Answer

Verified

484.2k+ views

Hint: Write the expression of f(x) for x>0 and x<0. Check if f(x) is continuous, if it is continuous check for its differentiability. Then check if f’ and f’’ are continuous.

Complete step-by-step answer:

We know that \[\left| x \right|\] is equal to x for x \[ \geqslant \] 0 and -x for x < 0. Then, we can express f(x) as follows:

\[f(x) = \left\{ \begin{gathered}

(x - x)( - x){\text{ }},x < 0 \\

(x + x)x{\text{ }},x \geqslant 0 \\

\end{gathered} \right.\]

\[f(x) = \left\{ \begin{gathered}

0{\text{ }},x < 0 \\

2{x^2}{\text{ }},x \geqslant 0 \\

\end{gathered} \right.............(1)\]

We know that constant and polynomial functions are continuous in their domain, so f(x) is continuous for all x > 0 and x < 0. Let us check the continuity of f(x) at x = 0.

The left-hand limit of f(x) at x=0 is as follows:

\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f(x)\]

\[LHL = \mathop {\lim }\limits_{x \to 0} 0\]

\[LHL = 0............(2)\]

Hence, the LHL of f(x) at x = 0 is zero.

Now, the right-hand limit of f(x) at x = 0 is as follows:

\[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x)\]

\[RHL = \mathop {\lim }\limits_{x \to 0} 2{x^2}\]

\[RHL = 0............(3)\] The value of f(x) at x = 0 is as follows:

\[f(0) = 2{(0)^2}\]

\[f(0) = 0..........(4)\]

From equations (2), (3) and (4), we have:

\[LHL = RHL = f(0)\]

Hence, f(x) is continuous at x = 0.

Therefore, f(x) is continuous everywhere.

We now check the differentiability of f(x) at x=0.

The left-hand derivative of f(x) at x = 0 is as follows:

\[LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x - h) - f(x)}}{{ - h}}\]

Here, x = 0, then, we have:

\[LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f( - h) - f(0)}}{{ - h}}\]

Using equation (1) in the above equation, we have:

\[LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{0 - 2{{(0)}^2}}}{{ - h}}\]

\[LHD = \mathop {\lim }\limits_{h \to 0} 0\]

\[LHD = 0..........(5)\]

The right-hand derivative of f(x) is given as follows:

\[RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]

Here, x = 0, then, we have:

\[RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(h) - f(0)}}{h}\]

Using equation (1), in the above equation, we obtain:

\[RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{2{h^2} - 0}}{h}\]

Simplifying further, we get:

\[RHD = \mathop {\lim }\limits_{h \to 0} 2h\]

\[RHD = 0.........(6)\]

From equation (5) and equation (6), we have:

\[LHD = RHD\]

Hence, f(x) is differentiable everywhere.

We find the derivative of f(x) as follows:

\[f'(x) = \left\{ \begin{gathered}

0{\text{ }},x < 0 \\

4x{\text{ }},x \geqslant 0 \\

\end{gathered} \right.............(7)\]

Again, this is continuous for x > 0 and x < 0.

At x = 0, we have:

\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f'(x)\]

\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} 0\]

\[LHL = 0\]

\[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f'(x)\]

\[RHL = \mathop {\lim }\limits_{x \to 0} 4x\]

\[RHL = 0\]

\[f'(0) = 0\]

Hence, we have:

\[LHL = RHL = f'(0)\]

Therefore, f’(x) is continuous everywhere.

We now find f’’(x).

\[f'(x) = \left\{ \begin{gathered}

0{\text{ }},x < 0 \\

4{\text{ }},x \geqslant 0 \\

\end{gathered} \right.............(8)\]

We know clearly that at x = 0,

\[LHL = 0\]

\[RHL = 4\]

\[LHL \ne RHL\]

Hence, f’’(x) is not continuous.

Hence, the correct options are (a), (b) and (c).

Note: To check continuity, finding left-hand limit and right-hand limit and equating them is not sufficient, we need to also check the value of the function at that point. A function is differentiable only if the function is continuous.

Complete step-by-step answer:

We know that \[\left| x \right|\] is equal to x for x \[ \geqslant \] 0 and -x for x < 0. Then, we can express f(x) as follows:

\[f(x) = \left\{ \begin{gathered}

(x - x)( - x){\text{ }},x < 0 \\

(x + x)x{\text{ }},x \geqslant 0 \\

\end{gathered} \right.\]

\[f(x) = \left\{ \begin{gathered}

0{\text{ }},x < 0 \\

2{x^2}{\text{ }},x \geqslant 0 \\

\end{gathered} \right.............(1)\]

We know that constant and polynomial functions are continuous in their domain, so f(x) is continuous for all x > 0 and x < 0. Let us check the continuity of f(x) at x = 0.

The left-hand limit of f(x) at x=0 is as follows:

\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f(x)\]

\[LHL = \mathop {\lim }\limits_{x \to 0} 0\]

\[LHL = 0............(2)\]

Hence, the LHL of f(x) at x = 0 is zero.

Now, the right-hand limit of f(x) at x = 0 is as follows:

\[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x)\]

\[RHL = \mathop {\lim }\limits_{x \to 0} 2{x^2}\]

\[RHL = 0............(3)\] The value of f(x) at x = 0 is as follows:

\[f(0) = 2{(0)^2}\]

\[f(0) = 0..........(4)\]

From equations (2), (3) and (4), we have:

\[LHL = RHL = f(0)\]

Hence, f(x) is continuous at x = 0.

Therefore, f(x) is continuous everywhere.

We now check the differentiability of f(x) at x=0.

The left-hand derivative of f(x) at x = 0 is as follows:

\[LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x - h) - f(x)}}{{ - h}}\]

Here, x = 0, then, we have:

\[LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f( - h) - f(0)}}{{ - h}}\]

Using equation (1) in the above equation, we have:

\[LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{0 - 2{{(0)}^2}}}{{ - h}}\]

\[LHD = \mathop {\lim }\limits_{h \to 0} 0\]

\[LHD = 0..........(5)\]

The right-hand derivative of f(x) is given as follows:

\[RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]

Here, x = 0, then, we have:

\[RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(h) - f(0)}}{h}\]

Using equation (1), in the above equation, we obtain:

\[RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{2{h^2} - 0}}{h}\]

Simplifying further, we get:

\[RHD = \mathop {\lim }\limits_{h \to 0} 2h\]

\[RHD = 0.........(6)\]

From equation (5) and equation (6), we have:

\[LHD = RHD\]

Hence, f(x) is differentiable everywhere.

We find the derivative of f(x) as follows:

\[f'(x) = \left\{ \begin{gathered}

0{\text{ }},x < 0 \\

4x{\text{ }},x \geqslant 0 \\

\end{gathered} \right.............(7)\]

Again, this is continuous for x > 0 and x < 0.

At x = 0, we have:

\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f'(x)\]

\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} 0\]

\[LHL = 0\]

\[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f'(x)\]

\[RHL = \mathop {\lim }\limits_{x \to 0} 4x\]

\[RHL = 0\]

\[f'(0) = 0\]

Hence, we have:

\[LHL = RHL = f'(0)\]

Therefore, f’(x) is continuous everywhere.

We now find f’’(x).

\[f'(x) = \left\{ \begin{gathered}

0{\text{ }},x < 0 \\

4{\text{ }},x \geqslant 0 \\

\end{gathered} \right.............(8)\]

We know clearly that at x = 0,

\[LHL = 0\]

\[RHL = 4\]

\[LHL \ne RHL\]

Hence, f’’(x) is not continuous.

Hence, the correct options are (a), (b) and (c).

Note: To check continuity, finding left-hand limit and right-hand limit and equating them is not sufficient, we need to also check the value of the function at that point. A function is differentiable only if the function is continuous.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which places in India experience sunrise first and class 9 social science CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE