Lagrange's Mean Value Theorem MCQ Quiz in தமிழ் - Objective Question with Answer for Lagrange's Mean Value Theorem - இலவச PDF ஐப் பதிவிறக்கவும்

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) = \(\rm \frac{f(b)-f(a)}{b-a}\).

Calculation:

The given function f(x) = \(\rm x+\frac1x\) is both differentiable and continuous in the interval [1, 3].

f'(x) = \(\rm 1-\frac1{x^2}\)

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

f'(c) = \(\rm \frac{f(b)-f(a)}{b-a}\)

⇒ \(\rm 1-\frac1{c^2}=\frac{f(3)-f(1)}{3-1}\)

⇒ \(\rm 1-\frac1{c^2}=\frac{\left(3+\frac13\right)-\left(1+\frac11\right)}{2}\)

⇒ \(\rm 1-\frac1{c^2}=\frac{\frac43}{2}=\frac23\)

⇒ \(\rm \frac1{c^2}=1-\frac23=\frac13\)

⇒ 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:

\(\frac{f(b) - f(a)}{b - a} = f'(C)\)

Calculation

Given: \(f(x) = x + x^{-1}\), \(x \in [1, 3]\)

Here, \(a = 1\), \(b = 3\)

\(f(a) = f(1) = 1 + 1^{-1} = 1 + 1 = 2\)

\(f(b) = f(3) = 3 + 3^{-1} = 3 + \frac{1}{3} = \frac{10}{3}\)

\(f'(x) = 1 - x^{-2} = 1 - \frac{1}{x^2}\)

Now, \(\frac{f(3) - f(1)}{3 - 1} = f'(C)\)

\(\frac{\frac{10}{3} - 2}{3 - 1} = 1 - \frac{1}{C^2}\)

\(\frac{\frac{4}{3}}{2} = 1 - \frac{1}{C^2}\)

\(\frac{2}{3} = 1 - \frac{1}{C^2}\)

\(\frac{1}{C^2} = 1 - \frac{2}{3} = \frac{1}{3}\)

\(C^2 = 3\)

\(C = \pm \sqrt{3}\)

Since \(C \in (1, 3)\), we take the positive value.

\(C = \sqrt{3}\)

Hence option 1 is correct.

Calculation

Given:

For all \(x \in [0, 2024]\), \(f(x)\) is differentiable.

\(f(0) = -2\) and \(f'(x) \ge 5\).

By the Mean Value Theorem, there exists \(c \in (0, 2024)\) such that

\(f'(c) = \frac{f(2024) - f(0)}{2024 - 0}\)

⇒ \(f'(c) = \frac{f(2024) - (-2)}{2024}\)

⇒ \(f(2024) = 2024 f'(c) - 2\)

Since \(f'(x) \ge 5\), we have \(f'(c) \ge 5\).

⇒ \(f(2024) \ge 2024 \times 5 - 2\)

⇒ \(f(2024) \ge 10120 - 2\)

⇒ \(f(2024) \ge 10118\)

∴ The least possible value of \(f(2024)\) 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

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)

Calculation:

Given, f ''(x) < 0

⇒ f '(x) is decreasing in [1, 3]

Let C₁ ∈ (1, 2). Using LMVT:

f '(C₁) = \(\displaystyle \frac{f(2)-f(1)}{2-1}\) = f(2) - f(1)

Let C2 ∈ (2, 3). Using LMVT:

f '(C2) = \(\displaystyle \frac{f(3)-f(2)}{3-2}\) = f(3) - f(2)

Since, f '(x) is a decresing function

⇒ f '(C2) < f '(C1) 

⇒ f(3) - f(2) < f(2) - f(1)

⇒ f(3) + f(1) < 2f(2)

∴ f(3) + f(1) < 2f(2).

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 \(f'(c) = \frac{f(b) - f(a)}{b - a}\)

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

\(f'(c) = \frac{f(e) - f(1)}{e - 1}\)

⇒ \(\frac{1}{c} = \frac{1 - 0}{e - 1}\)

⇒ \(\frac{1}{c} = \frac{1}{e - 1}\)

⇒ c = e - 1

∴ The value of c is e - 1.

Hence option 3 is correct

Calculation:

Using Lagrange's mean value theorem,

f '(c) = \(\frac{f(b)-f(a)}{b-a}\)

⇒ \(\frac{-c}{\sqrt{25 - c^2}} = \frac{\sqrt{25 - 5^2} - \sqrt{25 - 1^2}}{5 - 1}\)

⇒ \(\frac{-c}{\sqrt{25 - c^2}} = \frac{\sqrt{25 - 25} - \sqrt{25 - 1}}{4}\)

⇒ \(\frac{-c}{\sqrt{25 - c^2}} = \frac{-\sqrt{24}}{4}\)

⇒ \(4c = \sqrt{24} \cdot \sqrt{25 - c^2}\)

⇒ 16c² = 24(25 - c²)

⇒ 16c² + 24c² - 600 = 0

⇒ 40c² = 600

⇒ c² = 15

⇒ c = ± \(\sqrt{15}\)

⇒ c = \(\sqrt{15}\) {∵ - \(\sqrt{15}\) ∈ [1, 5]} 

∴ The value of c is \(\sqrt{15}\).

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)

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

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)

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, \(f'(x) = \frac{{f(2) - f(0)}}{{2 - 0}}\)

\( \Rightarrow 3{x^2} - 4x + 1 = \frac{{2 - 0}}{{2 - 0}} = 1 \)

\(\Rightarrow 3{x^2} - 4x = 0\)

\( \Rightarrow x(3x - 4) = 0 \)

\(\Rightarrow x = 0,x = 4/3.\)

We will take only x = 4/3 ϵ (0, 2)