Continuity of a function MCQ Quiz in தமிழ் - Objective Question with Answer for Continuity of a function - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

For a function say f,  exists

, where l is a finite value.

Any function say f is said to be continuous at point say a if and only if , where l is a finite value.

Calculation:

Given: 

Here, we have to find the value of x for which f(x) is not continuous.

So, if any function is not continuous at x = a then 

So, for the function f(x) if denominator is 0 at x = a then we can say that f(a) is infinite and limit cannot exist.

Let's find the value of x for which the denominator of f(x) is 0.

⇒ x² + 5x - 6 = 0

⇒ (x + 6) (x - 1) = 0.

⇒ x = -6, 1.

Hence, option 3 is correct.

Concept:

Continuity:

For a function say f,  exists

where l is a finite value.

Any function say f is said to be continuous at a point say 'a', if and only if:

where l is a finite value.

Calculation:

On substituting h = x – 2, we get:

This can be rearranged as:

= 1 ⋅ 1 ⋅ 1

= 1 and f(2) = k

∴ For the function to be continuous the value of the function f(x) at x = 2 must equal the limiting value of 1, i.e. k = 1

Concept:

Limit exists if LHL = RHL ⇔  

Calculation:

1. 

⇒ -1 ≤ sin  ≤ 1

⇒ -1 ≤  ≤ 1

Here, LHL ≠ RHL, so limit doesn't exist.

2. We have, 

Let, x = 1/t, if x → 0, t → 1/x = ∞ 

        (∵ something divide by ∞ = 0)

Here, LHL = RHL, so the limit exists

Hence, option (3) is correct.

The correct answer is option 4.

Given: 

Calculation:

⇒ 

For the value of x = 2

The function f(2) =  = 0

For the value of x = 0; f(0) =  = Impossible value

For the value of x = -2; f(-2) =  = 2

So, the function has some definite solution for all the values of x except x = 0.

Hence, the function is a continuous function for all the values of x except x = 0. 

Concept:

Differentiable Functions:

• If a graph has a sharp corner at a point, then the function is not differentiable at that point.
• If a graph has a break at a point, then the function is not differentiable at that point.
• If a graph has a vertical tangent line at a point, then the function is not differentiable at that point.

Calculation:

Givne that,

f(x) = |x| - 1     ----(1)

Step: 1

Step: 2 

Step: 3 

Clearly, we can see that,

f(x) is continuous at x = 1 and

f(x) is not differentiable at x = 0 because there is a corner at x = 0.

∴ Only statement 1 is correct.

Additional Information 

• Differentiable functions are those functions whose derivatives exist.
• If a function is differentiable, then it is continuous.
• If a function is continuous, then it is not necessarily differentiable.
• The graph of a differentiable function does not have breaks, corners, or cusps.

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. 

II. 

Then we can apply L-Hospital Rule ⇔  

Note: We have to differentiate both the numerator and denominator with respect to x unless and until  where l is a finite value.

A function f(x) is said to be continuous at a point x = a, in its domain if  exists 

Calculation:

Given: 

To Find: f(0)

Function is continuous at x = 0

Therefore, f(0) = 

          (Form 0/0)

Apply L-Hospital Rule,

Concept:

f(x) is Continuous at x = a, if  OR if derivative of function is defined 

If f(x) = |x| ⇔ f(x) = -x, for x  0,

Calculation:

1 f(x) = ex

The derivative of the function is ex and it is defined everywhere from negative infinity to positive infinity and it doesn't take zero value at any point. So, ex is a continuous function.

2.

So, g(x) is continuous

Hence, option Both 1 and 2 correct.

Explanation:

Here, it is given that

Continuity at x = a, f(a) = 0

Hence, LHL =  f(x) =  f(a - h) =  

and RHL =  f(x) =  

Hence, LHL = f(a) = RHL

Hence, f(x) is continuous.

Hence, f(x) is continuous on ] 0, ∞[

Now, we can check the differentiability at x = a

Lf'(a) =  

and 

Hence, Lf' (a) = Rf' (a)

Hence, f(x) is differentiable at x = a.

Therefore, f(x) is differentiable at (0, ∞).

Concept:

• A function f(x) is continuous at x = a if f(x)= f(x)= f(a)
•  =1

Explanation:

f(x)=0\end{array}\right.\)

Function f(x) is continuous at x= 0 if 

 f(x) = f(x) = f(0)  ---- (1)

Now f(x)=  

 =  +  

=  + 1   ---- (2)

f(x)=  

= × 

=     ----- [∵ (a+b) (a- b) = a² - b²)]

=  =  =  ----- (3)

f(0) = b  ----- (4)

From equations 1, 2, 3, and 4 we get 

⇒ b =  and  

⇒  = ⇒ a = -2

∴ a = -2 and b = 

so a +  b = 

The correct option is option (4).

Concept:

The function f(x) is continuous at an if all the below condition follows:

1. f(a) is real and finite.

2.  are real and finite.

3. 

Calculation:

f(x) =  2} \end{array}} \right.\) 

For x = 2,

⇒ 

⇒ 

For x = 2,

⇒ 

⇒ 

f(x) = f(2) = 1

∴ The function is not continuous at x = 2