Let g: R → R be a differentiable function with g (0) = 0, g ' (0) = 0 and g ' (1) \neq 0. \text{Let}\textrm{ }\text{f}\textrm{ } \left(\right. \text{x} \left.\right) \textrm{ } = \left{\right. \frac{x}{\left|\right. x \left|\right.} g \left(\right. x \left.\right) , \textrm{ }\textrm{ } x \neq 0 \\ 0 , x = 0 and h (x) = e|x| for all x \in R. Let (f o h)(x) denote f(h(x)) and (h o f ) (x) denote h(f(x)) . Then which of the following is(are) true?

Engineering

Mathematics

Important Points on Derivability

Question

Let g: R → R be a differentiable function with g (0) = 0, g ' (0) = 0 and g ' (1) $\neq$ 0.

$Let f (x) = {\begin{cases} \frac{x}{| x |} g (x), x \neq 0 \\ 0, x = 0 \end{cases}$

and h (x) = e^|x| for all x $\in$ R. Let (f o h)(x) denote f(h(x)) and (h o f ) (x) denote h(f(x)) .
Then which of the following is(are) true?

h is differentiable at x = 0

h o f is differentiable at x = 0

f is differentiable at x = 0

f o h is differentiable at x = 0

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(A) $f' (0^{+}) = \underset{h \to 0}{Lim} \frac{\frac{h}{h} g (h) - 0}{h} = g' (0) = 0$

$f' (0^{-}) = \underset{h \to 0}{Lim} \frac{\frac{h}{h} g (- h) - 0}{h} = - g' (0) = 0$

$∴$ f (x) is divisible at x = 0

(B) Obviously h (x) is non-derivable at x = 0

$f o h (0 +) = \underset{h \to 0}{Lim} \frac{g (e^{h}) - g (1)}{h} \underset{h \to 0}{= Lim} \frac{g (e^{h}) - g (1)}{h} = g' (1)$

$f o h (0 h) = \underset{h \to 0}{Lim} \frac{g (e^{h}) - g (1)}{- h} = - g' (1)$

(D) h o f (x) = $e^{| \frac{x}{| x |} g (x) |}$ = e^{| g (x) |}, since g ' (0) = 0 and g (0) = 0

⇒ | g (x) | is differentiable is equal to zero.

⇒ h o f (x) is derivable at x = 0.