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

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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) $\neq$ (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} 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} h^3 - 0/h = \lim_{h \to 0} 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.

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