Continuity and Differentiability Test

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Continuity and Differentiability Test - 2

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Question 1
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A
Only in some cases.

B
Cannot be determined

C
Continuous

D
Not continuous

Solution

= 1 x 2
= 2 ≠f(0)
Hence, given function is not continuous at x = 0
Question 2
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What is/are conditions for a function to be continuous on (a,b)?

A
The function is continuous at each point of (a,b).

B
The function is right continuous.

C
The function is left continuous.

D
Right continuous, left continuous, continuous at each point of (a,b).

Solution

The three conditions required for a function f is said to be continuous on (a,b) if f is continuous at each point of (a,b), f is right continuous at x = a, f is left continuous at x = b.
Question 3
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A
f(x) is continuous at x = 2.

B
f(x) is discontinuous at x = 2.

C

D
None of these.

Solution

and

and
f(2) = 1.
Therefore , f(x) is discontinuous at x = 2
Question 4
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Which of the following functions are not continuous.

A

B

C
e^x

D

Solution

Here, f(x) = [X] could be expressed graphically as:
From the graph it is clear that the function is discontinuous.
Question 5
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Examine the continuity of the function

A
Discontinuous at x = -2.

B
Continuous everywhere.

C
Discontinuous at x = 2.

D
Discontinuous at x = 4.

Solution

Set the denominator in:
equal to 0 to find where the expression is undefined.
Subtract 2 from both sides of the equation.
The domain is all values of x that make the expression defined.
Interval Notation:
Set-Builder Notation:
Since the domain is not all real numbers, f(x) is not continuous over all real numbers.
f(x) is discontinuous at x = -2.
Question 6
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For what values of a and b, f is a continuous function.

A
a = 2,b = 0

B
a = 1,b = 0

C
a = 0,b = 2

D
a = 0,b = 0

Solution

For continuity: LHL = RHL
At x=2,
LHL: x <2 ⇒f(x) = 2*a
RHL: x ≥ 2 ⇒f(x) = 4
For continuity: LHL = RHL
⇒2a = 4 ⇒a = 2
At x = 0,
LHL: x <0 ⇒f(x) = b
RHL: x ≥0 ⇒f(x) = 0 * a
For continuity: LHL = RHL
⇒b = 0
Question 7
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Discuss the continuity of function f(x) = |x-1| + |x+1|, x R

A
Discontinuous at x = 1.

B
Discontinuous at x = ±1.

C
Continuous everywhere.

D
Discontinuous at x = -1.

Solution

f(x) = |x-1| + |x+1| , x €R

►For x ≥1,
f(x) = x-1 + x+1
f(x) = 2x

►For -1 ≤x ≤1,
f(x) = -x + 1 + x+1
f(x) = 2

►For x ≤-1,
f(x) = -x + 1 - x-1
f(x) = -2x

As the graph of f(x) shows, the function is continuous throughout its domain.
Question 8
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Examine the continuity of function f(x) = (x-1)(x-2)

A
Discontinuous at x = 1,2.

B
Discontinuous at x = 1.

C
Continuous everywhere.

D
Discontinuous at x = 2.

Solution

Lim f (x) = lim (x-1)(x-2) at x tend to k
►So it get k² -3k+2
►Now f (k) = k² -3k+2
►So f (x) =f (k) so continous at everywhere
Question 9
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What is the point of discontinuity for signum function?

A
x = 1

B
x = -1

C
x = 0

D
Function is continuous on R.

Solution

Sgn(x) = |x|/x
► for x >0, sgn(x) = 1
► for x = 0, sgn(x) = 0
► for x <0, sgn(x) = -1
► Now see graphically or theoretically sgn(x) is discontinuous at x = 0
Question 10
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Function f(x) = log x + is continuous at

A
(0,1)

B
[-1,1]

C
(0,∞)

D
(0,1]
Solution
[-1,1] cannot be continuous interval because log is not defined at 0.
The value of x cannot be greater than 1 because then the function will become complex.
(0,1) will not be considered because its continuous at 1 as well. Hence D is the correct option.