Worked Solutions
C3 Worksheet 1 - Worked Solutions
1.
3x2 +16x+6
x2 +3x−4
B
C
can be expressed in the form A + x−1
+ x+4
. Find the values of A, B
and C.
B
C
In order to express A + x−1
+ x+4
as a single fraction, we must give each individual
fraction (including A on its own) a common denominator:
A+
B
C
A(x − 1)(x + 4)
B(x + 4)
C(x − 1)
+
=
+
+
x−1 x+4
(x − 1)(x + 4)
(x − 1)(x + 4) (x − 1)(x + 4)
2
A (x + 3x − 4) + Bx + 4B + Cx − C
=
x2 + 3x − 4
2
Ax + (3A + B + C)x + (−4A + 4B − C)
=
x2 + 3x − 4
3x2 + 16x + 6
≡
x2 + 3x − 4
Equating like terms we have:
A = 3,
⇒
⇒
⇒
⇒
⇒
3A + B + C = 16,
9 + B + C = 16,
B + C = 7,
C = 7 − B,
5B = 25
−4A + 4B − C = 6
−12 + 4B − C = 6
4B − C = 18
4B − (7 − B) = 18
A = 3, B = 5, C = 2
Hence,
3x2 + 16x + 6
5
2
=3+
+
2
x + 3x − 4
x−1 x+4
One must always check this answer by working backwards from the solution obtained.
2
2. (a) State the domain and range of the function p(x) = x−3
.
The domain of a function refers to the set of values for which the function is
defined, i.e. the set of values that can be successfully put into the function. All
values of x can be put into p(x) without any problems apart from x = 3. When
x = 3 we have division by 0 which is undefined. Hence, the domain is given by
{x ∈ R, x 6= 3}. x ∈ R reads ‘x is a member of the real numbers’. We add x 6= 3
to tell the reader that x can be anything except 3. We can observe this from the
2
graph of p(x) = x−3
:
www.studywell.com
c StudyWell Publications Ltd. 2013
Worked Solutions
p(x) =
2
x−3
y = 0, asymptote
3
− 23
x = 3, asymptote
We can also see from the graph the values that can come out of the function, i.e.
the range of the function. We see that all values can be obtained except p(x) = 0,
hence the range of the function is {p(x) ∈ R, p(x) 6= 0}. The domain and range
of p(x) are given by
DOMAIN = {x ∈ R, x 6= 3} , RANGE = {p(x) ∈ R, p(x) 6= 0}
√
(b) State the range of the function q(x) = 3x − 2 given that the domain is
restricted to {x ∈ R, x ≥ 2}.
√
If x ≥ 2, then 3x − 2 ≥ 4 and it follows that 3x − 2 ≥ 2, i.e.
RANGE = {q(x) ∈ R, q(x) ≥ 2}
3. Find the exact solutions of the following equations:
(a) 9 ln(2x) + 5 = 23
9 ln 2x + 5 = 23
⇒
9 ln 2x = 18
⇒
ln 2x = 2
⇒
⇒
2x = e2
1 2
x =
e
2
(b) 2e3x−5 = 8
2e3x−5 = 8
⇒
⇒
e3x−5 = 4
3x − 5 = ln 4
⇒
⇒
www.studywell.com
3x = ln 4 + 5
ln 4 + 5
x =
3
c StudyWell Publications Ltd. 2013
Worked Solutions
4. Show that ex − x3 has a root somewhere between x = 1 and x = 2.
Let f (x) = ex − x3 , then f (1) = e1 − 13 = 1.718 and f (2) = e2 − 23 = −0.611 to 3
decimal places. Since f (1) is positive and f (2) is negative, f (x) will be 0 somewhere in
between. Hence, the value of x which gives 0, i.e. the root of the function, is somewhere
between 1 and 2.
5. (a) Sketch the graph of f (x) = x2 + 6x − 7, labelling the coordinates of the
vertex and any intersection of axes.
The coordinates of the vertex of the curve can be found by completing the square:
f (x) = x2 + 6x − 7
= (x + 3)2 − 9 − 7
= (x + 3)2 − 16
Hence, the vertex of the curve is at (−3, −16).
The curve crosses the y-axis when x = 0, i.e. at y = −7. The curve crosses the
x-axis when y = 0, i.e. when
⇒
⇒
x2 + 6x − 7 = 0
(x + 7)(x − 1) = 0
x = −7, x = 1
y
−7
1
x
−7
(−3, −16)
(b) Sketch |f (x)|.
|f (x)| takes the parts of the curve of f (x) that are negative and makes them
positive.
y
(−3, 16)
7
−7
1
x
(c) Sketch f (|x|).
f (|x|) makes x coordinates positive before evaluating f at those x coordinates.
www.studywell.com
c StudyWell Publications Ltd. 2013
Worked Solutions
This means that f is never evaluated at a negative x value and the curve in the
positive x region is copied across the y-axis.
y
−1
x
1
−7
6. Show that 1 + cot2 (θ) = cosec2 (θ).
1 + cot2 (θ) = 1 +
= 1+
1
tan2 (θ)
1
sin2 (θ)
cos2 (θ)
2
cos (θ)
sin2 (θ)
sin2 (θ) cos2 (θ)
+
sin2 (θ) sin2 (θ)
sin2 (θ) + cos2 (θ)
sin2 (θ)
1
sin2 (θ)
cosec2 (θ)
= 1+
=
=
=
=
7. Using the formula cos(x + y) = cos(x) cos(y) − sin(x) sin(y), find the exact value
of cos(75).
Let x + y = 75. By choosing x and y to be angles that we know how to evaluate the
cosine for, i.e. x = 45, y = 30, we can use the formula to find cos(75).
cos(75) = cos(45) cos(30) − sin(45) sin(30)
√
√
√
2
3
2 1
×
−
×
=
2
√2 √ 2 √ 2
2 3
2
=
−
4
√
√ 4
6− 2
=
4
8. Differentiate the following standard trigonometric functions:
www.studywell.com
c StudyWell Publications Ltd. 2013
Worked Solutions
(a) sin(x)
(b) cos(x)
(c) tan(x)
(a)
d
dx
sin(x) = cos(x)
(b)
d
dx
cos(x) = − sin(x)
(c) We can use the quotient rule to differentiate tan(x):
d
d sin(x)
tan(x) =
dx
dx cos(x)
d
d
sin(x) − sin(x) dx
cos(x)
cos(x) dx
=
2
cos (x)
2
cos (x) + sin2 (x)
=
cos2 (x)
1
=
cos2 (x)
= sec2 (x)
www.studywell.com
c StudyWell Publications Ltd. 2013