Question

Let f (x) = |x| and g (x) = |x³|, then

पर्याय

•  f (x) and g (x) both are continuous at x = 0

• f (x) and g (x) both are differentiable at x = 0

• f (x) is differentiable but g (x) is not differentiable at x = 0

•  f (x) and g (x) both are not differentiable at x = 0

Answer 1

Option (a) f (x) and g (x) both are continuous at x = 0 

Given: 

$$f\left(x\right)=\left|x\right|,g\left(x\right)=\left|x^3\right|$$

We know  

$$\left|x\right|$$  is continuous at x=0 but not differentiable at x = 0 as (LHD at x = 0) ≠ (RHD at x = 0).
Now, for the function 

${\displaystyle g\left(x\right)=\left|x^3\right|=\{\begin{array}{ll}{x}^3& x\ge 0\\ -x^3& x<0\end{array}}$

Continuity at x = 0:

$$\underset{x\to 0^{-}}{lim}g\left(x\right)=\underset{h\to 0}{lim}g(0-h)=\underset{h\to 0}{lim}-(-h^3)=\underset{h\to 0}{lim}{h}^3=0.$$

(RHL at x = 0) =  

$$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}{h}^3=0.$$

and 

$$g\left(0\right)=0.$$

Thus,   

$$\underset{x\to 0^{-}}{lim}g\left(x\right)=\underset{x\to 0^{+}}{lim}g\left(x\right)=g\left(0\right)$$

Hence, 

$$g(x)$$ is continuous at x = 0.

Differentiability at x = 0:

(LHD at x = 0) = 

$$\underset{x\to 0^{-}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{0-h-0}=\underset{h\to 0}{lim}\frac{h^3-0}{-h}=0.$$

(RHD at x = 0) = 

$$\underset{x\to c^{+}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{0+h-0}=\underset{h\to 0}{lim}\frac{h^3-0}{h}=\underset{h\to 0}{lim}\frac{h^3}{h}=0$$

 Thus, (LHD at x = 0) = (RHD at x = 0). 
Hence, the function

$$g\left(x\right)$$  is differentiable at x = 0.