Q. For which value of a is the function f(x) = { x^2, x < 1; ax + 1, x >= 1 } continuous at x = 1?
Solution
Setting the two pieces equal at x = 1 gives 1 = a(1) + 1, leading to a = 0.
Correct Answer: C — 2
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Q. For which value of b is the function f(x) = { 2x + 1, x < 1; b, x = 1; x^2 + 1, x > 1 continuous at x = 1?
Solution
Setting the left limit (2(1) + 1 = 3) equal to the right limit (1^2 + 1 = 2), we find b = 3.
Correct Answer: B — 2
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Q. For which value of b is the function f(x) = { x^2 - 1, x < 1; b, x = 1; 3x - 2, x > 1 continuous at x = 1?
Solution
Setting limit as x approaches 1 gives b = 2 for continuity.
Correct Answer: C — 2
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Q. For which value of b is the function f(x) = { x^2 - 4, x < 2; bx + 2, x >= 2 } continuous at x = 2?
Solution
Setting the two pieces equal at x = 2 gives us 0 = 2b + 2. Solving for b gives b = -1.
Correct Answer: B — 4
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Q. For which value of b is the function f(x) = { x^3 - 3x + b, x < 1; 2x + 1, x >= 1 continuous at x = 1?
Solution
Setting 1 - 3 + b = 2 gives b = 4 for continuity.
Correct Answer: A — 0
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Q. For which value of c is the function f(x) = { 3x + c, x < 1; 2x^2, x >= 1 continuous at x = 1?
Solution
Setting 3(1) + c = 2(1)^2 gives c = -1 for continuity.
Correct Answer: B — 0
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Q. For which value of c is the function f(x) = { x^2 - 4, x < c; 3x - 5, x >= c } continuous at x = c?
Solution
Setting the two pieces equal at x = c: c^2 - 4 = 3c - 5. Solving gives c = 3.
Correct Answer: C — 3
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Q. For which value of c is the function f(x) = { x^2 - c, x < 1; 2x + 1, x >= 1 continuous at x = 1?
Solution
Setting x^2 - c = 2x + 1 at x = 1 gives c = 2.
Correct Answer: C — 2
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Q. For which value of c is the function f(x) = { x^2, x < c; 2x + 1, x >= c continuous at x = c?
Solution
Setting x^2 = 2x + 1 at x = c gives c = 2.
Correct Answer: C — 2
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Q. For which value of k is the function f(x) = kx^2 + 2x + 1 differentiable at x = -1?
Solution
f'(x) = 2kx + 2; f'(-1) = -2k + 2 must exist for any k.
Correct Answer: A — 0
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Q. For which value of m is the function f(x) = { 3x + m, x < 1; 2, x = 1; mx + 1, x > 1 continuous at x = 1?
Solution
Setting 3 + m = 2 and 2 = m + 1 gives m = 1 for continuity.
Correct Answer: B — 0
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Q. For which value of p is the function f(x) = { x^2 + p, x < 0; 3x - 1, x >= 0 } continuous at x = 0?
Solution
Setting the two pieces equal at x = 0 gives us p = -1.
Correct Answer: B — 0
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Q. If f(x) = 1/(x-1), what is the point of discontinuity?
-
A.
x = 0
-
B.
x = 1
-
C.
x = -1
-
D.
x = 2
Solution
The function is discontinuous at x = 1 because it leads to division by zero.
Correct Answer: B — x = 1
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Q. If f(x) = 2x^3 - 9x^2 + 12x, find the intervals where f(x) is increasing.
-
A.
(-∞, 1)
-
B.
(1, 3)
-
C.
(3, ∞)
-
D.
(0, 2)
Solution
Find f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. Testing intervals, f(x) is increasing on (1, 3).
Correct Answer: B — (1, 3)
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Q. If f(x) = 5x^2 + 3x, what is f'(1)?
Solution
f'(x) = 10x + 3; f'(1) = 10*1 + 3 = 13.
Correct Answer: B — 10
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Q. If f(x) = e^(2x), what is f'(x)?
-
A.
2e^(2x)
-
B.
e^(2x)
-
C.
2x*e^(2x)
-
D.
e^(x)
Solution
Using the chain rule, f'(x) = 2e^(2x).
Correct Answer: A — 2e^(2x)
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Q. If f(x) = e^x - x^2, find the x-coordinate of the local maximum.
Solution
Find f'(x) = e^x - 2x. Setting f'(x) = 0 gives a local maximum at x = 1.
Correct Answer: B — 1
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Q. If f(x) = e^x, then f'(0) is equal to?
Solution
f'(x) = e^x; f'(0) = e^0 = 1.
Correct Answer: B — 1
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Q. If f(x) = e^x, what is f''(0)?
Solution
f''(x) = e^x, thus f''(0) = e^0 = 1.
Correct Answer: A — 1
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Q. If f(x) = ln(x) + x^2, then the function is increasing for:
-
A.
x > 0
-
B.
x < 0
-
C.
x > 1
-
D.
x < 1
Solution
The derivative f'(x) = 1/x + 2x. For f'(x) > 0, we need 1/x + 2x > 0, which holds for x > 0.
Correct Answer: A — x > 0
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Q. If f(x) = ln(x) for x > 0, is f differentiable at x = 1?
-
A.
Yes
-
B.
No
-
C.
Only continuous
-
D.
Only left differentiable
Solution
f'(x) = 1/x; f'(1) = 1, hence f is differentiable at x = 1.
Correct Answer: A — Yes
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Q. If f(x) = ln(x), what is f'(x)?
-
A.
1/x
-
B.
x
-
C.
ln(x)
-
D.
0
Q. If f(x) = sin(x) + cos(x), then the critical points in the interval [0, 2π] are:
-
A.
π/4, 5π/4
-
B.
π/2, 3π/2
-
C.
0, π
-
D.
π/3, 2π/3
Solution
To find critical points, we set f'(x) = cos(x) - sin(x) = 0. This gives tan(x) = 1, leading to x = π/4 and x = 5π/4 in the interval [0, 2π].
Correct Answer: A — π/4, 5π/4
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Q. If f(x) = x^2 + 2x + 1 for x < 0 and f(x) = kx + 1 for x >= 0, find k such that f is differentiable at x = 0.
Solution
Setting the left-hand derivative equal to the right-hand derivative at x = 0 gives k = 2.
Correct Answer: A — -1
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Q. If f(x) = x^2 + 2x + 1, find f'(1).
Solution
f'(x) = 2x + 2, thus f'(1) = 2(1) + 2 = 4.
Correct Answer: C — 3
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Q. If f(x) = x^2 + 2x + 1, what is f'(1)?
Solution
Calculating the derivative f'(x) = 2x + 2, we find f'(1) = 4.
Correct Answer: B — 3
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Q. If f(x) = x^2 + 2x + 3, find f'(1).
Solution
f'(x) = 2x + 2. Therefore, f'(1) = 2(1) + 2 = 4.
Correct Answer: C — 4
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Q. If f(x) = x^2 - 4, what is the limit of f(x) as x approaches 2?
-
A.
0
-
B.
2
-
C.
4
-
D.
Undefined
Solution
The limit as x approaches 2 is f(2) = 2^2 - 4 = 0.
Correct Answer: C — 4
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Q. If f(x) = x^2 - 4x + 4, find f'(2).
Solution
f'(x) = 2x - 4. Thus, f'(2) = 2(2) - 4 = 0.
Correct Answer: A — 0
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Q. If f(x) = x^2 for x < 1 and f(x) = 2x - 1 for x ≥ 1, is f differentiable at x = 1?
-
A.
Yes
-
B.
No
-
C.
Only continuous
-
D.
Only left differentiable
Solution
f'(1) from left = 2 and from right = 2; hence f is differentiable at x = 1.
Correct Answer: B — No
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