College Algebra (MATH 107) Final Exam Solutions, Summer 2019

This document presents the solutions to a College Algebra final examination (MATH 107) administered in the Summer of 2019. The exam covers a range of topics, starting with multiple-choice questions on domains and ranges of functions, solving equations, and identifying increasing intervals. Further questions address function symmetry, inequalities, and linear equations, including slope-intercept form and parallel lines. The solutions also cover function properties such as the vertical line test and logarithmic properties, along with exponential functions and their graphical representations. The document provides detailed answers to all multiple-choice questions and showcases the step-by-step solutions for each question. It is designed to help students prepare for exams and improve their understanding of core algebraic concepts.

Q 1.
Answer:Option C
Figure 1:Domain of function Figure 2:Range of function
Domain:It is the interval over which the function is defined.
As seen in figure-1, the function is defined over the interval [-1,3], which is
therefore the domain.
Range:It is the range of values which the function assumes over the do-
main.
As seen in figure-2, the function takes values in-between -3 and 1, therefore,
the range is [-3,1].
Q 2.
Answer:Option C
√ 24 − 5x = x
[Square both sides]
24 − 5x = x2
[Subtract x2 from both sides]
−x2 − 5x + 24 = 0
[Multiply throughout by -1]
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x2 + 5x − 24 = 0
=⇒ x 2 + 8x − 3x − 24 = 0
x(x + 8) − 3(x + 8) = 0
(x + 8)(x − 3) = 0
∴ x = −8 or 3
Q 3.
Answer:Option C
Minima of the function occurs at the stationary point x = −2 and Maxima
occurs at the stationary point x = 4, as seen in the figure above.
The funciton decreases upto the minima and then increases from minima
to maxima. It again decreases once maxima is achieved.Therefore,the
function increases only in the interval (−2, 4).
Q 4.
Answer:Option D
A function f (x) is symmetric:
About y- axis if:f (x) = f (−x)
About x- axis if:f (x) = −f (x)
About origin if:f(x) = −f(−x)
For f (x) = |3 + x|,
f (−x) = |3 − x| 6= f (x)
−f (x) = −|3 + x| 6= f (x)
2

−f (−x) = −|3 − x| 6= f (x)
Therefore, the function is not symmetric about any of the above.
Q 5.
Answer:Option B
|7 − 4x| ≥ 9
=⇒ 7 − 4x ≥ 9 or 7 − 4x ≤ −9
1.) 7 − 4x ≥ 9
[Subtract 7 from both sides]
−4x ≥ 2
[Divide throughout by −4]
=⇒ x ≤ −1
2
2.) 7 − 4x ≤ −9
[Subtract 7 from both sides]
−4x ≤ −16
[Divide throughout by −4]
x ≥ 4
Therefore, the solution interval is
(−∞, −0.5] ∪ [4, ∞)
Q 6.
Answer:Option A
4x − 7y = 28
Arrange the equation of line in slope-intercept form
y = 4
7x − 4
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The y- intercept of the line is −4 which is the case for graph in option A.
Q 7.
Answer:Option B
x + 4y = 7
Re-arrange in slope-intercept form
4y = 7 − x =⇒ y = −
1
4x + 7
4
Line parallel to x + 4y = 7 will have slope m = −1
4.
Equation of line passing through point (x1, y1) = (12, −7), with slope m =
−1
4, is
y − y1 = m(x − x1) =⇒ y + 7 = −
1
4(x − 12)
Re-arranging in slope-intercept form
y = −1
4x − 4
Q 8.
Answer:Option B
Figure 1:Vertical line test Figure 2:Horizontal line test
Function :A graph is a function if each input has only one output.
Alternately, if a vertical line drawn over a graph intersects it only once, then
the graph is said to be a function.
One-to-one:A graph is said to be a one-to-one function if each input pro-
duces a unique output.
Alternately,a graph is said to be a one-to-one function if a horizontalline
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drawn over the graph intercepts it only once.
As it can be seen form figures 1 and 2, the graph passes the vertical test but
not the horizontal test, hence it is a function but not one-to-one.
Q 9.
Answer:Option B
log (x + 2) + log (1) − 3 log (y)
Properties of logarithms:
1. log (ab) = log a + log b
2. loga
b = log a − log b
3. n log a = log an
log (x + 2) + log (1)
| {z }
1
−3 log (y)
log ((x + 2) · 1) − 3 log y
| {z }
3
log (x + 2) − log (y3)
| {z }
2
answer
−−−−→ log
x + 2
y3
Q 10.
Answer:Option A
The graph intersects the y- axis at point (0, 3), implies f (x) = 3 when x = 0.
f (x) = e−x + 2
is the only equation that satisfies the above condition.
Q 11.
Answer:Option D
Let us consider an arbitrary function which satisfies the condition that it
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intersects the x- axis only once.
The function need not be linear,it need not be one-to-one (horizontalline
test) which means it is not necessarily invertible and it can cross the x- axis
from below so it is not necessarily always positive.
But the function will always have only one positive real root which the point
of intersection with the x- axis.
Q 12.
Answer:Option A
The points (−1, 4),(0, 2) and (2, −2) on f (x) are shifted downwards to
(−1, 2), (0, 0) and (2, −4) respectively.
This indicates a vertical shift of 2 units downwards.
This vertical shift is represented by the relation:
g(x) = f (x) − 2
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