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)
= x2 – 3x – 2x + 6 + x2 – 3x – x + 3 + x2 – 2x – x + 2
= 3x2 – 12x + 11
f’ (c) = 3c2 – 12c + 11
by Rolle’s Theorem f’ (c ) = 0
3c2 – 12c + 11 = 0
4. Find all the value s of ‘c’ in Rolle’s theorem for the function
y = f(x) = x2 + 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) = x2 – 1 on [– 1, 1]
Sol: Given function is f(x) = x2 – 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) = x3 + bx2 + ax on [1, 3] with c = 2 +
. Find the values of a and b.
Sol: Given function is f(x) = x3 + bx2 + 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 x2 + 2b x + a
f’ (c) = 0
⟹ 3 c2 + 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 cos2 c – 1) = 0
cos c – 4 cos2 c – 2 = 0
4 cos2 c – cos c + 2 = 0
c = ∈ (0, π)
∴ Rolle’s theorem is verified
8. Verify Rolle’s theorem for the function f(x) = (x2 – 1) (x – 2) on [ – 1, 2]
Sol: Given function is f(x) = (x2 – 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) = (x2 – 1) (1 – 0) + (2x – 0) (x – 2)
f’ (x) = (x2 – 1) (1) + (2x) (x – 2)
f’ (x) = x2 – 1 + 2x2 – 4x
= 3x2 – 4x – 1
f’ (c) = 0
3c2 – 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
c2 – c – 6 = 0
c2 – 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 x2 – 3x + k has two distinct roots in [0, 1].
Sol: given function is f (x) = x2 – 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 (x2 + 2) – log 3 on [ – 1, 1]
Sol: Given function is f(x) = log (x2 + 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) = x2 – 3x – 1 , a =
, b =
; f(x) = x2 – 3x – 1 , a =
13. Find ‘c’ so that f’ (c) = ; f(x) = ex, a =0, b = 1
Sol: Given f(x) = ex, a =0, b = 1
f (b) = f(1) = e
f (a) = f(0) = e0 = 1
f’ (x) = ex
f’ (c) = 
ec =
ec = e – 1
c =
14. Verify Lagrange’s Mean value theorem for the function f(x) = x2 – 1 on [2, 3]
Sol: Given function is f(x) = x2 – 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 cos2c – 1) = 0
cos c – 4cos2c + 2 = 0
4 cos2 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 = x3 where the tangent is parallel to the chord joining the points (1, 1) and (3, 27)
Sol: Given curve is y = x3
f’ (x) = 3x2
let A = (1, 1) and B = (3, 27)
slope of AB= 
= 13
given slope of AB = slope of the tangent
3x2 = 13
∴ the point is 
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