Question

The set of points where the function f (x) = x |x| is differentiable is 

 

पर्याय

• \[\left( - \infty , \infty \right)\]

• \[\left( - \infty , 0 \right) \cup \left( 0, \infty \right)\]

• \[\left( 0, \infty \right)\]

• \[\left[ 0, \infty \right]\]

Answer 1

(a) \[\left( - \infty , \infty \right)\]

\[\text{ We have }, \]
\[f\left( x \right) = x\left| x \right|\]

`⇒ f(x) {(-x^2, x<0),(0,x=0),(x^2 , x>0):}`
\[\text{ When, x < 0, we have }\]
\[ f\left( x \right) = - x^2 \text{which being a polynomial function is continuous and differentiable in} \left( - \infty , 0 \right)\]
\[\text{ When, x > 0, we have }\]
\[ f\left( x \right) = x^2 \text{which being a polynomial function is continuous and differentiable in} \left( 0, \infty \right)\]
\[\text{Thus possible point of non - differentiability of} f\left( x \right) is x = 0\]
\[\text{ Now , LHD} \left( at x = 0 \right) = \lim_{x \to 0^-} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0}\]
\[ = \lim_{x \to 0^-} \frac{- x^2 - 0}{x}\]
\[ = \lim_{h \to 0} \frac{- \left( - h \right)^2}{- h}\]
\[ = \lim_{h \to 0} h\]
\[ = 0\]
\[\text{ And RHD} \left( \text{ at } x = 0 \right) = \lim_{x \to 0^+} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0}\]
\[ = \lim_{x \to 0^+} \frac{x^2 - 0}{x}\]
\[ = \lim_{h \to 0} \frac{h^2}{h}\]
\[ = \lim_{h \to 0} h\]
\[ = 0\]
\[ \therefore \text { LHD } \left( \text { at x } = 0 \right) =\text {  RHD } \left(\text {  at x } = 0 \right)\]
\[{\text{ So }, f\left( x \right) \text{ is also differentiable at } x} = 0\]
\[\text{i . e . }f\left( x \right) \text { is differentiable in }\left( - \infty , \infty \right)\]