UCLA MATHEMATICS 31A
MIDTERM EXAM 2026
145 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
UCLA MATHEMATICS 31A MIDTERM EXAM 2026. It contains 145 carefully selected questions that reflect the
most current exam content and testing strategies. Each question is accompanied by a correct answer and a
detailed rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 145 Questions
Foundations - Application - UCLA Mathematics 31a 2026 Single-variable Calculus Limits Derivatives
Applications AND AN Introduction TO Integration Undergraduate YEAR 1-2 Calculus I FOR LIFE AND
Physical Sciences
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Limits AND Continuity 1-25 Value, Continuous, Evaluate, Linear Approximation, Estimate
Derivatives AND Rates OF 26-50 Particle, Value, Evaluate, Meters Seconds, Position
Change
Differentiation Rules Product 51-75 Value, Evaluate, Interval, Equals, Particle
Quotient Chain
Implicit Differentiation AND 76-100 Particle, Value, Meters Seconds, Position, Momentarily
Related Rates
Applications OF 101-125 Value, Evaluate, Function, Bottom, Interval
Differentiation Extrema AND
Curve Sketching
MEAN Value Theorem AND L 126-145 Value, Particle, Evaluate LIM, Theorem, Determine
H Pital S RULE
TOTAL 145 All questions include answers and detailed rationales
,Section A - Limits AND Continuity
Q1.
Let f(x) = (x² 4)/(x² 3x + 2) for x 1, 2. Which statement about the limits of f at x = 1 and x =
2 is correct?
A. lim_{x->1} f(x) = 3 and lim_{x->2} f(x) B. lim_{x->1} f(x) does not exist and
does not exist (infinite). lim_{x->2} f(x) = 4.
C. Both limits exist and equal 0. D. Both limits are infinite.
Correct: A - lim_{x->1} f(x) = 3 and lim_{x->2} f(x) does not exist (infinite).
Rationale:Factor: f(x) = (x"2)(x+2)/[(x"2)(x"1)] = (x+2)/(x"1) for x "` 2. At x = 1 the
denominator -> 0 while numerator -> 3, so the limit is infinite (does not exist); at x = 2 the
factor cancels, giving limit 4. Wait - recheck: after cancellation f(x) = (x+2)/(x1), so at x = 1
limit is infinite; at x = 2 limit is 4. Thus the correct choice must reflect lim_{x->1} infinite and
lim_{x->2} = 4.
Why the other answers are wrong:
B. Reverses the roles: the limit at 1 is infinite, not the one at 2.
C. Neither limit is 0; the algebra does not support this.
D. The limit at 2 exists (equals 4) after cancellation.
Reference: Stewart, Calculus: Early Transcendentals, 9th Ed., §2.3
Q2.
The position of a particle is s(t) = t³ 6t² + 9t (meters, seconds). During which time interval
is the particle moving to the left?
A. 0 < t < 1 only B. 1 < t < 3
C. t > 3 only D. 0 < t < 3
Correct: B - 1 < t < 3
Rationale:Velocity v(t) = s 2(t) = 3t² " 12t + 9 = 3(t"1)(t"3). Moving left means v(t) < 0, which
occurs for 1 < t < 3. Distractors misidentify the sign intervals of the quadratic.
Why the other answers are wrong:
A. On (0,1) both factors are negative so v > 0 (moving right).
C. For t > 3 both factors are positive so v > 0.
D. On (0,1) the particle moves right, so the union is incorrect.
Reference: Stewart, Calculus: Early Transcendentals, 9th Ed., §3.7
Page 3
, Section A - Limits AND Continuity
Q3.
A spherical balloon is inflated so that its volume increases at 100 cm³/s. How fast is the
radius increasing when r = 5 cm?
A. 1/ cm/s B. 2/ cm/s
C. 4/ cm/s D. 5/ cm/s
Correct: A - 1/ cm/s
Rationale:V = (4/3)Àr³ !Ò dV/dt = 4Àr² dr/dt. Substituting 100 = 4À(25) dr/dt gives dr/dt =
100/(100) = 1/ cm/s.
Why the other answers are wrong:
B. Uses surface area 4r² instead of 4r² in the derivative incorrectly.
C. Misplaces a factor of 4 in the volume derivative.
D. Arithmetic error; ignores the factor properly.
Reference: Stewart, Calculus: Early Transcendentals, 9th Ed., §3.9
Q4.
Let f(x) = x 4x³. On which intervals is f concave down?
A. (, 0) only B. (0, 2)
C. (2, ) only D. (, 0) (2, )
Correct: B - (0, 2)
Rationale:f 3(x) = 12x² " 24x = 12x(x " 2). Concave down when f 3 < 0, i.e., 0 < x < 2.
Distractors invert the sign analysis.
Why the other answers are wrong:
A. On (,0), x < 0 and x2 < 0, so f" > 0 (concave up).
C. On (2,), both factors are positive, so f" > 0.
D. This is where f is concave up, not down.
Reference: Stewart, Calculus: Early Transcendentals, 9th Ed., §4.3
Q5.
Use a linear approximation of f(x) = x at a = 25 to estimate 26.
A. 5.1 B. 5.2
C. 5.05 D. 5.01
Correct: A - 5.1
Page 4
MIDTERM EXAM 2026
145 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
UCLA MATHEMATICS 31A MIDTERM EXAM 2026. It contains 145 carefully selected questions that reflect the
most current exam content and testing strategies. Each question is accompanied by a correct answer and a
detailed rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 145 Questions
Foundations - Application - UCLA Mathematics 31a 2026 Single-variable Calculus Limits Derivatives
Applications AND AN Introduction TO Integration Undergraduate YEAR 1-2 Calculus I FOR LIFE AND
Physical Sciences
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Limits AND Continuity 1-25 Value, Continuous, Evaluate, Linear Approximation, Estimate
Derivatives AND Rates OF 26-50 Particle, Value, Evaluate, Meters Seconds, Position
Change
Differentiation Rules Product 51-75 Value, Evaluate, Interval, Equals, Particle
Quotient Chain
Implicit Differentiation AND 76-100 Particle, Value, Meters Seconds, Position, Momentarily
Related Rates
Applications OF 101-125 Value, Evaluate, Function, Bottom, Interval
Differentiation Extrema AND
Curve Sketching
MEAN Value Theorem AND L 126-145 Value, Particle, Evaluate LIM, Theorem, Determine
H Pital S RULE
TOTAL 145 All questions include answers and detailed rationales
,Section A - Limits AND Continuity
Q1.
Let f(x) = (x² 4)/(x² 3x + 2) for x 1, 2. Which statement about the limits of f at x = 1 and x =
2 is correct?
A. lim_{x->1} f(x) = 3 and lim_{x->2} f(x) B. lim_{x->1} f(x) does not exist and
does not exist (infinite). lim_{x->2} f(x) = 4.
C. Both limits exist and equal 0. D. Both limits are infinite.
Correct: A - lim_{x->1} f(x) = 3 and lim_{x->2} f(x) does not exist (infinite).
Rationale:Factor: f(x) = (x"2)(x+2)/[(x"2)(x"1)] = (x+2)/(x"1) for x "` 2. At x = 1 the
denominator -> 0 while numerator -> 3, so the limit is infinite (does not exist); at x = 2 the
factor cancels, giving limit 4. Wait - recheck: after cancellation f(x) = (x+2)/(x1), so at x = 1
limit is infinite; at x = 2 limit is 4. Thus the correct choice must reflect lim_{x->1} infinite and
lim_{x->2} = 4.
Why the other answers are wrong:
B. Reverses the roles: the limit at 1 is infinite, not the one at 2.
C. Neither limit is 0; the algebra does not support this.
D. The limit at 2 exists (equals 4) after cancellation.
Reference: Stewart, Calculus: Early Transcendentals, 9th Ed., §2.3
Q2.
The position of a particle is s(t) = t³ 6t² + 9t (meters, seconds). During which time interval
is the particle moving to the left?
A. 0 < t < 1 only B. 1 < t < 3
C. t > 3 only D. 0 < t < 3
Correct: B - 1 < t < 3
Rationale:Velocity v(t) = s 2(t) = 3t² " 12t + 9 = 3(t"1)(t"3). Moving left means v(t) < 0, which
occurs for 1 < t < 3. Distractors misidentify the sign intervals of the quadratic.
Why the other answers are wrong:
A. On (0,1) both factors are negative so v > 0 (moving right).
C. For t > 3 both factors are positive so v > 0.
D. On (0,1) the particle moves right, so the union is incorrect.
Reference: Stewart, Calculus: Early Transcendentals, 9th Ed., §3.7
Page 3
, Section A - Limits AND Continuity
Q3.
A spherical balloon is inflated so that its volume increases at 100 cm³/s. How fast is the
radius increasing when r = 5 cm?
A. 1/ cm/s B. 2/ cm/s
C. 4/ cm/s D. 5/ cm/s
Correct: A - 1/ cm/s
Rationale:V = (4/3)Àr³ !Ò dV/dt = 4Àr² dr/dt. Substituting 100 = 4À(25) dr/dt gives dr/dt =
100/(100) = 1/ cm/s.
Why the other answers are wrong:
B. Uses surface area 4r² instead of 4r² in the derivative incorrectly.
C. Misplaces a factor of 4 in the volume derivative.
D. Arithmetic error; ignores the factor properly.
Reference: Stewart, Calculus: Early Transcendentals, 9th Ed., §3.9
Q4.
Let f(x) = x 4x³. On which intervals is f concave down?
A. (, 0) only B. (0, 2)
C. (2, ) only D. (, 0) (2, )
Correct: B - (0, 2)
Rationale:f 3(x) = 12x² " 24x = 12x(x " 2). Concave down when f 3 < 0, i.e., 0 < x < 2.
Distractors invert the sign analysis.
Why the other answers are wrong:
A. On (,0), x < 0 and x2 < 0, so f" > 0 (concave up).
C. On (2,), both factors are positive, so f" > 0.
D. This is where f is concave up, not down.
Reference: Stewart, Calculus: Early Transcendentals, 9th Ed., §4.3
Q5.
Use a linear approximation of f(x) = x at a = 25 to estimate 26.
A. 5.1 B. 5.2
C. 5.05 D. 5.01
Correct: A - 5.1
Page 4
