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 

`g(x) = |x^3| = {(x^3, xge 0),(-x^3 , x<0):}`

Continuity at x = 0:

\[\lim_{x \to 0^-} g\left( x \right) = \lim_{h \to 0} g\left( 0 - h \right) = \lim_{h \to 0} - \left( - h^3 \right) = \lim_{h \to 0} h^3 = 0 .\]

(RHL at x = 0) =  

\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} h^3 = 0 .\]

and 

\[g\left( 0 \right) = 0 .\]

Thus,   

\[\lim_{x \to 0^-} g\left( x \right) = \lim_{x \to 0^+} g\left( x \right) = g\left( 0 \right)\]

Hence, 

\[g(x)\] is continuous at x = 0.

Differentiability at x = 0:

(LHD at x = 0) = 

\[\lim_{x \to 0^-} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0} = \lim_{h \to 0} \frac{f\left( 0 - h \right) - f\left( 0 \right)}{0 - h - 0} = \lim_{h \to 0} \frac{h^3 - 0}{- h} = 0 .\]

(RHD at x = 0) = 

\[\lim_{x \to c^+} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0} = \lim_{h \to 0} \frac{f\left( 0 + h \right) - f\left( 0 \right)}{0 + h - 0} = \lim_{h \to 0} \frac{h^3 - 0}{h} = \lim_{h \to 0} \frac{h^3}{h} = 0\]

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

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