# Let \[f\left( x \right)=\left\{ \begin{align}

& 1-\sqrt{1-{{x}^{2}}}\text{ for }-1\le x\le 1 \\

& 1+\log \dfrac{1}{x}\text{ for }x>1 \\

\end{align} \right.\]

Then

(a) \[f\]is continuous at \[x=1\]

(b) \[f\]is not differentiable at \[x=1\]

(c) \[f\]is continuous and differentiable at \[x=1\]

(d) \[{{f}^{'}}\left( x \right)\]exists for all \[x\in \left( 0,1 \right)\]

Answer

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Hint: If the value of limit of the function at a point \[x=a\] is equal to the value of the function at \[x=a\] , the function is said to be continuous at \[x=a\]. A function is differentiable at \[x=a\] , if the left-hand derivative of the function is equal to the right hand derivative of the function at \[x=a\].

Complete step-by-step answer:

The given function is \[f\left( x \right)=\left\{ \begin{align}

& 1-\sqrt{1-{{x}^{2}}}\text{ for }-1\le x\le 1 \\

& 1+\log \dfrac{1}{x}\text{ for }x>1 \\

\end{align} \right.\]

We will check if the function is continuous or differentiable at critical points of the function .

A function is differentiable at \[x=a\] , if the left-hand derivative of the function is equal to the right hand derivative of the function at \[x=a\].

We know , the left hand derivative of \[f\left( x \right)\] at \[x=a\] is given as \[{{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}\] and the right hand derivative of \[f\left( x \right)\] at \[x=a\] is given as \[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}\].

First , we will check the differentiability of the function \[f\left( x \right)\] at \[x=1\].

The left-hand derivative of \[f\left( x \right)\]at \[x=1\]is given by

\[{{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1-h \right)-f\left( 1 \right)}{-h}\]

\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1-\sqrt{1-{{\left( 1-h \right)}^{2}}} \right)-\left( 1-\sqrt{1-{{1}^{2}}} \right)}{-h}\]

\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1-\sqrt{1-1+2h-{{h}^{2}}}-1}{-h}\]

\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{-\sqrt{2h-{{h}^{2}}}}{-h}\]

Now , on substituting \[h=0\] in the limit , we can see that it gives an indeterminate value \[\dfrac{0}{0}\]. In such conditions, we apply L’ Hopital’s rule to evaluate the limit.

L’ Hopital’s Rule states that “ if \[L=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}\] and \[f(x)=g(x)=0\] or \[\infty \], then \[L=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)}\] .”

So , to find the value of the limit , we must differentiate the numerator and the denominator with respect to \[x\].

\[{{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{2\sqrt{2h-{{h}^{2}}}}.2-2h}{-1}\]

\[=\dfrac{2}{0}=\infty \]

The right-hand derivative of \[f\left( x \right)\]at \[x=1\]is given by

\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}\]

\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1+\log \dfrac{1}{\left( 1+h \right)}-\left( 1+\log 1 \right)}{h}\]

\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1+\log \dfrac{1}{\left( 1+h \right)}-1-0}{h}\]

\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\log \dfrac{1}{\left( 1+h \right)}}{h}\]

Applying L’ Hopital’s Rule , we get

\[\begin{align}

& {{R}^{'}}=\dfrac{\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1}{\dfrac{1}{1+h}}\times \dfrac{-1}{{{(1+h)}^{2}}}}{1} \\

& =-1 \\

\end{align}\]

Since the left hand derivative is not equal to the right hand derivative, hence , the function is not differentiable at \[x=1\].

Now, we will check the continuity of the function at \[x=1\].

A function is said to be continuous at \[x=a\] if the value of limit of the function at a point \[x=a\] is equal to the value of the function at \[x=a\] .

The right-hand limit of \[f\left( x \right)\]at \[x=1\] is given by

\[\underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1+\log \dfrac{1}{1+h} \right)\]

\[\begin{align}

& =1+0 \\

& =1 \\

\end{align}\]

The left-hand limit of \[f\left( x \right)\]at \[x=1\]is given by

\[\begin{align}

& \underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1-\sqrt{1-{{\left( 1-h \right)}^{2}}} \right) \\

& =1 \\

\end{align}\]

Value of function at \[x=1\]is given as

\[\begin{align}

& f\left( 1 \right)=1-\sqrt{1-{{1}^{2}}} \\

& =1 \\

\end{align}\]

Clearly , left hand limit \[=\]right hand limit\[=f\left( 1 \right)\]

So , the function is continuous at \[x=1\].

Now , in the interval \[\left( 0,1 \right)\], \[f\left( x \right)=1-\sqrt{1-{{x}^{2}}}\]. On differentiating \[f(x)\] with respect to \[x\] , we get

\[\dfrac{d}{dx}(f(x))=\dfrac{-1}{2\sqrt{1-{{x}^{2}}}}\times (-2x)\]

\[=\dfrac{x}{\sqrt{1-{{x}^{2}}}}\]

Clearly , \[f'(x)\] exists \[\forall x\in (-1,1)\]. Hence , \[{{f}^{'}}\left( x \right)\]exists for all \[x\in \left( 0,1 \right)\].

Answer is (a),(b),(d)

Note: If a function is differentiable at a, it should necessarily be continuous. But if a function is continuous, it is not necessary that it is differentiable at \[x=a\].

Complete step-by-step answer:

The given function is \[f\left( x \right)=\left\{ \begin{align}

& 1-\sqrt{1-{{x}^{2}}}\text{ for }-1\le x\le 1 \\

& 1+\log \dfrac{1}{x}\text{ for }x>1 \\

\end{align} \right.\]

We will check if the function is continuous or differentiable at critical points of the function .

A function is differentiable at \[x=a\] , if the left-hand derivative of the function is equal to the right hand derivative of the function at \[x=a\].

We know , the left hand derivative of \[f\left( x \right)\] at \[x=a\] is given as \[{{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}\] and the right hand derivative of \[f\left( x \right)\] at \[x=a\] is given as \[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}\].

First , we will check the differentiability of the function \[f\left( x \right)\] at \[x=1\].

The left-hand derivative of \[f\left( x \right)\]at \[x=1\]is given by

\[{{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1-h \right)-f\left( 1 \right)}{-h}\]

\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1-\sqrt{1-{{\left( 1-h \right)}^{2}}} \right)-\left( 1-\sqrt{1-{{1}^{2}}} \right)}{-h}\]

\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1-\sqrt{1-1+2h-{{h}^{2}}}-1}{-h}\]

\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{-\sqrt{2h-{{h}^{2}}}}{-h}\]

Now , on substituting \[h=0\] in the limit , we can see that it gives an indeterminate value \[\dfrac{0}{0}\]. In such conditions, we apply L’ Hopital’s rule to evaluate the limit.

L’ Hopital’s Rule states that “ if \[L=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}\] and \[f(x)=g(x)=0\] or \[\infty \], then \[L=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)}\] .”

So , to find the value of the limit , we must differentiate the numerator and the denominator with respect to \[x\].

\[{{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{2\sqrt{2h-{{h}^{2}}}}.2-2h}{-1}\]

\[=\dfrac{2}{0}=\infty \]

The right-hand derivative of \[f\left( x \right)\]at \[x=1\]is given by

\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}\]

\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1+\log \dfrac{1}{\left( 1+h \right)}-\left( 1+\log 1 \right)}{h}\]

\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1+\log \dfrac{1}{\left( 1+h \right)}-1-0}{h}\]

\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\log \dfrac{1}{\left( 1+h \right)}}{h}\]

Applying L’ Hopital’s Rule , we get

\[\begin{align}

& {{R}^{'}}=\dfrac{\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1}{\dfrac{1}{1+h}}\times \dfrac{-1}{{{(1+h)}^{2}}}}{1} \\

& =-1 \\

\end{align}\]

Since the left hand derivative is not equal to the right hand derivative, hence , the function is not differentiable at \[x=1\].

Now, we will check the continuity of the function at \[x=1\].

A function is said to be continuous at \[x=a\] if the value of limit of the function at a point \[x=a\] is equal to the value of the function at \[x=a\] .

The right-hand limit of \[f\left( x \right)\]at \[x=1\] is given by

\[\underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1+\log \dfrac{1}{1+h} \right)\]

\[\begin{align}

& =1+0 \\

& =1 \\

\end{align}\]

The left-hand limit of \[f\left( x \right)\]at \[x=1\]is given by

\[\begin{align}

& \underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1-\sqrt{1-{{\left( 1-h \right)}^{2}}} \right) \\

& =1 \\

\end{align}\]

Value of function at \[x=1\]is given as

\[\begin{align}

& f\left( 1 \right)=1-\sqrt{1-{{1}^{2}}} \\

& =1 \\

\end{align}\]

Clearly , left hand limit \[=\]right hand limit\[=f\left( 1 \right)\]

So , the function is continuous at \[x=1\].

Now , in the interval \[\left( 0,1 \right)\], \[f\left( x \right)=1-\sqrt{1-{{x}^{2}}}\]. On differentiating \[f(x)\] with respect to \[x\] , we get

\[\dfrac{d}{dx}(f(x))=\dfrac{-1}{2\sqrt{1-{{x}^{2}}}}\times (-2x)\]

\[=\dfrac{x}{\sqrt{1-{{x}^{2}}}}\]

Clearly , \[f'(x)\] exists \[\forall x\in (-1,1)\]. Hence , \[{{f}^{'}}\left( x \right)\]exists for all \[x\in \left( 0,1 \right)\].

Answer is (a),(b),(d)

Note: If a function is differentiable at a, it should necessarily be continuous. But if a function is continuous, it is not necessary that it is differentiable at \[x=a\].

Last updated date: 30th Sep 2023

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