NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability Ex 5.8
Page No 186:
Question 1:
Verify Rolle’s Theorem for the function![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_2bcdc671.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_2ba55237.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_2bcdc671.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_2ba55237.gif)
Answer:
The given function,
, being a polynomial function, is continuous in [−4, 2] and is differentiable in (−4, 2).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_2bcdc671.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_m7310293a.gif)
∴ f (−4) = f (2) = 0
⇒ The value of f (x) at −4 and 2 coincides.
Rolle’s Theorem states that there is a point c ∈ (−4, 2) such that![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_m1c568ee8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_m1c568ee8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_7b1d23d4.gif)
Hence, Rolle’s Theorem is verified for the given function.
Question 2:
Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples?
(i) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_1e3439b6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_1e3439b6.gif)
(ii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_97a178a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_97a178a.gif)
(iii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_7ad1b82a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_7ad1b82a.gif)
Answer:
By Rolle’s Theorem, for a function
, if
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_m4d43444b.gif)
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
(c) f (a) = f (b)
then, there exists some c ∈ (a, b) such that ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_5faadb82.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_5faadb82.gif)
Therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.
(i) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_1e3439b6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_1e3439b6.gif)
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9
⇒ f (x) is not continuous in [5, 9].
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_7c07b588.gif)
The differentiability of f in (5, 9) is checked as follows.
Let n be an integer such that n ∈ (5, 9).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_332814a3.gif)
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_1e3439b6.gif)
(ii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_97a178a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_97a178a.gif)
It is evident that the given function 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].
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_4dac9e05.gif)
The differentiability of f in (−2, 2) is checked as follows.
Let n be an integer such that n ∈ (−2, 2).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_332814a3.gif)
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (−2, 2).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_97a178a.gif)
(iii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_7ad1b82a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_7ad1b82a.gif)
It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_1291b89d.gif)
∴f (1) ≠ f (2)
It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_7ad1b82a.gif)
Question 3:
If
is a differentiable function and if
does not vanish anywhere, then prove that
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7010/Chapter%205_html_5f981c35.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7010/Chapter%205_html_21ad67ba.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7010/Chapter%205_html_m335959d3.gif)
Answer:
It is given that
is a differentiable function.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7010/Chapter%205_html_5f981c35.gif)
Since every differentiable function is a continuous function, we obtain
(a) f is continuous on [−5, 5].
(b) f is differentiable on (−5, 5).
Therefore, by the Mean Value Theorem, there exists c ∈ (−5, 5) such that
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7010/Chapter%205_html_413b28d7.gif)
It is also given that
does not vanish anywhere.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7010/Chapter%205_html_4525ddca.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7010/Chapter%205_html_m53324de.gif)
Hence, proved.
Question 4:
Verify Mean Value Theorem, if
in the interval
, where
and
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_5f926d23.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_m6b18dd4b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_m4a09daae.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_4f3fcf.gif)
Answer:
The given function is![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_5f926d23.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_5f926d23.gif)
f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x − 4.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_26322854.gif)
Mean Value Theorem states that there is a point c ∈ (1, 4) such that![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_2e12a26b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_2e12a26b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_74f0b6d.gif)
Hence, Mean Value Theorem is verified for the given function.
Question 5:
Verify Mean Value Theorem, if
in the interval [a, b], where a = 1 and b = 3. Find all
for which ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_m87c1540.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_245fb984.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_m24c12d16.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_m87c1540.gif)
Answer:
The given function f is![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_245fb984.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_245fb984.gif)
f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3x2 − 10x − 3.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_m6eec5660.gif)
Mean Value Theorem states that there exist a point c ∈ (1, 3) such that![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_33edd1e0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_33edd1e0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_23b1964d.gif)
Hence, Mean Value Theorem is verified for the given function and
is the only point for which![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_m87c1540.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_m1e82b740.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_m87c1540.gif)
Question 6:
Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.
Answer:
Mean Value Theorem states that for a function
, if
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_m4d43444b.gif)
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
then, there exists some c ∈ (a, b) such that ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_10edd9c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_10edd9c.gif)
Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.
(i) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_1e3439b6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_1e3439b6.gif)
It is evident that the given function f (x) is not continuous at every integral point.
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 an integer such that n ∈ (5, 9).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_332814a3.gif)
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is not applicable for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_1e3439b6.gif)
(ii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_97a178a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_97a178a.gif)
It is evident that the given function 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 an integer such that n ∈ (−2, 2).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_332814a3.gif)
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (−2, 2).
It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is not applicable for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_97a178a.gif)
(iii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_7ad1b82a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_7ad1b82a.gif)
It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).
It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is applicable for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_7ad1b82a.gif)
It can be proved as follows.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_m6ac1a3cd.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_7f2b4507.gif)