Question
A function $$f:R \to R$$ is defined as $$f\left( x \right) = {x^2}$$ for and for $$x \geqslant 0$$ and $$f\left( x \right) = - x$$ for $$x < 0.$$
Consider the following statements in respect of the above function :
1. The function is continuous at $$x = 0.$$
2. The function is differentiable at $$x = 0.$$
Which of the above statements is/are correct ?
A.
1 only
B.
2 only
C.
Both 1 and 2
D.
Neither 1 nor 2
Answer :
1 only
Solution :
\[f:R \to R,\,f\left( x \right) = \left\{ \begin{array}{l}
\,\,\,{x^2},\,\,\,\,\,x \ge 0\\
- x,\,\,\,\,\,x < 0
\end{array} \right.\]
For continuity at $$x = 0$$
$$\eqalign{
& f\left( {0 - 0} \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 - h} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{h \to 0} \left[ {\left( {0 - h} \right)} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{h \to 0} \,h = 0 \cr
& f\left( {0 + 0} \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{h \to 0} {\left( {0 + h} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0 \cr
& {\text{and }}f\left( 0 \right) = 0 \cr
& f\left( x \right){\text{ is continuous at }}x = 0 \cr
& {\text{For differentiability at }}x = 0 \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{ - \left( { - h} \right) - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{{ - h}} = - 1 \cr
& {\text{and }}\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \,h = 0 \cr
& f\left( x \right){\text{ is not differentiable at }}x = 0 \cr} $$