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 1

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) =Rolles and Langranges Theorem 1

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

Rolles and Langranges Theorem 2

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 +rolles th 3 . 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 + rolles th 1 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 +rolles th 1

   3  + 2b ( ) + a = 0

   3  + 4b ) + a = 0

3  + 4b ) + a = 0

  +  + 4b +  + a = 0

13 +  + 4b +  + a = 0 ———- (2)

Equation (2) – Equation (1)

rolles th 2

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

rolles th 4

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  

Rolles theorem 1

  ∴ Rolle’s theorem is verified

9. Verify Rolle’s theorem for the function f(x) = x (x + 3) Rolles theorem 2 on [ – 3, 0]

Sol:  Given function is f(x) = x (x + 3) Rolles theorem 8   

       f(x) is continuous on [ – 3, 0]

        f(x) is differentiable on (– 3, 0)

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

f (0) = (0) (0 + 3)Rolles theorem 4   = 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) Rolles theorem 8Rolles theorem 5  

        Rolles theorem 6  

       f’ (c) = 0

  Rolles theorem 7

       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) =Rolles theorem 9 (2x) = Rolles theorem 10

     by Rolle’s theorem f’ (c) = 0

           Rolles theorem 11= 0 ⟹ 2c = 0 ⟹ c = 0∈ (– 1, 1)

     ∴ Rolle’s theorem is verified

12. Find ‘c’ so that f’ (c) = Rolles theorem 15  ; f(x) = x2 – 3x – 1 , a =Rolles theorem 12 , b =Rolles theorem 13
Rolles theorem 14  
13. Find ‘c’ so that f’ (c) =Rolles theorem 15  ; 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) = Rolles theorem 16

            ec =Rolles theorem 17

            ec = e – 1

            c =Rolles theorem 18

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) = Rolles theorem 16

   2c  Rolles theorem 19= 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) = Rolles theorem 16

cos c – 2cos 2c = Rolles theorem 20 = 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 = Rolles theorem 21

          =  Rolles theorem 22

 c = (Rolles theorem 22)∈ (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) = 

Rolles theorem 23

1/c = log 2

c =Rolles theorem 24  ∈ (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= Rolles theorem 25  = 13

          given slope of AB = slope of the tangent

               3x2 = 13

            Rolles theorem 26

∴ the point is Rolles theorem 27

 

 


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