Question

Examine if Rolle's theorem is applicable to the function f(x) = [x] for xe 1-2, 2]. What
can you say about the converse of Rolle's theorem?
(NCERT)
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Answer 1

Answer:

Rolle's theorem holds for a function f:[a,b]→R, if following three conditions holds-

(1) f is continuous on [a,b]

(2) f is differentiable on (a,b)

(3) f(a)=f(b)

Then, there exists some c∈(a,b) such that f′(c)=0.

Rolle's Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis. 

(i)

Given f(x)=[x] for x∈ [5, 9]

Since, the greatest integer function is not continuous at integral values. 

So, f(x) is not continuous at x=5,6,7,8,9

So, f(x)  is not continuous in [5,9]. 

Since, condition (1) does not holds ,so need to check the other conditions.

Hence, Rolle's theorem is not applicable on given function

(ii)

Given function f(x)=[x] for x∈[−2,2]

Since, the greatest integer function is not continuous at integral points.

So, f(x) is not continuous at x=−2,−1,0,1,2 

f(x) is not continuous in [−2,2]. 

(iii)

f(x)=x2−1 for x∈[1,2]

It is evident that f, being a polynomial function, is continuous in [1,2] and is differentiable in (1,2).

Also f(1)=(1)2−1=0

and f(2)=(2)2−1=3

∴f(1)=f(2)

It is observed that f does not satisfy a condition of the hypothesis of Rolles Theorem.

Hence, Rolles Theorem is not applicable for f(x)=x2−1 for x∈[1,2].

Step-by-step explanation:

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