Question

Let g : $R\to R$ be a differentiable function with $g(0)=0,g^{\prime }(0)=0$ and $g^{\prime }(1)\ne 0$. Let
$f\left(x\right)=\left(\begin{array}{cc}\frac{x}{|x|}g\left(x\right),& x\ne 0\\ 0,& x=0\end{array}\right)$and $h\left(x\right)=e^{|x|}$ for all $x\in R$. Let $\left(foh\right)\left(x\right)$ denote $f\left(h\left(x\right)\right)$ and $\left(hof\right)\left(x\right)$ denote $h\left(f\left(x\right)\right)$. Then which of the following is (are) true ?

• A. f is differentiable at x = 0
• B. h is differentiable at x = 0
• C. f o h is differentiable at x = 0
• D. h o f is differentiable at x = 0

Answer 1

Answer: (A), (D)

$g(0)=0,g\prime (0)=0,g\prime (1)\ne 0$
$f(x)=\{\begin{array}{ll}g(x)& x>0\\ 0& x=0\\ -g(x)& x<0\end{array}$

$f^{\prime }(x)=\{\begin{array}{ll}{g}^{\prime }(x)& x>0\\ 0& x=0\\ -g^{\prime }(x)& x<0\end{array}$
f(x) is diff. at $x=0$, as g′(0) = 0
So option (A) is true.
$h(x)=e^{|x|}$
not diff. at $x=0$
So option (B) is not true
$foh(x)=f(h(x))ash(x)>0$
So $f(h(x))=g(h(x))$
$\frac{df\left(h\left(x\right)\right)}{dx}=g\prime \left(h\left(x\right)\right)h\prime \left(x\right)$ since $g\prime \left(1\right)\ne 0$
at $x=0,f(h(x))$ is not differentiable
So option (C) is not true
$hof(x)=h(f(x))$
$=e^{|f(x)|}$
Differentiable at $x=0$
So option (D) is true.