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

Calculation

Given:

Function: $f(x)=x(x+3)e^{-x/2}$

Interval: $[-3,0]$

$f^{\prime }(x)=(2x+3)e^{-x/2}+x(x+3)e^{-x/2}(-1/2)$

$f^{\prime }(x)=e^{-x/2}[(2x+3)-(x^2+3x)/2]$

$f^{\prime }(x)=e^{-x/2}[2x+3-x^2/2-3x/2]$

$f^{\prime }(x)=e^{-x/2}[-x^2/2+x/2+3]$

Set $f^{\prime }(c)=0$:

$e^{-c/2}[-c^2/2+c/2+3]=0$

$-c^2/2+c/2+3=0$

$c^2-c-6=0$

$(c-3)(c+2)=0$

⇒ $c=3$ or $c=-2$

Since $c\in (-3,0)$, we have $c=-2$.

∴ The value of $c$ is -2.

Hence option 3 is correct

Concept:

Lagrange's Mean Value Theorem (MVT):

• The Lagrange's Mean Value Theorem states that for a function f(x) continuous on [a, b] and differentiable on (a, b), there exists at least one point c in (a, b) such that:
• f'(c) = (f(b) - f(a)) / (b - a).
• This theorem helps find a point c where the slope of the tangent to the function equals the average rate of change over the interval [a, b].
• In this case, we are given the function f(x) = √(x² - 4) and the interval [2, 4], and we need to find the value of c where the slope of the tangent is equal to the average rate of change over the interval.

Calculation:

Given,

• f(x) = √(x² - 4)
• Interval: [2, 4]

According to Lagrange's Mean Value Theorem, we have:

f'(c) = (f(4) - f(2)) / (4 - 2)

First, calculate f(4) and f(2):

• f(4) = √(4² - 4) = √(16 - 4) = √12
• f(2) = √(2² - 4) = √(4 - 4) = √0 = 0

Now, substitute into the equation for the average rate of change:

f'(c) = (√12 - 0) / (4 - 2) = √12 / 2 = √3

Next, find the derivative of f(x) = √(x² - 4):

f'(x) = (1/2)(x² - 4)^(-1/2) * 2x = x / √(x² - 4)

Now, set f'(c) = √3:

c / √(c² - 4) = √3

Squaring both sides:

c² / (c² - 4) = 3

c² = 3(c² - 4)

c² = 3c² - 12

-2c² = -12

c² = 6

c = √6 or c = -√6

∴ The value of c is √6.

Concept: To find the average value of a function f(x) over the interval [a, b], we use the formula:

$\text{Average value of}f(x)\text{over}[a,b]=\frac{1}{b-a}{\int }_a^{b}f(x)dx$

Given the function: f(x) = 4x² and the interval [1, 3], we substitute f(x) into the formula for average value.

Explanation:

The average value of the function f(x) = 4x² over the interval [1, 3] is calculated as follows:

$\frac{1}{3-1}{\int }_1^{3}4x^{2}dx$

First, we compute the definite integral:

${\int }_1^{3}4x^{2}dx$

We find the antiderivative of 4x²:

$\int 4x^{2}dx=\frac{4x^3}{3}+C$

Now we evaluate this antiderivative from 1 to 3:

${\left[\frac{4x^3}{3}\right]}_1^3=\frac{4(3{)}^3}{3}-\frac{4(1{)}^3}{3}$

Simplifying the expression:

$\frac{4\cdot 27}{3}-\frac{4\cdot 1}{3}=\frac{108}{3}-\frac{4}{3}=\frac{108-4}{3}=\frac{104}{3}$

Next, we calculate the average value by dividing the integral by the length of the interval (3 - 1 = 2):

$\frac{1}{2}\left(\frac{104}{3}\right)=\frac{104}{6}=\frac{52}{3}$

The correct answer is option 4.

Explanation:

Lagrange’s mean value theorem as f(b) - f(a) = (b - a)f'(c), a < c < b.

Here f(x) = x(x - 1), a = 0, $b=\frac{1}{2}$

⇒ f'(x) = x + x - 1  =2x - 1

So, f(0) = 0, f($\frac{1}{2}$) = $\frac{1}{2}$(- $\frac{1}{2}$) = - $\frac{1}{4}$ and f'(c) = 2c - 1

Then using Lagrange's theorem

f($\frac{1}{2}$) - f(0) = ($\frac{1}{2}$ - 0)f'(c), 0 < c < $\frac{1}{2}$

⇒ - $\frac{1}{4}$ - 0 = $\frac{1}{2}$(2c - 1)

⇒ c = - $\frac{1}{4}$ + $\frac{1}{2}$ ⇒ c = $\frac{1}{4}$

(1) is correct

For Lagrange’s mean value theorem, there exists at least one real number ‘c’ in (a, b) such that

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$     ---(1)

Now,

f(a) = f(0) = 0

$f\left(b\right)=f\left(\frac{1}{2}\right)=\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)=\frac{3}{8}$

$f^{\prime }\left(x\right)=x\left(x^2-3x+2\right)=x^3-3x^2+2x$

f'(x) = 3x² – 6x + 2

f’(c) = 3c² – 6c + 2

Put in equation (1)

$3c^2-6c+2=\frac{\frac{3}{8}-0}{\frac{1}{2}-0}$

$3c^2-6c+2=\frac{3}{4}$

12c² – 24c + 8 = 3

12c² – 24c + 5 = 0

$c=\frac{24\pm \sqrt{{24}^2-12\times 5\times 4}}{2\times 12}$

c = 1 ± 0.764 = 1.764 or 0.236

But, c = 0.236, since it only lies between 0 and 1/2

Concept:

Lagrange’s Mean Value Theorem:

If f(x) is real valued function such that –

• f(x) is continuous in the closed interval [a,b]
• (f(x) is differentiable in the open interval (a,b)
• f(a) ≠ f(b)

Then there exist at least one value x, c (a,b) such that –

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

Geometrical Interpretation:

• Between two points a and b, f(a) ≠ f(b) of the graph of f(x) then there exists one point where the tangent is parallel to the chord $\overline{AB}$

Explanation:

(a) is true with reference to geometrical interpretations.

(b) is false, if f(x) has same value for every value of x, it will violate f(a) ≠ f(b).

Given:

f(-1) = 0

|f’(x)| ≤ 2

-2 ≤ f ’(x) ≤ 2

Using Lagrange mean value theorem:

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

Since value of f (-1) = 0 is given, we take the interval [-1, 2], i.e.

-2 ≤ f’(x) ≤ 2

$-2\le \frac{f\left(2\right)-f\left(-1\right)}{2-\left(-1\right)}\le 2$

$-2\le \frac{f\left(2\right)}{3}\le 2$

-6 ≤ f(2) ≤ 6

We observe that the option 2 satisfies this condition.

By Lagrange’s mean value theorem we have

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

$\begin{array}{l}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{\mathrm{f}\left(1\right)-\mathrm{f}\left(-1\right)}{1-\left(-1\right)}\end{array}$

f(1) = 1 - 1 + 1 = 1

f(-1) = 1 - 1 - 1 = -1

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

0 - 2x + 3x² = 1 ⇒ 3x² - 2x - 1 = 0

$$\begin{gathered} \therefore \mathrm{x}=1\mathrm{a}\mathrm{n}\mathrm{d}-\frac{1}{3} \\ \therefore \mathrm{x}=-\frac{1}{3}\mathrm{o}\mathrm{n}\mathrm{l}\mathrm{y}\mathrm{l}\mathrm{i}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{n}\left(-1,1\right) \end{gathered}$$

Concept:

Let f(x) is a function define on [a ,b] such that, 

1. f(x) is a Continuous on [a , b]
2. f(x) is Differentiable on [a , b]

Then, there exist a real number C ∈ (a , b) such that, According to Lagrangian Mean Value Theorem,

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

Calculation:

Given:

f(x) = $\sqrt{X^2-4}$    , and f'(x) = $\frac{1}{2\sqrt{X^2-4}}\times 2X$

The function satisfies Lagrange's Mean Value Theorem that means it satisfies two condition given above 1 and 2

Therefore for the value of C we can write down above formula 

f'(c) = $\frac{f\left(b\right)-f\left(a\right)}{b-a}$  = $\frac{\sqrt{16-4}-\sqrt{4-4}}{4-2}$   = $\frac{\sqrt{12}}{2}$  = √3 

$\frac{c}{\sqrt{c^2-4}}$ = √3

Squaring on both sides,

3c² - 12 = c² 

2c² = 12

c² = 6

c = √6

Additional Information

• (a, b) means an open interval.
• The value of C is obtained in the open interval (a, b) which means between a and b.  

Concept:

By lagrangian mean value theorem,

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a},\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{c}ϵ\left(\mathrm{a},\mathrm{b}\right)$

Calculation:

Given:

f(x) = x² – 2x + 2, x ϵ [1, 3]

By lagrangian mean value theorem

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

$f^{\prime }\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$

$2c-2=\frac{\left(9-6+2\right)-\left(1-2+2\right)}{3-1}$

$2c-2=\frac{5-1}{2}$

c = 2

Explanation:

Lagrange’s mean value theorem as f(b) - f(a) = (b - a)f'(c), a < c < b.

Here f(x) = x(x - 1), a = 0, $b=\frac{1}{2}$

⇒ f'(x) = x + x - 1  =2x - 1

So, f(0) = 0, f($\frac{1}{2}$) = $\frac{1}{2}$(- $\frac{1}{2}$) = - $\frac{1}{4}$ and f'(c) = 2c - 1

Then using Lagrange's theorem

f($\frac{1}{2}$) - f(0) = ($\frac{1}{2}$ - 0)f'(c), 0 < c < $\frac{1}{2}$

⇒ - $\frac{1}{4}$ - 0 = $\frac{1}{2}$(2c - 1)

⇒ c = - $\frac{1}{4}$ + $\frac{1}{2}$ ⇒ c = $\frac{1}{4}$

(1) is correct

y = f(x) = 5x2 + 10x in the internal x = (1, 2)

Since, the function y is continuous in the interval (1, 2), as well as is differentiable at each point so, from Lagrange mean value theorem there exist at least a point where:

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

Here, we have

a = 1, b = 2

So, for x = a = 1, we obtain:

y = f(a) = f(1) = 5(1)2 + 10(1) = 15

and for x = b = 2

y = f(b) = f(2) = 5(2)2 + 10(2) = 40

Therefore:

$f^{\prime }\left(c\right)=\frac{40-15}{2-1}=25$

Let$\mathrm{f}(\mathrm{x})=\mathrm{a}\mathrm{x}+\mathrm{b}$

$\therefore {\mathrm{f}}^{\prime }(\mathrm{x})=\mathrm{a}$

Given that, 

$6{\mathrm{f}}^{{}^{\prime }}(\mathrm{c})+\mathrm{f}(-3)=\mathrm{f}(3).$

$\Rightarrow 6\mathrm{a}=3\mathrm{a}+\mathrm{b}–(-3\mathrm{a}+\mathrm{b})$

$\Rightarrow \mathrm{a}=\mathrm{a}$

∴ C can be any value $(-3,3)$

So, no condition on c

Concept:

Lagrange’s Mean Value Theorem:

If f(x) is real valued function such that –

• f(x) is continuous in the closed interval [a,b]
• (f(x) is differentiable in the open interval (a,b)
• f(a) ≠ f(b)

Then there exist at least one value x, c (a,b) such that –

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

Geometrical Interpretation:

• Between two points a and b, f(a) ≠ f(b) of the graph of f(x) then there exists one point where the tangent is parallel to the chord $\overline{AB}$

Explanation:

(a) is true with reference to geometrical interpretations.

(b) is false, if f(x) has same value for every value of x, it will violate f(a) ≠ f(b).

Since given function is continuous and differentiable then by Lagrange’s Mean-Value Theorem.

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

$a=-\frac{11}{7},b=\frac{13}{7}$

$f\left(x\right)=x^3-3x-1$

$f^{\prime }\left(x\right)=3x^2-3$

$f^{{}^{\prime }}\left(c\right)=3c^2-3$

$f\left(b\right)={\left(-\frac{11}{7}\right)}^3-3\left(-\frac{11}{7}\right)-1=-\frac{57}{343}$

$f\left(a\right)={\left(\frac{13}{7}\right)}^3-3\left(\frac{13}{7}\right)-1=-\frac{57}{343}$

$3c^2-3=\frac{-\frac{57}{343}-\left(-\frac{57}{343}\right)}{-\frac{11}{7}-\frac{13}{7}}$

$3c^2-3=0$

$c^2=1$