At a point MCQ Quiz in தமிழ் - Objective Question with Answer for At a point - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Definition:

• A function f(x) is said to be continuous at a point x = a in its domain, if  exists or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ .

Calculation:

For x ≠ 0, the given function can be re-written as:

Since the equation of the function is same for x 0, we have:

For the function to be continuous at x = 0, we must have:

⇒ K = .

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:  is continuous at x =3.

AS we know that, if a function f is continuous at point say a then 

⇒ LHL = lim_x→3^- (5x - 9) = (5 ⋅ 3) - 9 = 6.

⇒ RHL =  limx→3+ (bx - 3) = 3b - 3.

As function is continuous at x = 3 so, LHL = RHL.

⇒ 3b - 3 = 6 ⇒ 3b = 9

⇒ b = 3

Hence, option 2 is correct.

Concept:

A function f(x) is continuous at a certain point x = a only if it follows the following conditions. 

• f(a) exists
•  exists
•  = f(a)

The second condition can also be written as 

All polynomial functions are continuous functions.

Calculation:

Statement I: f(x) is continuous only and only if  exists.

We know that function f(x) is continuous at a certain point x = a only if it follows the following conditions. 

• f(a) exists
•  exists
•  = f(a)

The second condition can also be written as 

So, Statement I is incorrect.

Statement II: f(x) = x3 - x2 + 3x + 6 is not continuous at x = 0.

All polynomial functions are continuous functions.

So, Statement II is incorrect.

∴ None of the statements are correct.

Concept:

Graph of f(x) = a^-x , a > 1

• Domain: 
• Range: 
• y-intercept: (0, 1)
• Decreasing
• Continuous

Solution:

Statement I: The function f(x) = 2-x continuous at x = 0

Graph of 2^-x is

By graph the function f(x) = 2^-x is continuous at x = 0

∴ Statement I is correct.

Statement II: The function f(x) = \(\frac{5}{x^{4}-16}\) is continuous at all points of its domain.

Given function is f(x) = 

Domain of the function is 

Given function is continuous at all points of its domain.

∴ Statement II is correct.

So, The correct option is (3)

Concept:

If f(x) =  such that, 

,

{ form }

then, by L's hospital Rule,

Calculation:

Given, 

{  form}

By L's Hospital Rule,

⇒ 

⇒ 

⇒ 

∴ The correct answer is option (4).

Concept:

Let y = f(x) be a function. Then,

The function is continuous if it satisfies the following conditions:

Calculation:

 f(x) = 

Since, f(x) is continuous at x  =1 

⇒  lim_x→112x + 3λ = 0

⇒ 12(1) + 3λ = 0 ⇒ 12 + 3λ = 0 

⇒ λ = -4

∴ Option 1 is correct">limx→a−⁡f(x)=limx→a+⁡f(x)=f(a)" id="MathJax-Element-3-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">

Concept:

The function f(x) is continuous at x = a if

 f(a^-) = f(a) = f(a⁺)

Calculation:

Given, 

⇒ f(0^-) = 2(0) = 0

f(0) = 2(0) + 1 = 1

and f(0⁺) = 2(0) + 1 = 1

So, f(0-) ≠ f(0) = f(0+)

⇒ f(x) is discontinuous at x = 0.

∴ The correct answer is option (3).

Concept:

• If, , then,  exists or the function is continuous at x = a.

• If, , then,  does not exist or the function is discontinuous at x = a.

Calculation:

LHL of f(x) at  x = 0 = 

⇒ 1/5

RHL of f(x) at  x = 0 = 

\( ⇒\lim_{h\rightarrow 0}f(0+h)\)

⇒ -1/5 

Hence, LHL ≠ RHL

⇒  does not exist.

Hence, f(x) is not continuous at x = 0 because  does not exist.

Concept:

Let y = f(x) be a function. Then for a function, we say,

The function is continuous if it satisfies the following conditions.

Calculation:

Given that,

           ----(1)

Statement: 1

⇒      ----(2)

We know that, 

sin θ ∈  [ -1 , 1] 

⇒ sin (∞) is a definite value.

Similarly,

   ----(3)

According to the question,

f(0) = 0        ----(4)

From equations (2), (3), (4),

Statement 1 is incorrect.

Statement: 2

⇒ LHL =     ----(5)

Similarly, 

RHL =         -----(6)

And,

f() = sin 

⇒ f() =        ----(7)

From equation (5), (6), (7)

We can say that f(x) is continuous at x = 

∴  Statement 2 is incorrect.

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: y = |2x - 4| 

⇒ 

⇒ LHL=  lim_x→2^- (4 - 2x) = 4 - 2 ⋅ 2 = 0.

⇒ RHL =  limx→2+ (2x - 4) = 2 ⋅ 2 - 4 = 0.

Hence limit exists at x = 2

y(2) = 2x - 4 = 2 ⋅ 2 - 4 = 0

⇒ LHL = RHL = y(2)

Hence function is continuous at x = 2.

Hence, option 1 is correct.