Rolle's theorem is not applicable in which one of the following cases?

  1. f(x) = x2 - 4x + 5 in [1, 3]
  2. f(x) = x2 - x in [0, 1]
  3. f(x) = |x| in [-2, 2]
  4. f(x) = [x] in [2.5, 2.7], Where [.] is GIF function.

Answer (Detailed Solution Below)

Option 3 : f(x) = |x| in [-2, 2]
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NDA 01/2025: English Subject Test
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Detailed Solution

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Concept:

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

 

Calculations:

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

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

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

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

 f(1) = (1)2 - 4(1) + 5 = 2

f(3) = (3)- 4(3) + 5 = 2

Here, f(1) = f(3) 

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

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

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

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

 f(0) = (0)2 - 0 = 0

f(0) = (1)2 - 1 = 0

Here, f(0) = f(1) 

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

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

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

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

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

 f(- 2) = |- 2| = 2

f(2) = | 2 |= 2

Here, f(- 2) = f(2) 

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

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

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

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

f(2.5) = [x]  = [2.5] = 2

f(2.7) = [x]  = [2.7] = 2

Here, f(2.5) = f(2.7) 

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

Hence option(3)  is correct.

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