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 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 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 maximum value of f(x) = -x^2 + 4x + 1.
Solution
The vertex occurs at x = 2. f(2) = -2^2 + 4(2) + 1 = 5, which is the maximum value.
Correct Answer: B — 5
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Q. Determine the minimum value of the function f(x) = x^2 - 4x + 5.
Solution
The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1. Thus, the minimum value is 1.
Correct Answer: A — 1
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Q. Determine the moment of inertia of a solid sphere of mass M and radius R about an axis through its center.
-
A.
2/5 MR^2
-
B.
3/5 MR^2
-
C.
4/5 MR^2
-
D.
MR^2
Solution
The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.
Correct Answer: A — 2/5 MR^2
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Q. Determine the nature of the lines represented by the equation 7x^2 + 2xy + 3y^2 = 0.
-
A.
Parallel
-
B.
Intersecting
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C.
Coincident
-
D.
Perpendicular
Solution
The discriminant indicates that the lines intersect at two distinct points.
Correct Answer: B — Intersecting
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Q. Determine the point at which the function f(x) = x^3 - 3x^2 + 4 has a local minimum.
-
A.
(1, 2)
-
B.
(2, 1)
-
C.
(0, 4)
-
D.
(3, 4)
Solution
Find f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(x - 2) = 0, so x = 0 or x = 2. f''(2) = 6 > 0, so (2, 1) is a local minimum.
Correct Answer: A — (1, 2)
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Q. Determine the point at which the function f(x) = |x - 1| is not differentiable.
-
A.
x = 0
-
B.
x = 1
-
C.
x = 2
-
D.
x = -1
Solution
The function |x - 1| is not differentiable at x = 1 due to a cusp.
Correct Answer: B — x = 1
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Q. Determine the point at which the function f(x) = |x - 3| is not differentiable.
-
A.
x = 1
-
B.
x = 2
-
C.
x = 3
-
D.
x = 4
Solution
The function f(x) = |x - 3| is not differentiable at x = 3 because it has a sharp corner.
Correct Answer: C — x = 3
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Q. Determine the point at which the function f(x) = |x^2 - 4| is differentiable.
-
A.
x = -2
-
B.
x = 0
-
C.
x = 2
-
D.
x = -4
Solution
f(x) is not differentiable at x = -2 and x = 2, but is differentiable everywhere else.
Correct Answer: A — x = -2
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Q. Determine the point of inflection for the function f(x) = x^4 - 4x^3 + 6.
-
A.
(1, 3)
-
B.
(2, 2)
-
C.
(3, 1)
-
D.
(0, 6)
Solution
f''(x) = 12x^2 - 24x. Setting f''(x) = 0 gives x = 0 and x = 2. The point of inflection is at (1, 3).
Correct Answer: A — (1, 3)
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Q. Determine the point of inflection for the function f(x) = x^4 - 4x^3 + 6x^2.
-
A.
(1, 3)
-
B.
(2, 2)
-
C.
(3, 1)
-
D.
(0, 0)
Solution
Find f''(x) = 12x^2 - 24x + 12. Setting f''(x) = 0 gives x = 1 and x = 2. Testing intervals shows a change in concavity at x = 1.
Correct Answer: A — (1, 3)
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Q. Determine the point of intersection of the lines y = 2x + 1 and y = -x + 4.
-
A.
(1, 3)
-
B.
(2, 5)
-
C.
(3, 7)
-
D.
(4, 9)
Solution
Setting 2x + 1 = -x + 4 gives 3x = 3, hence x = 1. Substituting back gives y = 3.
Correct Answer: A — (1, 3)
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Q. Determine the points where f(x) = x^3 - 3x is not differentiable.
-
A.
x = 0
-
B.
x = 1
-
C.
x = -1
-
D.
Nowhere
Solution
The function is a polynomial and is differentiable everywhere, hence nowhere.
Correct Answer: D — Nowhere
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Q. Determine the points where the function f(x) = x^4 - 4x^3 is not differentiable.
-
A.
x = 0
-
B.
x = 1
-
C.
x = 2
-
D.
None
Solution
The function is a polynomial and is differentiable everywhere. Thus, there are no points where it is not differentiable.
Correct Answer: D — None
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Q. Determine the scalar product of the vectors (0, 1, 2) and (3, 4, 5).
Solution
Scalar product = 0*3 + 1*4 + 2*5 = 0 + 4 + 10 = 14.
Correct Answer: B — 11
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Q. Determine the scalar product of the vectors A = (1, 1, 1) and B = (2, 2, 2).
Solution
A · B = 1*2 + 1*2 + 1*2 = 2 + 2 + 2 = 6.
Correct Answer: C — 6
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Q. Determine the scalar product of the vectors A = (2, 2, 2) and B = (3, 3, 3).
Solution
A · B = 2*3 + 2*3 + 2*3 = 6 + 6 + 6 = 18.
Correct Answer: A — 12
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Q. Determine the solution for the inequality -2x + 6 > 0.
-
A.
x < 3
-
B.
x > 3
-
C.
x < -3
-
D.
x > -3
Solution
-2x + 6 > 0 => -2x > -6 => x < 3.
Correct Answer: B — x > 3
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Q. Determine the solution for the inequality -2x + 6 ≥ 0.
-
A.
x ≤ 3
-
B.
x ≥ 3
-
C.
x ≤ -3
-
D.
x ≥ -3
Solution
-2x + 6 ≥ 0 => -2x ≥ -6 => x ≤ 3.
Correct Answer: B — x ≥ 3
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Q. Determine the solution for the inequality -3x + 1 ≤ 4.
-
A.
x ≥ -1
-
B.
x ≤ -1
-
C.
x ≥ 1
-
D.
x ≤ 1
Solution
-3x + 1 ≤ 4 => -3x ≤ 3 => x ≥ -1.
Correct Answer: B — x ≤ -1
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Q. Determine the solution for the inequality -3x + 4 ≤ 1.
-
A.
x ≥ 1
-
B.
x ≤ 1
-
C.
x ≥ -1
-
D.
x ≤ -1
Solution
-3x + 4 ≤ 1 => -3x ≤ -3 => x ≥ 1.
Correct Answer: B — x ≤ 1
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Q. Determine the solution for the inequality 2x + 3 ≤ 7.
-
A.
x ≤ 2
-
B.
x ≥ 2
-
C.
x ≤ 3
-
D.
x ≥ 3
Solution
2x + 3 ≤ 7 => 2x ≤ 4 => x ≤ 2.
Correct Answer: A — x ≤ 2
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Q. Determine the solution for the inequality 6 - x > 2.
-
A.
x < 4
-
B.
x > 4
-
C.
x < 6
-
D.
x > 6
Solution
6 - x > 2 => -x > -4 => x < 4.
Correct Answer: A — x < 4
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Q. Determine the solution for the inequality 6 - x ≤ 3.
-
A.
x ≥ 3
-
B.
x ≤ 3
-
C.
x ≥ 6
-
D.
x ≤ 6
Solution
6 - x ≤ 3 => -x ≤ -3 => x ≥ 3.
Correct Answer: B — x ≤ 3
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Q. Determine the solution for the inequality 7 - 3x < 1.
-
A.
x < 2
-
B.
x > 2
-
C.
x ≤ 2
-
D.
x ≥ 2
Solution
7 - 3x < 1 => -3x < -6 => x > 2.
Correct Answer: A — x < 2
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Q. Determine the solution for the inequality 7x - 2 ≤ 5x + 6.
-
A.
x ≤ 4
-
B.
x ≥ 4
-
C.
x ≤ 3
-
D.
x ≥ 3
Solution
7x - 2 ≤ 5x + 6 => 2x ≤ 8 => x ≤ 4.
Correct Answer: A — x ≤ 4
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Q. Determine the solution set for the inequality 2(x - 1) ≥ 3.
-
A.
x ≤ 2
-
B.
x ≥ 2
-
C.
x ≤ 3
-
D.
x ≥ 3
Solution
2(x - 1) ≥ 3 => x - 1 ≥ 1.5 => x ≥ 2.
Correct Answer: B — x ≥ 2
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