Embibe Experts Solutions for Chapter: Limits, Exercise 1: JEE Advanced Paper 1 - 2016

Let $F:R\to R$ be a function. We say that $f$ has:
 PROPERTY $1$  if $\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{\sqrt{\left|h\right|}}$ exists and is finite, and 
PROPERTY $2$  if $\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h^2}$ exists and is finite.
Then which of the following options is/are correct?

Let $f\left(x\right)=\frac{1-x(1+\left|1-x\right|)}{|1-x|}\cos \left(\frac{1}{1-x}\right) \mathrm{for} x\ne 1,$ then

Let $f\mathbb{ }:\mathbb{R}\to \left(0, \infty \right)$ and $g\mathbb{ }:\mathbb{R}\to \mathbb{R}$ be twice differentiable functions such that $f^{″}$ and $g^{″}$ are continuous functions on $\mathbb{R}$ . Suppose $f^{\prime }\left(2\right)=g\left(2\right)=0, f^{″}\left(2\right)\ne 0$ and $g^{\text{'}}\left(2\right)\ne 0$, $\underset{x\to 2}{\lim }\frac{f\left(x\right) g\left(x\right)}{f^{\text{'}} \left(x\right) g^{\text{'}}\left(x\right)}=1$, then

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right-hand limit $\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{x}}}-e^{-1}}{x^a}$ is equal to a non-zero real number, is