Q. Calculate the limit: lim (x -> 0) (ln(1 + x)/x) (2023)
A.
1
B.
0
C.
Undefined
D.
Infinity
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Solution
Using L'Hôpital's Rule, we differentiate the numerator and denominator to find lim (x -> 0) (1/(1 + x)) = 1.
Correct Answer:
A
— 1
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Q. Calculate the limit: lim (x -> 0) (x^2 sin(1/x))
A.
0
B.
1
C.
∞
D.
Undefined
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Solution
Since |sin(1/x)| ≤ 1, we have |x^2 sin(1/x)| ≤ |x^2|. Thus, lim (x -> 0) x^2 sin(1/x) = 0.
Correct Answer:
A
— 0
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Q. Calculate the limit: lim (x -> 0) (x^3)/(sin(x)) (2023)
A.
0
B.
1
C.
∞
D.
Undefined
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Solution
Using the fact that sin(x) ~ x as x approaches 0, we find that lim (x -> 0) (x^3)/(sin(x)) = 0.
Correct Answer:
A
— 0
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Q. Calculate the limit: lim (x -> 2) (x^3 - 8)/(x - 2)
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Solution
Factoring gives lim (x -> 2) ((x - 2)(x^2 + 2x + 4))/(x - 2) = lim (x -> 2) (x^2 + 2x + 4) = 12.
Correct Answer:
A
— 4
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Q. Calculate the limit: lim (x -> ∞) (3x^2 + 2)/(5x^2 - 4) (2023)
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Solution
Dividing numerator and denominator by x^2 gives lim (x -> ∞) (3 + 2/x^2)/(5 - 4/x^2) = 3/5.
Correct Answer:
A
— 3/5
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Q. Calculate the limit: lim (x -> ∞) (3x^2 + 2)/(5x^2 - 4x + 1) (2023)
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Solution
Dividing numerator and denominator by x^2 gives lim (x -> ∞) (3 + 2/x^2)/(5 - 4/x + 1/x^2) = 3/5.
Correct Answer:
A
— 3/5
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Q. Determine the continuity of the function f(x) = { x^2, x < 1; 2, x = 1; x + 1, x > 1 } at x = 1.
A.
Continuous
B.
Not continuous
C.
Depends on the limit
D.
Only left continuous
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Solution
The left limit as x approaches 1 is 1, the right limit is 2, and f(1) = 2. Since the left and right limits do not match, f(x) is not continuous at x = 1.
Correct Answer:
B
— Not continuous
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Q. Determine the continuity of the function f(x) = { x^2, x < 1; 2x - 1, x ≥ 1 } at x = 1.
A.
Continuous
B.
Discontinuous
C.
Only left continuous
D.
Only right continuous
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Solution
At x = 1, f(1) = 2(1) - 1 = 1 and lim x→1- f(x) = 1, lim x→1+ f(x) = 1. Thus, f(x) is continuous at x = 1.
Correct Answer:
A
— Continuous
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Q. Determine the continuity of the function f(x) = |x| at x = 0. (2020)
A.
Continuous
B.
Not continuous
C.
Depends on the limit
D.
Only left continuous
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Solution
The function f(x) = |x| is continuous at x = 0 since both the left-hand limit and right-hand limit equal f(0) = 0.
Correct Answer:
A
— Continuous
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Q. Determine the critical points of the function f(x) = x^2 - 4x + 4. (2022)
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Solution
f'(x) = 2x - 4; Setting f'(x) = 0 gives x = 2 as the critical point.
Correct Answer:
C
— 2
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Q. Determine the derivative of f(x) = x^3 - 4x + 7. (2023)
A.
3x^2 - 4
B.
3x^2 + 4
C.
x^2 - 4
D.
3x^2 - 7
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Solution
Using the power rule, f'(x) = 3x^2 - 4.
Correct Answer:
A
— 3x^2 - 4
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Q. Determine the derivative of f(x) = x^5 - 3x^3 + 2x. (2023)
A.
5x^4 - 9x^2 + 2
B.
5x^4 - 9x + 2
C.
5x^4 - 3x^2 + 2
D.
5x^4 - 3x^3
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Solution
Using the power rule, f'(x) = 5x^4 - 9x^2 + 2.
Correct Answer:
A
— 5x^4 - 9x^2 + 2
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Q. Determine the local maxima and minima of f(x) = x^2 - 4x + 3.
A.
Maxima at x=2
B.
Minima at x=2
C.
Maxima at x=1
D.
Minima at x=1
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Solution
f'(x) = 2x - 4. Setting f'(x) = 0 gives x = 2. f''(x) = 2 > 0 indicates a local minimum at x = 2.
Correct Answer:
B
— Minima at x=2
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Q. Determine the local maxima and minima of f(x) = x^4 - 8x^2 + 16. (2023)
A.
Maxima at x = 0
B.
Minima at x = 2
C.
Maxima at x = 2
D.
Minima at x = 0
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Solution
f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f''(x) = 12x^2 - 16. Minima at x = 0.
Correct Answer:
D
— Minima at x = 0
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Q. Determine the local maxima of f(x) = -x^2 + 4x. (2022)
A.
(2, 4)
B.
(0, 0)
C.
(4, 0)
D.
(1, 1)
Show solution
Solution
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f(2) = -2^2 + 4(2) = 4.
Correct Answer:
A
— (2, 4)
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Q. Determine the local maxima or minima of f(x) = -x^2 + 4x. (2019)
A.
Maxima at x=2
B.
Minima at x=2
C.
Maxima at x=4
D.
Minima at x=4
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Solution
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. Since f''(x) = -2 < 0, it is a maxima.
Correct Answer:
A
— Maxima at x=2
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Q. Determine the maximum value of f(x) = -2x^2 + 4x + 1. (2023)
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Solution
The vertex is at x = -4/(2*(-2)) = 1. The maximum value is f(1) = -2(1)^2 + 4(1) + 1 = 3.
Correct Answer:
C
— 3
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Q. Determine the maximum value of f(x) = -x^2 + 4x. (2020)
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Solution
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f(2) = -2^2 + 4(2) = 8.
Correct Answer:
A
— 4
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Q. Determine the maximum value of the function f(x) = -x^2 + 6x - 8. (2022)
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Solution
The vertex is at x = -6/(2*(-1)) = 3. The maximum value is f(3) = -3^2 + 6*3 - 8 = 1.
Correct Answer:
B
— 4
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Q. Determine the minimum value of f(x) = x^2 - 6x + 10. (2019)
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Solution
The minimum occurs at x = -b/(2a) = 6/(2*1) = 3. f(3) = 3^2 - 6(3) + 10 = 3.
Correct Answer:
B
— 3
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Q. Determine the minimum value of the function f(x) = x^2 - 4x + 6. (2020)
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Solution
The function is a upward-opening parabola. The minimum occurs at x = -b/(2a) = 4/(2*1) = 2. f(2) = 2^2 - 4(2) + 6 = 2.
Correct Answer:
A
— 2
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Q. Determine the slope of the tangent line to f(x) = x^2 at x = 3. (2023)
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Solution
f'(x) = 2x; thus, f'(3) = 2(3) = 6.
Correct Answer:
B
— 6
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Q. Differentiate f(x) = 4x^2 * e^x. (2022)
A.
4e^x + 4x^2e^x
B.
4x^2e^x + 4xe^x
C.
4e^x + 2x^2e^x
D.
8xe^x
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Solution
Using the product rule, f'(x) = 4e^x + 4x^2e^x.
Correct Answer:
A
— 4e^x + 4x^2e^x
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Q. Differentiate f(x) = 4x^2 + 3x - 5. (2019)
A.
8x + 3
B.
4x + 3
C.
2x + 3
D.
8x - 3
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Solution
Using the power rule, f'(x) = 8x + 3.
Correct Answer:
A
— 8x + 3
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Q. Differentiate f(x) = 4x^5 - 2x^3 + x. (2022)
A.
20x^4 - 6x^2 + 1
B.
20x^4 - 6x^2
C.
4x^4 - 2x^2 + 1
D.
5x^4 - 2x^2
Show solution
Solution
Using the power rule, f'(x) = 20x^4 - 6x^2 + 1.
Correct Answer:
A
— 20x^4 - 6x^2 + 1
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Q. Differentiate f(x) = ln(x^2 + 1). (2022)
A.
2x/(x^2 + 1)
B.
1/(x^2 + 1)
C.
2x/(x^2 - 1)
D.
x/(x^2 + 1)
Show solution
Solution
Using the chain rule, f'(x) = 2x/(x^2 + 1).
Correct Answer:
A
— 2x/(x^2 + 1)
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Q. Differentiate f(x) = x^2 * e^x. (2022)
A.
x^2 * e^x + 2x * e^x
B.
2x * e^x + x^2 * e^x
C.
x^2 * e^x + e^x
D.
2x * e^x
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Solution
Using the product rule, f'(x) = x^2 * e^x + 2x * e^x.
Correct Answer:
A
— x^2 * e^x + 2x * e^x
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Q. Differentiate f(x) = x^2 * ln(x).
A.
2x * ln(x) + x
B.
x * ln(x) + 2x
C.
2x * ln(x)
D.
x^2/x
Show solution
Solution
Using the product rule, f'(x) = 2x * ln(x) + x.
Correct Answer:
A
— 2x * ln(x) + x
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Q. Differentiate the function f(x) = ln(x^2 + 1).
A.
2x/(x^2 + 1)
B.
2/(x^2 + 1)
C.
1/(x^2 + 1)
D.
x/(x^2 + 1)
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Solution
Using the chain rule, f'(x) = (1/(x^2 + 1)) * (2x) = 2x/(x^2 + 1).
Correct Answer:
A
— 2x/(x^2 + 1)
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Q. Differentiate the function f(x) = x^2 * e^x.
A.
x^2 * e^x + 2x * e^x
B.
2x * e^x + x^2 * e^x
C.
x^2 * e^x + e^x
D.
2x * e^x + e^x
Show solution
Solution
Using the product rule, f'(x) = (x^2)' * e^x + x^2 * (e^x)' = 2x * e^x + x^2 * e^x.
Correct Answer:
A
— x^2 * e^x + 2x * e^x
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