# NCERT Solutions for Exercise 5.8 Class 12 Maths Chapter 5 - Continuity and Differentiability

In the previous exercises of this Class 12 NCERT syllabuschapter, you have already learned about the first-order derivatives and second-order derivatives. In NCERT solutions for Class 12 Maths chapter 5 exercise 5.8, you will learn about the two important theorems called Rolle's theorem and the Mean value theorem. These theorems are used to prove the inequality of derivatives, study the properties of the derivatives. The proof of these theorems also given in the Class 12 Maths ch 5 ex 5.8. You can go through the proof to get in-depth knowledge of these theorems.

There are some examples given in the NCERT book before the Class 12th Maths chapter 5 exercise 5.8. You should be thorough with the exercise 5.8 Class 12 Maths as one question from this exercise is generally asked in the board exams. If you need the NCERT solutions for other classes, you can click on NCERT Solutions link.

Also, see

• Continuity and Differentiability Exercise 5.1
• Continuity and Differentiability Exercise 5.2
• Continuity and Differentiability Exercise 5.3
• Continuity and Differentiability Exercise 5.4
• Continuity and Differentiability Exercise 5.5
• Continuity and Differentiability Exercise 5.6
• Continuity and Differentiability Exercise 5.7
• Continuity and Differentiability Miscellaneous Exercise

## Continuity and Differentiability Exercise: 5.8

Question:1 Verify Rolle’s theorem for the function$f (x) = x^2 + 2x - 8, x \epsilon [- 4, 2].$

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a $c \ \epsilon \ (x,y)$ such that $f^{'}(c)= 0$
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
$f (x) = x^2 + 2x - 8$
Now, being a polynomial function, $f (x) = x^2 + 2x - 8$ is both continuous in [-4,2] and differentiable in (-4,2)
Now,
$f (-4) = (-4)^2 + 2(-4) - 8= 16-8-8=16-16=0$
Similalrly,
$f (2) = (2)^2 + 2(2) - 8= 4+4-8=8-8=0$
Therefore, value of $f (-4) = f(2)=0$ and value of f(x) at -4 and 2 are equal
Now,
According to roll's theorem their is point c , $c \ \epsilon (-4,2)$ such that $f^{'}(c)=0$
Now,
$f^{'}(x)=2x+2\\ f^{'}(c)=2c+2\\ f^{'}(c)=0\\ 2c+2=0\\ c = -1$
And $c = -1 \ \epsilon \ (-4,2)$
Hence, Rolle's theorem is verified for the given function $f (x) = x^2 + 2x - 8$

Question:2 (1) 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 example?
$f (x) = [x] \: \: for \: \: x \epsilon [ 5,9]$

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a $c \ \epsilon \ (x,y)$ such that $f^{'}(c)= 0$
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
$f (x) = [x]$
It is clear that Given function $f (x) = [x]$ is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n , $n \ \epsilon \ [5,9]$
$\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}$$= -\infty$
$( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)
Now,
R.H.L. at x = n , $n \ \epsilon \ [5,9]$
$\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0$
$( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)$
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, the function is not differential in (5,9)
Hence, Rolle's theorem is not applicable for given function $f (x) = [x]$ , $x \ \epsilon \ [5,9]$

Question:2 (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 example?

$f (x) = [x] \: \:for \: \: x \epsilon [ -2,2]$

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a $c \ \epsilon \ (x,y)$ such that $f^{'}(c)= 0$
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
$f (x) = [x]$
It is clear that Given function $f (x) = [x]$ is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , $n \ \epsilon \ [-2,2]$
$\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}$$= -\infty$
$( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)
Now,
R.H.L. at x = n , $n \ \epsilon \ [-2,2]$
$\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0$
$( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)$
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Rolle's theorem is not applicable for given function $f (x) = [x]$ , $x \ \epsilon \ [-2,2]$

Question:2 (3) 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 example?
$f (x) = x^2 - 1 \: \:for \: \: x \epsilon [ 1,2]$

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then there exist a $c \ \epsilon \ (x,y)$ such that $f^{'}(c)= 0$
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
$f (x) = x^2-1$
Now, being a polynomial , function $f (x) = x^2-1$ is continuous in [1,2] and differentiable in(1,2)
Now,
$f(1)=1^2-1 = 1-1 = 0$
And
$f(2)=2^2-1 = 4-1 = 3$
Therefore, $f(1)\neq f(2)$
Therefore, All conditions are not satisfied
Hence, Rolle's theorem is not applicable for given function $f (x) = [x]$ , $x \ \epsilon \ [-2,2]$

Question:3 If $f ; [ -5 ,5] \rightarrow R$ is a differentiable function and if $f ' (x)$ does not vanish
anywhere, then prove that $f (-5) \neq f(5)$

It is given that
$f ; [ -5 ,5] \rightarrow R$ is a differentiable function
Now, f is a differential function. So, f is also a continuous function
We obtain the following results
a ) f is continuous in [-5,5]
b ) f is differentiable in (-5,5)
Then, by Mean value theorem we can say that there exist a c in (-5,5) such that
$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$
$f^{'}(c) = \frac{f(5)-f(-5)}{5-(-5)}\\ f^{'}(c)= \frac{f(5)-f(-5)}{10}\\ 10f^{'}(c)= f(5)-f(-5)$
Now, it is given that $f ' (x)$ does not vanish anywhere
Therefore,
$10f^{'}(c)\neq 0\\ f(5)-f(-5) \neq 0\\ f(5)\neq f(-5)$
Hence proved

Question:4 Verify Mean Value Theorem, if $f (x) = x^2 - 4x - 3$in the interval [a, b], where
a = 1 and b = 4.

Condition for M.V.T.
If $f ; [ a ,b] \rightarrow R$
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, there exist a c in (a,b) such that
$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$
It is given that
$f (x) = x^2 - 4x - 3$ and interval is [1,4]
Now, f is a polynomial function , $f (x) = x^2 - 4x - 3$ is continuous in[1,4] and differentiable in (1,4)
And
$f(1)= 1^2-4(1)-3= 1-7= -6$
and
$f(4)= 4^2-4(4)-3= 16-16-3= 16-19=-3$
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$
$f^{'}(c) = \frac{f(4)-f(1)}{4-1}\\ f^{'}(c)= \frac{-3-(-6)}{3}\\ f^{'}(c)= \frac{3}{3}\\ f^{'}(c)= 1$
Now,
$f^{'}(x) =2x-4\\ f^{'}(c)-2c-4\\ 1=2c-4\\ 2c=5\\ c=\frac{5}{2}$
And $c=\frac{5}{2} \ \epsilon \ (1,4)$
Hence, mean value theorem is verified for the function $f (x) = x^2 - 4x - 3$

Question:5 Verify Mean Value Theorem, if$f (x) = x^3 - 5x^2- 3x$in the interval [a, b], where
a = 1 and b = 3. Find all $c \epsilon (1,3)$ for which f '(c) = 0.

Condition for M.V.T.
If $f ; [ a ,b] \rightarrow R$
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, their exist a c in (a,b) such that
$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$
It is given that
$f (x) = x^3 - 5x^2- 3x$ and interval is [1,3]
Now, f being a polynomial function , $f (x) = x^3 - 5x^2- 3x$ is continuous in[1,3] and differentiable in (1,3)
And
$f(1)= 1^3-5(1)^2-3(1)= 1-5-3=1-8=-7$
and
$f(3)= 3^3-5(3)^2-3(3)= 27-5.9-9= 18-45=-27$
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$
$f^{'}(c) = \frac{f(3)-f(1)}{3-1}\\ f^{'}(c)= \frac{-27-(-7)}{2}\\ f^{'}(c)= \frac{-20}{2}\\ f^{'}(c)= -10$
Now,
$f^{'}(x) =3x^2-10x-3\\ f^{'}(c)=3c^2-10c-3\\ -10=3c^2-10c-3\\ 3c^2-10c+7=0\\ 3c^2-3c-7c+7=0\\ (c-1)(3c-7)=0\\ c = 1 \ \ \ and \ \ \ c = \frac{7}{3}$
And $c=1,\frac{7}{3} \ and \ \frac{7}{3}\ \epsilon \ (1,3)$
Hence, mean value theorem is varified for following function $f (x) = x^3 - 5x^2- 3x$ and $c=\frac{7}{3}$ is the only point where f '(c) = 0

Question:6 Examine the applicability of Mean Value Theorem for all three functions given in
the above exercise 2.

According to Mean value theorem function
$f:[a,b]\rightarrow R$ must be
a ) continuous in given closed interval say [a,b]
b ) differentiable in given open interval say (a,b)
Then their exist a $c \ \epsilon \ (x,y)$ such that
$f^{'}(c)= \frac{f(b)-f(a)}{b-a}$
If all these conditions are satisfies then we can verify mean value theorem
Given function is
$f (x) = [x]$
It is clear that Given function $f (x) = [x]$ is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n , $n \ \epsilon \ [5,9]$
$\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}$$= -\infty$
$( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)
Now,
R.H.L. at x = n , $n \ \epsilon \ [5,9]$
$\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0$
$( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)$
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (5,9)
Hence, Mean value theorem is not applicable for given function $f (x) = [x]$ , $x \ \epsilon \ [5,9]$

Similaly,
Given function is
$f (x) = [x]$
It is clear that Given function $f (x) = [x]$ is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , $n \ \epsilon \ [-2,2]$
$\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}$$= -\infty$
$( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)
Now,
R.H.L. at x = n , $n \ \epsilon \ [-2,2]$
$\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0$
$( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)$
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Mean value theorem is not applicable for given function $f (x) = [x]$ , $x \ \epsilon \ [-2,2]$

Similarly,
Given function is
$f (x) = x^2-1$
Now, being a polynomial , function $f (x) = x^2-1$ is continuous in [1,2] and differentiable in(1,2)
Now,
$f(1)=1^2-1 = 1-1 = 0$
And
$f(2)=2^2-1 = 4-1 = 3$
Now,
$f^{'}(c)= \frac{f(b)-f(a)}{b-a}$
$f^{'}(c)= \frac{f(2)-f(1)}{2-1}\\ f^{'}(c)=\frac{3-0}{1}\\ f^{'}(c)= 3$
Now,
$f^{'}(x)= 2x\\ f^{'}(c)=2c\\ 3=2c\\ c=\frac{3}{2}$
And $c=\frac{3}{2} \ \epsilon \ (1,2)$
Therefore, mean value theorem is applicable for the function $f (x) = x^2-1$

## More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.8:-

In Class 12 Maths chapter 5 exercise 5.8 solutions, you will get 6 questions related to verifying the two theorems called the mean value theorem and Rolle's theorem. The three examples given before this exercise are also related to the same also. All the questions in the Class 12th Maths chapter 5 exercise 5.8 are very similar but you must solve all the problems by yourself to get familiar with these types of questions.

Also Read| Continuity and Differentiability Class 12th Chapter 5 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.8:-

• In Class 12 Maths chapter 5 exercise 5.8 solutions you will get some new ways to approach the problem.
• NCERT solutions for Class 12 Maths Chapter 5 exercise 5.8 are designed by subject matter experts in a descriptive manner that you can understand very easily.
• Class 12th Maths chapter 5 exercise 5.8 can be used as revision notes.

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