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

Calculation

Given:

Function: \(f(x) = x(x+3)e^{-x/2}\)

Interval: \([-3, 0]\)

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

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

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

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

Set \(f'(c) = 0\):

\(e^{-c/2} [-c^2/2 + c/2 + 3] = 0\)

\(-c^2/2 + c/2 + 3 = 0\)

\(c^2 - c - 6 = 0\)

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

⇒ \(c = 3\) 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:

Rolle's Theorem:

Let f(x) be a real valued function defined on a closed interval [a, b] such that

(i) It is continuous on the closed interval [a, b]

(ii) It is differentiable on the open interval (a, b) and 

(iii) f(a) = f(b)

Then, There exists at least one point c \(\in \) (a, b)  such that f'(c)=0

Explanation:

We have , f(x) = |x| = \( |x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases} \)

since |x| is not differential at \(x = 0 \in (-3, 3) \)

⇒ f(x) is not differential at x=0

⇒ Rolle's theorem is not applicable to the given function.

Hence Option(2) is the correct answer.

Concept:

Rolle's Theorem:

Let f(x) be a real valued function defined on a closed interval [a, b] such that

(i) It is continuous on the closed interval [a, b]

(ii) It is differentiable on the open interval (a, b) and 

(iii) f(a) = f(b)

Then, There exists at least one point  \(c \in (a, b) \) such that \(f'(c) = 0 \).

Explanation:

Let  \(f(x) = \frac{a x^3}{3} + \frac{bx^2}{2} +cx \) be a Polynomial. Then

f(0) = 0

f(2) = \(\frac{8a}{3} \)+ \(\frac{4b}{2} \)+2c  ⇒ \(\frac{8a+6b+6c}{6} \) =\(\frac{4a+3b+3c}{3} \)= 0

Hence 0 and 2 are the root of f(x).

Since f(0)=f(2)

and f(x) is continuous in [0, 2] and differential in (0, 2)

⇒ All conditions of Rolle's theorem satisfied 

Hence  by Rolle's Theorem:  

At least one root exists.

Hence Option(1) is the correct answer.

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

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 –

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

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).

Concept:

Mean value theorem for integrals:

Let f be continuous on [a, b]. Then there is a point x_o in (a, b) such that

\(f\left( {{x_o}} \right) = \frac{1}{{b - a}}{\rm{\;}}\mathop \smallint \limits_a^b f\left( x \right)dx\)

Calculation:

\(f\left( \xi \right) = \frac{1}{{b - a}}{\rm{\;}}\mathop \smallint \limits_a^b f\left( x \right)dx\)

\(\mathop \smallint \limits_a^b f\left( x \right)dx = f\left( \xi \right)\left( {b - a} \right)\)

Given:

f(-1) = 0

|f’(x)| ≤ 2

-2 ≤ f ’(x) ≤ 2

Using Lagrange mean value theorem:

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

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

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

\(\begin{array}{l} {{\rm{f}}'}\left( {\rm{x}} \right) = \frac{{{\rm{f}}\left( 1 \right) - {\rm{f}}\left( { - 1} \right)}}{{1 - \left( { - 1} \right)}}\\ \end{array}\)

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

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

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

0 - 2x + 3x² = 1 ⇒ 3x² - 2x - 1 = 0

\(\therefore {\rm{\;x\;}} = {\rm{\;}}1{\rm{\;and\;}} - \frac{1}{3}\\ \therefore {\rm{\;x}} = - \frac{1}{3}{\rm{\;only\;lies\;in\;}}\left( { - 1,{\rm{\;}}1} \right) \)

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.  

Explanation:

Try to solve options.

Let us take option 3

Then we can say f(x) = e^{sin x}

F(0) = 1

f’(x) = e^{sin x} cos x

f’(0) = 1

f”(x) = e^{sin x} cos² x – e^{sin x} sin x

f”(0) = 1

Similarly f’’’ (0) = 0

f^iv (0) = -3

Now from Maclaurin’s series of expansion –

\(f\left( x \right)=f\left( 0 \right)+\frac{x}{1!}~{f}'\left( 0 \right)+\frac{{{x}^{2}}}{2!}~{f}''\left( 0 \right)+\frac{{{x}^{3}}}{3!}~{f}'''\left( 0 \right)+\frac{{{x}^{4}}}{4!}f'''\left( 0 \right)\)

\(=1+x\cdot 1+\frac{{{x}^{2}}}{2!}~\left( 1 \right)+\frac{{{x}^{3}}}{3!}~\left( 0 \right)+\frac{{{x}^{4}}}{4!}\left( -3 \right)\)

Concept:

By lagrangian mean value theorem,

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\;,{\rm{\;where\;c \epsilon \;}}{\rm{\;}}\left( {{\rm{a}},{\rm{\;b}}} \right)\;\)

Calculation:

Given:

f(x) = x² – 2x + 2, x ϵ [1, 3]

By lagrangian mean value theorem

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

\(f'\left( c \right) = \frac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}}\)

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

\(2c - 2 = \frac{{5 - 1}}{2}\)

c = 2

Rolle's theorem.

To apply Rolle's theorem following 3 conditions should be satisfied:

f(x) should be continuous in interval [a, b],

f(x) should be differentiable in interval (a, b), and

f(a)=f(b)

If these 3 conditions are satisfied simultaneously then, there exists at least one ′x′ such that f′(x)=0

Explanation:

Statement I: There exists \(\theta \in \left( {\frac{\pi }{6},\frac{\pi }{3}} \right)\) such that f’(θ) = 0

For the given question, it satisfies all the three conditions, therefore there exists at least one θ that gives f′(θ)=0

Therefore, statement II is true

Statement II: There exists \(\theta \in \left( {\frac{\pi }{6},\frac{\pi }{3}} \right)\) such that f’(θ) ≠ 0  

The given function is also not a constant function therefore for some θ, f′(θ) ≠ 0

Concept:

At x = 0, the Taylor series expansion of sin x and cos x is given by

\(\sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \ldots \)

\(\cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - \ldots \)

Application:

1) \(\sin \left( {{x^3}} \right) = {x^3} - \frac{{{x^9}}}{{3!}} + \frac{{{x^{15}}}}{{5!}} - \ldots \)

2) \(\sin \left( {{x^2}} \right) = {x^2} - \frac{{{x^6}}}{{3!}} + \frac{{{x^{10}}}}{{5!}} - \ldots \)

3) \(\cos \left( {{x^3}} \right) = 1 - \frac{{{x^6}}}{{2!}} + \frac{{{x^{12}}}}{{4!}} + \ldots \)

4) \(\cos \left( {{x^2}} \right) = 1 - \frac{{{x^4}}}{{2!}} + \frac{{{x^8}}}{{4!}} + \ldots \)

Concept:

Taylor’s series expansion is given by,

\(f\left( x \right) = f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + \frac{{f''\left( {{x_0}} \right)}}{{2!}}{\left( {x - {x_0}} \right)^2}\)

Calculation:

f(x) = e^x + sin x, f(π) = e^π

f’(x) = e^x + cos x, f’(π) = e^π – 1

f’’(π) = e^x – sin x, f’’(π) = e^π

\(f\left( \pi \right) = {e^\pi } + \left( {{e^\pi } - 1} \right)\left( {x - \pi } \right) + \frac{{{e^\pi }}}{{2!}}{\left( {x - \pi } \right)^2} + \ldots \)