Lagrange's Mean Value Theorem MCQ Quiz - Objective Question with Answer for Lagrange's Mean Value Theorem - Download Free PDF

Concept:

Lagrange’s Mean Value Theorem:
If a function f is defined on the closed interval [a, b] satisfying:

• The function f is continuous on the closed interval [a, b].
• The function f is differentiable on the open interval (a, b).

Then there exists a value x = c such that f'(c) = .

Calculation:

The given function f(x) =  is both differentiable and continuous in the interval [1, 3].

f'(x) = 

By the Mean Value Theorem, there exists a c ∈ [1, 3], such that:

f'(c) = 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ c = √3.

Concept

Mean Value Theorem: If a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number C in the interval (a, b) such that:

Calculation

Given: ,

Here, ,

Now,

Since , we take the positive value.

Hence option 1 is correct.

Calculation

Given:

For all ,  is differentiable.

 and .

By the Mean Value Theorem, there exists such that

⇒

⇒

Since , we have .

⇒

⇒

⇒

∴ The least possible value of is 10118.

Hence option 2 is correct

Concept Used:

Lagrange's Mean Value Theorem (LMVT): If f(x) is continuous on [a, b] and differentiable on (a, b), then there exists a c in (a, b) such that

Calculation

f(x) = log x

⇒ f'(x) = 1/x

a = 1, b = e

f(a) = f(1) = log 1 = 0

f(b) = f(e) = log e = 1

⇒

⇒

⇒ c = e - 1

∴ The value of c is e - 1.

Hence option 3 is correct

Calculation:

Using Lagrange's mean value theorem,

f '(c) = 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 16c² = 24(25 - c²)

⇒ 16c² + 24c² - 600 = 0

⇒ 40c² = 600

⇒ c² = 15

⇒ c = ± 

⇒ c =  {∵ -  ∈ [1, 5]} 

∴ The value of c is .

The correct answer is Option 1.

Concept:

Lagrange’s Mean Value Theorem:
If a function f is defined on the closed interval [a, b] satisfying:

• The function f is continuous on the closed interval [a, b].
• The function f is differentiable on the open interval (a, b).

Then there exists a value x = c such that f'(c) = .

Calculation:

The given function f(x) =  is both differentiable and continuous in the interval [1, 3].

f'(x) = 

By the Mean Value Theorem, there exists a c ∈ [1, 3], such that:

f'(c) = 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ c = √3.

Concept:

Mean value theorem states that if f(x) is continuous over the interval [a, b] and differentiable over the interval (a, b), then there exists at least one c in (a, b) such that f'(c) = 

Calculation:

Given: f(x) = x(x - 2), x ϵ [1, 2]

f'(x) = x - 2 + x

f'(x) = 2x - 2

f'(x) =2(x - 1)

f'(c) = 2(c - 1)

The mean value theorem states that there exists at least one c in (a, b) such that f'(c) = 

Here b = 2 ⇒  f(2) = 2(2 - 2) = 0

a = 1 ⇒  f(1) = 1(1 - 2) = - 1

⇒ 2 (c - 1) = 

⇒ 2(c - 1) = 1

⇒ c - 1 = 

∴ c =  ∈ (1, 2)

The correct answer is 3/2.

Concept:

Mean value theorem states that if f(x) is continuous over the interval [a, b] and differentiable over the interval (a, b) then there exists at least one c in (a, b) such that

f'(c) = 

Calculation:

Given: f(x) = x + , x ϵ [1, 3]

 f'(x) = 1 -   

The mean value theorem states there exists at least one c in (a, b) such that

f'(c) = 

Here b = 3 ⇒ f(3) = 3 +  = 

a = 1 ⇒  f(1) = 1 + 1 = 2

Now, f'(c) = 

    =  

   =  

⇒ c² = 3

⇒ c = ± √3

∴  c = √3 ∈ (1, 3)

The correct answer is √3.

Concept:

Mean Value Theorem: Let f  : [a, b] → R be a continuous function on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that

f'(c) =  

Explanation:

Given that

According to the mean value theorem, f'(x) will be defined for some value a t 

Hence, correct answer is a t 

Concept:

For mean value theorem in [a, b]

Calculation:

f(x) = log_e x

Explanation -

f(-7) = -3 and f’(x) ≤ 2

Applying LMVT in [-7, 0], we get

(f(-7) – f(0))/-7 = f’(c) ≤ 2

(-3-f(0))/-7 ≤ 2

f(0) + 3 ≤ 14

f(0) ≤ 11

Applying LMVT in [-7, -1], we get

(f(-7) – f(-1))/(-7 +1) = f’(c) ≤ 2

-3 – f(-1))/-6 = f’(c) ≤ 2

f(-1) + 3 = ≤ 12

f(-1) ≤ 9

Therefore f(-1) + f(0) ≤ 20

Hence Option (2) is correct.

Concept:

Lagrange’s Mean Value Theorem:
If a function f is defined on the closed interval [a, b] satisfying:

• The function f is continuous on the closed interval [a, b].
• The function f is differentiable on the open interval (a, b).

Then there exists a value x = c such that f'(c) = .

Calculation:

The given function f(x) =  is both differentiable and continuous in the interval [1, 3].

f'(x) = 

By the Mean Value Theorem, there exists a c ∈ [1, 3], such that:

f'(c) = 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ c = √3.

Concept

Lagrange mean value theorem (LMVT)

Let f(x) be a function defined in [a, b] such that

f (x) is continuous in [a, b] and differentiable in (a, b)

Then there exists at least one point such that c ∈ (a, b) such that

Calculation

Given: The function f be a twice differentiable function on (1, 6). 

f(2) = 8, f'(2) = 5, f'(x) ≥ 1, f"(x) ≥ 4, ∀ x ∈ (1, 6)

       ----(i)

⇒ f'(5) ≥ 17

       ----(ii)

⇒ f(5) ≥ 11

∴ f'(5) + f(5) ≥ 28

Concept

Lagrange mean value theorem (LMVT)

Let f(x) be a function defined in [a, b] such that

f (x) is continuous in [a, b] and differentiable in (a, b)

Then there exists at least one point such that C ∈ (a, b) such that

Calculation

Given:

f(x) = x(x -1)2 

⇒ f(0) = 0

f(2) = 2 

⇒  f(0) ≠ f(2)

Thus mean value theorem is applicable

Then, 

We will take only x = 4/3 ϵ (0, 2) 

Concept:

Differentiation formula

Calculation:

We have f(a) = f(0) = 0 and f(b) = f(​) = 

⇒ 

Now we have f(x) = x³ −3x²+ 2x

⇒ f′(x) = 3x²− 6x + 2

⇒ f′(c) = 3c² − 6c + 2

Putting all these value in lagrange's mean value theorem

,(a

⇒  = 3c² − 6c + 2

⇒ c = ​​

Hence, c = ​​.