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

Calculation: 

${\mathrm{f}}^{\prime \prime }(\mathrm{x})={\mathrm{g}}^{\prime \prime }(\mathrm{x})+6\mathrm{x}...(\mathrm{i})$

${\mathrm{f}}^{\prime }(1)=4{\mathrm{g}}^{\prime }(1)-3=9...(\mathrm{i}\mathrm{i})$

$\mathrm{f}(2)=3\mathrm{g}(2)=12...(\mathrm{i}\mathrm{i}\mathrm{i})$

By integrating (1) 

⇒ f'(x) = g'(x) + 6$\frac{{\mathrm{x}}^2}{2}$ + 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) + $\frac{3{\mathrm{x}}^2}{3}$ + 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, 

⇒ $\mathrm{g}(-2)-\mathrm{f}(-2)=20\text{ (Option (1) is true) }$

Now, for – 1 < x , 2 

⇒ h (x) = f(x) - g(x) = x3 + 3x - 6 

⇒ h'(x) = 3x² + 3 

⇒ h(x)↑ 

So, h(–1) < h(x) < h(2) 

⇒ – 10 < h(x) < 8 

⇒ $|\mathrm{h}(\mathrm{x})|<10$ 

Option 1 is True

Now, h'(x) = f'(x) – g'(x) = 3x² + 3 

If |h'(x)| < 6 ⇒ |3x² + 3| < 6 

⇒ 3x² + 3 < 6 

⇒ x² < 1 

⇒ \(\rm -1

Option 3 is True 

If x ∈ (–1, 1) |f'(x) – g'(x)| < 6 

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$\left(\frac{3}{2}\right)$ = +ve

So there exists x₀ ∈$(1,\frac{3}{2})$ 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) = $\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}$.

Calculation:

The given function f(x) = $\mathrm{x}+\frac{1}{\mathrm{x}}$ is both differentiable and continuous in the interval [1, 3].

f'(x) = $1-\frac{1}{{\mathrm{x}}^2}$

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

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

⇒ $1-\frac{1}{{\mathrm{c}}^2}=\frac{\mathrm{f}(3)-\mathrm{f}(1)}{3-1}$

⇒ $1-\frac{1}{{\mathrm{c}}^2}=\frac{(3+\frac{1}{3})-(1+\frac{1}{1})}{2}$

⇒ $1-\frac{1}{{\mathrm{c}}^2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}$

⇒ $\frac{1}{{\mathrm{c}}^2}=1-\frac{2}{3}=\frac{1}{3}$

⇒ 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:

$\frac{f(b)-f(a)}{b-a}=f^{\prime }(C)$

Calculation

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

Here, $a=1$, $b=3$

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

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

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

Now, $\frac{f(3)-f(1)}{3-1}=f^{\prime }(C)$

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

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

$\frac{2}{3}=1-\frac{1}{C^2}$

$\frac{1}{C^2}=1-\frac{2}{3}=\frac{1}{3}$

$C^2=3$

$C=\pm \sqrt{3}$

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 $f^{\prime }(x)\ge 5$.

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

$f^{\prime }(c)=\frac{f(2024)-f(0)}{2024-0}$

⇒ $f^{\prime }(c)=\frac{f(2024)-(-2)}{2024}$

⇒ $f(2024)=2024f^{\prime }(c)-2$

Since $f^{\prime }(x)\ge 5$, we have $f^{\prime }(c)\ge 5$.

⇒ $f(2024)\ge 2024\times 5-2$

⇒ $f(2024)\ge 10120-2$

⇒ $f(2024)\ge 10118$

∴ The least possible value of $f(2024)$ is 10118.

Hence option 2 is correct

Concept 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 $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

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

$f^{\prime }(c)=\frac{f(e)-f(1)}{e-1}$

⇒ $\frac{1}{c}=\frac{1-0}{e-1}$

⇒ $\frac{1}{c}=\frac{1}{e-1}$

⇒ c = e - 1

∴ The value of c is e - 1.

Hence option 3 is correct

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

Calculation:

The given function f(x) = $\mathrm{x}+\frac{1}{\mathrm{x}}$ is both differentiable and continuous in the interval [1, 3].

f'(x) = $1-\frac{1}{{\mathrm{x}}^2}$

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

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

⇒ $1-\frac{1}{{\mathrm{c}}^2}=\frac{\mathrm{f}(3)-\mathrm{f}(1)}{3-1}$

⇒ $1-\frac{1}{{\mathrm{c}}^2}=\frac{(3+\frac{1}{3})-(1+\frac{1}{1})}{2}$

⇒ $1-\frac{1}{{\mathrm{c}}^2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}$

⇒ $\frac{1}{{\mathrm{c}}^2}=1-\frac{2}{3}=\frac{1}{3}$

⇒ 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) = $\frac{f(b)-f(a)}{b-a}$

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) = $\frac{f(b)-f(a)}{b-a}$

Here b = 2 ⇒  f(2) = 2(2 - 2) = 0

a = 1 ⇒  f(1) = 1(1 - 2) = - 1

⇒ 2 (c - 1) = $\frac{0-(-1)}{2-1}$

⇒ 2(c - 1) = 1

⇒ c - 1 = $\frac{1}{2}$

∴ c = $\frac{3}{2}$ ∈ (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) = $\frac{f(b)-f(a)}{b-a}$

Calculation:

Given: f(x) = x + $\frac{1}{\mathrm{x}}$, x ϵ [1, 3]

 f'(x) = 1 - $\frac{1}{x^2}$  

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

f'(c) = $\frac{f(b)-f(a)}{b-a}$

Here b = 3 ⇒ f(3) = 3 + $\frac{1}{3}$ = $\frac{10}{3}$

a = 1 ⇒  f(1) = 1 + 1 = 2

Now, f'(c) = $\frac{\frac{10}{3}-2}{3-1}$

 $1-\frac{1}{c^2}$   = $\frac{\frac{10-6}{3}}{2}$ 

 $1-\frac{1}{c^2}=\frac{1}{2}\times \frac{4}{3}$

 $1-\frac{1}{c^2}=\frac{2}{3}$

 $\frac{1}{c^2}=1-\frac{2}{3}$

  $\frac{1}{c^2}$ =  $\frac{1}{3}$

⇒ 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) = $\frac{[\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})]}{(\mathrm{b}-\mathrm{a})}$ 

Explanation:

Given that

$f^{\prime }(x)=\frac{f(b)-f(a)}{b-a}$

According to the mean value theorem, f'(x) will be defined for some value a < xt < b, where f(x) is differentiable.

Hence, correct answer is a < xt < b.

Concept:

For mean value theorem in [a, b]

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

f(x) = log_e x

$f^{\prime }\left(x\right)=\frac{1}{x}\therefore f^{\prime }\left(c\right)=\frac{1}{c}$

$\frac{1}{c}=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{{\log }_{e}3-{\log }_{e}1}{3-1}$

$\frac{1}{c}=\frac{{\log }_{e}3}{2}$

$c=\frac{2}{{\log }_{e}3}=2{\log }_{3}e$

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

Calculation:

The given function f(x) = $\mathrm{x}+\frac{1}{\mathrm{x}}$ is both differentiable and continuous in the interval [1, 3].

f'(x) = $1-\frac{1}{{\mathrm{x}}^2}$

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

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

⇒ $1-\frac{1}{{\mathrm{c}}^2}=\frac{\mathrm{f}(3)-\mathrm{f}(1)}{3-1}$

⇒ $1-\frac{1}{{\mathrm{c}}^2}=\frac{(3+\frac{1}{3})-(1+\frac{1}{1})}{2}$

⇒ $1-\frac{1}{{\mathrm{c}}^2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}$

⇒ $\frac{1}{{\mathrm{c}}^2}=1-\frac{2}{3}=\frac{1}{3}$

⇒ 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

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

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)

$f^{″}(x)=\frac{f^{\prime }\left(5\right)-f^{\prime }\left(2\right)}{5-2}\ge 4$       ----(i)

⇒ f'(5) ≥ 17

$f^{\prime }\left(x\right)=\frac{f\left(5\right)-f\left(2\right)}{5-2}\ge 1$       ----(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

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

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, $f^{\prime }(x)=\frac{f(2)-f(0)}{2-0}$

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

$\Rightarrow 3x^2-4x=0$

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

$\Rightarrow x=0,x=4/3.$

We will take only x = 4/3 ϵ (0, 2) 

Concept:

Differentiation formula

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}{\mathrm{x}}^{\mathrm{n}}=\mathrm{n}{\mathrm{x}}^{\mathrm{n}-1}$

Calculation:

We have f(a) = f(0) = 0 and f(b) = f($\frac{1}{2}$​) = $\frac{3}{8}$

⇒ $\frac{f(b)-f(a)}{b-a}=\frac{\frac{3}{8}-0}{\frac{1}{2}-0}=\frac{3}{4}$

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

$\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}={\mathrm{f}}^{\prime }(\mathrm{c})$,(a < c, b) We get

⇒ $\frac{3}{4}$ = 3c² − 6c + 2

⇒ c = $1\pm \frac{\sqrt{21}}{6}$​​

Hence, c = ​​$1-\frac{\sqrt{21}}{6}$.