respect to x: Calculate the second
derivatives (Don’t do any that involve the quotient rule!). Find the turning
points and classify as maxima, minima or points of inflexion.
1. y = x2
Turning points are where
dy
dx
dy
= 2x
dx
d2 y
=2
dx2
= 0, so in this case:
2x = 0
x = 0
A minimum has a positive (or zero) second dervative; a maximum has a
negative (or zero) second derivative and a point of inflexion has zero as its
second derivative. In this case, the second derivative is positive (always
2) so it is a minimum.
2. y = x3
dy
= 3x2
dx
d2 y
= 6x
dx2
3x2
= 0
x
= 0
Turnining point is at x = 0 in this case, so we substitute x into second
derivative at this point to find if it is a maximum or minimum.
d2 y
= 6×0
dx2
d2 y
= 0
dx2
So this could be a point of inflexion, maximum or minumum. We must
calculate the gradient just before and just after the turning point to see
how it changes. Take, x = −0.1 and 0.1.
dy
= 3 × (−0.1)2
dx x=−0.1
dy
= 3 × (0.1)2
dx x=0.1
The gradient is positive on both sides of the turning point, so the turning
point can be neither a maximum or a minimum so it must be a point of
inflexion.
16
3. y = x7
dy
= 7x6
dx
d2 y
= 42x5
dx2
7x6
x
= 0
= 0
Once again second derivative is zero at turning point so we must substitute
values at either side to classify it.
dy
dx x=−0.1
dy
dx x=0.1
= 7 × (−0.1)6
= 7 × (0.1)6
Again gradient is positive either side of turning point so it is a point of
inflexion.
4. y = 4x5
dy
= 20x4
dx
d2 y
= 80x3
dx2
Turning point at x = 0, point of inflexion.
5. y = 6x2 + 12x
dy
= 12x + 12
dx
d2 y
= 12
dx2
12x + 12 = 0
12x = −12
x = −1
Minimum at x = −1.
6. y = 3x3 + 7x2
dy
= 9x2 + 14x
dx
d2 y
= 18x + 14
dx2
17
9x2 + 14x = 0
x(9x + 14) = 0
x = 0, −
14
9
Two turning points now:
d2 y
dx2 x=0
d2 y
dx2 x=0
2
d y
dx2 x=− 14
9
d2 y
dx2 x=− 14
9
d2 y
dx2 x=− 14
9
= 18x + 14
= 14
= 18x + 14
= −28 + 14
= −14
So there is a minimum at x = 0 and a maximum at x = − 14
9 .
7. y = 5x3 + 2x2 + 3x + 5
dy
= 15x2 + 4x + 3
dx
d2 y
= 30x + 4
dx2
15x2 + 4x + 3 = 0
This has no real roots! So we cannot solve this (yet).
8. y = ax3 + bx2 + cx + d
dy
= 3ax2 + 2bx + c
dx
d2 y
= 6ax + 2b
dx2
This is purely algebraic, without some values of a,b and c we cannot do
much more. We could write down that the turning points are solutions of
the quadratic formula. Note the multipliers.
p
−2b ± (2b)2 − 4 × 3a × c
x=
2 × 3a
18
9. y = axn + bxn−1 + cxn−2
dy
= naxn−1 + (n − 1)bxn−2 + (n − 2)cxn−3
dx
d2 y
= n(n − 1)axn−2 + (n − 1)(n − 2)bxn−3 + (n − 2)(n − 3)cxn−4
dx2
No more to do.
10. y = (x + 2)2
dy
= 2x + 4
dx
d2 y
=2
dx2
2x + 4 = 0
x = −2
Minimum at x = −2
11. y = (x + 1)3
dy
= 3x2 + 6x + 3
dx
d2 y
= 6x + 6
dx2
3x2 + 6x + 3 = 0
x2 + 2x + 1 = 0
(x + 1)(x + 1) = 0
x = −1
d2 y
dx2 x=−1
d2 y
dx2 x=−1
= 6 × (−1) + 6
= 0
So we must find gradient at either side of turning point. It turns out it is
positive before and after x = −1 so there is a point of inflexion at x = −1.
12. y = (x + 1)(x + 2)2
dy
= 3x2 + 10x + 8
dx
d2 y
= 6x + 10
dx2
19
3x2 + 10x + 8 = 0
4
x = −2, −
3
d2 y
dx2 x=−2
d2 y
dx2 x=−2
d2 y
dx2 x=− 34
d2 y
dx2 x=− 34
= 6 × (−2) + 10
= −2
4
= 6 × (− ) + 10
3
= 2
Maximum at -2, minimum at − 34 .
13. y = (x + 3)2 (x − 2)2
dy
= 4x3 + 6x2 − 22x − 12
dx
d2 y
= 12x2 + 12x − 22
dx2
4x3 + 6x2 − 22x − 12 = 0
We do not know how to solve this, so we leave it at this. (There are
methods for solving cubics, but it is quite complicated).
14. y = sin(x)
dy
= cos(x)
dx
d2 y
= − sin(x)
dx2
cos(x)
= 0
We know from the fact that trigonometric functions are periodic, that
there are an infinite number of solutions to the above. You should find on
your calculator though that there is, for example, a maximum at 90◦ and
a minimum at −90◦ and 270◦ .
15. y = cos(x)
dy
= − sin(x)
dx
d2 y
= − cos(x)
dx2
Again an infinite number of solutions, you can verify that there is a maximum at 0◦ and a minumum at 180◦ .
20
16. y = cos(2x)
dy
= −2 sin(2x)
dx
d2 y
= −4 cos(2x)
dx2
−2 sin(2x) = 0
Maximum at 0◦ and a minumum at 90◦ .
17. y = − cos(2x)
dy
= 2 sin(2x)
dx
d2 y
= 4 cos(2x)
dx2
Minimum at 0◦ and a maximum at 90◦ .
18. y =
1
2
sin(−x)
dy
1
= − cos(−x)
dx
2
1
d2 y
= sin(−x)
dx2
2
◦
A minimum at 90 and a maximum at −90◦ and 270◦ .
19. y =
1
3
sin( 12 x)
dy
1
1
= cos( x)
dx
6
2
d2 y
1
1
= − sin( x)
dx2
12
2
◦
◦
Maximum at 0 and a minumum at 360 .
20. y = sin(sin(x))
dy
= cos(x) cos(sin(x))
dx
d2 y
= − cos2 (x) sin(sin(x)) − sin(x) cos(sin(x))
dx2
cos(x) cos(sin(x))
= 0
So either cos(x) = 0, ie x = 90◦ , or cos(sin(x)) = 0. In the latter case we
require a case where, eg sin(x) = 90 or sin(x) = 270; this isn’t possible, so
the only solutions are where cos(x) = 0.
21
21. y = ex
dy
= ex
dx
d2 y
= ex
dx2
ex
x
= 0
= ln(0)
This is an error, so there are no values of x for which ex = 0 so there are
no turning points.
22. y = e2 x
dy
= e2
dx
d2 y
=0
dx2
e2
= 0
This is clearly nonsense, so there are no turning points. This is a straight
line with gradient e2 .
23. y = e3x
2
2
dy
= 6xe3x
dx
2
d2 y
= 6e3x (6x2 + 1)
2
dx
6xe3x
2
= 0
2
We know that the second factor (e3x ) has no x for which it is equal to 0,
so
6x = 0
x = 0
d2 y
dx2 x=0
= e0 × (6)
This is a positive number, so there is a minimum at x = 0.
22
24. y = e2x
3
3
dy
= 6x2 e2x
dx
3
d2 y
= 12xe2x (3x2 + 1)
2
dx
6x2 e2x
3
6x2
x
= 0
= 0
= 0
d2 y
=0
dx2 x=0
So this could be any of the possible turning points, evaluating the gradient
before and after the point shows it to be a point of inflexion (at x = 0).
25. y = 2e4x
2
2
dy
= 16xe4x
dx
2
d2 y
= 16e4x (8x + 1)
dx2
2
16xe4x
x
= 0
= 0
d2 y
= 16 × e0 × 1
dx2 x=0
d2 y
= 16
dx2 x=0
So we have a minimum at x = 0.
2
26. y = 2e4x + 3x2
2
dy
= 16xe4x + 6x
dx
2
d2 y
= 16e4x (8x + 1) + 6
dx2
2
16xe4x + 6x = 0
2
x(16e4x + 6) = 0
The term in brackets cannot be equal to zero, so x = 0 and the turning
point is a minimum.
23
2
27. y = ex + sin(x2 ) + 2x
2
dy
= 2xex + 2x cos(x2 ) + 2
dx
2
d2 y
= 2ex (4x2 + 1) − 4x2 sin(x2 )
dx2
2
2xex + 2x cos(x2 ) + 2 = 0
This is more or less impossible. So we cannot easily determine turning
points.
28. y = ex+3
dy
= ex+3
dx
d2 y
= ex+3
dx2
ex+3
= 0
No turning points.
29. y = x2 + sin( 12 x) + x3
dy
1
1
= 2x + cos( ) + 3x2
dx
2
2
d2 y
1
1
= 2 + sin( x) + 6x
2
dx
4
2
2x +
1
1
cos( ) + 3x2
2
2
= 0
As above, too hard to continue with.
30. y = sin((x + 3)2 ) + cos( 12 x) + 4
1
1
dy
= (2x + 6) cos((x + 3)2 ) − sin( x)
dx
2
2
d2 y
1
1
= −(2x + 6)2 sin((x + 3)2 ) − cos( x)
dx2
4
2
Not possible to determine turning points.
31. y = sin(ex )
dy
= ex cos(ex )
dx
d2 y
= ex (cos(ex ) − ex sin(ex ))
dx2
Again much too hard to solve.
24
2
32. y = sin(ex )
2
2
dy
= 2xex cos(ex )
dx
2
2
2
2
d2 y
= 4x2 ex (cos(ex ) − ex sin(ex ))
2
dx
33. y = 2xex
dy
= 2ex (1 + x)
dx
d2 y
= 2ex (2 + x)
dx2
2ex (1 + x)
(1 + x)
x
= 0
= 0
= −1
d2 y
= 2e−1
dx2 x=−1
This is a positive number, so there is a minimum at x = −1.
34. y = 2xe2x
dy
= 2e2x (1 + 2x)
dx
d2 y
= 8e2x (1 + x)
dx2
2e2x (1 + 2x) = 0
1 + 2x = 0
x = −
d2 y
dx2 x=− 12
d2 y
dx2 x=− 12
1
2
1
= 8e1 ( )
2
= 4e
This is a positive number, so there is a minimum at x = − 21 .
35. y = 2xe2x + sin(x)
dy
= 2e2x (1 + 2x) + cos(x)
dx
d2 y
= 8e2x (1 + x) − sin(x)
dx2
This is too hard to go further with.
25
36. y = cos(x2 )
dy
= −2x sin(x2 )
dx
d2 y
= −2 sin(x2 ) − 4x2 cos(x2 )
dx2
You can work out that there is a maximum at x = 0, but beyond that it
is somewhat impossible.
37. y = cos(x + 2)
dy
= − sin(x + 2)
dx
d2 y
= − cos(x + 2)
dx2
Your answer should be the same as for y = cos(x), but offset by 2. These
will vary depending on whether you use degrees or radians.
38. y = e−x
2
2
dy
= −2xe−x
dx
2
d2 y
= 2e−x (2x2 − 1)
dx2
−2xe−x
2
= 0
−2x = 0
x = 0
d2 y
dx2 x=0
d2 y
dx2 x=0
= −2e0 (0 + 1)
= −2e
This is a negative number, so there is a maximum at x = 0.
39. y = −2xe−x
2
2
dy
= 2e−x (2x2 − 1)
dx
2
d2 y
= 4xe−x (3 − 2x2 )
dx2
2
2e−x (2x2 − 1) = 0
1
x = ±√
2
26
d2 y
dx2 x= √12
1
1
1
= 4 √ e− 2 (3 − 2 )
2
2
d2 y
dx2 x= √12
1
1
= 8 √ e− 2
2
This is a positive number, so there is a minimum at x =
d2 y
dx2 x=− √12
1
1
1
= −4 √ e− 2 (3 − 2 )
2
2
d2 y
dx2 x=− √12
1
1
= −8 √ e− 2
2
√1
2
This is a negative number, so there is a maximum at x = − √12
40. y = x−4
dy
= −4x−5
dx
d2 y
= 20x−6
dx2
−4x−5
x−5
x5
= 0
= 0
1
=
0
This is an error so there are no turning points.
41. y = sin(x) cos(x)
dy
= cos2 (x) − sin2 (x)
dx
d2 y
= −4 sin(x) cos(x)
dx2
cos2 (x) − sin2 (x)
2
cos (x)
cos(x)
= 0
= sin2 (x)
= sin(x)
This is satisfied by, for instance, x = sin−1 ( 21 ) = 45◦ which is a maximum,
and x = sin−1 (− 21 ) = −45◦ which is a minimum.
42. y = −2 sin(x) sin(x)
dy
= −4 sin(x) cos(x)
dx
27
d2 y
= −4(cos2 (x) − sin2 (x))
dx2
−4 sin(x) cos(x)
sin(x) cos(x)
= 0
= 0
So there are maxima and minima at points where sin(x) = 0 and cos(x) =
0.
43. y = cos( 12 x) cos(x)
1
1
1
dy
= sin( x) cos(x) − cos( x) sin(x)
dx
2
2
2
3
1
1
1
d2 y
= − cos( x) cos(x) + sin( x) sin(x)
2
dx
4
2
4
2
A bit impossible.
44. y = cos(2x) sin( x2 )
dy
1
1
1
= −2 sin(2x) sin( x) + cos(2x) cos( x)
dx
2
2
2
17
x
x
d2 y
= − cos(2x) sin( ) − 2 sin(2x) cos( )
dx2
4
2
2
Too hard to take further.
45. y = cos2 (x)
dy
= −2 sin(x) cos(x)
dx
d2 y
= −2(cos2 (x) − sin2 (x))
dx2
This has turning points where sin(x) = 0 or cos(x) = 0.
46. y = sin(x) + 2x2 + ex
dy
= cos(x) + 4x + ex
dx
d2 y
= − sin(x) + 4
dx2
Not solvable.
2
2
47. y = sin(xex ) + ex + ex
2
2
2
dy
= ex (2x2 + 1) cos(xex ) + 2xex + ex
dx
28
d2 y
dx2
2
2
= −ex (2x2 + 1)2 sin(xex )
2
2
+ 2xex (2x2 + 3) cos(xex )
2
+ +2ex (2x2 + 1) + ex
Probably. We won’t take this one any further.
48. y =
1
x
1
dy
=− 2
dx
x
d2 y
2
= 3
2
dx
x
This has no turning points.
49. y =
1
x2
dy
2
=− 3
dx
x
6
d2 y
= 4
2
dx
x
This has no turning points.
50. y =
3
x5
dy
15
=− 6
dx
x
90
d2 y
= 7
dx2
x
This has no turning points.
51. y =
x2
x4
2
dy
=− 3
dx
x
d2 y
6
= 4
2
dx
x
= 0
52. y =
x+1
x+2
dy
1
=
dx
(x + 2)2
Although we can go on to find and classify turning points, the question
asked us not to.
53. y =
x−1
x+2
dy
3
=
dx
(x + 2)2
29
54. y =
55. y =
56. y =
57. y =
58. y =
59. y =
2x+1
3x−4
2x+1
x2 +2
3x3 +1
4x+2
3x3 +2x+5
3x2 +2
(x+1)(x+4)
x−1
dy
−11
=
dx
(3x − 4)2
dy
−6x2 − 2x + 4
=
dx
(x2 + 2)2
dy
24x3 + 18x2 − 4
=
dx
(4x + 2)2
dy
9x4 + 12x2 + 26
=
dx
(3x2 + 2)2
dy
x2 − 2x − 4
=
dx
(x − 1)2
(2x+4)(x−4)
(x−4)2
−12
dy
=
dx
(x − 4)2
60. y = ln(x)
dy
1
=
dx
x
d2 y
1
=− 2
dx2
x
No turning points
61. y = ln(2x)
dy
1
=
dx
x
1
d2 y
=− 2
2
dx
x
No turning points
62. y = ln(ex )
dy
=1
dx
d2 y
=0
dx2
This is just the straight line with constant gradient of 1 y = x.
30
63. y = sin(ln(x))
cos(ln(x))
dy
=
dx
x
− cos(ln(x)) − sin(ln(x))
d2 y
=
dx2
x2
Not solvable.
64. y =
1
ln(x)
dy
−1
=
dx
x ln(x)
65. y = tan(x)
dy
= 1 + tan2 (x)
dx
66. y = 2x2 sin(x2 )
dy
= 4x(sin(x2 ) + x2 cos(x2 ))
dx
d2 y
= 4(sin(x2 ) + x2 cos(x2 )) + 8x2 (2 cos(x2 ) − x2 sin(x2 ))
dx2
Maybe. . .
67. y = 2x2 sin(x2 ) + ex sin(3x)
dy
= 4x(sin(x2 ) + x2 cos(x2 )) + ex (sin(3x) + 3 cos(3x))
dx
d2 y
dx2
= 4(sin(x2 ) + x2 cos(x2 )) + 8x2 (2 cos(x2 ) − x2 sin(x2 ))
+ ex (6 cos(3x) − 8 sin(3x))
Assuming last question was correct.
2
68. y = x2 ex sin(x)
2
dy
= ex (2x sin(x) + 2x3 sin(x) + x2 cos(x)))
dx
d2 y
dx2
2
= 2xex (2x sin(x) + 2x3 sin(x) + x2 cos(x)))
2
+ ex ((2x cos(x))(2 + x2 ) + (2 + 5x2 ) sin(x))
Another horribly difficult one to go any further with.
69. y = x2 tan(x)
dy
= 2x tan(x) + x2 (1 + tan2 (x))
dx
31
70. y = ln(x)ex sin(x2 )
1
dy
= ln(x)(ex (sin(x2 ) + 2x cos(x2 ))) + (ex sin(x2 ))
dx
x
d2 y
= daunting
dx2
71. y = ln(x) + ex sin(x2 )
1
dy
= + ex (sin(x2 ) + 2x cos(x2 ))
dx
x
d2 y
dx2
−1
+ ex (sin(x2 ) + 2x cos(x2 ))
x2
+ ex (2x cos(x2 ) + 2 cos(x2 ) − 4x2 sin(x2 )))
=
72. y = ln(sin(x)) − ex sin(x−2 )
dy
cos(x)
2 cos(x−2 )
=
− ex (sin(x−2 ) +
)
dx
sin(x)
x−3
73. y = x4 + 3x3 + 6x2 + sin(x)ex + 4
dy
= 4x3 + 9x2 + 12x + ex (sin(x) + cos(x))
dx
d2 y
= 12x2 + 18x + 12 + 2ex cos(x)
dx2
74. y = tan(2x)
dy
= 2(1 + tan2 (2x))
dx
75. y = 5
dy
=0
dx
d2 y
=0
dx2
No turning points - a straight line.
32
6
Integration
Integrate the following expressions:
1.
2.
3.
4.
5.
6.
Z
1 3
x +c
3
Z
x2 dx =
Z
x3 dx =
Z
x5 dx =
Z
xn dx =
Z
1
1
x dx = x2 + c
3
6
1
4
1
5
1
n
x4 + c
x5 + c
xn + c
5 7
1
2
5
x +c
x + x6 dx = x2 +
3
7
3
49
7.
Z
Z
2
(x + 2) dx =
1
=
3
x2 + 4x + 4 dx
x3 + 2x2 + 4x + c
8.
Z
(x + 1)(x + 4) dx =
Z
x2 + 5x + 4 dx
=
1 3
5
x + x2 + 4x + c
3
2
=
Z
9.
Z
(2x + 4)(3x + 6) dx
6x2 + 24x + 24 dx
= 2x3 + 12x2 + 24x + c
33
10.
Z
(ax + b)(cx + d) dx
Z
=
acx2 + (bc + ad)x + bd dx
ac
=
3
x3 +
bc + ad
2
x2 + bdx + C
I used C for the constant here because c has already been used
to represent something else.
11.
12.
13.
14.
Z
Z
Z
y 2 dx = y 2 x + c
Z
y 2 dy =
1
3
y3 + c
cos(x) dx = sin(x) + c
sin(x) + cos(x) dx = − cos(x) + sin(x) + c
15.
Z
ex dx = ex + c
16.
Z
2
x2 dx
=
0
=
1 3
x
3
8
3
x=2
−
1 3
x
3
x=0
17.
Z
4
2
x dx
=
2
=
=
1 3
1 3
x
− x
3
3
x=4
x=2
64 8
−
3
3
56
3
18.
Z
0
−2
1 3
x
3
8
= −
3
x2 dx =
34
x=0
−
1 3
x
3
x=−2
19.
Z
2
x2 dx =
−2
=
=
1 3
x
3
x=2
−
−
1 3
x
3
x=−2
8
8
−−
3
3
16
3
20.
Z
2
x3 dx
0
1 4
x
4
16
=
4
= 4
=
x=2
1 4
x
4
x=0
21.
Z
4
3
x dx
=
2
=
=
=
1 4
1 4
− x
x
4
4
x=4
x=2
256 16
−
4
4
240
4
60
22.
Z
0
1 4
x dx =
x
4
−2
16
= −
4
= −4
3
1
− x4
4
x=0
x=−2
23.
Z
2
−2
1 4
1
x
− x4
4
4
x=2
x=−2
16 16
=
−
4
4
= 0
x3 dx =
Is this reasonable? How could we get a better answer?
35
24.
Z
π
2
cos(x) dx
0
= [sin(x)]x= π − [sin(x)]x=0
2
= 1−0
= 1
25.
Z
π
2
0
sin(x) dx = [− cos(x)]x= π − [− cos(x)]x=0
2
= 0−1
= −1
Find the following areas
26. Under the curve y = 21 x3 between 0 and 5.
The answer is the result of the integral
Z 5
1 4
1
1 3
x dx =
x
− x4
8
8
0 2
x=5
x=0
625
=
8
27. Under the curve y = cos(x) between 0 and π.
Z π
sin(x) dx = [sin(x)]x=π − [sin(x)]x=0
0
= 0−0
= 0
Again, we have to consider if this answer is reasonable.
28. Under the curve y = sin(x) between 0 and π.
Z π
sin(x) dx = [− cos(x)]x=π − [− cos(x)]x=0
0
= −1 − (−1)
= 0
And once more we have to consider if this answer is reasonable.
29. Under the curve y = ex between 1 and 2.
Z 2
ex dx = [ex ]x=2 − [ex ]x=1
1
= e2 − e
= 4.67
36
30. Under the curve y = x1 between
Z e
1
dx =
1 x
=
=
7
1 and e.
[ln(x)]x=e − [ln(x)]x=1
1−0
1
Sums, factorials and what is e?
No problems here. A mostly optional section. Just remember what n! ans
mean.
37
P
8
Complex Numbers
Work out the following:
1.
(3 + 2i) + (5 − 6i) = 8 − 4i
2.
(1 + 5i) − (6 − i) = −5 + 6i
3.
(3 + 12i) + (2 − 5i) = 5 + 7i
4.
(4 − 2i) − (5 + 2i) = −1
5.
(7 + 9i) + (24 + 11i) = 31 + 20i
6.
(7 + 9i)(24 + 11i) = 168 + 77i + 216i + 99i2
= 168 − 99 + 293i
= 69 + 293i
7.
(5 + 2i)(6 + i) = 30 + 5i + 12i + 2i2
= 30 − 2 + 17i
= 28 + 17i
8.
i(3 + 12i) = 3i + 12i2
= −12 + 3i
9.
(2i + 5)(7 − 15i) = 14i − 30i2 + 35 − 75i
35 + 30 − 54i
= 65 − 61i
38
10.
(2i + 5)
(7 − 15i)
=
=
=
=
5 + 2i
7 − 15i
(5 × 7 + 2 × −15) + (2 × 7 − 5 × −15)i
72 + (−15)2
(35 − 30) + (14 + 75)i
49 + 225
5 + 89i
274
= 0.018 + 0.325i
11. Write down the above answers in phasor form.
The phasor form is the complex number z = a + bi written as:
z = |z|eiθ
where
|z| =
and
p
a2 + b2
θ = tan
−1
b
a
So the answers are:
p
−1 −4
8 − 4i =
82 + (−4)2 ei tan ( 8 )
= 8.94e−0.464i
−5 + 6 = 7.81e2.26i
You have to think carefully about θ here. Draw the diagram and
don’t just trust your calculator.
5 + 7i = 8.60e0.951i
−1 = eπi
Again you have to think carefully about θ here. Draw the diagram, θ is not zero.
31 + 20i = 36.9e0.573i
69 + 293i = 301e1.34i
28 + 17i = 32.8e0.546i
39
−12 + 3i = 12.4e2.90i
65 − 61i = 89.1e−0.754i
5 + 89i
274
= 0.325e1.51i
12. Write down the original factors in the questions which involve multiplication in the form z = |z|(cos(θ) + i sin(θ)). Work out the product in this
form and make sure it agrees with your original answer written in this
form.
Lest us consider the product (7 + 9i)(24 + 11i)
7 + 9i = 11.4(cos(0.909) + i sin(0.909))
24 + 11i = 26.4(cos(0.429) + i sin(0.429))
(7 + 9i)(24 + 11i) = 11.4 ∗ 26.4(cos(0.909 + 0.429) + i sin(0.909 + 0.429))
= 301(cos(1.34) + i sin(1.34))
(We work out |z| and θ in same way as when writing down phasor form,
but then calculate sine and cosine).
69 + 293i = 301(cos(1.34) + i sin(1.34))
So the two ways of calculating give the same answer. (The answers to 7,8
and 9 can be checked in the same way).
13. Work out (4 − 2i)2 both by multiplying brackets and by using the power
rule. Your answers should agree.
Firstly by multiplying brackets.
(4 − 2i)2
= (4 − 2i)(4 − 2i)
= 16 − 16i + 4i2
= 16 − 16i − 4
= 12 − 16i
For second part we write the complex number in phasor form:
√ −0.464i
20e
4 − 2i =
√
(4 − 2i)2 = ( 20e−0.464i )2
√
(4 − 2i)2 = ( 20)2 e2×−0.464i )
= 20e−0.927
Then we write our original answer in phasor form:
p
−1 −16
12 − 16i =
122 + (−16)2 etan ( 12 )
√
=
400e−0.927
= 20e−0.927
So our answers are the same.
40
9
Statistics
The concentration of Pb2+ in river water near a factory was measured on 15
consecutative days. The measured values in parts per million is given below
12, 15, 7, 18, 19, 20, 10, 11, 13, 15, 16, 20, 22, 9, 8
Calculate the mean of the distribution of measurements.
Calculate the varience of the distribution.
Calculate the standard deviation of the distribution.
Regulations determine that the mean [Pb2+ ] must not exceed 15 parts per million, and also that the standard deviation must be not be greater that 5 parts per
million in any period of 10 days or more. Further there must be no more than 4
days per 15 days in which the measured concentration is more than 1.5s from 15.
Are the pollution levels legal?
Given that the factory has no way of reducing its minimum level pollution
below current levels, how low would the maximum level have to drop if the laws
were changed such that the mean must not exceed 12 parts per million and that
the standard deviation must drop to 4 parts per million over a 10 day or longer
period?
The mean of the set of numbers
12, 15, 7, 18, 19, 20, 10, 11, 13, 15, 16, 20, 22, 9, 8
is 14.79
The variance is 21.41.
The standard deviation is 4.63.
The mean is legal over the range of days, the standard deviation is
also legal. 1.5s is 6.94 only 7 is more than 1.5s from 15 so overall the
pollution levels are legal.
The next part is a bit horrible. If all the data were 12 we would be
OK (but question doesn’t allow this!) There are many sets of numbers which would fix criteria, but try setting maximum value to 10
or 11. Nothing like this last part will be in the test.
41