Mathematics
Q. Find the distance between the parallel planes x + 2y + 3z = 4 and x + 2y + 3z = 10. (2023)
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Solution
Distance = |d1 - d2| / √(a² + b² + c²) = |4 - 10| / √(1² + 2² + 3²) = 6 / √14.
Correct Answer: A — 2
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Q. Find the equation of the line parallel to y = 3x + 2 and passing through (4, 5).
A.
y = 3x - 7
B.
y = 3x + 5
C.
y = 3x + 2
D.
y = 3x - 2
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Solution
Since the line is parallel, it has the same slope. Using point-slope form: y - 5 = 3(x - 4) gives y = 3x - 7.
Correct Answer: A — y = 3x - 7
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Q. Find the equation of the line parallel to y = 3x + 2 that passes through the point (4, 1).
A.
y = 3x - 11
B.
y = 3x + 1
C.
y = 3x + 2
D.
y = 3x - 2
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Solution
Since the line is parallel, it has the same slope (3). Using point-slope form: y - 1 = 3(x - 4) gives y = 3x - 11.
Correct Answer: A — y = 3x - 11
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Q. Find the equation of the line parallel to y = 5x - 2 that passes through the point (2, 3).
A.
y = 5x - 7
B.
y = 5x + 2
C.
y = 5x - 5
D.
y = 5x + 1
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Solution
Since the slope is the same (5), using point-slope form: y - 3 = 5(x - 2) gives y = 5x - 7.
Correct Answer: A — y = 5x - 7
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Q. Find the equation of the line parallel to y = 5x - 3 that passes through the point (2, 1).
A.
y = 5x - 9
B.
y = 5x + 1
C.
y = 5x - 7
D.
y = 5x + 3
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Solution
Since the slope is the same (5), using point-slope form: y - 1 = 5(x - 2) gives y = 5x - 9.
Correct Answer: A — y = 5x - 9
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Q. Find the equation of the line that passes through (0, 0) and has a slope of 5.
A.
y = 5x
B.
y = x/5
C.
y = 5/x
D.
y = 1/5x
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Solution
Using the slope-intercept form y = mx + b, with m = 5 and b = 0, we get y = 5x.
Correct Answer: A — y = 5x
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Q. Find the equation of the line that passes through the origin and has a slope of -3.
A.
y = -3x
B.
y = 3x
C.
y = -x/3
D.
y = 1/3x
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Solution
Using the slope-intercept form, the equation is y = -3x.
Correct Answer: A — y = -3x
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Q. Find the equation of the line that passes through the point (4, -1) and is perpendicular to the line y = 3x + 2.
A.
y = -1/3x + 5/3
B.
y = 3x - 13
C.
y = -3x + 11
D.
y = 1/3x - 5/3
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Solution
The slope of the given line is 3, so the slope of the perpendicular line is -1/3. Using point-slope form, we get y + 1 = -1/3(x - 4), which simplifies to y = -1/3x + 11/3.
Correct Answer: C — y = -3x + 11
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Q. Find the equation of the line that passes through the point (4, 5) and is perpendicular to the line y = 1/3x + 2.
A.
y = -3x + 17
B.
y = 3x - 7
C.
y = -3x + 5
D.
y = 1/3x + 5
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Solution
The slope of the given line is 1/3, so the slope of the perpendicular line is -3. Using point-slope form, we get y - 5 = -3(x - 4), which simplifies to y = -3x + 17.
Correct Answer: A — y = -3x + 17
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Q. Find the equation of the line that passes through the points (2, 3) and (4, 7).
A.
y = 2x - 1
B.
y = 2x + 1
C.
y = 3x - 3
D.
y = x + 1
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Solution
The slope m = (7 - 3) / (4 - 2) = 2. Using point-slope form: y - 3 = 2(x - 2) gives y = 2x + 1.
Correct Answer: B — y = 2x + 1
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Q. Find the general solution of dy/dx = 3x^2. (2020)
A.
y = x^3 + C
B.
y = 3x^3 + C
C.
y = x^2 + C
D.
y = 3x + C
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Solution
Integrating 3x^2 gives y = x^3 + C.
Correct Answer: A — y = x^3 + C
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Q. Find the general solution of the equation y' = 3x^2y.
A.
y = Ce^(x^3)
B.
y = Ce^(3x^3)
C.
y = C/x^3
D.
y = Cx^3
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Solution
This is a separable equation. Integrating gives y = Ce^(x^3).
Correct Answer: A — y = Ce^(x^3)
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Q. Find the integral of (2x + 1)^3 dx. (2019)
A.
(1/4)(2x + 1)^4 + C
B.
(1/3)(2x + 1)^4 + C
C.
(1/5)(2x + 1)^4 + C
D.
(1/2)(2x + 1)^4 + C
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Solution
Using substitution, the integral is (1/4)(2x + 1)^4 + C.
Correct Answer: A — (1/4)(2x + 1)^4 + C
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Q. Find the integral of (2x + 3)dx. (2022)
A.
x^2 + 3x + C
B.
x^2 + 3x + 1
C.
x^2 + 3 + C
D.
2x^2 + 3x + C
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Solution
Integrating term by term: ∫2xdx = x^2 and ∫3dx = 3x. Thus, ∫(2x + 3)dx = x^2 + 3x + C.
Correct Answer: A — x^2 + 3x + C
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Q. Find the integral of cos(2x)dx. (2023)
A.
(1/2)sin(2x) + C
B.
sin(2x) + C
C.
(1/2)cos(2x) + C
D.
2sin(2x) + C
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Solution
The integral of cos(kx) is (1/k)sin(kx) + C. Here, k=2, so the integral is (1/2)sin(2x) + C.
Correct Answer: A — (1/2)sin(2x) + C
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Q. Find the integral of cos(x). (2023)
A.
sin(x) + C
B.
-sin(x) + C
C.
cos(x) + C
D.
-cos(x) + C
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Solution
The integral of cos(x) is sin(x) + C.
Correct Answer: A — sin(x) + C
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Q. Find the integral of cos(x)dx. (2023)
A.
sin(x) + C
B.
-sin(x) + C
C.
cos(x) + C
D.
-cos(x) + C
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Solution
The integral of cos(x) is sin(x) + C.
Correct Answer: A — sin(x) + C
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Q. Find the integral of e^x dx. (2022)
A.
e^x + C
B.
e^x
C.
x e^x + C
D.
ln(e^x) + C
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Solution
The integral of e^x is e^x + C.
Correct Answer: A — e^x + C
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Q. Find the integral of sin(x). (2020)
A.
-cos(x) + C
B.
cos(x) + C
C.
sin(x) + C
D.
-sin(x) + C
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Solution
The integral of sin(x) is -cos(x) + C.
Correct Answer: A — -cos(x) + C
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Q. Find the integral of sin(x)dx. (2020)
A.
-cos(x) + C
B.
cos(x) + C
C.
sin(x) + C
D.
-sin(x) + C
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Solution
The integral of sin(x) is -cos(x) + C.
Correct Answer: A — -cos(x) + C
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Q. Find the integral of x^5 dx. (2020)
A.
(1/6)x^6 + C
B.
(1/5)x^6 + C
C.
(1/4)x^6 + C
D.
(1/7)x^6 + C
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Solution
The integral is (1/6)x^6 + C.
Correct Answer: B — (1/5)x^6 + C
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Q. Find the length of the diagonal of a rectangular box with dimensions 2, 3, and 6 units. (2022)
A.
√49
B.
√45
C.
√36
D.
√50
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Solution
Diagonal = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7 units.
Correct Answer: A — √49
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Q. Find the length of the diagonal of a rectangular box with dimensions 2, 3, and 6. (2023)
A.
√49
B.
√36
C.
√45
D.
√50
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Solution
Diagonal = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7.
Correct Answer: A — √49
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Q. Find the limit: lim (x -> 1) (x^4 - 1)/(x - 1) (2023)
A.
0
B.
1
C.
4
D.
Undefined
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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. Thus, lim (x -> 1) = 4.
Correct Answer: A — 0
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Q. Find the limit: lim (x -> 3) (x^2 - 9)/(x - 3) (2023)
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Solution
The expression can be factored as ((x - 3)(x + 3))/(x - 3). For x ≠ 3, this simplifies to x + 3. Thus, lim (x -> 3) (x + 3) = 6.
Correct Answer: A — 0
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Q. Find the magnitude of the vector A = 3i - 4j. (2020)
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Solution
|A| = √(3^2 + (-4)^2) = √(9 + 16) = √25 = 5.
Correct Answer: A — 5
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Q. Find the maximum area of a triangle with a base of 10 m and height varying. (2020)
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Solution
Area = 1/2 * base * height. Max area occurs when height is maximized, thus Area = 1/2 * 10 * 10 = 50.
Correct Answer: B — 50
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Q. Find the maximum area of a triangle with a base of 10 units and height as a function of the base. (2021)
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Solution
Area = 1/2 * base * height. Max area occurs when height is maximized at 10 units, giving Area = 50.
Correct Answer: B — 50
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Q. Find the maximum height of the projectile modeled by h(t) = -16t^2 + 32t + 48. (2020)
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Solution
The maximum occurs at t = -b/(2a) = -32/(2*-16) = 1. h(1) = 64.
Correct Answer: A — 48
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Q. Find the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48. (2020)
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Solution
The maximum occurs at t = -b/(2a) = 64/(2*16) = 2. h(2) = -16(2^2) + 64(2) + 48 = 80.
Correct Answer: B — 64
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