Q. Find the solution set for the inequality 6 - 3x ≤ 0.
-
A.
x ≥ 2
-
B.
x < 2
-
C.
x ≤ 2
-
D.
x > 2
Solution
6 - 3x ≤ 0 => -3x ≤ -6 => x ≥ 2.
Correct Answer: A — x ≥ 2
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Q. Find the solution set for the inequality 7 - 3x > 1.
-
A.
x < 2
-
B.
x > 2
-
C.
x < 3
-
D.
x > 3
Solution
7 - 3x > 1 => -3x > -6 => x < 2.
Correct Answer: B — x > 2
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Q. Find the solution set for the inequality 8x + 1 ≤ 5.
-
A.
x ≤ 0.5
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B.
x < 0.5
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C.
x ≥ 0.5
-
D.
x > 0.5
Solution
8x + 1 ≤ 5 => 8x ≤ 4 => x ≤ 0.5.
Correct Answer: A — x ≤ 0.5
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Q. Find the solution set for the inequality x + 2 > 3.
-
A.
x > 1
-
B.
x < 1
-
C.
x > -1
-
D.
x < -1
Solution
x + 2 > 3 => x > 1.
Correct Answer: A — x > 1
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Q. Find the solutions of the equation 2sin(x) + √3 = 0.
-
A.
x = 5π/6
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B.
x = 7π/6
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C.
x = π/6
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D.
x = 11π/6
Solution
Solving gives sin(x) = -√3/2, so x = 7π/6 and 11π/6.
Correct Answer: B — x = 7π/6
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Q. Find the solutions of the equation 2sin(x) - 1 = 0 in the interval [0, 2π].
-
A.
π/6, 5π/6
-
B.
π/4, 3π/4
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C.
π/3, 2π/3
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D.
π/2, 3π/2
Solution
The solutions are x = π/6 and x = 5π/6.
Correct Answer: A — π/6, 5π/6
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Q. Find the solutions of the equation 2sin(x) - 1 = 0.
-
A.
π/6
-
B.
5π/6
-
C.
7π/6
-
D.
11π/6
Solution
The solutions are x = π/6 and x = 5π/6.
Correct Answer: A — π/6
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Q. Find the sum of the roots of the equation 3x^2 - 12x + 9 = 0.
Solution
The sum of the roots is given by -b/a = 12/3 = 4.
Correct Answer: B — 4
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Q. Find the unit vector in the direction of the vector (3, 4).
-
A.
(0.6, 0.8)
-
B.
(0.8, 0.6)
-
C.
(1, 1)
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D.
(0.5, 0.5)
Solution
Magnitude = √(3^2 + 4^2) = 5. Unit vector = (3/5, 4/5) = (0.6, 0.8).
Correct Answer: A — (0.6, 0.8)
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Q. Find the unit vector in the direction of the vector (3, 4, 0).
-
A.
(0.6, 0.8, 0)
-
B.
(0.3, 0.4, 0)
-
C.
(1, 1, 0)
-
D.
(0, 0, 1)
Solution
Magnitude = √(3^2 + 4^2) = 5. Unit vector = (3/5, 4/5, 0) = (0.6, 0.8, 0).
Correct Answer: A — (0.6, 0.8, 0)
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Q. Find the unit vector in the direction of the vector (4, 3).
-
A.
(4/5, 3/5)
-
B.
(3/5, 4/5)
-
C.
(1, 0)
-
D.
(0, 1)
Solution
Unit vector = (4, 3) / √(4^2 + 3^2) = (4, 3) / 5 = (4/5, 3/5).
Correct Answer: A — (4/5, 3/5)
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Q. Find the unit vector in the direction of the vector (6, 8).
-
A.
(0.6, 0.8)
-
B.
(0.8, 0.6)
-
C.
(1, 1)
-
D.
(0.5, 0.5)
Solution
Magnitude = √(6^2 + 8^2) = √(36 + 64) = √100 = 10. Unit vector = (6/10, 8/10) = (0.6, 0.8).
Correct Answer: A — (0.6, 0.8)
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Q. Find the unit vector in the direction of the vector v = (4, -3).
-
A.
(4/5, -3/5)
-
B.
(3/5, 4/5)
-
C.
(4/3, -3/4)
-
D.
(3/4, 4/3)
Solution
Magnitude |v| = √(4^2 + (-3)^2) = √(16 + 9) = 5. Unit vector = (4/5, -3/5).
Correct Answer: A — (4/5, -3/5)
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Q. Find the value of (1 + 2)^4 using the binomial theorem.
Solution
Using the binomial theorem, (1 + 2)^4 = C(4,0) * 1^4 * 2^0 + C(4,1) * 1^3 * 2^1 + C(4,2) * 1^2 * 2^2 + C(4,3) * 1^1 * 2^3 + C(4,4) * 1^0 * 2^4 = 1 + 8 + 24 + 32 + 16 = 81.
Correct Answer: A — 16
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Q. Find the value of (1 + i)^2.
Solution
(1 + i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i.
Correct Answer: B — 2
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Q. Find the value of (1 + i)^4.
Solution
(1 + i)^4 = (√2 e^(iπ/4))^4 = 4 e^(iπ) = 4(-1) = -4.
Correct Answer: C — 8
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Q. Find the value of (1 + x)^10 at x = 1. (2048)
-
A.
10
-
B.
11
-
C.
1024
-
D.
2048
Solution
Using the binomial theorem, (1 + 1)^10 = 2^10 = 1024.
Correct Answer: C — 1024
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Q. Find the value of (1 + x)^10 at x = 2.
-
A.
1024
-
B.
2048
-
C.
512
-
D.
256
Solution
Using the binomial theorem, (1 + 2)^10 = 3^10 = 59049.
Correct Answer: B — 2048
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Q. Find the value of a for which the function f(x) = { ax + 1, x < 1; 2, x = 1; x^2 + a, x > 1 is continuous at x = 1.
Solution
Setting ax + 1 = 2 and x^2 + a = 2 at x = 1 gives a = 0.
Correct Answer: A — 0
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Q. Find the value of a for which the function f(x) = { ax + 1, x < 1; 3, x = 1; 2x + a, x > 1 is continuous at x = 1.
Solution
Setting ax + 1 = 3 and 2x + a = 3 at x = 1 gives a = 2.
Correct Answer: A — 1
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Q. Find the value of a for which the function f(x) = { ax + 1, x < 2; 3x - 5, x >= 2 } is continuous at x = 2.
Solution
Setting the two pieces equal at x = 2 gives us 2a + 1 = 1. Solving for a gives a = 0.
Correct Answer: C — 3
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Q. Find the value of a for which the function f(x) = { ax + 1, x < 2; x^2 - 3, x >= 2 } is continuous at x = 2.
Solution
Setting the two pieces equal at x = 2: 2a + 1 = 2^2 - 3. Solving gives a = 2.
Correct Answer: C — 3
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Q. Find the value of a for which the function f(x) = { ax + 1, x < 2; x^2 - 4, x >= 2 } is differentiable at x = 2.
Solution
Set the left-hand limit equal to the right-hand limit and their derivatives at x = 2.
Correct Answer: B — 1
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Q. Find the value of a for which the function f(x) = { x^2 + a, x < 1; 3, x = 1; 2x + 1, x > 1 is continuous at x = 1.
Solution
Setting the left limit (1 + a) equal to the right limit (3), we find a = 2.
Correct Answer: A — -1
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Q. Find the value of b for which the function f(x) = { x^2 + b, x < 1; 2x + 3, x >= 1 is continuous at x = 1.
Solution
Setting 1 + b = 2 + 3 gives b = 4.
Correct Answer: C — 2
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Q. Find the value of b for which the function f(x) = { x^2 + b, x < 1; 3x - 1, x >= 1 is continuous at x = 1.
Solution
Setting 1 + b = 2 gives b = 1.
Correct Answer: A — -1
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Q. Find the value of c such that the function f(x) = { x^2 + c, x < 1; 2x + 1, x >= 1 } is differentiable at x = 1.
Solution
Setting the left-hand limit equal to the right-hand limit gives c = 1.
Correct Answer: B — 1
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Q. Find the value of c such that the function f(x) = { x^2 + c, x < 2; 4, x >= 2 } is continuous at x = 2.
Solution
Setting the two pieces equal at x = 2 gives 4 = 4 + c, hence c = 0.
Correct Answer: B — 2
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Q. Find the value of c such that the function f(x) = { x^3 - 3x + 2, x < 1; c, x = 1; x^2 + 1, x > 1 is continuous at x = 1.
Solution
To ensure continuity at x = 1, we set the left limit (1 - 3 + 2 = 0) equal to the right limit (1 + 1 = 2), leading to c = 2.
Correct Answer: C — 2
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Q. Find the value of c such that the function f(x) = { x^3 - 3x + 2, x < c; 4, x = c; 2x - 1, x > c is continuous at x = c.
Solution
Setting limit as x approaches c from left equal to 4 and from right gives c = 1.
Correct Answer: A — 1
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