Continuity in Education for Competitive Exam Success

Continuity

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.

A. 0
B. 1
C. 2
D. 3

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.

A. -1
B. 0
C. 1
D. 2

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 < 2; 4, x >= 2 } is continuous at x = 2.

A. 0
B. 2
C. 4
D. 6

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.

A. 0
B. 1
C. 2
D. 3

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.

A. 1
B. 2
C. 3
D. 4

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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Q. Find the value of k such that the function f(x) = { kx + 1, x < 1; 2x - 1, x >= 1 } is continuous at x = 1.

A. 0
B. 1
C. 2
D. 3

Solution

Setting k(1) + 1 = 2(1) - 1 gives k + 1 = 1, so k = 0.

Correct Answer: B — 1

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Q. Find the value of k such that the function f(x) = { kx + 1, x < 1; 3, x = 1; x^2 + 1, x > 1 is continuous at x = 1.

A. 1
B. 2
C. 3
D. 4

Solution

Setting k(1) + 1 = 3 gives k = 2 for continuity.

Correct Answer: C — 3

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Q. Find the value of k such that the function f(x) = { kx + 1, x < 2; x^2 - 3, x >= 2 } is continuous at x = 2.

A. 1
B. 2
C. 3
D. 4

Solution

Setting k(2) + 1 = 2^2 - 3 gives 2k + 1 = 1, leading to k = 0.

Correct Answer: B — 2

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Q. Find the value of k such that the function f(x) = { kx + 2, x < 1; 3, x = 1; 2x + 1, x > 1 } is continuous at x = 1.

A. 1
B. 2
C. 3
D. 4

Solution

Setting k(1) + 2 = 3 gives k = 1.

Correct Answer: A — 1

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Q. Find the value of k such that the function f(x) = { kx, x < 0; 0, x = 0; x^2 + k, x > 0 is continuous at x = 0.

A. -1
B. 0
C. 1
D. 2

Solution

Setting k = 0 for continuity at x = 0 gives f(0) = 0.

Correct Answer: B — 0

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Q. Find the value of k such that the function f(x) = { kx, x < 0; x^2 + 1, x >= 0 is continuous at x = 0.

A. 0
B. 1
C. 2
D. 3

Solution

Setting k(0) = 0^2 + 1 gives k = 1.

Correct Answer: A — 0

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Q. Find the value of m for which the function f(x) = { 2x + m, x < 1; mx + 3, x >= 1 is continuous at x = 1.

A. 1
B. 2
C. 3
D. 4

Solution

Setting 2(1) + m = m(1) + 3 gives m = 1.

Correct Answer: B — 2

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Q. Find the value of m for which the function f(x) = { 2x + m, x < 1; x^2 + 1, x >= 1 is continuous at x = 1.

A. 0
B. 1
C. 2
D. 3

Solution

Setting 2(1) + m = 1^2 + 1 gives m = 0.

Correct Answer: A — 0

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Q. Find the value of m for which the function f(x) = { 2x + m, x < 3; x^2 - 3, x >= 3 } is continuous at x = 3.

A. 0
B. 1
C. 2
D. 3

Solution

Setting the two pieces equal at x = 3 gives us 6 + m = 6. Thus, m = 0.

Correct Answer: C — 2

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Q. Find the value of m for which the function f(x) = { 3x + m, x < 1; 2x^2, x >= 1 is continuous at x = 1.

A. -1
B. 0
C. 1
D. 2

Solution

Setting 3(1) + m = 2(1)^2 gives m = -1.

Correct Answer: B — 0

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Q. Find the value of m such that the function f(x) = { x^2 + m, x < 1; 4 - x, x >= 1 } is continuous at x = 1.

A. 0
B. 1
C. 2
D. 3

Solution

Setting the two pieces equal at x = 1: 1^2 + m = 4 - 1. Solving gives m = 2.

Correct Answer: B — 1

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Q. Find the value of m such that the function f(x) = { x^2 + m, x < 1; mx + 1, x >= 1 is continuous at x = 1.

A. 0
B. 1
C. 2
D. 3

Solution

Setting x^2 + m = mx + 1 at x = 1 gives m = 1 for continuity.

Correct Answer: A — 0

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Q. For the function f(x) = { x^2 + 1, x < 0; 2x + b, x = 0; 3 - x, x > 0 to be continuous at x = 0, what is b?

A. -1
B. 0
C. 1
D. 2

Solution

Setting the left limit (0 + 1 = 1) equal to the right limit (3 - 0 = 3), we find b = 1.

Correct Answer: B — 0

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Q. For the function f(x) = { x^2, x < 1; 3, x = 1; 2x, x > 1 }, what is the value of f(1)?

A. 1
B. 2
C. 3
D. 4

Solution

By definition, f(1) = 3, as given in the piecewise function.

Correct Answer: C — 3

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Q. For the function f(x) = { x^2, x < 3; 9, x = 3; 3x, x > 3 } to be continuous at x = 3, the value of f(3) must be:

A. 6
B. 9
C. 3
D. 12

Solution

For continuity, f(3) must equal the limit as x approaches 3, which is 9.

Correct Answer: B — 9

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Q. For what value of b is the function f(x) = { x^3 - 3x + b, x < 1; 2x + 1, x >= 1 continuous at x = 1?

A. 0
B. 1
C. 2
D. 3

Solution

Setting 1^3 - 3(1) + b = 2(1) + 1 gives b = 2.

Correct Answer: B — 1

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Q. For which value of a is the function f(x) = { 2x + a, x < 0; x^2 + 1, x >= 0 continuous at x = 0?

A. -1
B. 0
C. 1
D. 2

Solution

Setting a = 1 gives continuity at x = 0.

Correct Answer: B — 0

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Q. For which value of a is the function f(x) = { 3x + a, x < 2; 4x - 1, x >= 2 continuous at x = 2?

A. -1
B. 0
C. 1
D. 2

Solution

Setting 3(2) + a = 4(2) - 1 gives a = 1.

Correct Answer: A — -1

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Q. For which value of a is the function f(x) = { ax + 1, x < 0; 2, x = 0; 3x - 1, x > 0 } continuous at x = 0?

A. -1
B. 0
C. 1
D. 2

Solution

Setting ax + 1 = 2 at x = 0 gives a = 2.

Correct Answer: A — -1

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Q. For which value of a is the function f(x) = { ax + 1, x < 0; 2x + a, x >= 0 } continuous at x = 0?

A. -1
B. 0
C. 1
D. 2

Solution

Setting the two pieces equal at x = 0 gives 1 = a, hence a = 1.

Correct Answer: A — -1

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Q. For which value of a is the function f(x) = { ax + 2, x < 1; 3, x >= 1 } continuous at x = 1?

A. 1
B. 2
C. 3
D. 4

Solution

Setting ax + 2 = 3 at x = 1 gives a = 1.

Correct Answer: B — 2

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Q. For which value of a is the function f(x) = { x^2 + a, x < 1; 3, x >= 1 } continuous at x = 1?

A. 1
B. 2
C. 3
D. 0

Solution

To ensure continuity at x = 1, we set limit as x approaches 1 from left (1 + a) equal to f(1) = 3, thus a = 2.

Correct Answer: B — 2

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Q. For which value of a is the function f(x) = { x^2 - a, x < 0; 2x + 1, x >= 0 } continuous at x = 0?

A. -1
B. 0
C. 1
D. 2

Solution

Setting the two pieces equal at x = 0 gives -a = 1, so a = -1.

Correct Answer: A — -1

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Q. For which value of a is the function f(x) = { x^2 - a, x < 1; 3x - 2, x >= 1 } continuous at x = 1?

A. 0
B. 1
C. 2
D. 3

Solution

Setting the two pieces equal at x = 1 gives 1 - a = 1. Thus, a = 0.

Correct Answer: C — 2

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Q. For which value of a is the function f(x) = { x^2, x < 1; ax + 1, x >= 1 } continuous at x = 1?

A. 0
B. 1
C. 2
D. 3

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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Showing 31 to 60 of 124 (5 Pages)

Continuity

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