GATE CSE 2007 | Question: 1

Question

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?

• A. $P$ is true and $Q$ is false.
• B. $P$ is false and $Q$ is true.
• C. Both $P$ and $Q$ are true.
• D. Both $P$ and $Q$ are false.

Answer 1

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

Comments

at x= 0, Left hand derivative is not equal to right hand derivative. So it is not differentiable.

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at x=0 Left hand limit and Right hand limit are equal , for derivability we check LHD(left hand derivative) and RHD(Right hand derivative) , those are not equal so function at x=0 is not differentiable.

$LHD $ = $f'(a^{-})$ = $\frac{f(a+h)-f(a)}{h}$   where $h->0^{-}$

$RHD$ = $f'(a^{+})$ = $\frac{f(a+h)-f(a)}{h}$   where $h->0^{+}$

LHD = $f'(0^{-})$ = $\frac {f(0+h)-f(0)}{h}$ = $\frac {\left | 0+h \right | -\left | 0 \right |}{h}$ = $\lim_{h->0^{-}} \frac{\left | h \right |}{h} = \frac{-h}{h}=-1$

RHD = $f'(0^{+})$ = $\frac {f(0+h)-f(0)}{h}$=$\frac {\left | 0+h \right | -\left | 0 \right |}{h}$ = $\lim_{h->0^{+}} \frac{\left | h \right |}{h} = \frac{h}{h}=1$

$LHD\neq RHD$

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it should be LHD!=RHD instead of LHL!=RHL in the above answer,,,writing LHL!=RHL is completely wrong for mod(x) at x=0

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The Answer needs to be edited, As LHL != RHL is wrong statement in the answer. It should be LHD != RHD. As the function is not Differentiable at x = 0. But its Continuous at x = 0.

If I am wrong. Please do correct me.

Thanks.

Answer 2

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

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

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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