NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.8


Chapter 5 Ex.5.8 Question 1

Verify Rolle’s Theorem for the function \(f\left( x \right) = {x^2} + 2x - 8,x \in \left[ { - 4,2} \right]\)

 

Solution

Video Solution

 

Given, \(f\left( x \right) = {x^2} + 2x - 8\), being polynomial function is continuous in \(\left[ { - 4,2} \right]\) and also differentiable in \(\left( { - 4,2} \right)\).

\[\begin{align}f\left( { - 4} \right) &= {\left( { - 4} \right)^2} + 2.\left( { - 4} \right) - 8\\ &= 16 - 8 - 8\\ &= 0\end{align}\]

\[\begin{align}f\left( 2 \right) &= {\left( 2 \right)^2} + 2 \times 2 - 8\\ &= 4 + 4 - 8\\ &= 0\end{align}\]

Therefore, \(f\left( { - 4} \right) = f\left( 2 \right) = 0\)

The value of \(f\left( x \right)\)at \( - 4\) and 2 coincides.

Rolle’s Theorem states that there is a point \(c \in \left( { - 4,2} \right)\) such that \(f'\left( c \right) = 0\)

\(f\left( x \right) = {x^2} + 2x - 8\)

Therefore, \(f'\left( x \right) = 2x + 2\)

Hence,

\[\begin{align}f'\left( c \right) &= 0\\2c + 2 &= 0\\c &= - 1\end{align}\]

Thus, \(c = - 1 \in \left( { - 4,2} \right)\)

Hence, Rolle’s Theorem is verified.

Chapter 5 Ex.5.8 Question 2

Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say something about the converse of Rolle’s Theorem from these examples?

(i) \(f\left( x \right) = \left[ x \right]{\;\rm{ for\; }} x \in \left[ {5,9} \right]\)

(ii) \(f\left( x \right) = \left[ x \right]{\;\rm{ for\; }}x \in \left[ { - 2,2} \right]\)

(iii) \(f\left( x \right) = {x^2} - 1{\;\rm{ for\;}}x \in \left[ {1,2} \right]\)

 

Solution

Video Solution

 

By Rolle’s Theorem, \(f:\left[ {a,b} \right] \to {\bf{R}}\),

If

         (a) \(f\) is continuous on \(\left[ {a,b} \right]\)

         (b) \(f\) is continuous on \(\left( {a,b} \right)\)

         (c) \(f\left( a \right) = f\left( b \right)\)

Then, there exists some \(c \in \left( {a,b} \right)\) such that \(f'\left( c \right) = 0\)

Thus, Rolle’s Theorem is not applicable to those functions that do not satisfy any of three conditions of the hypothesis.

(i) \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]\)

     Since, the given function \(f\left( x \right)\) is not continuous at every integral point.

     In general, \(f\left( x \right)\) is not continuous at \(x = 5\) and \(x = 9\)

             Therefore, \(f\left( x \right)\) is not continuous in \(\left[ {5,9} \right]\)

     Also, \(f\left( 5 \right) = \left[ 5 \right] = 5\) and \(f\left( 9 \right) = \left[ 9 \right] = 9\)

     Thus, \(f\left( 5 \right) \ne f\left( 9 \right)\)

     The differentiability of \(f\) in \(\left( {5,9} \right)\) is checked as follows.

     Let \(n\) be an integer such that \(n \in \left( {5,9} \right)\)

     The LHD of \(f\) at \(x = n\) is

                           \(\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - 1}}{h} = \infty \)

     The RHD of \(f\) at \(x = n\) is

                           \(\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{n - n}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0\)

     Since LHD and RHD of \(f\) at \(x = n\) are not equal, f is not differentiable at \(x = n\)

             Therefore, \(f\) is not differentiable in \(\left( {5,9} \right)\).

     It is observed that \(f\) does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

     Thus, Rolle’s Theorem is not applicable for \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]\).

(ii) \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]\)

     Since, the given function \(f\left( x \right)\) is not continuous at every integral point.

     In general, \(f\left( x \right)\) is not continuous at \(x = - 2\) and \(x = 2\)

                   Therefore, \(f\left( x \right)\) is not continuous in \(\left[ { - 2,2} \right]\)

     Also, \(f\left( { - 2} \right) = \left[ { - 2} \right] = - 2\) and \(f\left( 2 \right) = \left[ 2 \right] = 2\)

     Thus, \(f\left( { - 2} \right) \ne f\left( 2 \right)\)

     The differentiability of \(f\) in \(\left( { - 2,2} \right)\) is checked as follows.

     Let \(n\) be an integer such that \(n \in \left( { - 2,2} \right)\)

     The LHD of \(f\) at \(x = n\) is

                        \(\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - 1}}{h} = \infty \)

     The RHD of \(f\) at \(x = n\) is

                        \(\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0\)

     Since LHD and RHD of \(f\) at \(x = n\) are not equal, f is not differentiable at \(x = n\)

                Therefore, \(f\) is not differentiable in \(\left( { - 2,2} \right)\).

     It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

     Thus, Rolle’s Theorem is not applicable for \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]\)

(iii) \(f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]\)

      Since, \(f\) being a polynomial function is continuous in \(\left[ {1,2} \right]\) and is differentiable in \(\left( {1,2} \right)\).

      Thus,

              \[\begin{align}f\left( 1 \right)& = {\left( 1 \right)^2} - 1 = 0\\f\left( 2 \right)& = {\left( 2 \right)^2} - 1 = 3\end{align}\]

      Therefore, \(f\left( 1 \right) \ne f\left( 2 \right)\)

      Since, \(f\) does not satisfy a condition of the hypothesis of Rolle’s Theorem.

      Hence, Rolle’s Theorem is not applicable for \(f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]\).

Chapter 5 Ex.5.8 Question 3

If \(f:\left[ { - 5,5} \right] \to {\bf{R}}\) is a differentiable function and if \(f'\left( x \right)\) does not vanish anywhere, then prove that \(f\left( { - 5} \right) \ne f\left( 5 \right)\).

 

Solution

Video Solution

 

Given, \(f:\left[ { - 5,5} \right] \to {\bf{R}}\) is a differentiable function.

Since every differentiable function is a continuous function, we obtain

\(f\) is continuous on \(\left[ { - 5,5} \right]\)

\(f\) is continuous on \(\left( { - 5,5} \right)\)

Thus, by the Mean Value Theorem, there exists \(c \in \left( { - 5,5} \right)\) such that

\[\begin{align} &\Rightarrow \; f'\left( c \right) = \frac{{f\left( 5 \right) - f\left( { - 5} \right)}}{{5 - \left( { - 5} \right)}}\\& \Rightarrow \; 10f'\left( c \right) = f\left( 5 \right) - f\left( { - 5} \right)\end{align}\]

It is also given that \(f'\left( x \right)\) does not vanish anywhere.

Therefore, \(f'\left( c \right) \ne 0\)

Thus,

\[\begin{align}& \Rightarrow \; 10f'\left( c \right) \ne 0\\ &\Rightarrow \; f\left( 5 \right) - f\left( { - 5} \right) \ne 0\\ &\Rightarrow \; f\left( 5 \right) \ne f\left( { - 5} \right)\end{align}\]

Hence proved.

Chapter 5 Ex.5.8 Question 4

Verify Mean Value Theorem, if \(f\left( x \right) = {x^2} - 4x - 3\) in the integral \(\left[ {a,b} \right]\), where \(a = 1\) and \(b = 4\).

 

Solution

Video Solution

 

Given, \(f\left( x \right) = {x^2} - 4x - 3\)

\(f\), being a polynomial function, is continuous in \(\left[ {1,4} \right]\) and is differentiable in \(\left( {1,4} \right)\), whose derivative is \(2x - 4\).

Thus,

\[\begin{align}f\left( 1 \right) &= {1^2} - 4 \times 1 - 3 = - 6\\f\left( 4 \right) &= {4^2} - 4 \times 4 - 3 = - 3\end{align}\]

Therefore,

\[\begin{align}\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}& = \frac{{f\left( 4 \right) - f\left( 1 \right)}}{{4 - 1}}\\ &= \frac{{ - 3 - \left( { - 6} \right)}}{3}\\& = \frac{3}{3}\\& = 1\end{align}\]

Mean Value Theorem states that there is a point \(c \in \left( {1,4} \right)\) such that \(f'\left( c \right) = 1\)

Hence,

\[\begin{align}& \Rightarrow \; f'\left( c \right) = 1\\& \Rightarrow \; 2c - 4 = 1\\& \Rightarrow \; c = \frac{5}{2}\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{where }}c = \frac{5}{2} \in \left( {1,4} \right)} \right]\end{align}\]

Thus, mean value theorem is verified for the given function.

Chapter 5 Ex.5.8 Question 5

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

 

Solution

Video Solution

 

Given, f is \(f\left( x \right) = {x^3} - 5{x^2} - 3x\)

\(f\), being a polynomial function, is continuous in \(\left[ {1,3} \right]\) and is differentiable in \(\left( {1,3} \right)\), whose derivative is \(3{x^2} - 10x - 3\)

Thus,

\[\begin{align}f\left( 1 \right) &= {1^3} - 5 \times {1^2} - 3 \times 1 = - 7\\f\left( 3 \right) &= {3^3} - 5 \times {3^2} - 3 \times 3 = - 27\end{align}\]

Therefore,

\[\begin{align}\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} &= \frac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}}\\ &= \frac{{ - 27 - \left( { - 7} \right)}}{{3 - 1}}\\& = - 10\end{align}\]

Mean Value Theorem states that there exists a point \(c \in \left( {1,3} \right)\) such that \(f'\left( c \right) = - 10\)

Hence,

\[\begin{align}& \Rightarrow \; f'\left( c \right) = - 10\\& \Rightarrow \; 3{c^2} - 10c - 3 = - 10\\ &\Rightarrow \; 3{c^2} - 10c + 7 = 0\\ &\Rightarrow \; 3{c^2} - 3c - 7c + 7 = 0\\& \Rightarrow \; 3c\left( {c - 1} \right) - 7\left( {c - 1} \right) = 0\\& \Rightarrow \; \left( {c - 1} \right)\left( {3c - 7} \right) = 0\\ &\Rightarrow \; c = 1,\frac{7}{3} \qquad \left[ {{\text{where }}c = \frac{7}{3} \in \left( {1,3} \right)} \right]\end{align}\]

Thus, Mean Value Theorem is verified for the given function and \(c = \frac{7}{3} \in \left( {1,3} \right)\) is the only point for which \(f'\left( c \right) = 0\).

Chapter 5 Ex.5.8 Question 6

Examine the applicability of Mean Value Theorem for all three functions given

(i) \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]\)

(ii) \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]\)

(iii) \(f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]\)

 

Solution

Video Solution

 

Mean Value Theorem states that for a function \(f:\left[ {a,b} \right] \to {\bf{R}}\), if

\(f\) is continuous on \(\left[ {a,b} \right]\)

\(f\) is continuous on \(\left( {a,b} \right)\)

Then there exists some \(c \in \left( {a,b} \right)\) such that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)

Thus, Mean Value Theorem is not applicable to those functions that do not satisfy any of three conditions of the hypothesis.

(i) \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]\)

     Since, the given function \(f\left( x \right)\) is not continuous at every integral point.

     In general, \(f\left( x \right)\) is not continuous at \(x = 5\) and \(x = 9\)

              Therefore, \(f\left( x \right)\) is not continuous in \(\left[ {5,9} \right]\)

      The differentiability of \(f\) in \(\left( {5,9} \right)\) is checked as follows.

      Let \(n\) be an integer such that \(n \in \left( {5,9} \right)\)

          The LHD of \(f\) at \(x = n\) is

                       \(\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - 1}}{h} = \infty \)

          The RHD of \(f\) at \(x = n\) is

                       \(\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{n - n}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0\)

   Since LHD and RHD of \(f\) at \(x = n\) are not equal, \(f\) is not differentiable at \(x = n\)

              Therefore, \(f\) is not differentiable in \(\left( {5,9} \right)\).

    It is observed that \(f\) does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

   Thus, Mean Value Theorem is not applicable for \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]\)

(ii) \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]\)

     Since, the given function \(f\left( x \right)\) is not continuous at every integral point.

      In general, \(f\left( x \right)\) is not continuous at \(x = - 2\) and \(x = 2\)

                Therefore, \(f\left( x \right)\) is not continuous in \(\left[ { - 2,2} \right]\)

      The differentiability of \(f\) in \(\left( { - 2,2} \right)\) is checked as follows.

      Let n be an integer such that \(n \in \left( { - 2,2} \right)\)

          The LHD of \(f\) at \(x = n\) is

                     \(\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - 1}}{h} = \infty \)

           The RHD of \(f\) at \(x = n\) is

                     \(\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{n - n}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0\)

      Since LHD and RHD of \(f\) at \(x = n\) are not equal, \(f\) is not differentiable at \(x = n\)

                    Therefore, \(f\) is not differentiable in \(\left( { - 2,2} \right)\) .

      It is observed that \(f\) does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

   Thus, Mean Value Theorem is not applicable for \(f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]\).

(iii) \(f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]\)

      Since, \(f\) being a polynomial function is continuous in \(\left[ {1,2} \right]\) and is differentiable in \(\left( {1,2} \right)\)

      It is observed that \(f\) satisfies all the conditions of the hypothesis of Mean Value Theorem.

      Hence, Mean Value Theorem is applicable for \(f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]\).

      It can be proved as follows.

      We have, \(f\left( x \right) = {x^2} - 1\)

      Then,

        \[\begin{align}f\left( 1 \right) &= {\left( 1 \right)^2} - 1 = 0,\\f\left( 2 \right) &= {\left( 2 \right)^2} - 1 = 3\end{align}\]

      Therefore,

       \[\begin{align}\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} &= \frac{{f\left( 2 \right) - f\left( 1 \right)}}{{2 - 1}} = \frac{{3 - 0}}{1}\\& = 3\end{align}\]

       Hence, \(f'\left( x \right) = 2x\)

      Thus,

      \[\begin{align} &\Rightarrow \; f'\left( c \right) = 3\\& \Rightarrow \; 2c = 3\\ &\Rightarrow \; c = \frac{3}{2}\\ &\Rightarrow \; c = 1.5 \qquad \left[ {{\text{where }}1.5 \in \left[ {1,2} \right]} \right]\end{align}\]

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