Q. Determine the family of curves represented by the equation y = ax^2 + bx + c.
-
A.
Parabolas
-
B.
Circles
-
C.
Ellipses
-
D.
Straight lines
Solution
The equation y = ax^2 + bx + c represents a family of parabolas with varying coefficients a, b, and c.
Correct Answer:
A
— Parabolas
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Q. Determine the family of curves represented by the equation y = ax^3 + bx.
-
A.
Cubic functions
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B.
Quadratic functions
-
C.
Linear functions
-
D.
Exponential functions
Solution
The equation y = ax^3 + bx represents a family of cubic functions where a and b are constants.
Correct Answer:
A
— Cubic functions
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Q. Determine the family of curves represented by the equation y = ax^3 + bx^2 + cx + d.
-
A.
Cubic functions
-
B.
Quadratic functions
-
C.
Linear functions
-
D.
Exponential functions
Solution
The equation y = ax^3 + bx^2 + cx + d represents a family of cubic functions.
Correct Answer:
A
— Cubic functions
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Q. Determine the family of curves represented by the equation y = e^(kx) for varying k.
-
A.
Exponential curves
-
B.
Linear functions
-
C.
Quadratic functions
-
D.
Logarithmic functions
Solution
The equation y = e^(kx) represents a family of exponential curves with varying growth rates determined by k.
Correct Answer:
A
— Exponential curves
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Q. Determine the family of curves represented by the equation y = k/x, where k is a constant.
-
A.
Hyperbolas
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B.
Circles
-
C.
Ellipses
-
D.
Parabolas
Solution
The equation y = k/x represents a family of hyperbolas with varying values of 'k'.
Correct Answer:
A
— Hyperbolas
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Q. Determine the family of curves represented by the equation y = kx^2, where k is a constant.
-
A.
Circles
-
B.
Ellipses
-
C.
Parabolas
-
D.
Hyperbolas
Solution
The equation y = kx^2 represents a family of parabolas that open upwards or downwards depending on the sign of k.
Correct Answer:
C
— Parabolas
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Q. Determine the focus of the parabola defined by the equation x^2 = 12y.
-
A.
(0, 3)
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B.
(0, -3)
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C.
(3, 0)
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D.
(-3, 0)
Solution
The equation x^2 = 4py gives 4p = 12, hence p = 3. The focus is at (0, p) = (0, 3).
Correct Answer:
A
— (0, 3)
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Q. Determine the focus of the parabola given by the equation x^2 = 8y.
-
A.
(0, 2)
-
B.
(0, 4)
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C.
(2, 0)
-
D.
(4, 0)
Solution
The standard form of the parabola is x^2 = 4py. Here, 4p = 8, so p = 2. The focus is at (0, p) = (0, 2).
Correct Answer:
B
— (0, 4)
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Q. Determine the hybridization of the central atom in BF3.
-
A.
sp
-
B.
sp2
-
C.
sp3
-
D.
dsp3
Solution
Boron in BF3 is sp2 hybridized, forming three equivalent sp2 hybrid orbitals.
Correct Answer:
B
— sp2
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Q. Determine the hybridization of the central atom in O3.
-
A.
sp
-
B.
sp2
-
C.
sp3
-
D.
dsp3
Solution
The central atom in ozone (O3) is sp2 hybridized, forming a resonance structure.
Correct Answer:
B
— sp2
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Q. Determine the hybridization of the central atom in PCl5.
-
A.
sp
-
B.
sp2
-
C.
sp3
-
D.
dsp3
Solution
Phosphorus in PCl5 is dsp3 hybridized, allowing it to form five bonds.
Correct Answer:
D
— dsp3
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Q. Determine the intervals where f(x) = -x^2 + 4x is concave up. (2023)
-
A.
(-∞, 0)
-
B.
(0, 2)
-
C.
(2, ∞)
-
D.
(0, 4)
Solution
f''(x) = -2, which is always negative, indicating concave down everywhere.
Correct Answer:
C
— (2, ∞)
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Q. Determine the intervals where f(x) = x^3 - 3x is increasing. (2021)
-
A.
(-∞, -1)
-
B.
(-1, 1)
-
C.
(1, ∞)
-
D.
(-∞, 1)
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = -1, 1. f'(x) > 0 for x > 1.
Correct Answer:
C
— (1, ∞)
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Q. Determine the intervals where f(x) = x^4 - 4x^3 has increasing behavior. (2023)
-
A.
(-∞, 0)
-
B.
(0, 2)
-
C.
(2, ∞)
-
D.
(0, 4)
Solution
f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). f'(x) > 0 for x in (0, 3).
Correct Answer:
B
— (0, 2)
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Q. Determine the intervals where f(x) = x^4 - 4x^3 has local minima. (2020)
-
A.
(0, 2)
-
B.
(1, 3)
-
C.
(2, 4)
-
D.
(0, 1)
Solution
f'(x) = 4x^3 - 12x^2. Setting f'(x) = 0 gives x = 0, 3. Testing intervals shows local minima at (0, 2).
Correct Answer:
A
— (0, 2)
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Q. Determine the intervals where the function f(x) = x^3 - 3x is increasing.
-
A.
(-∞, -1)
-
B.
(-1, 1)
-
C.
(1, ∞)
-
D.
(-∞, 1)
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f'(x) > 0 for x > 1, so f(x) is increasing on (1, ∞).
Correct Answer:
C
— (1, ∞)
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Q. Determine the intervals where the function f(x) = x^4 - 4x^3 has increasing behavior.
-
A.
(-∞, 0) U (2, ∞)
-
B.
(0, 2)
-
C.
(0, ∞)
-
D.
(2, ∞)
Solution
f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). The function is increasing where f'(x) > 0, which is in the intervals (-∞, 0) and (3, ∞).
Correct Answer:
A
— (-∞, 0) U (2, ∞)
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Q. Determine the length of the latus rectum of the parabola y^2 = 16x.
Solution
The length of the latus rectum for the parabola y^2 = 4px is given by 4p. Here, p = 4, so the length is 16.
Correct Answer:
B
— 8
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Q. Determine the limit: lim (x -> 0) (tan(5x)/x) (2022)
-
A.
0
-
B.
1
-
C.
5
-
D.
Undefined
Solution
Using the standard limit lim (x -> 0) (tan(kx)/x) = k, we have k = 5. Thus, lim (x -> 0) (tan(5x)/x) = 5.
Correct Answer:
C
— 5
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Q. Determine the limit: lim (x -> 1) (x^3 - 1)/(x - 1) (2020)
Solution
Factoring gives (x - 1)(x^2 + x + 1)/(x - 1). For x ≠ 1, this simplifies to x^2 + x + 1. Evaluating at x = 1 gives 3.
Correct Answer:
C
— 3
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Q. Determine the limit: lim (x -> 1) (x^4 - 1)/(x - 1) (2021)
-
A.
0
-
B.
1
-
C.
4
-
D.
Undefined
Solution
Factoring gives (x - 1)(x^3 + x^2 + x + 1)/(x - 1). For x ≠ 1, this simplifies to x^3 + x^2 + x + 1. Evaluating at x = 1 gives 4.
Correct Answer:
D
— Undefined
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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
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C.
Maxima at x=1
-
D.
Minima at x=1
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^3 - 3x.
-
A.
Maxima at (1, -2)
-
B.
Minima at (0, 0)
-
C.
Maxima at (0, 0)
-
D.
Minima at (1, -2)
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f''(1) = 6 > 0 (min), f''(-1) = 6 > 0 (min). Local maxima at (0, 0).
Correct Answer:
A
— Maxima at (1, -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
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 and minima of the function f(x) = x^3 - 6x^2 + 9x.
-
A.
(0, 0)
-
B.
(2, 0)
-
C.
(3, 0)
-
D.
(1, 0)
Solution
f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1, 3. f''(1) > 0 (min), f''(3) < 0 (max).
Correct Answer:
C
— (3, 0)
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Q. Determine the local maxima and minima of the function f(x) = x^4 - 4x^3 + 4x.
-
A.
Maxima at (0, 0)
-
B.
Minima at (2, 0)
-
C.
Maxima at (2, 0)
-
D.
Minima at (0, 0)
Solution
f'(x) = 4x^3 - 12x^2 + 4. Setting f'(x) = 0 gives x = 0 and x = 2. f''(0) = 4 > 0 (min), f''(2) = -8 < 0 (max).
Correct Answer:
B
— Minima at (2, 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)
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 of f(x) = -x^3 + 3x^2 + 1. (2021)
-
A.
(0, 1)
-
B.
(1, 3)
-
C.
(2, 5)
-
D.
(3, 4)
Solution
f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x = 0 or x = 2. f(2) = 5 is a local maximum.
Correct Answer:
B
— (1, 3)
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Q. Determine the local maxima of f(x) = x^4 - 8x^2 + 16. (2021)
-
A.
(0, 16)
-
B.
(2, 12)
-
C.
(4, 0)
-
D.
(1, 9)
Solution
Find f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 12 is a local maximum.
Correct Answer:
B
— (2, 12)
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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
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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