## Rolles and Langranges Theorem vsaq’s

Rolles and Langranges Theorem

## Rolles and Langranges Theorem vsaq’s

### Rolle’s theorem

If f: [a, b] ⟶ R be a function satisfying the following conditions

- f is continuous on [a, b]
- f is differentiable (a, b)
- f(a) = f(b)

then there exists at least one cϵ (a, b) such that f’(c) = 0

### Langrage’s theorem

If f: [a, b] ⟶ R be a function satisfying the following conditions

- f is continuous on [a, b]
- f is differentiable (a, b)

then there exists at least one cϵ (a, b) such that f’(c) =

Rolles and Langranges Theorem vsaq’s

###### 1. State Rolle’s theorem

Sol:

If f: [a, b] ⟶ R be a function satisfying the following conditions

- f is continuous on [a, b]
- f is differentiable (a, b)
- f(a) = f(b)

then there exists at least one cϵ (a, b) such that f’(c) = 0

###### 2. State Langrage’s theorem

Sol: If f: [a, b] ⟶ R be a function satisfying the following conditions

- f is continuous on [a, b]
- f is differentiable (a, b)

then there exists at least one cϵ (a, b) such that f’(c) =

###### 3. If f(x) = (x – 1) (x – 2) (x – 3), prove that there is more than ‘c’

###### in (1, 3) such that f’ (c) = 0

Sol: Given function is f(x) = (x – 1) (x – 2) (x – 3)

(i)f(x) is continuous on [1, 3]

(ii) f(x) is differentiable on (1, 3)

(iii) f(1) = (1 – 1) (1 – 2) (1 – 3)

= 0 (– 1)( – 2)

= 0

f(3) = (3 – 1) (3 – 2) (3 – 3)

= (2) (1) (0)

= 0

f(1) = f(3)

f’(x) = (1 – 0) (x – 2) (x – 3) + (x – 1) (1 – 0) (x – 3) + (x – 1) (x – 2) (1 – 0)

= (x – 2) (x – 3) + (x – 1) (x – 3) + (x – 1) (x – 2)

= x^{2} – 3x – 2x + 6 + x^{2} – 3x – x + 3 + x^{2} – 2x – x + 2

= 3x^{2} – 12x + 11

f’ (c) = 3c^{2} – 12c + 11

by Rolle’s Theorem f’ (c ) = 0

3c^{2} – 12c + 11 = 0

###### 4. Find all the value s of ‘c’ in Rolle’s theorem for the function

###### y = f(x) = x^{2 } + 4 on [– 3, 3]

Sol: Given function is f(x) = (x – 1) (x – 2) (x – 3)

f(x) is continuous on [– 3, 3]

f(x) is differentiable on (– 3, 3)

f (– 3) = (– 3)^{2} + 4 = 9 + 4 = 13

f (3) = (3)^{2} + 4 = 9 + 4 = 13

f (– 3) = f(3)

By Rolle’s theorem there exist c ∈ (– 3, 3) such that f’ (c) = 0

f’ (x) = 2x

f’ (c) = 2c

⟹ 2c = 0

C = 0 ∈ (– 3, 3)

###### 5. Find the value of ‘c’ from Rolle’s theorem for the function f(x) = x^{2} – 1 on [– 1, 1]

Sol: Given function is f(x) = x^{2} – 1

f(x) is continuous on [– 1, 1]

f(x) is differentiable on (– 1, 1)

f (– 1) = (– 1)^{2} – 1 = 1 – 1 = 0

f (– 1) = (– 1)^{2} – 1 = 1 – 1 = 0

f (– 1) = f(1)

By Rolle’s theorem there exist c ∈ (– 1, 1) such that f’ (c) = 0

f’ (x) = 2x

f’ (c) = 2c

⟹ 2c = 0

c = 0 ∈ (– 1, 1)

6. It is given that Rolle’s theorem holds for the function f(x) = x^{3} + bx^{2} + ax on [1, 3] with c = 2 + . Find the values of a and b.

Sol: Given function is f(x) = x^{3} + bx^{2} + ax on [1, 3] satisfying Rolle’s theorem

⟹ f(x) is continuous on [ 1, 3]

f(x) is differentiable on (1, 3)

f (1) = f (3)

and there exists at least one c = 2 + ϵ (1, 3) such that f’ (c) = 0

now f(1) = f(3)

1 + b + a = 27 + 9b + 3a

2a + 8b + 26 = 0

a + 4b + 13 = 0 ———- (1)

f’ (x) = 3 x^{2} + 2b x + a

f’ (c) = 0

⟹ 3 c^{2} + 2bc + a = 0

Since c = 2 +

3 + 2b ( ) + a = 0

3 + 4b ) + a = 0

3 + 4b ) + a = 0

+ + 4b + + a = 0

13 + + 4b + + a = 0 ———- (2)

Equation (2) – Equation (1)

from (1) a + 4(– 6) + 13 = 0

a – 24 + 13 = 0

a = 11

∴ a = 11, b = – 6

###### 7. Verify Rolle’s theorem for the function f(x) = sin x – sin 2x on [0, π]

Sol: Given function is f(x) = sin x – sin 2x

f(x) is continuous on [0, π]

f(x) is differentiable on (0, π)

f(0) = sin (0) – sin 2(0) = 0

f( π) = sin ( π) – sin 2(π) = 0

f(0) = f( π)

now f’ (x) = cos x – 2 cos 2x

f ‘ (c) = 0

cos c – 2 cos 2c = 0

cos c – 2(2 cos^{2} c – 1) = 0

cos c – 4 cos^{2} c – 2 = 0

4 cos^{2} c – cos c + 2 = 0

c = ∈ (0, π)

∴ Rolle’s theorem is verified

**8. Verify Rolle’s theorem for the function f(x) = (x**^{2} – 1) (x – 2) on [ – 1, 2]

^{2}– 1) (x – 2) on [ – 1, 2]

Sol: Given function is f(x) = (x^{2} – 1) (x – 2)

f(x) is continuous on [ – 1, 2]

f(x) is differentiable on (– 1, 2)

f (– 1) = (1 – 1) (– 1 – 2) = 0 (– 3) = 0

f (2) = (4 – 1) (2 – 2) = 3 (0) = 0

f (– 1) = f (2)

f(x) satisfies all the conditions of Rolle’s theorem

f’ (x) = (x^{2} – 1) (1 – 0) + (2x – 0) (x – 2)

f’ (x) = (x^{2} – 1) (1) + (2x) (x – 2)

f’ (x) = x^{2} – 1 + 2x^{2} – 4x

= 3x^{2} – 4x – 1

f’ (c) = 0

3c^{2} – 4c – 1 = 0

∴ Rolle’s theorem is verified

**9. Verify Rolle’s theorem for the function f(x) = x (x + 3)** ** on [ – 3, 0]**

Sol: Given function is f(x) = x (x + 3)

f(x) is continuous on [ – 3, 0]

f(x) is differentiable on (– 3, 0)

f (– 3) = (– 3) (– 3 + 3) = (– 3) (0) = 0

f (0) = (0) (0 + 3) = 0

f (– 3) = f (0)

f(x) satisfies all the conditions of Rolle’s theorem

f’ (x) = 1 (x + 3) + x (1 + 0) + x (x + 3)

f’ (c) = 0

c^{2} – c – 6 = 0

c^{2} – 3c + 2c – 6 = 0

c (c – 3) + 2 (c – 3) = 0

(c – 3) (c + 2) = 0

c = 3 or c = – 2

c = 3 ∉(– 3, 0) and c = – 2 ∈ (– 3, 0)

∴ Rolle’s theorem is verified

**10. Show that there is no real number ‘k’ for which the equation x**^{2} – 3x + k has two distinct roots in [0, 1].

^{2}– 3x + k has two distinct roots in [0, 1].

Sol: given function is f (x) = x^{2} – 3x + k

Let α, β be the two distinct roots of f (x) and 0 < α < β <1

f’ (α) = 0 and f’ (β) = 0

f(x) is continuous on [ α, β]

f(x) is differentiable on (α, β)

f(α) = f(β)

f’ (x) = 2x – 3

f’ ( c ) = 0

2c – 3 = 0

2c = 3

c = 3/2

**11. Verify Rolle’s theorem for the function f(x) = log (x**^{2} + 2) – log 3 ** ****on [ – 1, 1]**

^{2}+ 2) – log 3

Sol: Given function is f(x) = log (x^{2} + 2) – log 3

f(x) is continuous on [ – 1, 1]

f(x) is differentiable on (– 1, 1)

f(– 1) = log (1 + 2) – log 3 = log 3 – log 3 = 0

f(1) = log (1 + 2) – log 3 = log 3 – log 3 = 0

f(– 1) = f(1)

f(x) satisfies all the conditions of Rolle’s theorem

f’ (x) = (2x) =

by Rolle’s theorem f’ (c) = 0

= 0 ⟹ 2c = 0 ⟹ c = 0∈ (– 1, 1)

∴ Rolle’s theorem is verified

**12. Find ‘c’ so that f’ (c) = ** ** ****; f(x) = x**^{2} – 3x – 1 , a = **, b =**

^{2}– 3x – 1 , a =

**13. Find ‘c’ so that f’ (c) =** ** ****; f(x) = e**^{x}, a =0, b = 1

^{x}, a =0, b = 1

Sol: Given f(x) = e^{x}, a =0, b = 1

f (b) = f(1) = e

f (a) = f(0) = e^{0} = 1

f’ (x) = e^{x}

f’ (c) =

e^{c} =

e^{c} = e – 1

c =

**14. Verify Lagrange’s Mean value theorem for the function f(x) = x**^{2} – 1 on [2, 3]

^{2}– 1 on [2, 3]

Sol: Given function is f(x) = x^{2} – 1

f(x) is continuous on [ 2, 3]

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

f’(x) = 2x

By Lagrange’s mean value theorem f’ (c) =

2c = 5

c = 5/2 ∈ (2, 3)

∴ Lagrange’s Mean value theorem is verified

**15. Verify the Lagrange’s Mean value theorem for the function f(x) = sin x – sin 2x on [0, π]**

Sol: Given function is f(x) = sin x – sin 2x** **

f(x) is continuous on [0, π]

f(x) is differentiable on (0, π)

f(π) = sin π – sin 2 π = 0 – 0 = 0

f(0) = sin 0 – sin 2(0) = 0 – 0 = 0

f’(x) = cos x – 2cos 2x

By Lagrange’s mean value theorem f’ (c) =

cos c – 2cos 2c = = 0

cos c – 2cos 2c = 0

cos c – 2 (2 cos^{2}c – 1) = 0

cos c – 4cos^{2}c + 2 = 0

4 cos^{2} c – cos c – 2 = 0

cos c =

=

c = ()∈ (2, 3)

∴ Lagrange’s Mean value theorem is verified

**16. Verify Lagrange’s Mean value theorem for the function f(x) = log x on [1, 2]**

Sol: Given function is f(x) = log x** **

f(x) is continuous on [1, 2]

f(x) is differentiable on (1, 2)

f(2) = log 2; f(1) = log 1 = 0

f’ (x) =

By Lagrange’s mean value theorem f’ (c) =

1/c = log 2

c = ∈ (1, 2)

∴ Lagrange’s Mean value theorem is verified

**17. Find the point on the graph of the curve y = x ^{3} where the tangent is parallel to the chord joining the points (1, 1) and (3, 27)**

Sol: Given curve is y = x^{3}

f’ (x) = 3x^{2}

let A = (1, 1) and B = (3, 27)

slope of AB= = 13

given slope of AB = slope of the tangent

3x^{2 }= 13

∴ the point is