Download Sample Midterm 2 Solutions

Solutions to the second sample midterm
Instructor: F. Balogh
March 2008
1.
Solve the equation
32x − 3x + 2 = 0.
[4]
2
Solution. Since 32x = (3x ) , the equation
2
(3x ) − 3x + 2 = 0
is a quadratic equation in 3x . Therefore if we define a new variable A = 3x it
has to be a solution of the quadratic equation
A2 − A + 2 = 0.
The discriminant (the expession under the square root in the Quadratic Formula) of this quadratic equation is
(−1)2 − 4 · 1 · 2 = 1 − 8 = −7.
This means that the quadratic equation has no solution because its discriminant
is negative. Therefore the original equation has no solution.
2.
Solve
x2 · 5x − 4 · 5x = 0.
[4]
Solution. The expression x2 · 5x − 4 · 5x on the left side can be factorised:
x2 · 5x − 4 · 5x = (x2 − 4)5x .
Then
(x2 − 4)5x = 0
if and only if
x2 − 4 = 0
or
5x = 0.
The equation x2 − 4 = 0 can be factorised further:
(x + 2)(x − 2) = 0.
Hence we get two solutions from this branch: x = −2 and x = 2.
On the other hand, the equation
5x = 0
has no solution because the exponential function 5x is positive for all x ∈ R.
To sum up, the two solutions of the original equation are x = −2 and x = 2.
3.
[5]
Find all solutions of the equation
log5 (x − 3) + log5 (x + 1) = 1.
1
Solution. The domain of definition of the expression on the left is given by the
condition
log5 (x − 3) + log5 (x + 1) is defined if and only if
x−3 > 0
and
x + 1 > 0.
This means that x > 3 and x > −1 has to be satisfied by x simultaneously (we
need both logarithmic functions to be defined) which is equivalent to requiring
that x > 3.
Using the laws of logarithms we have
log5 (x − 3) + log5 (x + 1)
=
1
log5 (x − 3)(x + 1)
=
1
log5 (x − 3)(x + 1)
=
log5 5.
The exponential form of this equation is
(x − 3)(x + 1) = 5
x2 − 2x − 3 = 5
x2 − 2x − 8
=
0.
This quadratic equation can be solved using the Quadratic Formula:
p
√
2 ± (−2)2 − 4 · 1 · (−8)
2 ± 4 + 32
2±6
=
=
= 1 ± 3.
x=
2·1
2
2
The first solution x = 1 + 3 = 4 solves the original equation because 4 > 3 (4 is
in the domain of the logarithmic expression).
Although x = 1 − 3 = −2 solves the derived quadratic equation, it is not a
solution of the original equation because −2 < 3 (log5 (−2 − 3) is not defined).
So the only solution of the original equation is x = 4 and we can verify that this
is solution by substitution:
LHS
=
log5 (4 − 3) + log5 (4 + 1) = log5 1 + log5 5 = 0 + 1 = 1,
RHS
=
1.
[6]
4.
Find the values√of the trigonometric functions of the angle θ
given that tan θ = − 3 and θ is in the second quadrant.
Solution. If P (x, y) is a point sitting on the terminal side of the angle θ in
standard position, then the tangent of θ is defined to be
tan θ =
y
.
x
So first√we fix a point P (x, y) which belongs to the second quadrant satisfying
y
x = − 3. One such point is given by the coordinates
x
y
= −1
√
=
3
(The signs of the coordinates are not arbitrary : x must be negative and y must
be positive because the point comes from the second quadrant.)
2
There is a natural
ambiguity in the choice of the point P : one can take x = −2
√
and y = −2 3, for example, this different choice has no effect on the values of
the trigonometric functions.
The distance of P from the origin is
q
√ 2 √
√
r = (−1)2 + 3 = 1 + 3 = 4 = 2.
Using the definitions of the trigonometric functions we get
√
3
y
sin θ =
=
r
2
x
−1
1
cos θ =
=
=−
r
2
√2
√
3
y
tan θ =
=
=− 3
x
−1
r
2
csc θ =
=√
y
3
2
r
=
= −2
sec θ =
x
−1
x
1
−1
cot θ =
= √ = −√ .
y
3
3
It is easy to see that θ is coterminal to 120◦ .
[3]
5. Express sin θ in terms of cos θ for an angle θ positioned in the
third quadrant.
Solution. First of all, we have to find an identity which relates the trigonometric functions sin θ and cos θ. The Pythagorean-type identity
cos2 θ + sin2 θ = 1
seems to be useful. To express sin θ in terms of cos θ we try to solve the equation
above for sin θ:
cos2 θ + sin2 θ
2
sin θ
sin θ
=
1
1 − cos2 θ
p
= ± 1 − cos2 θ
=
There is an ambiguity in choosing the sign because sin θ is the solution of a
quadratic equation. To fix the sign, we note that the sine function is negative
in the third quadrant since it is defined as
sin θ =
y
r
for some point P (x, y) on the terminal side of θ which has negative y-coordinate.
Therefore we have
p
sin θ = − 1 − cos2 θ
for θ with terminal side located in the third quadrant.
3
6.
Solve the triangle ABC4 given the following data:
(a)
a = 3m,
[8]
b = 5m,
c = 7m
(b)
c = 3m,
∠A = 32◦ ,
∠B = 41◦ .
Solution.
Part (a)
This is a problem of the type SSS: we know the sides and we have to determine
the unknown angles.
We use the Law of Cosines to find the cosines of the angles first and the angles
can be found approximately using the inverse of the cosine function.
b2 + c2 − a2
25 + 49 − 9
65
=
=
2bc
2·5·7
70
c2 + a2 − b2
49 + 9 − 25
33
cos β =
=
=
2ca
2·7·3
42
a2 + b2 − c2
9 + 25 − 49
15
cos γ =
=
=− .
2ab
2·3·5
30
The angles can be approximated using the calculator:
cos α
=
α
=
β
=
γ
=
65
≈ 21.7868◦
70
33
≈ 38.2132◦
cos−1
42
15
1
cos−1 −
= cos−1 −
= 120◦
30
2
cos−1
(The cosine of γ has the special value − 21 which corresponds to the special angle
γ = 120◦ so γ is exact.) It is easy to check that the decimal approximations
give
α + β + γ = 180◦ .
Part (b)
We have c = 3m, α = 32◦ and β = 41◦ which is data of ASA-type.
The third angle γ is determined by
γ = 180◦ − α − β = 180◦ − 32◦ − 41◦ = 107◦ .
To calculate a, we use the Law of Sines:
sin γ
sin α
=
a
c
and therefore
a=c
sin α
sin 32◦
= 3m
≈ 1.6624m.
sin γ
sin 107◦
Similarly,
sin β
sin γ
=
b
c
so
b=c
sin 41◦
sin β
= 3m
≈ 2.0581m.
sin γ
sin 107◦
4