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?
$f(x)=[x]forxϵ[5,9]$

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 $cϵ(x,y)$ such that $f^{{}^{\prime }}(c)=0$
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 , $nϵ[5,9]$
$\underset{h\to 0^{-}}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h\to 0^{-}}{lim}\frac{[n+h]-[n]}{h}=\underset{h\to 0^{-}}{lim}\frac{n-1-n}{h}=\underset{h\to 0^{-}}{lim}\frac{-1}{h}$$=-\mathrm{\infty }$
$([n+h]=n-1\because h<0therefore(n+h)<n)$
Now,
R.H.L. at x = n , $nϵ[5,9]$
$\underset{h\to 0^{+}}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h\to 0^{+}}{lim}\frac{[n+h]-[n]}{h}=\underset{h\to 0^{+}}{lim}\frac{n-n}{h}=\underset{h\to 0^{-}}{lim}\frac{0}{h}=0$
$([n+h]=n\because h>0therefore(n+h)>n)$
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 $f(x)=[x]$ , $xϵ[5,9]$

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?

$f(x)=[x]forxϵ[-2,2]$

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 $cϵ(x,y)$ such that $f^{{}^{\prime }}(c)=0$
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 , $nϵ[-2,2]$
$\underset{h\to 0^{-}}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h\to 0^{-}}{lim}\frac{[n+h]-[n]}{h}=\underset{h\to 0^{-}}{lim}\frac{n-1-n}{h}=\underset{h\to 0^{-}}{lim}\frac{-1}{h}$$=-\mathrm{\infty }$
$([n+h]=n-1\because h<0therefore(n+h)<n)$
Now,
R.H.L. at x = n , $nϵ[-2,2]$
$\underset{h\to 0^{+}}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h\to 0^{+}}{lim}\frac{[n+h]-[n]}{h}=\underset{h\to 0^{+}}{lim}\frac{n-n}{h}=\underset{h\to 0^{-}}{lim}\frac{0}{h}=0$
$([n+h]=n\because h>0therefore(n+h)>n)$
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 $f(x)=[x]$ , $xϵ[-2,2]$

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?
$f(x)=x^2-1forxϵ[1,2]$

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 $cϵ(x,y)$ such that $f^{{}^{\prime }}(c)=0$
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, $f(1)\ne f(2)$
Therefore, All conditions are not satisfied
Hence, Rolle's theorem is not applicable for given function $f(x)=[x]$ , $xϵ[-2,2]$

Question:3 If $f;[-5,5]\to R$ is a differentiable function and if $f^{\prime }(x)$ does not vanish
anywhere, then prove that $f(-5)\ne f(5)$

Answer:

It is given that
$f;[-5,5]\to R$ 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
$f^{{}^{\prime }}(c)=\frac{f(b)-f(a)}{b-a}$
$$\begin{gathered} f^{{}^{\prime }}(c)=\frac{f(5)-f(-5)}{5-(-5)} \\ {f}^{{}^{\prime }}(c)=\frac{f(5)-f(-5)}{10} \\ 10f^{{}^{\prime }}(c)=f(5)-f(-5) \end{gathered}$$
Now, it is given that $f^{\prime }(x)$ does not vanish anywhere
Therefore,
$$\begin{gathered} 10f^{{}^{\prime }}(c)\ne 0 \\ f(5)-f(-5)\ne 0 \\ f(5)\ne f(-5) \end{gathered}$$
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 $f;[a,b]\to R$
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, there exist a c in (a,b) such that
$f^{{}^{\prime }}(c)=\frac{f(b)-f(a)}{b-a}$
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
$f^{{}^{\prime }}(c)=\frac{f(b)-f(a)}{b-a}$
$$\begin{gathered} f^{{}^{\prime }}(c)=\frac{f(4)-f(1)}{4-1} \\ {f}^{{}^{\prime }}(c)=\frac{-3-(-6)}{3} \\ {f}^{{}^{\prime }}(c)=\frac{3}{3} \\ {f}^{{}^{\prime }}(c)=1 \end{gathered}$$
Now,
$$\begin{gathered} f^{{}^{\prime }}(x)=2x-4 \\ {f}^{{}^{\prime }}(c)-2c-4 \\ 1=2c-4 \\ 2c=5 \\ c=\frac{5}{2} \end{gathered}$$
And $c=\frac{5}{2}ϵ(1,4)$
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 $cϵ(1,3)$ for which f '(c) = 0.

Answer:

Condition for M.V.T.
If $f;[a,b]\to R$
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, their exist a c in (a,b) such that
$f^{{}^{\prime }}(c)=\frac{f(b)-f(a)}{b-a}$
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
$f^{{}^{\prime }}(c)=\frac{f(b)-f(a)}{b-a}$
$$\begin{gathered} f^{{}^{\prime }}(c)=\frac{f(3)-f(1)}{3-1} \\ {f}^{{}^{\prime }}(c)=\frac{-27-(-7)}{2} \\ {f}^{{}^{\prime }}(c)=\frac{-20}{2} \\ {f}^{{}^{\prime }}(c)=-10 \end{gathered}$$
Now,
$$\begin{gathered} f^{{}^{\prime }}(x)=3x^2-10x-3 \\ {f}^{{}^{\prime }}(c)=3c^2-10c-3 \\ -10=3c^2-10c-3 \\ 3c^2-10c+7=0 \\ 3c^2-3c-7c+7=0 \\ (c-1)(3c-7)=0 \\ c=1andc=\frac{7}{3} \end{gathered}$$
And $c=1,\frac{7}{3}and\frac{7}{3}ϵ(1,3)$
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
$f:[a,b]\to R$ must be
a ) continuous in given closed interval say [a,b]
b ) differentiable in given open interval say (a,b)
Then their exist a $cϵ(x,y)$ such that
$f^{{}^{\prime }}(c)=\frac{f(b)-f(a)}{b-a}$
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 , $nϵ[5,9]$
$\underset{h\to 0^{-}}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h\to 0^{-}}{lim}\frac{[n+h]-[n]}{h}=\underset{h\to 0^{-}}{lim}\frac{n-1-n}{h}=\underset{h\to 0^{-}}{lim}\frac{-1}{h}$$=-\mathrm{\infty }$
$([n+h]=n-1\because h<0therefore(n+h)<n)$
Now,
R.H.L. at x = n , $nϵ[5,9]$
$\underset{h\to 0^{+}}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h\to 0^{+}}{lim}\frac{[n+h]-[n]}{h}=\underset{h\to 0^{+}}{lim}\frac{n-n}{h}=\underset{h\to 0^{-}}{lim}\frac{0}{h}=0$
$([n+h]=n\because h>0therefore(n+h)>n)$
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 $f(x)=[x]$ , $xϵ[5,9]$

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 , $nϵ[-2,2]$
$\underset{h\to 0^{-}}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h\to 0^{-}}{lim}\frac{[n+h]-[n]}{h}=\underset{h\to 0^{-}}{lim}\frac{n-1-n}{h}=\underset{h\to 0^{-}}{lim}\frac{-1}{h}$$=-\mathrm{\infty }$
$([n+h]=n-1\because h<0therefore(n+h)<n)$
Now,
R.H.L. at x = n , $nϵ[-2,2]$
$\underset{h\to 0^{+}}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h\to 0^{+}}{lim}\frac{[n+h]-[n]}{h}=\underset{h\to 0^{+}}{lim}\frac{n-n}{h}=\underset{h\to 0^{-}}{lim}\frac{0}{h}=0$
$([n+h]=n\because h>0therefore(n+h)>n)$
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 $f(x)=[x]$ , $xϵ[-2,2]$

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,
$f^{{}^{\prime }}(c)=\frac{f(b)-f(a)}{b-a}$
$$\begin{gathered} f^{{}^{\prime }}(c)=\frac{f(2)-f(1)}{2-1} \\ {f}^{{}^{\prime }}(c)=\frac{3-0}{1} \\ {f}^{{}^{\prime }}(c)=3 \end{gathered}$$
Now,
$$\begin{gathered} f^{{}^{\prime }}(x)=2x \\ {f}^{{}^{\prime }}(c)=2c \\ 3=2c \\ c=\frac{3}{2} \end{gathered}$$
And $c=\frac{3}{2}ϵ(1,2)$
Therefore, mean value theorem is applicable for the function $f(x)=x^2-1$