MTH 130 - Pre-Calculus Mathematics Final Exam 2026/2027 | Complete Questions & Verified Answers | Calculus Preparation & Comprehensive Bridge Course Assessment

MTH 130 - Pre-Calculus Mathematics Final Exam
(2026/2027) | QUESTIONS AND ANSWERS
MTH 130: Pre-Calculus Mathematics Comprehensive Examination | Core Domains: Advanced
Algebraic Functions & Analysis, Trigonometric Functions & Analytic Trigonometry, Polar
Coordinates & Complex Numbers, Conic Sections & Parametric Equations, Sequences, Series &
Mathematical Induction, Binomial Theorem, Limits & Introduction to Continuity, and
Mathematical Modeling | Calculus Preparation Focus | Comprehensive Bridge Course Final
Exam Format


Exam Structure

The MTH 130 Pre-Calculus Mathematics Final Exam for the 2026/2027 academic cycle is a
100-question, multiple-choice question (MCQ) and extended problem-solving examination.

Introduction​
This MTH 130 Pre-Calculus Mathematics Final Exam guide for the 2026/2027 cycle assesses the
synthesis of advanced algebra, trigonometry, and analytic geometry necessary for success in
calculus. The content emphasizes deep function analysis, rigorous problem-solving, and the
foundational concepts of limits and continuity that bridge algebraic manipulation to differential
calculus.

Answer Format​
All correct answers and mathematical solutions must be presented in bold and green,
followed by step-by-step rationales that show advanced algebraic and trigonometric
manipulations, graphical interpretations, the use of mathematical induction or series formulas,
and the precise application of limit definitions and properties.



Questions and Solutions
Question 1: If \( f(x) = \frac{x^2 - 9}{x - 3} \), what is \( \lim_{x \to 3} f(x) \)?


●​ (A) 0
●​ (B) 3
●​ (C) 6
●​ (D) Undefined


(C) 6


Simplify: \( f(x) = \frac{(x - 3)(x + 3)}{x - 3} = x + 3 \) for \( x \ne 3 \). Thus, \( \lim_{x \to 3}
f(x) = 3 + 3 = 6 \).

,Question 2: Which of the following is equivalent to \( \sin\left(\frac{3\pi}{2} - x\right) \)?


●​ (A) \( \sin x \)
●​ (B) \( -\sin x \)
●​ (C) \( \cos x \)
●​ (D) \( -\cos x \)


(D) \( -\cos x \)


Use identity: \( \sin(a - b) = \sin a \cos b - \cos a \sin b \). \( \sin\left(\frac{3\pi}{2} - x\right)
= \sin\frac{3\pi}{2}\cos x - \cos\frac{3\pi}{2}\sin x = (-1)\cos x - (0)\sin x = -\cos x \).

Question 3: The complex number \( z = 1 - i \) expressed in polar form is:


●​ (A) \( \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \)
●​ (B) \( \sqrt{2} \left( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right) \)
●​ (C) \( 2 \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) \)
●​ (D) \( \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) \)


(B) \( \sqrt{2} \left( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right) \)


Magnitude: \( r = \sqrt{1^2 + (-1)^2} = \sqrt{2} \). Argument: \( \theta = \tan^{-1}(-1/1) =
-\frac{\pi}{4} \), which is coterminal with \( \frac{7\pi}{4} \) in \( [0, 2\pi) \). So polar form
is \( \sqrt{2} \left( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right) \).

Question 4: The graph of \( 4x^2 + 9y^2 - 16x + 18y = 11 \) represents a:


●​ (A) Circle
●​ (B) Ellipse
●​ (C) Hyperbola
●​ (D) Parabola


(B) Ellipse


Group terms: \( 4(x^2 - 4x) + 9(y^2 + 2y) = 11 \). Complete squares: \( 4[(x - 2)^2 - 4] + 9[(y +
1)^2 - 1] = 11 \) → \( 4(x - 2)^2 + 9(y + 1)^2 = 11 + 16 + 9 = 36 \). Divide by 36: \( \frac{(x -
2)^2}{9} + \frac{(y + 1)^2}{4} = 1 \), standard form of an ellipse.

,Question 5: What is the sum of the first 10 terms of the geometric sequence \( 3, 6, 12, 24,
\dots \)?


●​ (A) 3069
●​ (B) 1533
●​ (C) 3072
●​ (D) 1023


(A) 3069


First term \( a = 3 \), common ratio \( r = 2 \), \( n = 10 \). Sum formula: \( S_n = a \frac{r^n
- 1}{r - 1} = 3 \cdot \frac{2^{10} - 1}{2 - 1} = 3(1024 - 1) = 3 \cdot 1023 = 3069 \).

Question 6: Evaluate \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} \).


●​ (A) 0
●​ (B) \( \frac{1}{2} \)
●​ (C) 1
●​ (D) Undefined


(B) \( \frac{1}{2} \)


Use standard limit or L’Hôpital’s Rule (allowed as preview): \( \lim_{x \to 0} \frac{1 - \cos
x}{x^2} = \lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2} \lim_{x \to 0} \frac{\sin x}{x} =
\frac{1}{2} \cdot 1 = \frac{1}{2} \). Alternatively, use identity \( 1 - \cos x = 2\sin^2(x/2) \),
then simplify.

Question 7: If \( \log_2(x) + \log_2(x - 2) = 3 \), then \( x = \)?


●​ (A) -2
●​ (B) 1
●​ (C) 4
●​ (D) 6


(C) 4

, Combine logs: \( \log_2[x(x - 2)] = 3 \) → \( x(x - 2) = 2^3 = 8 \). So \( x^2 - 2x - 8 = 0 \) → \(
(x - 4)(x + 2) = 0 \) → \( x = 4 \) or \( x = -2 \). But domain requires \( x > 2 \), so only \( x = 4
\) is valid.

Question 8: The period of \( y = \tan(3x) \) is:


●​ (A) \( \pi \)
●​ (B) \( \frac{\pi}{3} \)
●​ (C) \( 3\pi \)
●​ (D) \( \frac{2\pi}{3} \)


(B) \( \frac{\pi}{3} \)


Standard period of \( \tan x \) is \( \pi \). For \( \tan(kx) \), period is \( \frac{\pi}{|k|} \).
Here \( k = 3 \), so period = \( \frac{\pi}{3} \).

Question 9: The binomial expansion of \( (x - 2)^5 \) has a constant term equal to:


●​ (A) -32
●​ (B) 32
●​ (C) -10
●​ (D) 0


(A) -32


General term: \( \binom{5}{k} x^{5 - k} (-2)^k \). Constant term when exponent of \( x \) is 0
→ \( 5 - k = 0 \) → \( k = 5 \). Term: \( \binom{5}{5} x^0 (-2)^5 = 1 \cdot 1 \cdot (-32) = -32
\).

Question 10: Which equation represents a function that is its own inverse?


●​ (A) \( f(x) = x + 1 \)
●​ (B) \( f(x) = \frac{1}{x}, x \ne 0 \)
●​ (C) \( f(x) = x^2 \)
●​ (D) \( f(x) = 2x \)


(B) \( f(x) = \frac{1}{x}, x \ne 0 \)