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

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

Answer:

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$
$( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)$
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?

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

Answer:

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$
$( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)$
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?
$f (x) = x^2 - 1 \: \:for \: \: x \epsilon [ 1,2]$

Answer:

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 $f (-5) \neq f(5)$

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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$
$( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)$
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$
$( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)$
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 $f (x) = x^2-1$