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Consider the following two statements about the function $f(x)=\left\vert x\right\vert$:

  • P. $f(x)$ is continuous for all real values of $x$.
  • Q. $f(x)$ is differentiable for all real values of $x$ .

Which of the following is TRUE?

  1. $P$ is true and $Q$ is false.
  2. $P$ is false and $Q$ is true.
  3. Both $P$ and $Q$ are true.
  4. Both $P$ and $Q$ are false.
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2 Answers

Best answer
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25 votes

Ans is A.

$f(x)=\mid x\mid.$ Here, for all values of $x, f(x)$ exists. Therefore, it is continuous for all real values of $x.$ 

At $x=0, f(x)$ is not differentiable. Because if we take the left hand derivative here, it is negative while the right hand derivative is positive making $\text{LHD} \neq \text{RHD}$  

Ref: http://math.stackexchange.com/questions/991475/why-is-the-absolute-value-function-not-differentiable-at-x-0

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$f(x) = \begin{cases} -x & \text{ if }  x < 0 \\ +x & \text{ if } x \geq 0  \end{cases}$

Continuity at $ x=0 $

$\begin{align*} \lim_{x \rightarrow 0^{-}}f(x)\ &= \lim_{x \rightarrow 0^{+}}f(x)\\ 0 &= 0\\ \end{align*}$

So $ f(x) = |x|$ is continous at x = 0


Differentiability at $x=0$

$\begin{align*} \lim_{h \rightarrow 0}\frac{f(0)-f(0-h)}{0-(0-h)} &= \lim_{h \rightarrow 0}\frac{f(0+h)-f(0)}{(0+h)-(0)}\\ \lim_{h \rightarrow 0}\frac{0-(-0+h)}{h)} &= \lim_{h \rightarrow 0}\frac{(0+h)-0}{h} \\ -1 &= 1\\ \end{align*}$

So $ f(x) = |x|$ is not differentiable at x = 0

P is true and Q is false

Option $A$ is the aswer

Answer:

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