Mean Value Theorems

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

(i) Given function is:

⇒ on [1,3]

Let us check the differentiability of the function f(x).

Let’s find the derivative of f(x),

⇒

⇒

⇒

⇒

⇒

Let’s the differentiability at the value of x = 2

⇒

⇒

⇒

⇒

∴ f is not differentiable at x = 2, so it is not differentiable in the closed interval (1,3).

So, Rolle’s theorem is not applicable for the function f on the interval [1,3].

Given function is:

⇒ f(x) = [x], – 1≤x≤1 where [x] denotes the greatest integer not exceeding x.

Let us check the continuity of the function ‘f’.

Here in the interval xϵ[ – 1,1], the function has to be Right continuous at x = 1 and left continuous at x = 1.

⇒

⇒ where h>0.

⇒

⇒ ......(1)

⇒

⇒ , where h>0

⇒

⇒ ......(2)

From (1) and (2), we can see that the limits are not the same so, the function is not continuous in the interval [ – 1,1].

∴ Rolle’s theorem is not applicable for the function f in the interval [ – 1,1].

Given function is:

⇒ for – 1≤x≤1

Let us check the continuity of the function ‘f’ at the value of x = 0.

We can not directly find the value of limit at x = 0, as the function is not valid at x = 0. So, we take the limit on either sides and x = 0, and we check whether they are equal or not.

Right – Hand Limit:

⇒

We assume that the limit , kϵ[ – 1,1].

⇒ , where h>0

⇒

⇒

⇒ ...... (1)

Left – Hand Limit:

⇒

⇒ , where h>0

⇒

⇒

⇒

⇒

⇒ ......(2)

From (1) and (2), we can see that the Right hand and left – hand limits are not equal, so the function ‘f’ is not continuous at x = 0.

∴ Rolle’s theorem is not applicable to the function ‘f’ in the interval [ – 1,1].

Given function is:

⇒ f(x) = 2x² – 5x + 3 on [1,3]

Since given function ‘f’ is a polynomial. So, it is continuous and differentiable every where.

Now, we find the values of function at the extremum values.

⇒ f(1) = 2(1)²–5(1) + 3

⇒ f(1) = 2 – 5 + 3

⇒ f(1) = 0 ......(1)

⇒ f(3) = 2(3)²–5(3) + 3

⇒ f(3) = 2(9)–15 + 3

⇒ f(3) = 18 – 12

⇒ f(3) = 6 ......(2)

From (1) and (2), we can say that,

f(1)≠f(3)

∴ Rolle’s theorem is not applicable for the function f in interval [1,3].

Given function is:

⇒ on [ – 1,1]

Let’s find the derivative of the given function:

⇒

⇒

⇒

⇒

Let’s check the differentiability of the function at x = 0.

⇒

⇒

⇒

Since the limit for the derivative is undefined at x = 0, we can say that f is not differentiable at x = 0.

∴ Rolle’s theorem is not applicable to the function ‘f’ on [ – 1,1].

Given function is:

⇒

Let’s check the continuity at x = 1 as the equation of function changes.

Left – Hand Limit:

⇒

⇒

⇒ ......(1)

Right – Hand Limit:

⇒

⇒

⇒ ......(2)

From (1) and (2), we can see that the values of both side limits are not equal. So, the function ‘f’ is not continuous at x = 1.

∴ Rolle’s theorem is not applicable to the function ‘f’ in the interval [0,2].

First let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = x² – 8x + 12 on [2,6]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

Let us find the values at extremums:

⇒ f(2) = 2² – 8(2) + 12

⇒ f(2) = 4 – 16 + 12

⇒ f(2) = 0

⇒ f(6) = 6² – 8(6) + 12

⇒ f(6) = 36 – 48 + 12

⇒ f(6) = 0

∴ f(2) = f(6), Rolle’s theorem applicable for function ‘f’ on [2,6].

Let’s find the derivative of f(x):

⇒

⇒

⇒

⇒ f’(x) = 2x – 8

We have f’(c) = 0 cϵ(2,6), from the definition given above.

⇒ f’(c) = 0

⇒ 2c – 8 = 0

⇒ 2c = 8

⇒

⇒ C = 4ϵ(2,6)

∴ Rolle’s theorem is verified.

First let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = x² – 4x + 3 on [1,3]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

Let us find the values at extremums:

⇒ f(1) = 1² – 4(1) + 3

⇒ f(1) = 1 – 4 + 3

⇒ f(1) = 0

⇒ f(3) = 3² – 4(3) + 3

⇒ f(3) = 9 – 12 + 3

⇒ f(3) = 0

∴ f(1) = f(3), Rolle’s theorem applicable for function ‘f’ on [1,3].

Let’s find the derivative of f(x):

⇒

⇒

⇒

⇒ f’(x) = 2x – 4

We have f’(c) = 0 cϵ(1,3), from the definition given above.

⇒ f’(c) = 0

⇒ 2c – 4 = 0

⇒ 2c = 4

⇒

⇒ C = 2ϵ(1,3)

∴ Rolle’s theorem is verified.

First let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = (x – 1)(x – 2)² on [1,2]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e, on R.

Let us find the values at extremums:

⇒ f(1) = (1 – 1)(1 – 2)²

⇒ f(1) = 0(1)²

⇒ f(1) = 0

⇒ f(2) = (2 – 1)(2 – 2)²

⇒ f(2) = 1.0²

⇒ f(2) = 0

∴ f(1) = f(2), Rolle’s theorem applicable for function ‘f’ on [1,2].

Let’s find the derivative of f(x):

⇒

Differentiating using UV rule

⇒

⇒ f’(x) = ((x – 2)²×1) + ((x – 1)×2×(x – 2))

⇒ f’(x) = x² – 4x + 4 + 2(x² – 3x + 2)

⇒ f’(x) = 3x² – 10x + 8

We have f’(c) = 0 cϵ(1,2), from the definition given above.

⇒ f’(c) = 0

⇒ 3c² – 10c + 8 = 0

⇒

⇒

⇒

⇒

⇒ ϵ(1,2) (neglecting the value 2)

∴ Rolle’s theorem is verified.

First let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = x(x – 1)² on [0,1]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e, on R.

Let us find the values at extremums:

⇒ f(0) = 0(0 – 1)²

⇒ f(0) = 0

⇒ f(1) = 1(1 – 1)²

⇒ f(1) = 0²

⇒ f(1) = 0

∴ f(0) = f(1), Rolle’s theorem applicable for function ‘f’ on [0,1].

Let’s find the derivative of f(x):

⇒

Differentiating using UV rule,

⇒

⇒ f’(x) = ((x – 1)²×1) + (x×2×(x – 1))

⇒ f’(x) = (x – 1)² + 2(x² – x)

⇒ f’(x) = x² – 2x + 1 + 2x² – 2x

⇒ f’(x) = 3x² – 4x + 1

We have f’(c) = 0 cϵ(0,1), from the definition given above.

⇒ f’(c) = 0

⇒ 3c² – 4c + 1 = 0

⇒

⇒

⇒

⇒

⇒ ϵ(0,1)

∴ Rolle’s theorem is verified.

First let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = (x² – 1)(x – 2) on [ – 1,2]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e, on R.

Let us find the values at extremums:

⇒ f( – 1) = (( – 1)² – 1)( – 1 – 2)

⇒ f( – 1) = (1 – 1)( – 3)

⇒ f( – 1) = (0)( – 3)

⇒ f( – 1) = 0

⇒ f(2) = (2² – 1)(2 – 2)

⇒ f(2) = (4 – 1)(0)

⇒ f(2) = 0

∴ f( – 1) = f(2), Rolle’s theorem applicable for function ‘f’ on [ – 1,2].

Let’s find the derivative of f(x):

⇒

Differentiating using UV rule,

⇒

⇒ f’(x) = ((x – 2)×2x) + ((x² – 1)×1)

⇒ f’(x) = 2x² – 4x + x² – 1

⇒ f’(x) = 2x² – 4x – 1

We have f’(c) = 0 cϵ( – 1,2), from the definition given above.

⇒ f’(c) = 0

⇒ 2c² – 4c – 1 = 0

⇒

⇒

⇒

⇒

⇒

⇒ ϵ( – 1,2)

∴ Rolle’s theorem is verified.

First let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = x(x – 4)² on [0,4]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

Let us find the values at extremums:

⇒ f(0) = 0(0 – 4)²

⇒ f(0) = 0

⇒ f(4) = 4(4 – 4)²

⇒ f(4) = 4(0)²

⇒ f(4) = 0

∴ f(0) = f(4), Rolle’s theorem applicable for function ‘f’ on [0,4].

Let’s find the derivative of f(x):

⇒

Differentiating using UV rule,

⇒

⇒ f’(x) = ((x – 4)²×1) + (x×2×(x – 4))

⇒ f’(x) = (x – 4)² + 2(x² – 4x)

⇒ f’(x) = x² – 8x + 16 + 2x² – 8x

⇒ f’(x) = 3x² – 16x + 16

We have f’(c) = 0 cϵ(0,4), from the definition given above.

⇒ f’(c) = 0

⇒ 3c² – 16c + 16 = 0

⇒

⇒

⇒

⇒

⇒ ϵ(0,4)

∴ Rolle’s theorem is verified.

First let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = x(x – 2)² on [0,2]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e, on R.

Let us find the values at extremums:

⇒ f(0) = 0(0 – 2)²

⇒ f(0) = 0

⇒ f(2) = 2(2 – 2)²

⇒ f(2) = 2(0)²

⇒ f(2) = 0

∴ f(0) = f(2), Rolle’s theorem applicable for function ‘f’ on [0,2].

Let’s find the derivative of f(x):

⇒

Differentiating using UV rule,

⇒

⇒ f’(x) = ((x – 2)²×1) + (x×2×(x – 2))

⇒ f’(x) = (x – 2)² + 2(x² – 2x)

⇒ f’(x) = x² – 4x + 4 + 2x² – 4x

⇒ f’(x) = 3x² – 8x + 4

We have f’(c) = 0 cϵ(0,1), from the definition given above.

⇒ f’(c) = 0

⇒ 3c² – 8c + 4 = 0

⇒

⇒

⇒

⇒

⇒ c = 1ϵ(0,2)

∴ Rolle’s theorem is verified.

First let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = x² + 5x + 6 on [ – 3, – 2]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

Let us find the values at extremums:

⇒ f( – 3) = ( – 3)² + 5( – 3) + 6

⇒ f( – 3) = 9 – 15 + 6

⇒ f( – 3) = 0

⇒ f( – 2) = ( – 2)² + 5( – 2) + 6

⇒ f( – 2) = 4 – 10 + 6

⇒ f( – 2) = 0

∴ f( – 3) = f( – 2), Rolle’s theorem applicable for function ‘f’ on [ – 3, – 2].

Let’s find the derivative of f(x):

⇒

⇒

⇒

⇒ f’(x) = 2x + 5

We have f’(c) = 0 cϵ( – 3, – 2), from the definition given above.

⇒ f’(c) = 0

⇒ 2c + 5 = 0

⇒ 2c = – 5

⇒

⇒ C = – 2.5ϵ( – 3, – 2)

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is :

⇒

We know that cosine function is continuous and differentiable on R.

Let’s find the values of the function at an extremum,

⇒

⇒

⇒

We know that cos( – x) = cosx

⇒ f(0) = 0

⇒

⇒

⇒

⇒

We got , so there exist a such that f’(c) = 0.

Let’s find the derivative of f(x)

⇒

⇒

⇒

We have f’(c) = 0,

⇒

⇒

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given fuction is:

⇒ f(x) = sin2x on

We know that sine function is continuous and differentiable on R.

Let’s find the values of function at extremum,

⇒ f(0) = sin2(0)

⇒ f(0) = sin0

⇒ f(0) = 0

⇒

⇒

⇒

We got , so there exist a such that f’(c) = 0.

Let’s find the derivative of f(x)

⇒

⇒

⇒ f’(x) = 2cos2x

We have f’(c) = 0,

⇒ 2cos2c = 0

⇒

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ cos2x on

We know that cosine function is continuous and differentiable on R.

Let’s find the values of the function at an extremum,

⇒

⇒

We know that cos( – x) = cosx

⇒ f(0) = 0

⇒

⇒

⇒

We got , so there exist a such that f’(c) = 0.

Let’s find the derivative of f(x)

⇒

⇒

⇒ f’(x) = – 2sin2x

We have f’(c) = 0,

⇒ – 2sin2c = 0

⇒ 2c = 0

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = e^xsinx on [0,]

We know that exponential and sine functions are continuous and differentiable on R.

Let’s find the values of the function at an extremum,

⇒ f(0) = e⁰sin(0)

⇒ f(0) = 1×0

⇒ f(0) = 0

⇒

⇒

⇒

We got , so there exist a such that f’(c) = 0.

Let’s find the derivative of f(x)

⇒

⇒

⇒ f’(x) = e^x(sinx + cosx)

We have f’(c) = 0,

⇒ e^c(sinc + cosc) = 0

⇒ sinc + cosc = 0

⇒

⇒

⇒

⇒

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = e^xcosx on

We know that exponential and cosine functions are continuous and differentiable on R.

Let’s find the values of the function at an extremum,

⇒

⇒

⇒

⇒

⇒

⇒

We got , so there exist a such that f’(c) = 0.

Let’s find the derivative of f(x)

⇒

⇒

⇒ f’(x) = e^x( – sinx + cosx)

We have f’(c) = 0,

⇒ e^c( – sinc + cosc) = 0

⇒ – sinc + cosc = 0

⇒

⇒

⇒

⇒

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = cos 2x on [0,]

We know that cosine function is continuous and differentiable on R.

Let’s find the values of function at extremum,

⇒ f(0) = cos2(0)

⇒ f(0) = cos(0)

⇒ f(0) = 1

⇒ f() = cos2()

⇒ f() = cos(2)

⇒ f() = 1

We got , so there exist a such that f’(c) = 0.

Let’s find the derivative of f(x)

⇒

⇒

⇒ f’(x) = – 2sin2x

We have f’(c) = 0,

⇒ – 2sin2c = 0

⇒ 2c = 0

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ on [0,]

This can be written as

⇒ f(x) = e ^{– x}sinx on [0,]

We know that exponential and sine functions are continuous and differentiable on R.

Let’s find the values of the function at an extremum,

⇒ f(0) = e ^{– 0}sin(0)

⇒ f(0) = 1×0

⇒ f(0) = 0

⇒

⇒

⇒

We got , so there exist a such that f’(c) = 0.

Let’s find the derivative of f(x)

⇒

⇒

⇒ f’(x) = sinx( – e ^{– x}) + e ^{– x}(cosx)

⇒ f’(x) = e ^{– x}( – sinx + cosx)

We have f’(c) = 0,

⇒ e ^{– c}( – sinc + cosc) = 0

⇒ – sinc + cosc = 0

⇒

⇒

⇒

⇒

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = sin3x on [0,]

We know that sine function is continuous and differentiable on R.

Let’s find the values of function at extremum,

⇒ f(0) = sin3(0)

⇒ f(0) = sin0

⇒ f(0) = 0

⇒ f() = sin3()

⇒ f() = sin(3)

⇒ f() = 0

We got , so there exist a such that f’(c) = 0.

Let’s find the derivative of f(x)

⇒

⇒

⇒ f’(x) = 3cos3x

We have f’(c) = 0,

⇒ 3cos3c = 0

⇒

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ on [ – 1,1]

We know that exponential function is continuous and differentiable over R.

Let’s find the value of function f at extremums,

⇒

⇒

⇒ f( – 1) = e⁰

⇒ f( – 1) = 1

⇒

⇒

⇒ f(1) = e⁰

⇒ f(1) = 1

We got f( – 1) = f(1) so, there exists a cϵ( – 1,1) such that f’(c) = 0.

Let’s find the derivative of the function f:

⇒

⇒

⇒

We have f’(c) = 0

⇒

⇒ 2c = 0

⇒ c = 0ϵ[ – 1,1]

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = log(x² + 2) – log3 on [ – 1,1]

We know that logarithmic function is continuous and differentiable in its own domain.

We check the values of the function at the extremum,

⇒ f( – 1) = log(( – 1)² + 2) – log3

⇒ f( – 1) = log(1 + 2) – log3

⇒ f( – 1) = log3 – log3

⇒ f( – 1) = 0

⇒ f(1) = log(1² + 2) – log3

⇒ f(1) = log(1 + 2) – log3

⇒ f(1) = log3 – log3

⇒ f(1) = 0

We have got f( – 1) = f(1). So, there exists a c such that cϵ( – 1,1) such that f’(c) = 0.

Let’s find the derivative of the function f,

⇒

⇒

⇒

We have f’(c) = 0

⇒

⇒ 2c = 0

⇒ c = 0ϵ( – 1,1)

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = sinx + cosx on

We know that sine and cosine functions are continuous and differentiable on R.

Let’s the value of function f at extremums:

⇒ f(0) = sin(0) + cos(0)

⇒ f(0) = 0 + 1

⇒ f(0) = 1

⇒

⇒

⇒

We have got . So, there exists a cϵ such that f’(c) = 0.

Let’s find the derivative of the function ‘f’.

⇒

⇒ f’(x) = cosx – sinx

We have f’(c) = 0

⇒ cosc – sinc = 0

⇒

⇒

⇒

⇒

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = 2sinx + sin2x on [0,]

We know that sine function continuous and differentiable over R.

Let’s check the values of function f at the extremums

⇒ f(0) = 2sin(0) + sin2(0)

⇒ f(0) = 2(0) + 0

⇒ f(0) = 0

⇒ f() = 2sin() + sin2()

⇒ f() = 2(0) + 0

⇒ f() = 0

We have got f(0) = f(). So, there exists a cϵ(0,) such that f’(c) = 0.

Let’s find the derivative of function ‘f’.

⇒

⇒

⇒ f’(x) = 2cosx + 2cos2x

⇒ f’(x) = 2cosx + 2(2cos²x – 1)

⇒ f’(x) = 4cos²x + 2cosx – 2

We have f’(c) = 0,

⇒ 4cos²c + 2cosc – 2 = 0

⇒ 2cos²c + cosc – 1 = 0

⇒ 2cos²c + 2cosc – cosc – 1 = 0

⇒ 2cosc(cosc + 1) – 1(cosc + 1) = 0

⇒ (2cosc – 1)(cosc + 1) = 0

⇒

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ on [ – 1,0]

We know that sine function is continuous and differentiable over R.

Let’s check the values of ‘f’ at an extremum

⇒

⇒

⇒

⇒ f( – 1) = 0

⇒

⇒

⇒ f(0) = 0 – 0

⇒ f(0) = 0

We have got f( – 1) = f(0). So, there exists a cϵ( – 1,0) such that f’(c) = 0.

Let’s find the derivative of the function ‘f’

⇒

⇒

⇒

We have f’(c) = 0

⇒

⇒

⇒

⇒

⇒

⇒

Cosine is positive between , for our convenience we take the interval to be , since the values of the cosine repeats.

We know that value is nearly equal to 1. So, the value of the c nearly equal to 0.

So, we can clearly say that cϵ( – 1,0).

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒

We know that sine function is continuous and differentiable over R.

Let’s check the values of function ‘f’ at the extremums,

⇒

⇒ f(0) = 0 – 4(0)

⇒ f(0) = 0

⇒

⇒

⇒

⇒

⇒ .

We got . So, there exists a cϵ such that f’(c) = 0.

Let’s find the derivative of function ‘f.’

⇒

⇒

⇒

⇒

⇒

We have f’(c) = 0

⇒

⇒

⇒

We know

⇒

⇒

⇒

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = 4^sinx on [0,]

We that sine function is continuous and differentiable over R.

Let’s check the values of function ‘f’ at extremums

⇒ f(0) = 4^sin(0)

⇒ f(0) = 4⁰

⇒ f(0) = 1

⇒ f() = 4^sinπ

⇒ f() = 4⁰

⇒ f() = 1

We got f(0) = f(). So, there exists a cϵ(0,) such that f’(c) = 0.

Let’s find the derivative of ‘f’

⇒

⇒

⇒

We have f’(c) = 0

⇒ 4^sinclog4cosc = 0

⇒ cosc = 0

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = x² – 5x + 4 on [1,4]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

Let us find the values at extremums:

⇒ f(1) = 1² – 5(1) + 4

⇒ f(1) = 1 – 5 + 4

⇒ f(1) = 0

⇒ f(4) = 4² – 5(4) + 4

⇒ f(4) = 16 – 20 + 4

⇒ f(4) = 0

∴ We got f(1) = f(4). So, there exists a cϵ(1,4) such that f’(c) = 0.

Let’s find the derivative of f(x):

⇒

⇒

⇒

⇒ f’(x) = 2x – 5

We have f’(c) = 0

⇒ f’(c) = 0

⇒ 2c – 5 = 0

⇒ 2c = 5

⇒

⇒ C = 2.5ϵ(1,4)

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = sin⁴x + cos⁴x on

We know that sine and cosine functions are continuous and differentiable functions over R.

Let’s find the value of function ‘f’ at extremums

⇒ f(0) = sin⁴(0) + cos⁴(0)

⇒ f(0) = (0)⁴ + (1)⁴

⇒ f(0) = 0 + 1

⇒ f(0) = 1

⇒

⇒

⇒

⇒

We got . So, there exists a cϵ such that f’(c) = 0.

Let’s find the derivative of the function ‘f’.

⇒

⇒

⇒ f’(x) = 4sin³xcosx–4cos³xsinx

⇒ f’(x) = 4sinxcosx(sin²x – cos²x)

⇒ f’(x) = 2(2sinxcosx)( – cos2x)

⇒ f’(x) = – 2(sin2x)(cos2x)

⇒ f’(x) = – sin4x

We have f’(c) = 0

⇒ – sin4c = 0

⇒ sin4c = 0

⇒ 4c = 0 or

⇒

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = sinx – sin2x on [0,]

We know that sine function is continuous and differentiable over R.

Let’s check the values of the function ‘f’ at the extremums.

⇒ f(0) = sin(0)–sin2(0)

⇒ f(0) = 0 – sin(0)

⇒ f(0) = 0

⇒ f() = sin() – sin2()

⇒ f() = 0 – sin(2)

⇒ f() = 0

We got f(0) = f(). So, there exists a cϵ such that f’(c) = 0.

Let’s find the derivative of the function ‘f’

⇒

⇒

⇒

⇒ f’(x) = cosx – 2(2cos²x – 1)

⇒ f’(x) = cosx – 4cos²x + 2

We have f’(c) = 0

⇒ cosc – 4cos²c + 2 = 0

⇒

⇒

⇒

We can see that cϵ(0,)

∴ Rolle’s theorem is verified.

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ y = 16 – x², xϵ[ – 1,1]

We know that polynomial function is continuous and differentiable over R.

Let’s check the values of ‘y’ at extremums

⇒ y( – 1) = 16 – ( – 1)²

⇒ y( – 1) = 16 – 1

⇒ y( – 1) = 15

⇒ y(1) = 16 – (1)²

⇒ y(1) = 16 – 1

⇒ y(1) = 15

We got y( – 1) = y(1). So, there exists a cϵ( – 1,1) such that f’(c) = 0.

We know that for a curve g, the value of the slope of the tangent at a point r is given by g’(r).

Let’s find the derivative of curve y

⇒

⇒ y’ = – 2x

We have y’(c) = 0

⇒ – 2c = 0

⇒ c = 0ϵ( – 1,1)

Value of y at x = 1 is

⇒ y = 16 – 0²

⇒ y = 16

∴ The point at which the curve y has a tangent parallel to x – axis (since the slope of x – axis is 0) is (0,16).

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ y = x² on [ – 2,2]

We know that polynomials are continuous and differentiable over R.

Let’s check the values of y at the extremums

⇒ y( – 2) = ( – 2)²

⇒ y( – 2) = 4

⇒ y(2) = (2)²
⇒ y(2) = 4

We got y( – 2) = y(2). So, there exists a c such that f’(c) = 0.

For a curve g to have a tangent parallel to x – axis at point r, the criteria to be satisfied is g’(r) = 0.

⇒ y’(x) = 0

⇒

⇒ 2x = 0

⇒ x = 0

The value of y is

⇒ y = (0)²

⇒ y = 0

The point at which the curve has tangent parallel to x – axis is (0,0).

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ on [ – 1,1]

We know that exponential functions are continuous and differentiable over R.

Let’s check the values of y at the extremums

⇒

⇒ y( – 1) = e^{1 – 1}

⇒ y( – 1) = e⁰
⇒ y( – 1) = 1

⇒

⇒ y(1) = e^{1 – 1}

⇒ y(1) = e⁰

⇒ y(1) = 1

We got y( – 1) = y(1). So, there exists a c such that f’(c) = 0.

For a curve g to have a tangent parallel to the x – axis at point r, the criteria to be satisfied is g’(r) = 0.

⇒ y’(x) = 0

⇒

⇒

⇒

⇒ 2x = 0

⇒ x = 0

The value of y is

⇒

⇒ y = e^{1 – 0}

⇒ y = e¹

⇒ y = e

The point at which the curve has a tangent parallel to the x – axis is (0,e).

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ y = 12(x + 1)(x – 2) on [ – 1,2]

We know that polynomials are continuous and differentiable over R.

Let’s check the values of y at the extremums

⇒ y( – 1) = 12( – 1 + 1)( – 1 – 2)

⇒ y( – 1) = 12(0)( – 3)

⇒ y( – 1) = 0

⇒ y(2) = 12(2 + 1)(2 – 2)
⇒ y(2) = 12(3)(0)

⇒ y(2) = 0

We got y( – 1) = y(2). So, there exists a c such that f’(c) = 0.

For a curve g to have a tangent parallel to the x – axis at point r, the criteria to be satisfied is g’(r) = 0.

⇒ y’(x) = 0

⇒

⇒

⇒ ((x + 1)×1) + ((x – 2)×1) = 0

⇒ x + 1 + x – 2 = 0

⇒ 2x – 1 = 0

⇒ 2x = 1

⇒

The value of y is

⇒

⇒

⇒ y = – 27

The point at which the curve has tangent parallel to x – axis is .

Given that f is continuous and differentiable in the interval [ – 5,5].

It is also given that f’(x) doesn’t vanish anywhere.

According to Rolle’s theorem for a differentiable function on [a,b] will have atleast one cϵ(a,b) such that f’(c) = 0, if the following condition had satisfied:

⇒ f(a) = f(b).

According to the problem it is given for any value of x, say r the values never equals to zero.

⇒ f’(r)≠0

This is possible when Rolle’s theorem is not applicable.

Let us Recap the Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

First, two conditions are satisfied according to the problem, so the only condition that cannot be satisfied is (c).

So, we can clearly say that f( – 5)≠f(5).

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = [x] for xϵ[5,9]

Let us check the continuity of the function ‘f’.

Here in the interval xϵ[5,9], the function has to be Right continuous at x = 5 and left continuous at x = 5.

Right Hand Limit:

⇒

⇒ where h>0.

⇒

⇒ ......(1)

Left Hand Limit:

⇒

⇒ , where h>0

⇒

⇒ ......(2)

From (1) and (2), we can clearly see that the limits are not same so, the function is not continuous in the interval [5,9].

∴ Rolle’s theorem is not applicable for the function f in the interval [5,9].

Given function is:

⇒ f(x) = [x] for xϵ[ – 2,2]

Let us check the continuity of the function ‘f’.

Here in the interval xϵ[ – 2,1], the function has to be Right continuous at x = 2 and left continuous at x = 2.

Right Hand Limit:

⇒

⇒ where h>0.

⇒

⇒ ......(1)

Left Hand Limit:

⇒

⇒ , where h>0

⇒

⇒ ......(2)

From (1) and (2), we can see that the limits are not the same so, the function is not continuous in the interval [ – 2,2].

∴ Rolle’s theorem is not applicable for the function f in the interval [ – 2,2].

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

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

Given function is:

⇒ f(x) = x³ + bx² + cx, xϵ[1,2]

According to the problem the Rolle’s theorem holds for the function ‘f’ at .

We can say that .

Let’s find the derivative of f(x)

⇒

⇒

⇒ f’(x) = 3x² + 2bx + c

We have

⇒

⇒

⇒

⇒ 8b + 3c = – 16 ...... (1)

We also have f(1) = f(2)

⇒ (1)³ + b(1)² + c(1) = (2)³ + b(2)² + c(2)

⇒ 1 + b(1) + c = 8 + b(4) + 2c

⇒ 3b + c = – 7 ......(2)

On solving (1) and (2), we get

⇒ b = – 5 and c = 8

∴ The values of b and c is – 5 and 8.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = x² – 1 on [2, 3]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [2, 3] and differentiable in (2, 3). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x² – 1

Differentiating with respect to x:

f’(x) = 2x

For f’(c), put the value of x=c in f’(x):

f’(c)= 2c

For f(3), put the value of x=3 in f(x):

f(3)= (3)² – 1

= 9 – 1

= 8

For f(2), put the value of x=2 in f(x):

f(2)= (2)² – 1

= 4 – 1

= 3

∴ f’(c) = f(3) – f(2)

⇒ 2c = 8 – 3

⇒ 2c = 5

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = x³ – 2x² – x + 3 on [0, 1]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [0, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x³ – 2x² – x + 3

Differentiating with respect to x:

f’(x) = 3x² – 2(2x) – 1

= 3x² – 4x – 1

For f’(c), put the value of x=c in f’(x):

f’(c)= 3c² – 4c – 1

For f(1), put the value of x=1 in f(x):

f(1)= (1)³ – 2(1)² – (1) + 3

= 1 – 2 – 1 + 3

= 1

For f(0), put the value of x=0 in f(x):

f(0)= (0)³ – 2(0)² – (0) + 3

= 0 – 0 – 0 + 3

= 3

∴ f’(c) = f(1) – f(0)

⇒ 3c² – 4c – 1 = 1 – 3

⇒ 3c² – 4c = 1 + 1 – 3

⇒ 3c² – 4c = – 1

⇒ 3c² – 4c + 1 = 0

⇒ 3c² – 3c – c + 1 = 0

⇒ 3c(c – 1) – 1(c – 1) = 0

⇒ (3c – 1) (c – 1) = 0

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = x (x – 1) on [1, 2]

= x² – x

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [1, 2] and differentiable in (1, 2). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x² – x

Differentiating with respect to x:

f’(x) = 2x – 1

For f’(c), put the value of x=c in f’(x):

f’(c)= 2c – 1

For f(2), put the value of x=2 in f(x):

f(2)= (2)² – 2

= 4 – 2

= 2

For f(1), put the value of x=1 in f(x):

f(1)= (1)² – 1

= 1 – 1

= 0

∴ f’(c) = f(2) – f(1)

⇒ 2c – 1 = 2 – 0

⇒ 2c = 2 + 1

⇒ 2c = 3

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = x² – 3x + 2 on [ – 1, 2]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [ – 1, 2] and differentiable in ( – 1, 2). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x² – 3x + 2

Differentiating with respect to x:

f’(x) = 2x – 3

For f’(c), put the value of x=c in f’(x):

f’(c)= 2c – 3

For f(2), put the value of x=2 in f(x):

f(2)= (2)² – 3(2) + 2

= 4 – 6 + 2

= 0

For f( – 1), put the value of x= – 1 in f(x):

f( – 1) = ( – 1)² – 3( – 1) + 2

= 1 + 3 + 2

= 6

⇒ 2c = – 2 + 3

⇒ 2c= – 1

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = 2x² – 3x + 1 on [1, 3]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [1, 3] and differentiable in (1, 3). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = 2x² – 3x + 1

Differentiating with respect to x:

f’(x) = 2(2x) – 3

= 4x – 3

For f’(c), put the value of x=c in f’(x):

f’(c)= 4c – 3

For f(3), put the value of x=3 in f(x):

f(3)= 2(3)² – 3(3) + 1

= 2(9) – 9 + 1

= 18 – 8 = 10

For f(1), put the value of x=1 in f(x):

f(1) = 2(1)² – 3(1) + 1

= 2(1) – 3 + 1

= 2 – 2 = 0

⇒ 4c = 5 + 3

⇒ 4c= 8

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = x² – 2x + 4 on [1, 5]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [1, 5] and differentiable in (1, 5). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x² – 2x + 4

Differentiating with respect to x:

f’(x) = 2x – 2

For f’(c), put the value of x=c in f’(x):

f’(c)= 2c – 2

For f(5), put the value of x=5 in f(x):

f(5)= (5)² – 2(5) + 4

= 25 – 10 + 4

= 19

For f(1), put the value of x=1 in f(x):

f(1) = (1)² – 2(1) + 4

= 1 – 2 + 4

= 3

⇒ 2c = 4 + 2

⇒ 2c= 6

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = 2x – x² on [0, 1]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [0, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

⇒ f’(c) = f(1) – f(0)

f(x) = 2x – x²

Differentiating with respect to x:

f’(x) = 2 – 2x

For f’(c), put the value of x=c in f’(x):

f’(c)= 2 – 2c

For f(1), put the value of x=1 in f(x):

f(1)= 2(1) – (1)²

= 2 – 1

= 1

For f(0), put the value of x=0 in f(x):

f(0) = 2(0) – (0)²

= 0 – 0

= 0

f’(c) = f(1) – f(0)

⇒ 2 – 2c = 1 – 0

⇒ – 2c = 1 – 2

⇒ – 2c = – 1

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = (x – 1) (x – 2)(x – 3) on [0, 4]

= (x² – x – 2x + 3) (x – 3)

= (x² – 3x + 3) (x – 3)

= x³ – 3x² + 3x – 3x² + 9x – 9

= x³ – 6x² + 12x – 9 on [0, 4]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x³ – 6x² + 12x – 9

Differentiating with respect to x:

f’(x) = 3x² – 6(2x) + 12

= 3x² – 12x + 12

For f’(c), put the value of x=c in f’(x):

f’(c)= 3c² – 12c + 12

For f(4), put the value of x=4 in f(x):

f(4)= (4)³ – 6(4)² + 12(4) – 9

= 64 – 96 + 48 – 9

= 7

For f(0), put the value of x=0 in f(x):

f(0)= (0)³ – 6(0)² + 12(0) – 9

= 0 – 0 + 0 – 9

= – 9

⇒ 3c² – 12c + 12 = 4

⇒ 3c² – 12c + 8 = 0

For quadratic equation, ax² + bx + c = 0

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

Here,

⇒ 25 – x² >0

⇒ x² < 25

⇒ – 5 < x < 5

∴ f(x) is continuous in [ – 3, 4]

Differentiating with respect to x:

Here also,

⇒ – 5 < x < 5

∴ f(x) is differentiable in ( – 3, 4)

So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

On differentiating with respect to x:

For f’(c), put the value of x=c in f’(x):

For f(4), put the value of x=4 in f(x):

⇒ f(4) = 3

For f( – 3), put the value of x= – 3 in f(x):

⇒ f( – 3) = 4

Squaring both sides:

⇒ 49c² = 25 – c²

⇒ 50c² = 25

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = tan ^{– 1} x on [0, 1]

tan ^{– 1} x has unique value for all x between 0 and 1.

∴ f(x) is continuous in [0, 1]

f(x) = tan ^{– 1} x

Differentiating with respect to x:

x² always has value greater than 0.

⇒ 1 + x² > 0

∴ f(x) is differentiable in (0, 1)

So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

⇒ f’(c)= f(1) – f(0)

f(x) = tan ^{– 1} x

Differentiating with respect to x:

For f’(c), put the value of x=c in f’(x):

For f(1), put the value of x=1 in f(x):

f(1) = tan ^{– 1} 1

For f(0), put the value of x=0 in f(x):

f(0) = tan ^{– 1} 0

⇒ f(0) = 0

f’(c)= f(1) – f(0)

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

∴ f(x) is continuous in [1, 3]

Differentiating with respect to x:

Here,

⇒ f’(x) exists for all values except 0

∴ f(x) is differentiable in (1, 3)

So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

On differentiating with respect to x:

For f’(c), put the value of x=c in f’(x):

For f(3), put the value of x=3 in f(x):

For f(1), put the value of x=1 in f(x):

⇒ f(1) = 2

⇒ 6(c² – 1) = 4c²

⇒ 6c² – 6 = 4c²

⇒ 6c² – 4c² = 6

⇒ 2c² = 6

⇒ c² = 3

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = x (x + 4)² on [0, 4]

= x [(x)²+2(4)(x)+(4)²]

= x(x² + 8x + 16)

= x³ + 8x² + 16x on [0, 4]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x³ + 8x² + 16x

Differentiating with respect to x:

f’(x) = 3x² + 8(2x) + 16

= 3x² + 16x + 16

For f’(c), put the value of x=c in f’(x):

f’(c)= 3c² + 16c + 16

For f(4), put the value of x=4 in f(x):

f(4)= (4)³ + 8(4)² + 16(4)

= 64 + 128 + 64

= 256

For f(0), put the value of x=0 in f(x):

f(0)= (0)³ + 8(0)² + 16(0)

= 0 + 0 + 0

= 0

⇒ 3c² + 16c + 16 = 64

⇒ 3c² + 16c + 16 – 64 = 0

⇒ 3c² + 16c – 48 = 0

For quadratic equation, ax² + bx + c = 0

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

Here,

⇒ x² – 4 >0

⇒ x² > 4

⇒ f(x) exists for all values expect ( – 2, 2)

∴ f(x) is continuous in [2, 4]

Differentiating with respect to x:

Here also,

⇒ f’(x) exists for all values of x except (2, – 2)

∴ f(x) is differentiable in (2, 4)

So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

On differentiating with respect to x:

For f’(c), put the value of x=c in f’(x):

For f(4), put the value of x=4 in f(x):

For f(2), put the value of x=2 in f(x):

⇒ f(2) = 0

Squaring both sides:

⇒ c² = 3(c² – 4)

⇒ c² = 3c² – 12

⇒ – 2c² = – 12

⇒ c² = 6

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = x² + x – 1 on [0, 4]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x² + x – 1

Differentiating with respect to x:

f’(x) = 2x + 1

For f’(c), put the value of x=c in f’(x):

f’(c)= 2c + 1

For f(4), put the value of x=4 in f(x):

f(4)= (4)² + 4 – 1

= 16 + 4 – 1

= 19

For f(0), put the value of x=0 in f(x):

f(0) = (0)² + 0 – 1

= 0 + 0 – 1

= – 1

⇒ 2c + 1 = 5

⇒ 2c = 5 – 1

⇒ 2c = 4

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = sin x – sin 2x – x on [0, π]

sin x and cos x functions are continuous everywhere on (−∞, ∞) and differentiable for all arguments.

So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = sin x – sin 2x – x

Differentiating with respect to x:

f(x) = sin x – sin 2x – x

For f’(c), put the value of x=c in f’(x):

f’(c) = cos c – 2cos 2c – 1

For f(π), put the value of x=π in f(x):

f(π) = sin π – sin 2π – π

= 0 – 0 – π

= – π

For f(0), put the value of x=0 in f(x):

f(0) = sin 0 – sin 2(0) – 0

= sin 0 – sin 0 – 0

= 0 – 0 – 0

= 0

⇒ cos c – 2cos 2c – 1 = – 1

⇒ cos c – 2(2cos² c – 1) = – 1 + 1

⇒ cos c – 4cos² c + 2 = 0

⇒ 4cos² c – cos c – 2 = 0

For quadratic equation, ax² + bx + c = 0

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = x³ – 5x² – 3x on [1, 3]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [1, 3] and differentiable in (1, 3). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x³ – 5x² – 3x

Differentiating with respect to x:

f’(x) = 3x² – 5(2x) – 3

= 3x² – 10x – 3

For f’(c), put the value of x=c in f’(x):

f’(c)= 3c² – 10c – 3

For f(3), put the value of x=3 in f(x):

f(3)= (3)³ – 5(3)² – 3(3)

= 27 – 45 – 9

= – 27

For f(1), put the value of x=1 in f(x):

f(1)= (1)³ – 5(1)² – 3(1)

= 1 – 5 – 3

= – 7

⇒ 3c² – 10c – 3 = – 10

⇒ 3c² – 10c – 3 + 10 = 0

⇒ 3c² – 10c + 7 = 0

⇒ 3c² – 7c – 3c + 7 = 0

⇒ c(3c – 7) – 1(3c – 7) = 0

⇒ (3c – 7) (c – 1) = 0

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = |x| on [ – 1, 1]

For differentiability at x=0,

{Since f(x)= – x, x<0}

= – 1

{Since f(x)= x, x>0}

= 1

⇒ f(x) is not differential at x=0

∴ Lagrange’s mean value theorem is not applicable for the function f(x) = |x| on [ – 1, 1].

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

Here,

⇒ f(x) exists for all values of x except 0

⇒ f(x) is discontinuous at x=0

∴ f(x) is not continuous in [ – 1, 1]

Hence the lagrange’s mean value theorem is not applicable to the

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

4x – 1>0

∴ f(x) is continuous in [1, 4]

Differentiating with respect to x:

Here,

⇒ 4x – 1>0

∴ f(x) is differentiable in (1, 4)

So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

On differentiating with respect to x:

For f’(c), put the value of x=c in f’(x):

For f(4), put the value of x=4 in f(x):

For f(1), put the value of x=1 in f(x):

⇒ (4c – 1)² = 45

Hence, Lagrange’s mean value theorem is verified.

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

Let f(x) = (x – 4)² on [4, 5]

This interval [a, b] is obtained by x – coordinates of the points of the chord.

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [4, 5] and differentiable in (4, 5). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = (x – 4)²

Differentiating with respect to x:

⇒ f’(x) = 2(x – 4)(1)

⇒ f’(x) = 2(x – 4)

For f’(c), put the value of x=c in f’(x):

f’(c)= 2(c – 4)

For f(5), put the value of x=5 in f(x):

f(5)= (5 – 4)²

= (1)²

= 1

For f(4), put the value of x=4 in f(x):

f(4) = (4 – 4)²

= (0)²

= 0

f’(c) = f(5) – f(4)

⇒ 2(c – 4) = 1 – 0

⇒ 2c – 8 = 1

⇒ 2c = 1 + 8

We know that, the value of c obtained in Lagrange’s Mean value Theorem is nothing but the value of x – coordinate of the point of the contact of the tangent to the curve which is parallel to the chord joining the points (4, 0) and (5, 1).

Now, Put this value of x in f(x) to obtain y:

y = (x – 4)²

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

Let f(x) = x² + x on [0, 1]

This interval [a, b] is obtained by x – coordinates of the points of the chord.

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [0, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x² + x

Differentiating with respect to x:

f’(x) = 2x + 1

For f’(c), put the value of x=c in f’(x):

f’(c)= 2c + 1

For f(1), put the value of x=1 in f(x):

f(1)= (1)² + 1

= 1 + 1

= 2

For f(0), put the value of x=0 in f(x):

f(0) = (0)² + 0

= 0 + 0

= 0

f’(c) = f(1) – f(0)

⇒ 2c + 1= 2 – 0

⇒ 2c = 2 – 1

⇒ 2c = 1

We know that the value of c obtained in Lagrange’s Mean value Theorem is nothing but the value of x – coordinate of the point of the contact of the tangent to the curve which is parallel to the chord joining the points (0, 0) and (1, 2).

Now, put this value of x in f(x) to obtain y:

y = x² + x

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

Let f(x) = (x – 3)² on [3, 4]

This interval [a, b] is obtained by x – coordinates of the points of the chord.

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [3, 4] and differentiable in (3, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = (x – 3)²

Differentiating with respect to x:

⇒ f’(x) = 2(x – 3)(1)

⇒ f’(x) = 2(x – 3)

For f’(c), put the value of x=c in f’(x):

f’(c)= 2(c – 3)

For f(4), put the value of x=4 in f(x):

f(4)= (4 – 3)²

= (1)²

= 1

For f(3), put the value of x=3 in f(x):

f(3) = (3 – 3)²

= (0)²

= 0

f’(c) = f(4) – f(3)

⇒ 2(c – 3) = 1 – 0

⇒ 2c – 6 = 1

⇒ 2c = 1 + 6

We know that, the value of c obtained in Lagrange’s Mean value Theorem is nothing but the value of x – coordinate of the point of the contact of the tangent to the curve which is parallel to the chord joining the points (3, 0) and (4, 1).

Now, Put this value of x in f(x) to obtain y:

y = (x – 3)²

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

Let f(x) = x³ – 3x on [1, 2]

This interval [a, b] is obtained by x – coordinates of the points of the chord.

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [1, 2] and differentiable in (1, 2). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x³ – 3x

Differentiating with respect to x:

f’(x) = 3x² – 3

For f’(c), put the value of x=c in f’(x):

f’(c)= 3c² – 3

For f(2), put the value of x=2 in f(x):

f(2)= (2)³ – 3(2)

= 8 – 6

= 2

For f(1), put the value of x=1 in f(x):

f(1) = (1)³ – 3(1)

= 1 – 3

= – 2

f’(c) = f(2) – f(1)

⇒ 3c² – 3 = 2 – ( – 2)

⇒ 3c² – 3 = 2 + 2

⇒ 3c² = 4 + 3

We know that, the value of c obtained in Lagrange’s Mean value Theorem is nothing but the value of x – coordinate of the point of the contact of the tangent to the curve which is parallel to the chord joining the points (1, – 2) and (2, 2).

Now, Put this value of x in f(x) to obtain y:

y = x³ – 3x

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

Let f(x) = x³ + 1 on [1, 3]

This interval [a, b] is obtained by x – coordinates of the points of the chord.

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [1, 3] and differentiable in (1, 3). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x³ + 1

Differentiating with respect to x:

f’(x) = 3x²

For f’(c), put the value of x=c in f’(x):

f’(c)= 3c²

For f(3), put the value of x=3 in f(x):

f(3)= (3)³ + 3

= 27 + 3

= 30

For f(1), put the value of x=1 in f(x):

f(1) = (1)³ + 3

= 1 + 3

= 4

⇒ 3c² = 13

We know that, the value of c obtained in Lagrange’s Mean value Theorem is nothing but the value of x – coordinate of the point of the contact of the tangent to the curve which is parallel to the chord joining the points (1, 2) and (3, 28).

Now, Put this value of x in f(x) to obtain y:

y = x³ + 1

sin x and cos x functions are continuous everywhere on (−∞, ∞) and differentiable for all arguments.

So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

x = a cos³ θ

y = a sin³ θ

We know that,

sin² θ + cos² θ = 1

x = acos³ θ

y = asin³ θ

For f’(c), put the value of x=c in f’(x):

f’(c) = – tan θ

For f(a), put the value of x=a in f(x):

= 0

For f(0), put the value of x=0 in f(x):

= a

⇒ – tan θ = – 1

⇒ tan θ = 1

Now put the value of θ in the function of x and y:

x = a cos³ θ

Similarly,

y = a sin³ θ

Let f(x) = tan x on [a, b]

We know that, tan x function is continuous and differentiable on

differentiable on (a, b).

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = tan x

Differentiating with respect to x:

f’(x) = sec² x

Since c lies between a and b

⇒ a < c < b

⇒ sec² a < sec² c < sec² b

⇒ sec² a (b – a)<tan b – tan a< sec² b (b – a)

Hence Proved

As the polynomial, na_nx^n–1 + (n–1) a_n–1x^n–2 + … + a₁ = 0 is a derivative of the polynomial a₀xⁿ + a_{n – 1}x^{n – 1} + a_{n – 2}x^{n – 2} +..............a₂x² + a₁x + a₀ = 0 --------- (i)

Putting x = 0 in equation (i),

f(0) = a₀ < 0, {Y – Intercept of the graph is negative}

On the other hand, an > 0 and ‘n’ is even, the leading term on xⁿ, is positive for only x.

For |x| to be large, the term anx_n will dominate, so

lim _x→-∞ f (x) = +∞

lim _x→+∞ f (x) = +∞

If lim _x→+∞ f (x) = +∞, there must exist

Same number α < 0, where f(α > 0)

f(0) = a₀ < 0,

α < β < 0, such that f(β) = 0

Also, there is some value 0 < α,

Where f(a) & so there exists,

0 < b < a with f(b) = 0

Additionally, the polynomial function (equation (i)) is continuous everywhere in R and consequently derivative in R.

a₀xⁿ + a_{n – 1}x^{n – 1} + a_{n – 2}x^{n – 2} +..............a₂x² + a₁x + a₀ = 0 is continuous on α,β and derivative on α,β.

Thus, it satisfies both the conditions of Rolle' s Theorem.

As per the Rolle’s Theorem, between any two roots of a function f(x), there exists at least one root of its derivative.

Thus, the equation na_nx^n–1 + (n–1) a_n–1x^n–2 + … + a₁ = 0 will have at least one root between α & β.

Hence, Option (C) is the answer.

Let f(x) = ax3 + bx2 + cx + d --------- (i)

f(0) = d

f(2) = a(2)³ + b(2)² + c(2) + d

= 8a + 4b + 2c + d

= 2(4a + 2b + c) + d

4a + 2b + c = 0 {Given}

= 2 (0) + d

= 0 + d

= d

f is continuous in closed interval [0, 2] and f is derivable in the open interval (0, 2).

Also, f(0) = f(2)

As per Rolle’s Theorem,

f’(α) = 0 for 0 < α < 2

f’(x) = 3ax² + 2bx + c

f’(α) = 3aα² + 2b(α) + c

3aα² + 2b(α) + c = 0

Hence equation (i) has at least one root in the interval (0, 2).

Thus, f’(x) must have one root in the interval (0, 2).

Hence, Option (C) is the answer.

It shows that f(x) is continuous on 1, 3 and derivable on 1, 3.

So, both the conditions of Lagrange’s Theorem are satisfied.

Consequently, there exists c Є 1, 3 such that

Hence, Є (1, 3) such that .

Hence, Option (B) is the answer.

In the Lagrange’s Mean Value Theorem, c Є (a, b) such that

f is continuous on [a, b] and differentiable on (a, b), then there exists a real number x Є (a, b)

So, in the case of x₁, , then x₁Є (a, b)

a <x₁ < b

Hence, Option (C) is the answer.

Φ(x) = a^{Sin x}, a > 0

Differentiating the above-mentioned function, with respect to ‘x’,

Φ’(x) = log a (cos x a ^{Sin x})

Φ’(c) = log a (cos c a ^{Sin c})

Let Φ’(c) = 0

log a (cos c a ^{Sin c}) = 0

cos c a ^{Sin c} = 0

cos c = 0

Also, the above-mentioned function, is derivable and continuous on the interval [0, π].

Thus, here Rolle’s Theorem is applicable on the above mentioned function in the interval [0,π].

Hence, Option (B) is the answer.

_{f(x) = 2x}³ – 5x² – 4x + 3

_{f’(x) = 6x}² – 10x – 4

_{f’(c) = 6c}² – 10c – 4

f’(c) = 0

6c² – 10c – 4 = 0

3c² – 5c – 2 = 0

3c² + c – 6c – 2 = 0

c (3c + 1) – 2 (3c + 1) = 0

(3c + 1) (c – 2) = 0

c = 2 or

c = 2 Є

Thus, as per Rolle’s Theorem, c = 2 Є .

So, the required value of c = 2

Hence, Option (A) is the answer.

y = x log x

Differentiating the function with respect to ‘x’,

Slope of tangent to the curve = 1 + log x

And, slope of the chord joining the points, (1, 0) & (e, e)

The tangent to the curve is parallel to the chord joining the points, (1, 0) & (e, e)

m = 1 + log x

Hence, Option (A) is the answer.

Differentiating the function with respect to ‘x’,

Using Sridharacharya Formula,

In a general equation, ax² + bx + c = 0

Є (-1, 0)

So the required value of Є (-1, 0).

Hence, Option (C) is the answer.

_{f(x) = x (x – 2)}

_{f(x) = x}² – 2x

Since, a polynomial function is always continuous and differentiable.

As f(x) is a polynomial function so it is always continuous on 1, 2 and differentiable on 1, 2.

f(x) satisfies both the conditions of Lagrange’s Theorem on 1, 2.

So, a real number has to exist c Є 1, 2, such that

f(x) = x² – 2x

f'(x) = 2x – 2

f(1) = 1² – 2(1)

= 1 – 2

= -1

f(2) = 2² – 2(2)

= 4 – 4

= 0

= 1

So, 2x – 2 = 1

2x = 3

Є (1, 2)

Hence, Option (D) is the answer.

_{f(x) = x}³ – 3x

_{The above mentioned polynomial function is continuous and derivable in R.}

the function is continous on [0, ] and derivable on [0, ].

Differentiating the function with respect to x,

_{f(x) = x}³ – 3x

_{f’(x) = 3x}² – 3

f’(c) = 3c² – 3

f’(c) = 0

3c² – 3 = 0

c² – 1 = 0

c² = 1

c = ± 1

Hence, c = 1 Є [0, ], as per the condition of Rolle’s Theorem.

The required value is c = 3.

Hence, Option (A) is the answer.

As, f(x) = e^x Sin x

Differentiating the function with respect to ‘x’,

f’(x) = e^x Cos x + Sin x e^x

f’(c) = e^c Cos c + Sin c e^c

As, e^x Cos x is continuous and derivable in R.

it is contionous on and derivable on .

f(0) = e⁰ Sin (0)

= 0

f’(c) = 0

e^c Cos c + Sin c e^c = 0

e^c (Cos c + Sin c) = 0

Cos c + Sin c = 0 -------- (i)

Cos c = - Sin c

c = Є (0, )

Hence, Option (D) is the answer.

f(x) = a^x

f’(x) = a^x log a

f(x) is increasing on R.

f’(x) > 0

a^x log a > 0

Logarithmic function is defined for positive values of a.

a > 0

a^x > 0

a^x log a > 0

It can be possible when log a^x > 0 & log a > 0 or a^x < 0 & log a < 0.

log a > 0

a > 1

Hence, f(x) is increasing when a > 1.

f(x) = Ax² + Bx + C

Differentiating the above-mentioned function with respect to ‘x’,

f’(x) = 2Ax + B

f’(c) = 2Ac + B

f’(c) = 0

2Ac + B = 0

-------- (i)

As f(a) = f(b),

f(a) = Aa² + Ba + C

f(b) = Ab² + Bb + C

Aa² + Ba + C = Ab² + Bb + C

Aa² + Ba = Ab² + Bb

A (a² – b²) + B (a – b) = 0

A (a + b) (a – b) + B (a – b) = 0

(a – b) {A (a + b) + B} = 0

a = b,

{As a ≠ b}

From equation (i)

Hence the required value is .

Rolle’s Theorem is stated as below: -

Let ‘f’, be a real valued function defined on the closed interval a, b such that

i. It is continuous in the closed interval [a, b].

ii. It is differentiable in the open interval (a, b).

iii. f(a) = f(b)

Then there exists a real number c Є (a, b) such that f’(c) = 0.

f(x) = a^x

f’(x) = a^x log a

f(x) is decreasing on R.

f’(x) < 0, ∀x ϵ R

a^x log a < 0, ∀x ϵ R

Logarithmic function is not defined for negative values of a.

a^x > 0

a^x log a < 0 can be possible when log a < 0, ∀x ϵ R.

1 > a > 0

Hence the function f(x), whose values are 1 > a > 0.

f(x) = log_ax

Let x₁, x₂ϵ (0, ∞) such that x₁ < x₂.

the function here is a logarithmic function, so either a > 1 or 1 > a > 0.

Case – 1

Let a > 1

x₁ < x₂

log_ax₁ < log_ax₂

f(x₁) < f(x₂)

x₁ < x₂ & f(x₁) < f(x₂), ∀ x₁, x₂ϵ (0, ∞)

Hence, f(x)is increasing on (0, ∞).

Case – 2

Let, 1 > a > 0

x₁ < x₂

log_ax₁ > log_ax₂

f(x₁) > f(x₂)

x₁ < x₂ & f(x₁) > f(x₂), ∀ x₁, x₂ϵ (0, ∞)

Thus, for a > 1, f(x) is increasing in its domain.

Lagrange’s Mean Value Theorem is stated as below :-

Let f(x) be a function defined on a, b such that

i. It is continuous on a, b and

ii. It is differentiable on a, b.

Then there exists a real number c Є a, b such that .

f(x) = 2x (x – 3)ⁿ

Differentiating the above-mentioned function with respect to ‘x’,

f’(x) = 2 [xn (x – 3)^{n – 1} + (x – 3)ⁿ]

-n + 3 = 0

-n = -3

n = 3

Hence, the required value of ‘n’ is 3.

f(x) = log_ax

Domain of the above mentioned function is (0, ∞)

Let x₁, x₂ϵ (0, ∞) such that x₁ < x₂.

the function here is a logarithmic function, so either a > 1 or 1 > a > 0.

Case – 1

Let a > 1

x₁ < x₂

log_ax₁ < log_ax₂

f(x₁) < f(x₂)

x₁ < x₂ & f(x₁) < f(x₂), ∀ x₁, x₂ϵ (0, ∞)

Hence, f(x)is increasing on (0, ∞).

Case – 2

Let 1 > a > 0

x₁ < x₂

log_ax₁ > log_ax₂

f(x₁) > f(x₂)

x₁ < x₂ & f(x₁) > f(x₂), ∀ x₁, x₂ϵ (0, ∞)

Hence, f(x)is decreasing on (0, ∞).

Thus, for 1 > a > 0, f(x) is decreasing in its domain.

f(x) = a (x + Sin x) + a

f’(x) = a (1 + Cos x) + 0

f’(x) = a (1 + Cos x)

For f(x), to be increasing, it must have,

f’(x) > 0

a (1 + Cos x) > 0 -------------- (i)

-1 ≤ Cos x ≤ 1, ∀x ϵ R

0 ≤ (1 + Cos x) ≤ 2, ∀x ϵ R

a > 0 {From eq. (i)}

a ϵ (0, ∞)

Hence the required set of values is a ϵ (0, ∞).

f(x) will exist, if

x² – 4 ≥ 0

x² ≥ 4

x ≤ -2 or x ≥ 2

for each x Є [2, 3], the function f(x) has a unique definite value, f(x) is continuous on (2, 3).

Exists for all x Є (2, 3).

So, f(x) is differentiable on (2,3).

Hence, both the conditions of Lagrange’s Theorem are satisfied.

Consequently, there exists c Є (2, 3) such that,

0

Squaring both sides,

5x² – 20 – x² = 0

4x² = 20

x² = 5

x = ± √5

Hence, c = √5 Є (2, 3) such that

Hence the above explanation verifies the Lagrange’s Theorem.

f(x) = Sin x – ax + 4

f’(x) = Cos x – a + 0

f’(x) = Cos x – a

f(x) is increasing on R.

f’(x) > 0

Cos x – a > 0

Cos x > a

Cos x ≥ -1, ∀x ϵ R

a < -1

a ϵ (-∞, -1)

Hence the required set of values is a ϵ (-∞, -1).

f(x) = b (x + Cos x) + 4

f’(x) = b (1 – Sin x) + 0

f’(x) = b (1 – Sin x)

f(x) is decreasing on R.

f’(x) < 0

b (1 – Sin x) < 0

Sin x ≤ 1

1 – Sin x ≥ 0

b (1 – Sin x) < 0 & 1 – Sin x ≥ 0

b < 0

b ϵ (-∞, 0)

Hence the required set of values is b ϵ (-∞, 0).

f(x) = x + Cos x + ax + b

f’(x) = 1 – Sin x + a + 0

f’(x) = 1 – Sin x + a

For, f(x) to be increasing, it must have

f’(x) > 0

1 – Sin x + a > 0

1 > Sin x - a

Sin x < a + 1

the maximum value of Sin x is 1.

Also, 1 < a + 1

a > 0

a ϵ (0, ∞)

Hence the required set of values is a ϵ (0, ∞).

f(x) = kx – Sin x

f’(x) = k – Cos x

For, f(x) to be increasing, it must have

f’(x) > 0

k – Cos x > 0

k > Cos x

Cos x < k

the minimum value of Xos x is 1.

Also, Cos x < k

The minimum value of k is 1.

k ϵ (1, ∞)

Hence the required set of values is k ϵ (1, ∞).

g(x) is decreasing on R.

f(x) = tan^-1x is an increasing function.

f o g(x) = f(g(x)) = tan^-1(g(x))

is a decreasing function

Composite of two functions,

Glossary –

↑ - Increasing

↓ - For decreasing f(x).

x₁ < x₂

g (x₁) > g (x₂)

Applying, tan^-1 on both the sides, of the mentioned equation,

tan^-1 {g (x₁)} > tan^-1 {g (x₂)}

f (x₁) > f (x₂)

Hence it is decreasing on R.

f(x) = ax + b

f’(x) = a + 0

f’(x) = a

For, f(x) to be decreasing, it must have

f’(x) < 0

a < 0

a ϵ (-∞, 0)

Hence the required set of values is a ϵ (-∞, 0).

f(x) = Sin x + Cos x, x ϵ [0, ]

f’(x) = Cos x – Sin x

For, f(x) to be increasing, it must have

f’(x) > 0

Cos x – Sin x > 0

Sin x < Cos x

tan x < 1

tan x < tan

x ϵ

Hence, the required interval is x ϵ .

f(x) = tan x – x

f’(x) = Sec²x – 1

Sec²x – 1 ≥ 0

Sec²x ≥ 1

1 + tan²x = Sec²x

1 + tan²x ≥ 1

tan²x ≥ 0

tan²x ≥ 0 ∀x ϵ [0, 2π]

f(x) is increasing its domain.

f(x) = Cos x + a²x + b

f’(x) = -Sin x + a² + 0

f’(x) = a² - Sin x

f(x) is strictly increasing on R

f’(x) > 0, ∀x ϵ R

a² – Sin x > 0, ∀x ϵ R

a² > Sin x, ∀x ϵ R

Maximum value of Sin x is 1.

a² > Sin x, a² is always greater than 1.

a² > 1

a² – 1 > 0

(a + 1) (a – 1) > 0

a ϵ (-∞, -1) ∪ (1, ∞)