Lagrange's Mean Value Theorem MCQ Quiz in मल्याळम - Objective Question with Answer for Lagrange's Mean Value Theorem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculation: 

By integrating (1) 

⇒ f'(x) = g'(x) + 6 + C

At x = 1, 

⇒ f'(1) = g'(1) + 3 + C

⇒ 9 = 4 + 3 + C ⇒ C = 3 

∴ f'(x) = g'(x) + 3x² + 3

Again by integrating,

⇒ f(x) = g(x) +  + 3x + D

At x = 2, 

⇒ f(2) = g(2) + 8 + 3(2) + D

⇒ 12 = 4 + 8 + 6 + D ⇒ D = –6

So, f(x) = g(x) + x³ + 3x - 6 

At x = –2, 

⇒ 

Now, for – 1

⇒ h (x) = f(x) - g(x) = x3 + 3x - 6 

⇒ h'(x) = 3x² + 3 

⇒ h(x)↑ 

So, h(–1)

⇒ – 10

⇒ 

Now, h'(x) = f'(x) – g'(x) = 3x² + 3 

If |h'(x)| ⇒ |3x² + 3|

⇒ 3x² + 3

⇒ x²

⇒  

If x ∈ (–1, 1) |f'(x) – g'(x)|

option (3) is true and now to solve 

⇒ f(x) – g(x) = 0 

⇒ x³ + 3x – 6 = 0

h(x) = x³ + 3x - 6

here, = h(1) −ve and h = +ve

So there exists x₀ ∈ such that f(x₀) = g(x₀)

(option (4) is true)   

Hence, the correct answer is Option 2. 

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.

Explanation -

LMVT is applicable on f(x) in [0,1], therefore it is continuous and differentiable in [0,1]

Now, f(0) = 11, f(1) = 16

f ′(x) = 3x2 − 8x + 8

f’(c) = (f(1) – f(0))/(1-0)

= (16 – 11)/1

⇒ 3c2 − 8c + 8 = 5

⇒ 3𝑐2 − 8𝑐 + 3 = 0

c = (8±2√7)/6 = (4±√7)/3

As c ∈ (0,1)

We get c = (4 – √7)/3

Hence the Correct option is (1)

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

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: 

Lagrange mean value theorem (LMVT)

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

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

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

Calculation:

Given, f ''(x)

⇒ f '(x) is decreasing in [1, 3]

Let C₁ ∈ (1, 2). Using LMVT:

f '(C₁) =  = f(2) - f(1)

Let C2 ∈ (2, 3). Using LMVT:

f '(C2) =  = f(3) - f(2)

Since, f '(x) is a decresing function

⇒ f '(C2) f '(C1) 

⇒ f(3) - f(2)

⇒ f(3) + f(1)

∴ f(3) + f(1)

The correct answer is Option 1.

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.

Explanation -

LMVT is applicable on f(x) in [0,1], therefore it is continuous and differentiable in [0,1]

Now, f(0) = 11, f(1) = 16

f ′(x) = 3x2 − 8x + 8

f’(c) = (f(1) – f(0))/(1-0)

= (16 – 11)/1

⇒ 3c2 − 8c + 8 = 5

⇒ 3𝑐2 − 8𝑐 + 3 = 0

c = (8±2√7)/6 = (4±√7)/3

As c ∈ (0,1)

We get c = (4 – √7)/3

Hence the Correct option is (1)