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

Given function f(x) = x^1/3

 f(x) is continuous in [-1, 1]

f(x) is not differentiable in [-1, 1] as f(x) is not differentiable at x = 0

Hence we cannot apply Lagrange’s mean value theorem for the given function.

Hence c does not exist.

f(x) being a polynomial function, it is continuous on [3, 4].

\(f'\left( x \right) = 2\left( {x - 3} \right)\), which exists for all x ∊ [3, 4].

f(x) is differentiable in [3, 4].

We are to find a point on the parabola whose slope equals the slope of the chord/line joining A(3, 0) and B(4, 1).

The slope of the line joining the two points is:

\(f'\left( c \right) = \frac{{f\left( 4 \right) - f\left( 3 \right)}}{{4 - 3}} = 1\)

We have to find the value of x which satisfies the following:

\(f'\left( x \right) = 2\left( {x - 3} \right)\) = 1

Solving the above we find that x = 7/2 is satisfying the above equation:

Solving for y, we get: \(y = {\left( {x - 3} \right)^2} \Rightarrow y = \frac{1}{4}\)

Thus required point is \(\left( {\frac{7}{2},\frac{1}{4}} \right)\)

Concept:

Lagrange's Mean Value Theorem:

Let f(x) be a function of x in interval [a, b] such that

1. f(x) is continuous in [a, b]
2. f(x) is differentiable in (a, b)

Then, there exist at least one point c ϵ (a, b)

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

Calculation:

In the given interval [2, 5] for f(x) = 2x

1. f(x) = 2x is continous.
2. f(x) = 2x is differentiable

f'(x) = (d/dx) 2x

f'(x) = 2 {replace x with c}

f'(c) = 2

f(a) = 2 × 2 = 4

f(b) = 2 × 5 = 10

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

2 = (10 - 4) / (5 - 2)

2 = 2

As, f'(c) has no variable 'c' in it.

There will be infinite many values of 'c' that will satisfy the condition.

So,  The number of values of C with reference to Lagrange's mean value theorem for the function f(x) = 2x in [2, 5] is "Infinite"

Given:

According to which theorem, there exists a real number C between a and b such that f'(C) = 0 whenever f'(a) f'(b) < 0 such that f be defined and derivable on [a, b].

Concept Used:

Dirichlet's theorem: there are infinitely many prime numbers contained in the collection of all numbers of the form na + b, in which the constants a and b are integers that have no common divisors except the number 1 (in which case the pair are known as being relatively prime) and the variable n is any natural number (1, 2, 3, …).

Milne theorem: In fluid dynamics the Milne-Thomson circle theorem or the circle theorem is a statement giving a new stream function for a fluid flow when a cylinder is placed into that flow.

Darboux's theorem: there exists a real number C between a and b such that f'(C) = 0 whenever f'(a) f'(b) < 0 such that f be defined and derivable on [a, b].

Riesz's theorem: [Riesz Representation Theorem, I]. Let Γ ∈ L p (X, µ) ∗ , where 1 ≤ p < ∞ and µ is σ-finite. Then if 1 p + 1 q = 1, there exists a unique g ∈ L q (X, µ) ∗ such that Γ(f) = Z X fg dµ = φg(f) Moreover, kΓk = kgkq.

Solution:

Dirichlet's theorem: there are infinitely many prime numbers contained in the collection of all numbers of the form na + b, in which the constants a and b are integers that have no common divisors except the number 1 (in which case the pair are known as being relatively prime) and the variable n is any natural number (1, 2, 3, …).

Milne theorem: In fluid dynamics the Milne-Thomson circle theorem or the circle theorem is a statement giving a new stream function for a fluid flow when a cylinder is placed into that flow.

Darboux's theorem: there exists a real number C between a and b such that f'(C) = 0 whenever f'(a) f'(b) < 0 such that f be defined and derivable on [a, b].

Riesz's theorem: [Riesz Representation Theorem, I]. Let Γ ∈ L p (X, µ) ∗ , where 1 ≤ p < ∞ and µ is σ-finite. Then if 1 p + 1 q = 1, there exists a unique g ∈ L q (X, µ) ∗ such that Γ(f) = Z X fg dµ = φg(f) Moreover, kΓk = kgkq.

\(\therefore\) Option 3 is correct.

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

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

Now,

f(a) = f(0) = 0

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

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

f'(x) = 3x² – 6x + 2

f’(c) = 3c² – 6c + 2

Put in equation (1)

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

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

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

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: 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.

Concept:

If \(f\left( t \right) = \;\mathop \smallint \nolimits_a^b g\left( x \right)dx,\) then, f’(x) = g(x)

Lagrange’s Mean value theorem,

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

Calculation:

\(f'\left( t \right) = \frac{1}{{{t^4}\; + \;5}}\)

Let, f(x) is defined on [3, 4]

\(f'\left( c \right) = \frac{{f\left( 4 \right) - f\left( 3 \right)}}{1}\)

⇒ \(\frac{1}{{{c^4} + 5}} = f\left( 4 \right)\)

3 < c < 4

3⁴ < c⁴ < 4⁴ ⇒ 3⁴ + 5 < c⁴ + 5 < 4⁴ + 5

\(\frac{1}{{{3^4}\; + \;5}} > \frac{1}{{{c^4}\; + \;5}} > \frac{1}{{{4^4}\; + \;5}}\)

⇒ \(\frac{1}{{86}}\left\langle {f\left( 4 \right)} \right\rangle \frac{1}{{261}}\)

f(x) = x³ – 3x² + 2x

⇒ f’(x) = 3x² – 6x + 2

⇒ f’(c) = 3c² – 6c + 2                                                                                     

by mean value theorem, we have

\(\begin{array}{l} {\rm{f'}}\left( {\rm{c}} \right){\rm{\;}} = {\rm{\;}}\frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}}\left( {\rm{a}} \right)}}\\ \Rightarrow \;f'\left( c \right)\; = \;\frac{{f\left( {\frac{1}{2}} \right) - f\left( 0 \right)}}{{\left( {\frac{1}{2} - 0} \right)}}\\ = \;\frac{{\left( {\frac{1}{8} - \frac{3}{4} + 1} \right) - \left( 0 \right)}}{{\frac{1}{2}}}\; = \;\frac{3}{4} \end{array}\)

⇒ 3c² – 6c + 2 = ¾

⇒ 12c² – 24c + 5 = 0

\(\begin{array}{l} \Rightarrow \;C\; = \;\frac{{6 \pm \sqrt {21} }}{6}\\ \Rightarrow \;C\; = \;\frac{{6 - \sqrt {21} }}{6}\; = \;0.236 \leftrightarrow \left[ {0,\frac{1}{2}} \right] \end{array}\)