Question: > 
> Set of all values of '$a$' for which $f(x)=\bigg(\dfrac{\sqrt{a+4}}{1-a}-1\bigg)x^5-3x+\log 5$ decreases for all real $x$
> 
> 
> 

$$
a\neq1\;\&\; a\geq -4\\
f'(x)=5\bigg(\dfrac{\sqrt{a+4}}{1-a}-1\bigg)x^4-3<0\\
\bigg(\dfrac{\sqrt{a+4}}{1-a}-1\bigg)x^4<\frac{3}{5}\text{ for all} x\in\mathcal{R}\\
\dfrac{\sqrt{a+4}}{1-a}-1=0\implies\sqrt{a+4}=1-a\\
a+4=1+a^2-2a\implies a^2-3a-3=0\implies a=\frac{3\pm\sqrt{21}}{2}
$$

My reference gives the solution $a\in\big[-4,\dfrac{3-\sqrt{21}}{2}\big]\cup(1,\infty)$, yet I am stuck here. Could anyone give a hint of whats the way forward ?

Answer: 
This has nothing to do with linearity and everything to do with independence. (Just because we can't apply linear operators in such a way in general doesn't mean that there isn't *some* linear operator in *some* situation that this rule applies to.) For instance, if the distributions are continuous $f\_U$ and $f\_V$ are the pdfs, we have $$ E(U/V) = \int (u/v) f\_U(u)f\_V(v) du dv =\left(\int uf\_U(u)du\right)\left(\int (1/v)f\_V(v)dv\right)=E(U)E(1/V).$$ It's just because the joint distribution factors as $f\_{U,V}(u,v) = f\_U(u)f\_V(v)$.
If $U$ and $V$ are independent random variables, then
$$\mathbb{E}[f(U) g(V)] = \mathbb{E}[f(U)] \mathbb{E}[g(V)]$$
for any reasonable functions $f$ and $g$. In particular, $\mathbb{E}[U/V] = \mathbb{E}[U] \mathbb{E}[1/V]$. These equalities do not hold in general (without the independence assumption).