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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.

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