Ex 5.8 class 12 maths ncert solutions | class 12 maths exercise 5.8 | class 12 maths ncert solutions chapter 5 exercise 5.8 | exercise 5.8 class 12 maths ncert solutions | continuity and differentiability class 12 ncert solutions
Looking for Ex 5.8 Class 12 Maths NCERT Solutions? You’ve found the perfect spot! This section offers thorough, step-by-step solutions for Exercise 5.8 Class 12 Maths, from Chapter 5 – Continuity and Differentiability. Designed according to the latest CBSE guidelines, these Class 12 Maths Chapter 5 Exercise 5.8 NCERT Solutions focus on deepening your understanding of advanced differentiation techniques, especially second-order derivatives and their practical applications. The exercise 5.8 class 12 maths NCERT solutions guide you through complex problems involving composite, implicit, and parametric differentiation, ensuring clarity and precision. Ideal for both board exam preparation and daily practice, this expert-curated resource helps boost your confidence in mastering the nuanced aspects of calculus!
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Exercise 5.8
1 . Verify Rolle's theorem for the function \( \mathrm{f}(x)=x^{2}+2 x-8, x \in[-4 \), 2].
Answer
The given function is \( \mathrm{f}(x)=x^{2}+2 x-8 \) and \( x \in[-4,2] \).
By Rolle's Theorem, for a function \( f:[a, b] \rightarrow R \), if
(a) \(f\) is continuous on \( [\mathrm{a}, \mathrm{b}] \)
(b) \( f \) is differentiable on \( (a, b) \)
(c) \( f(a)=f(b) \)
Then there exists some c in \( (\mathrm{a}, \mathrm{b}) \) such that \( \mathrm{f}^{\prime}(\mathrm{c})=0 \).
As \( f(x)=x^{2}+2 x-8 \) is a polynomial function,
(a) \( f(x) \) is continuous in \( [-4,2] \)
(b) \( f^{\prime}(x)=2 x+2 \)
So, \( f(x) \) is differentiable in \( (-4,2) \).
(c) \( f(a)=f(-4)=(-4)^{2}+2(-4)-8=16-8-8=16-16=0 \)
\( \mathrm{f}(\mathrm{b})=\mathrm{f}(2)=(2)^{2}+2(2)-8=4+4-8=8-8=0 \)
Hence, \( f(a)=f(b) \).
\( \therefore \) There is a point \( \mathrm{c} \in(-4,2) \) where \( \mathrm{f}^{\prime}(\mathrm{c})=0 \).
\(\mathrm{f}(x)=x^{2}+2 x-8\)
\(\mathrm{f}^{\prime}(x)=2 x+2\)
\(\mathrm{f}^{\prime}(\mathrm{c})=0\)
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=2 \mathrm{c}+2=0\)
\(\Rightarrow 2 \mathrm{c}=-2\)
\( \Rightarrow \mathrm{c}=-\frac{ 2 }{ 2 } \)
\( \Rightarrow \mathrm{c}=-1 \) where \( \mathrm{c}=-1 \in(-4,2) \)
Hence, Rolle's Theorem is verified.
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) \( \mathrm{f}(x)=[x] \) for \( x \in[5,9] \)
(ii) \( f(x)=[x] \) for \( x \in[-2,2] \)
(iii) \( f(x)=x^{2}-1 \) for \( x \in[1,2] \)
(i) \( \mathrm{f}(x)=[x] \) for \( x \in[5,9] \)
(ii) \( f(x)=[x] \) for \( x \in[-2,2] \)
(iii) \( f(x)=x^{2}-1 \) for \( x \in[1,2] \)
Answer
By Rolle's Theorem, for a function \( f:[a, b] \rightarrow R \), if
(a) \( f \) is continuous on \( [a, b] \)
(b) \( f \) is differentiable on \( (a, b) \)
(c) \( f(a)=f(b) \)
Then there exists some c in \( (\mathrm{a}, \mathrm{b}) \) such that \( \mathrm{f}^{\prime}(\mathrm{c})=0 \).
If a function does not satisfy any of the above conditions, then Rolle's Theorem is not applicable.
(i) \( \mathrm{f}(x)=[x] \) for \( x \in[5,9] \)
As the given function is a greatest integer function,
(a) \( f(x) \) is not continuous in \( [5,9] \)
(b) Let \(y\) be an integer such that \( y \in(5,9) \)
Left hand limit of \( \mathrm{f}(x) \) at \( x=y: \lim _{h \rightarrow 0^{-}} \frac{f(y+h)-f(y)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[y+h]-[y]}{h}= \) \( \lim _{h \rightarrow 0^{-}} \frac{y-1-y}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty \)
Right hand limit of \( \mathrm{f}(x) \) at \( x=y \) :
\(\lim _{h \rightarrow 0^{+}} \frac{f(y+h)-f(y)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[y+h]-[y]}{h}=\lim _{h \rightarrow 0^{+}} \frac{y-y}{h}=\lim _{h \rightarrow 0^{+}} \frac{0}{h}=0\)
Since, left and right hand limits of \( \mathrm{f}(x) \) at \( x=y \) is not equal, \( \mathrm{f}(x) \) is not differentiable at \( x=y \).
So, \( f(x) \) is not differentiable in \( [5,9] \)
(c) \( f( a )=f(5)=[5]=5 \)
\( f(b)=f(9)=[9]=9 \)
\( \mathrm{f}(\mathrm{a}) \neq \mathrm{f}(\mathrm{b}) \)
Here, \( \mathrm{f}(x) \) does not satisfy the conditions of Rolle's Theorem.
Rolle's Theorem is not applicable for \( \mathrm{f}(x)=[x] \) for \( x \in[5,9] \).
(ii) \( \mathrm{f}(x)=[x] \) for \( x \in[-2,2] \)
As the given function is a greatest integer function,
(a) \( f(x) \) is not continuous in \( [-2,2] \)
(b) Let \( y \) be an integer such that \( y \in(-2,2) \)
Left hand limit of \( \mathrm{f}(x) \) at \( x=y \) :
\(\lim _{h \rightarrow 0^{-}} \frac{f(y+h)-f(y)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[y+h]-[y]}{h}=\lim _{h \rightarrow 0^{-}} \frac{y-1-y}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty\)
Right hand limit of \( \mathrm{f}(x) \) at \( x=y \) :
\(\lim _{h \rightarrow 0^{+}} \frac{f(y+h)-f(y)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[y+h]-[y]}{h}=\lim _{h \rightarrow 0^{+}} \frac{y-y}{h}=\lim _{h \rightarrow 0^{+}} \frac{0}{h}=0\)
Since, left and right hand limits of \( \mathrm{f}(x) \) at \( x=y \) is not equal, \( \mathrm{f}(x) \) is not differentiable at \( x=y \).
So, \( f(x) \) is not differentiable in \( (-2,2) \)
(c) \( f(a)=f(-2)=[-2]=-2 \)
\( f(b)=f(2)=[2]=2 \)
\( \mathrm{f}(\mathrm{a}) \neq \mathrm{f}(\mathrm{b}) \)
Here, \( \mathrm{f}(x) \) does not satisfy the conditions of Rolle's Theorem.
Rolle's Theorem is not applicable for \( \mathrm{f}(x)=[x] \) for \( x \in[-2,2] \).
(iii) \( \mathrm{f}(x)=x^{2}-1 \) for \( x \in[1,2] \)
As the given function is a polynomial function,
(a) \( f(x) \) is continuous in \( [1,2] \)
(b) \( f^{\prime}(x)=2 x \)
So, \( \mathrm{f}(x) \) is differentiable in \( [1,2] \)
(c) \( \mathrm{f}(\mathrm{a})=\mathrm{f}(1)=1^{2}-1=1-1=0 \)
\( \mathrm{f}(\mathrm{b})=\mathrm{f}(2)=2^{2}-1=4-1=3 \)
\( \mathrm{f}(\mathrm{a}) \neq \mathrm{f}(\mathrm{b}) \)
Here, \( \mathrm{f}(x) \) does not satisfy a condition of Rolle's Theorem.
Rolle's Theorem is not applicable for \( f(x)=x^{2}-1 \) for \( x \in[1,2] \).
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3. If \( f:[-5,5] \rightarrow R \) is a differentiable function and if \( f^{\prime}(x) \) does not vanish anywhere, then prove that \( \mathrm{f}(-5) \neq \mathrm{f}(5) \).
Answer
Given: \( \mathrm{f}:[-5,5] \rightarrow \mathrm{R} \) is a differentiable function.
Mean Value Theorem states that for a function \( f:[a, b] \rightarrow R \), if
(a) \(f\) is continuous on \( [a, b] \)
(b) \(f\) is differentiable on ( \( \mathrm{a}, \mathrm{b} \) )
Then there exists some \( \mathrm{c} \in(\mathrm{a}, \mathrm{b}) \) such that
We know that a differentiable function is a continuous function.
So,
(a) \(f\) is continuous on \( [-5,5] \)
(b) \(f\) is differentiable on \( (-5,5) \)
\( \therefore \) By Mean Value Theorem, there exists \( \mathrm{c} \in(-5,5) \) such that
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=\frac{f(5)-f(-5)}{5-(-5)}\)
\(\Rightarrow 10 \mathrm{f}^{\prime}(\mathrm{c})=\mathrm{f}(5)-\mathrm{f}(-5)\)
It is given that \( f^{\prime}(x) \) does not vanish anywhere.
\(\therefore \mathrm{f}^{\prime}(\mathrm{c}) \neq 0\)
\(10 \mathrm{f}^{\prime}(\mathrm{c}) \neq 0\)
\(\mathrm{f}(5)-\mathrm{f}(-5) \neq 0\)
\(\mathrm{f}(5) \neq \mathrm{f}(-5)\)
Hence proved.
By Mean Value Theorem, it is proved that \( f(5) \neq f(-5) \).
4 . Verify Mean Value Theorem, if \( f(x)=x^{2}-4 x-3 \) in the interval \( [a \), b ], where \( \mathrm{a}=1 \) and \( \mathrm{b}=4 \).
Answer
Given: \( f(x)=x^{2}-4 x-3 \) in the interval \( [1,4] \)
Mean Value Theorem states that for a function \( f:[a, b] \rightarrow R \), if
(a) \(f\) is continuous on \([a, b]\)
(b) \(f\) is differentiable on \(( a, b )\)
Then there exists some \( \mathrm{c} \in(\mathrm{a}, \mathrm{b}) \) such that \( \mathrm{f}^{\prime}(\mathrm{c})=\frac{f(b)-f(a)}{b-a} \)
As \( f(x) \) is a polynomial function,
(a) \( f(x) \) is continuous in \( [1,4] \)
(b) \( f^{\prime}(x)=2 x-4 \)
So, \( f(x) \) is differentiable in \( (1,4) \).
\(\therefore \frac{f(b)-f(a)}{b-a}=\frac{f(4)-f(1)}{4-1}\)
\(f(4)=4^{2}-4(4)-3=16-16-3=-3\)
\(\mathrm{f}(1)=1^{2}-4(1)-3=1-4-3=-6\)
\(=\frac{f(4)-f(1)}{4-1}=\frac{-3-(-6)}{4-1}=\frac{3}{3}=1\)
\( \therefore \) There is a point \( \mathrm{c} \in(1,4) \) such that \( \mathrm{f}^{\prime}(\mathrm{c})=1 \)
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=1\)
\(\Rightarrow 2 \mathrm{c}-4=1\)
\(\Rightarrow 2 \mathrm{c}=1+4=5\)
\(\Rightarrow \mathrm{c}=\frac{ 5 }{ 2 } \text { where } \mathrm{c} \in(1,4)\)
The Mean Value Theorem is verified for the given \( f(x) \).
5 . Verify Mean Value Theorem, if \( f(x)=x^{3}-5 x^{2}-3 x \) in the interval \( [a, b] \), where \( a=1 \) and \( b=3 \). Find all \( c \in(1,3) \) for which \( f^{\prime}(c)=0 \).
Answer
Given: \( f(x)=x^{3}-5 x^{2}-3 x \) in the interval \( [1,3] \)
Mean Value Theorem states that for a function \( f:[a, b] \rightarrow R \), if
(a) \(f\) is continuous on \([a, b]\)
(b) \(f\) is differentiable on \( ( a, b ) \)
Then there exists some \( \mathrm{c} \in(\mathrm{a}, \mathrm{b}) \) such that \( \mathrm{f}^{\prime}(\mathrm{c})=\frac{f(b)-f(a)}{b-a} \)
As \( f(x) \) is a polynomial function,
(a) \( f(x) \) is continuous in \( [1,3] \)
(b) \( f^{\prime}(x)=3 x^2-10 x-3 \)
So, \( f(x) \) is differentiable in \( (1,3) \).
\(\therefore \frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}\)
\(f(3)=3^{3}-5(3)^{2}-3(3)=27-45-9=-27\)
\(f(1)=1^{3}-5(1)^{2}-3(1)=1-5-3=-7\)
\(\Rightarrow \frac{f(3)-f(1)}{3-1}=\frac{-27-(-7)}{3-1}=\frac{-20}{0}=-10\)
\( \therefore \) There is a point \( \mathrm{c} \in(1,4) \) such that \( \mathrm{f}^{\prime}(\mathrm{c})=-10 \)
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=-10\)
\(\Rightarrow 3 \mathrm{c}^{2}-10 \mathrm{c}-3=-10\)
\(\Rightarrow 3 \mathrm{c}^{2}-10 \mathrm{c}+7=0\)
\(\Rightarrow 3 \mathrm{c}^{2}-3 \mathrm{c}-7 \mathrm{c}+7=0\)
\(\Rightarrow 3 \mathrm{c}(\mathrm{c}-1)-7(\mathrm{c}-1)=0\)
\(\Rightarrow(\mathrm{c}-1)(3 \mathrm{c}-7)=0\)
\(\Rightarrow \mathrm{c}=1,\frac{ 7 }{ 3 } \text { where } \mathrm{c}=\frac{ 7 }{ 3 } \in(1,3)\)
The Mean Value Theorem is verified for the given \( f(x) \) and \( c=\frac{ 7 }{ 3 } \in(1 \), 3 ) is the only point for which \( f^{\prime}(c)=0 \).
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 \( f:[a, b] \rightarrow R \), if
(a) \( f \) is continuous on \( [a, b] \)
(b) \( f \) is differentiable on \( (a, b) \)
Then there exists some \( \mathrm{c} \in(\mathrm{a}, \mathrm{b}) \) such that \( \mathrm{f}^{\prime}(\mathrm{c})=\frac{f(b)-f(a)}{b-a} \)
If a function does not satisfy any of the above conditions, then Mean Value Theorem is not applicable.
(i) \( \mathrm{f}(x)=[x] \) for \( x \in[5,9] \)
As the given function is a greatest integer function,
(a) \( f(x) \) is not continuous in \( [5,9] \)
(b) Let \( y \) be an integer such that \( y \in(5,9) \)
Left hand limit of \( \mathrm{f}(x) \) at \( x=y \) :
\(\lim _{h \rightarrow 0^{-}} \frac{f(y+h)-f(y)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[y+h]-[y]}{h}=\lim _{h \rightarrow 0^{-}} \frac{y-1-y}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty\)
Right hand limit of \( \mathrm{f}(x) \) at \( x=y \) :
\(\lim _{h \rightarrow 0^{+}} \frac{f(y+h)-f(y)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[y+h]-[y]}{h}=\lim _{h \rightarrow 0^{+}} \frac{y-y}{h}=\lim _{h \rightarrow 0^{+}} \frac{0}{h}=0\)
Since, left and right hand limits of \( f(x) \) at \( x=y \) is not equal, \( f(x) \) is not differentiable at \( x=y \).
So, \( \mathrm{f}(x) \) is not differentiable in \( [5,9] \).
Here, \( \mathrm{f}(x) \) does not satisfy the conditions of Mean Value Theorem.
Mean Value Theorem is not applicable for \( \mathrm{f}(x)=[x] \) for \( x \in[5,9] \).
(ii) \( \mathrm{f}(x)=[x] \) for \( x \in[-2,2] \)
As the given function is a greatest integer function,
(a) \( \mathrm{f}(x) \) is not continuous in \( [-2,2] \)
(b) Let \( y \) be an integer such that \( y \in(-2,2) \)
Left hand limit of \( \mathrm{f}(x) \) at \( x=y \) :
\(\lim _{h \rightarrow 0^{-}} \frac{f(y+h)-f(y)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[y+h]-[y]}{h}=\lim _{h \rightarrow 0^{-}} \frac{y-1-y}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty\)
Right hand limit of \( f(x) \) at \( x=y \) :
\(\lim _{h \rightarrow 0^{+}} \frac{f(y+h)-f(y)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[y+h]-[y]}{h}=\lim _{h \rightarrow 0^{+}} \frac{y-y}{h}=\lim _{h \rightarrow 0^{+}} \frac{0}{h}=0\)
Since, left and right hand limits of \( f(x) \) at \( x=y \) is not equal, \( f(x) \) is not differentiable at \( x=y \).
So, \( \mathrm{f}(x) \) is not differentiable in \( (-2,2) \)
Here, \( \mathrm{f}(x) \) does not satisfy the conditions of Mean Value Theorem.
Mean Value Theorem is not applicable for \( \mathrm{f}(x)=[x] \) for \( x \in[-2,2] \).
(iii) \( \mathrm{f}(x)=x^2-1 \) for \( x \in[1,2] \)
As the given function is a polynomial function,
(a) \( f(x) \) is continuous in \( [1,2] \)
(b) \( f^{\prime}(x)=2 x \)
So, \( f(x) \) is differentiable in [1, 2].
Here, \( \mathrm{f}(x) \) satisfies the conditions of Mean Value Theorem.
So, Mean Value Theorem is applicable for \( \mathrm{f}(x) \).
\(\therefore \frac{f(b)-f(a)}{b-a}=\frac{f(2)-f(1)}{2-1}\)
\(\mathrm{f}(2)=2^{2}-1=4-1=3\)
\(\mathrm{f}(1)=1^{2}-1=1-1=0\)
\(\Rightarrow \frac{f(2)-f(1)}{2-1}=\frac{3-0}{2-1}=\frac{3}{1}=3\)
\( \therefore \) There is a point \( \mathrm{c} \in(1,2) \) such that \( \mathrm{f}^{\prime}(\mathrm{c})=3 \)
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=3\)
\(\Rightarrow 2 \mathrm{c}=3\)
\(\Rightarrow \mathrm{c}=\frac{3}{2} \text { where } \mathrm{c} \in(1,2)\)
Mean Value Theorem is applicable for \( f(x)=x^{2}-1 \) for \( x \in[1,2] \).
