Chapter 5 Continuity and Differentiability Ex 5.8 Solutions
Question - 1 : - Verify Rolle’s Theorem for the function
Answer - 1 : - The given function,
, being a polynomial function, iscontinuous in [−4, 2] and is differentiable in (−4, 2).
∴ f (−4) = f (2) = 0
⇒ The value of f (x) at −4 and 2 coincides.
Rolle’sTheorem states that there is a point c ∈ (−4, 2) such thatHence, Rolle’s Theorem isverified for the given function.
Examine if Rolle’s Theorem isapplicable to any of the following functions. Can you say some thing about theconverse of Rolle’s Theorem from these examples?
(i) 
(ii) 
(iii) 
Answer - 2 : - By Rolle’s Theorem, for a function
, if
(a) f iscontinuous on [a, b]
(b) f isdifferentiable on (a, b)
(c) f (a)= f (b)
then,there exists some c ∈ (a, b)such that 
Therefore, Rolle’s Theorem is notapplicable to those functions that do not satisfy any of the three conditionsof the hypothesis.
(i)
It is evident that the givenfunction f (x) is not continuous at every integralpoint.
In particular, f(x)is not continuous at x = 5 and x = 9
⇒ f (x)is not continuous in [5, 9].
The differentiability of f in(5, 9) is checked as follows.
Let n be aninteger such that n ∈ (5, 9).
Since the left and right handlimits of f at x = n are notequal, f is not differentiable at x = n
∴f is not differentiable in (5, 9).
It is observed that f doesnot satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence,Rolle’s Theorem is not applicable for
(ii)
It is evident that the givenfunction f (x) is not continuous at every integral point.
In particular, f(x)is not continuous at x = −2 and x = 2
⇒ f (x) is not continuous in [−2, 2].
The differentiability of f in(−2, 2) is checked as follows.
Let n be aninteger such that n ∈ (−2, 2).
Since the left and right handlimits of f at x = n are notequal, f is not differentiable at x = n
∴f is not differentiable in (−2, 2).
It is observed that f doesnot satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence,Rolle’s Theorem is not applicable for
(iii)
It is evident that f,being a polynomial function, is continuous in [1, 2] and is differentiable in(1, 2).
∴f (1) ≠ f (2)
It is observed that f doesnot satisfy a condition of the hypothesis of Rolle’s Theorem.
Hence,Rolle’s Theorem is not applicable for
is a differentiable function and if
does not vanish anywhere, then prove that
Answer - 3 : - It is given that is a differentiable function.
Since every differentiablefunction is a continuous function, we obtain
(a) f iscontinuous on [−5, 5].
(b) f isdifferentiable on (−5, 5).
Therefore, by the Mean ValueTheorem, there exists c ∈ (−5, 5) such that
It is also given that 
does not vanish anywhere.
Hence, proved.
Question - 4 : - Verify Mean Value Theorem, if 
in the interval
, where 
and
Answer - 4 : - The given function is
f, being apolynomial function, is continuous in [1, 4] and is differentiable in (1, 4)whose derivative is 2x − 4.
Mean Value Theorem states that there is a point c ∈ (1, 4) such that
Hence, Mean Value Theorem isverified for the given function.
in the interval [a, b],where a = 1 and b = 3. Find all 
for which 
Answer - 5 : - The given function f is
f, being apolynomial function, is continuous in [1, 3] and is differentiable in (1, 3)whose derivative is 3x2 − 10x − 3.
Mean Value Theorem states that there exist apoint c ∈(1, 3) such that
Hence, Mean Value Theorem is verified for the givenfunction and 
is the only point for which