Rolle's Theorem MCQ Quiz - Objective Question with Answer for Rolle's Theorem - Download Free PDF

Calculation:

Given, f(x) satisfies the Rolle's theorem in [3, 4] and [6, 8]

∴ In [3, 4], f(3) = f(4)

In [6, 8], f(6) = f(8)

∴ \(\int_3^8f'(x) \ dx\)

= \(\int_3^4 f'(x) \ dx\) + \(\int_4^6 f'(x) \ dx\) + \(\int_6^8 f'(x) \ dx\)

= [f(4) - f(3)] + \(\int_4^6 f'(x) \ dx\) + [f(8) - f(6)]

= 0 + \(\int_4^6 f'(x) \ dx\) + 0

= \(\int_4^6 f'(x) \ dx\)

∴ \(\int_3^8f'(x) \ dx\) is equal to \(\int_4^6 f'(x) \ dx\).

The correct answer is Option 1.

Given:

f(x) = (2x - 1)(3x + 2)(4x - 3)

Find the value of c in the interval (1/2, 3/4) such that f'(c) = 0

Concept Used:

Rolle's Theorem:

If a function f(x) is:

• Continuous on the closed interval [a, b]
• Differentiable on the open interval (a, b)
• f(a) = f(b)

Then there exists at least one c in the open interval (a, b) such that f'(c) = 0.

Calculation:

Given:

f(x) = (2x - 1)(3x + 2)(4x - 3)

Using the product rule:

\(f'(x) = 2(3x+2)(4x-3) + (2x-1)(3)(4x-3) + (2x-1)(3x+2)(4)\)

\(f'(x) = 2(12x^2 - x - 6) + 3(8x^2 - 10x + 3) + 4(6x^2 + x - 2)\)

\(f'(x) = 24x^2 - 2x - 12 + 24x^2 - 30x + 9 + 24x^2 + 4x - 8\)

\(f'(x) = 72x^2 - 28x - 11\)

3. Set f'(c) = 0 and solve for c:

\(72c^2 - 28c - 11 = 0\)

Using the quadratic formula:

\(c = \frac{(-b \pm \sqrt{b^2 - 4ac})}{2a}\)

\(c = \frac{(28 \pm \sqrt{(-28)^2 - 4 \cdot 72 \cdot -11})}{2 \cdot 72}\)

\(c = \frac{(28 \pm \sqrt{784 + 3168})}{144}\)

\(c = \frac{(28 \pm \sqrt{3952})}{144}\)

\(c = \frac{(28 \pm \sqrt{16 \cdot 247})}{144}\)

\(c = \frac{(28 \pm 4\sqrt{247})}{144}\)

\(c = \frac{(7 \pm \sqrt{247})}{36}\)

4. Check if c is in the interval (1/2, 3/4):

\(\frac{(7 - \sqrt{247})}{36} \approx \frac{(7 - 15.716)}{36} \approx \frac{-8.716}{36} \approx -0.242\)

\(\frac{(7 + \sqrt{247})}{36} \approx \frac{(7 + 15.716)}{36} \approx \frac{22.716}{36} \approx 0.631\)

Since \(\frac{(7 - \sqrt{247})}{36}\) is not in the interval (1/2, 3/4), and \(\frac{(7 + \sqrt{247})}{36}\) is in the interval (1/2, 3/4), there is one value of c that satisfies Rolle's Theorem.

∴ The value of c is \(\frac{(7 + \sqrt{247})}{36}\)

Hence option 4 is correct

Calculation:

Given, f(x) satisfies the Rolle's theorem in [3, 4] and [6, 8]

∴ In [3, 4], f(3) = f(4)

In [6, 8], f(6) = f(8)

∴ \(\int_3^8f'(x) \ dx\)

= \(\int_3^4 f'(x) \ dx\) + \(\int_4^6 f'(x) \ dx\) + \(\int_6^8 f'(x) \ dx\)

= [f(4) - f(3)] + \(\int_4^6 f'(x) \ dx\) + [f(8) - f(6)]

= 0 + \(\int_4^6 f'(x) \ dx\) + 0

= \(\int_4^6 f'(x) \ dx\)

∴ \(\int_3^8f'(x) \ dx\) is equal to \(\int_4^6 f'(x) \ dx\).

The correct answer is Option 1.

Calculation

Let \( h(f)=f(x)-2g(x) \)

as \( h(0)=h(1)=2 \)

Hence, using Rolle's theorem

\(h'(c)=0 \)

\( \Rightarrow f'(c)=2g'(c) \)

Hence option 4 is correct

In calculus, Rolle's theorem essentially states that any real-valued differentiable function that attains equal values at two distinct points must have a stationary point somewhere between them—that is, a point where the first derivative (the slope of the tangent line to the graph of the function) is zero

\( { f }^{ 1 }(c)=0 \)

\( \Rightarrow { e }^{ x }(\sin x - \cos x) + { e }^{ x }(\cos x + \sin x) = 0 \\ \Longrightarrow { e }^{ x }(2\sin x) = 0 \)

\( \Longrightarrow \sin x = 0 \)

From \( \left[ \frac { \pi }{ 4 } , \frac { 5\pi }{ 4 } \right] \) \( \sin x = 0 \) if \( x = \pi \)

\( \Rightarrow c = \pi \)

Concept:

Rolle's theorem:

Rolle's theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that,

f(a) = f(b), then, for some c ∈ [a, b]

f′(c) = 0 

Calculation:

The given function is f(x) = \(\rm \cos \dfrac{x}{2}\) on [π, 3π].

f(π) = \(\rm \cos \dfrac{\pi}{2}\) = 0 and f(3π) = \(\rm \cos \dfrac{3\pi}{2}\) = 0.

Since, f(π) = f(3π), there must exist a c ∈ [π, 3π] such that f'(c) = 0.

f'(x) = \(\rm \dfrac{d}{dx}\left (\cos \dfrac{x}{2} \right )=-\dfrac{1}{2}\left (\sin \dfrac {x}{2} \right )\)

⇒ f'(c) = \(\rm -\dfrac{1}{2}\left (\sin \dfrac {c}{2} \right )\) = 0

⇒ \(\rm \sin \dfrac {c}{2}\) = 0

⇒ \(\rm \dfrac {c}{2}\) = nπ

⇒ c = 2nπ, where n is an integer.

We want c ∈ [π, 3π], therefore c = 2π.

Concept:

Rolle's theorem states that a function f(x) is continuous over the interval [a, b] and differentiable over the interval (a, b) such that f(a) = f(b), then there exists c ϵ (a, b) such that f'(c) = 0.

Calculation:

Given:  f(x) = x³ - 3x, x ∈ [0, √3] 

∴ f'(x) = 3x² - 3 

f'(c) = 3c² - 3 = 0

⇒ 3c²  = 3

⇒ c² = 1

⇒ c = ± 1

∴ c = 1  ∈ (0, √3)

The correct answer is 1.

Concept:

Rolle's theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

Calculations:

Rolle's theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

(1) Consider the function f(x) = x² - 4x + 5 in [1, 3]

Here, f(x) is polynomail function so it is continuous on the closed interval [1, 3] and differentiable on the open interval (1, 3) 

Now, to find find f(1) and f(3), put x = 1 and x = 3 in f(x).

 f(1) = (1)² - 4(1) + 5 = 2

f(3) = (3)² - 4(3) + 5 = 2

Here, f(1) = f(3) 

Hence, Rolle's theorem is applicable for the function f(x) = x² - 4x + 5 in [1, 3]

(2) Consider the function f(x) = x² - x  in [0, 1]

Here, f(x) is polynomail function so it is continuous on the closed interval [0, 1] and differentiable on the open interval (0, 1) 

Now, to find find f(0) and f(1), put x = 0 and x = 1 in f(x).

 f(0) = (0)² - 0 = 0

f(0) = (1)² - 1 = 0

Here, f(0) = f(1) 

Hence, Rolle's theorem is applicable for the function f(x) = x² - x  in [0, 1]

(3) Consider the function f(x) = |x| in [-2, 2]

Here, f(x) is continuous on the closed interval [-2, 2] and not differentiable on the open interval (-2, 2) 

It makes sharp curve at x = 0 , hence it is not differentiable.

Now, to find find f(-2) and f(2), put x = - 2 and x = 2 in f(x).

 f(- 2) = |- 2| = 2

f(2) = | 2 |= 2

Here, f(- 2) = f(2) 

Hence, Rolle's theorem is not applicable for the function f(x) = |x| in [-2, 2]

(4) Consider the function f(x) = [x] in [2.5, 2.7]

Here, f(x) is continuous on the closed interval [2.5, 2.7] and differentiable on the open interval (2.5, 2.7) 

Now, to find find f(2.5) and f(2.7), put x = 2.5 and x = 2.7 in f(x).

f(2.5) = [x]  = [2.5] = 2

f(2.7) = [x]  = [2.7] = 2

Here, f(2.5) = f(2.7) 

Hence, Rolle's theorem is applicable for the function f(x) = [x] in [2.5, 2.7]

Hence option(3)  is correct.

Concept:

1) Rolle's theorem states that a function f(x) is continuous on [a, b] and differentiable in (a, b) such that f(a) = f(b), then there exists c ϵ (a, b) and f'(c) = 0.

2) If g(x) and h(x) are continuous at any real value c over the closed interval [a, b], then the following are also continuous at any real value c over the closed interval [a, b].

f(x) = g(x)⋅ h(x) 

f(x) = g(x) + h(x)

f(x) = g(x) - h(x)

f(x)= g(x)/h(x), as long as h(c) ≠ 0

Calculation:

Given:  f(x) = ex sin x, x ∈ [0, π] 

The function sin x is continuous in [0, π], as sin x is always continuous.

The function e^x is an exponential function and exponential functions are continuous at all real numbers.

Since both the functions e^x and sin x are continuous in [0, π].

Therefore, f(x) = ex sin x is continuous at x ∈ [0, π] 

Now, differentiate f(x) with respect to x, and we get

f'(x) = excos x + e^x sin x

f'(x) = e^x(cos x + sin x) 

Since the derivative of the function f(x) exists. Therefore, f(x) is differentiable in (0, π).

f(0) = e⁰ sin (0) = 0

f(π) = e^π sin (π ) = 0

⇒ f(0) = f(π)

All three conditions of Rolle's theorem are satisfied. Therefore, f'(c) = 0

⇒ e^c ( sin c + cos c) = 0

⇒ sin c + cos c = 0 (∵ e^c ≠ 0)

⇒ sin c = - cos c

⇒ tan c = -1

∴ c = 3π/ 4

The correct answer is 3π/4 .

Concept:

Rolle's theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(c) = 0 for some c ∈ [a, b].

Calculation:

The given function is f(x) = \(\rm \cos \dfrac{x}{2}\) on [π, 3π].

f(π) = \(\rm \cos \dfrac{\pi}{2}\) = 0 and f(3π) = \(\rm \cos \dfrac{3\pi}{2}\) = 0.

Since, f(π) = f(3π), there must exist a c ∈ [π, 3π] such that f'(c) = 0.

f'(x) = \(\rm \dfrac{d}{dx}\left (\cos \dfrac{x}{2} \right )=-\dfrac{1}{2}\left (\sin \dfrac {x}{2} \right )\)

⇒ f'(c) = \(\rm -\dfrac{1}{2}\left (\sin \dfrac {c}{2} \right )\) = 0

⇒ \(\rm \sin \dfrac {c}{2}\) = 0

⇒ \(\rm \dfrac {c}{2}\) = nπ

⇒ c = 2nπ, where n is an integer.

We want c ∈ [π, 3π], therefore c = 2π.

Given:

Y = f(x) = x(x - 3)² = x³ – 6x² + 9x ⇒ Polynomial

F(x) is

i) Continuous on [0, 3]

And ii) differentiable on (0, 3)

iii) f(0) = f(3) = 0

Thus, all three conditions of Rolle’s Theorem are satisfied.

By Rolle’s Theorem, C ϵ (0, 3) such that f’(c) = 0

f(C) = C (C - 3)²

f(C) = C [C² – 6c + 9]

f(C) = C³ – 6c² + 9C

Differentiating,

f’(C) = 3C² – 12C + 9 = 0

∴ C = 1, 3

But, C = 3 ∉ (0, 3)

∴ C = 1

If x = C = 1, then y (1) = f (1) = 1 (1 - 3)² = 4

∴ At a point (x, y) = (1, 4) on given curve, the tangent is parallel to x-axis.

Concept:

Rolle's theorem:

Rolle's theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that,

f(a) = f(b), then, for some c ∈ [a, b]

f′(c) = 0 

Calculation:

The given function is f(x) = \(\rm \cos \dfrac{x}{2}\) on [π, 3π].

f(π) = \(\rm \cos \dfrac{\pi}{2}\) = 0 and f(3π) = \(\rm \cos \dfrac{3\pi}{2}\) = 0.

Since, f(π) = f(3π), there must exist a c ∈ [π, 3π] such that f'(c) = 0.

f'(x) = \(\rm \dfrac{d}{dx}\left (\cos \dfrac{x}{2} \right )=-\dfrac{1}{2}\left (\sin \dfrac {x}{2} \right )\)

⇒ f'(c) = \(\rm -\dfrac{1}{2}\left (\sin \dfrac {c}{2} \right )\) = 0

⇒ \(\rm \sin \dfrac {c}{2}\) = 0

⇒ \(\rm \dfrac {c}{2}\) = nπ

⇒ c = 2nπ, where n is an integer.

We want c ∈ [π, 3π], therefore c = 2π.

Concept:

Rolle's theorem states that a function f(x) is continuous over the interval [a, b] and differentiable over the interval (a, b) such that f(a) = f(b), then there exists c ϵ (a, b) such that f'(c) = 0.

Calculation:

Given:  f(x) = x³ - 3x, x ∈ [0, √3] 

∴ f'(x) = 3x² - 3 

f'(c) = 3c² - 3 = 0

⇒ 3c²  = 3

⇒ c² = 1

⇒ c = ± 1

∴ c = 1  ∈ (0, √3)

The correct answer is 1.

Concept:

• If Rolle's theorem holds for the function f(x) for  x ∈ [1, 2] 

Also, f'(c) = 0

Where, c ∈ [1, 2] 

⇒ f(1) = f(2)     

Calculation:

Given: \(f'\left( {\frac{4}{3}} \right) = 0\)

⇒ f(1) = f(2)    

⇒ 1 - a + b - 4 = 8 - 4a + 2b - 4

⇒ 3a - b = 7      ----(i)

Also \(f'\left( {\frac{4}{3}} \right) = 0\) [Given]

\( ⇒ {\left( {3{x^2} - 2ax + b} \right)_{x = \frac{4}{3}}} = 0 ⇒ \frac{{16}}{3} - \frac{{8a}}{3} + b = 0\)

⇒ 8a - 3b -16 = 0

• Solving (i) and (ii), a = 5, b = 8 
• So option 1 is correct.

Concept:

Rolle’s Theorem,

• Rolle's theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that
• f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.
• Here, a = 3, b = 4 then f(3) = f(4)
• Also, f'(c) = 0

Calculation:

Given:

\(f\left( x \right) = {\log _e}\left( {\frac{{{x^2} + α }}{{7x}}} \right)\)​​

\(⇒ \frac{{9 + α }}{{21}} = \frac{{16 + α }}{{28}}\)

⇒ α = 12

Now after putting the value of  α, differentiate it with respect to x,

\(f\left( x \right) = {\log _e}\left( {\frac{{{x^2} + α }}{{7x}}} \right) \)

\(f'\left( x \right) =\frac{1}{(\frac{x^2+12}{7x})}\times\frac{1}{7}(\frac{2x \times x-(x^2+12)}{x^2})\)

\(⇒f'(x) = \frac{{7x}}{{{x^2} + 12}} \times \frac{{{x^2} - 12}}{{7{x^2}}} = \frac{{{x^2} - 12}}{{x\left( {{x^2} + 12} \right)}}\)

As per Rolle’s Theorem,

∴ f'(c) = 0

\( \frac{{{x^2} - 12}}{{x\left( {{x^2} + 12} \right)}}=0\)

x² -12 = 0

x2 =12

x = ±√12 

⇒ c = 2√3 

• So, the correct answer is option 2