# Let \[f(x) = (x + \left| x \right|)\left| x \right|\]. Which of the following is true for all x.

(a) f is continuous

(b) f is differentiable for some x

(c) f’ is continuous

(d) f’’ is continuous

Answer

Verified

361.8k+ 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.

Last updated date: 30th Sep 2023

•

Total views: 361.8k

•

Views today: 4.61k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

The poet says Beauty is heard in Can you hear beauty class 6 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

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

What is the past tense of read class 10 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE