Question: I have the \*expression
$$\frac{1}{\sqrt{4a^2-b^2}}$$

and I am being asked to evaluate the case when $2|a| \geq|b|$

Logically I know what this statement mean but don't know how it applies to this problem: Cases when $a$ is twice greater than the distance of $b$ is from $0$.
(Correct me if i'm wrong)

I also know that absolute values on both sides of an equation are inpracticle and it would be better to write the inequality as $|a/b| \geq 1/2$.

Which can also be written as $a/b \geq 1/2$ or $a/b \leq 1/2$ which doesn't make any sense.

here's a link to the problem (#3)
[http://imgur.com/a/GheWI](https://imgur.com/a/GheWI)

Thank you, I would appreciate help I've been struggling with this for a while

Answer: 
In addition to the things answered here, I just want to add this one. 
**Proposition** Let $G$ be a finite group and $H$ a subgroup of prime *index* $p$, with gcd$(|G|,p-1)=1$. Then $G' \subseteq H$.
 Note that this implies that $H \unlhd G$, and that it is in fact sufficient to prove that $H$ is normal, since then $G/H \cong C\_p$ is abelian.
**Proof** Firstly, we may assume by induction on $|G|$, that $H$ is core-free, that is core$\_G(H)=\bigcap\_{g \in G}H^g=1$. This means that $G$ can be homomorphically embedded in $S\_p$. Let $P \in Syl\_p(G)$ and note that because $|S\_p|=p \cdot (p-1) \cdots \cdot 1$, $|P|=p$. By the $N/C$-Theorem, $N\_G(P)/C\_G(P)$ embeds in Aut$(P) \cong C\_{p-1}$. By the assumption gcd$(|G|,p-1)=1$, we get that $N\_G(P)=C\_G(P)$. Since $P$ is abelian we have $P \subseteq C\_G(P)$, whence $P \subseteq Z(N\_G(P))$. We now can apply Burnside's Normal $p$-complement Theorem, which implies that $P$ has a normal complement $N$, that is $G=PN$ and $P \cap N=1$. Note that $|G/N|=p$. Look at the image of $H$ in $G/N$. Then $G=HN$, or $HN=N$. In the latter case $H \subseteq N$, and $|G:H|=|G:N|=p$, whence $H=N$ and we are done if we can refute the first case. If $G=HN$, then $|G:H \cap N|=|G:N|\cdot|N:H \cap N|=|G:N|\cdot |G:H|=p \cdot p=p^2$, contradicting the fact that $|G| \mid |S\_p|$. The proof is now complete.
**Corollary 1** Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$, the *smallest* prime dividing the order of $G$. Then $G' \subseteq H$. In particular, $H$ is normal.
**Corollary 2** Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$ and gcd$(|H|,p-1)=1$. Then $H$ is normal.
Observe that this last result renders a well-known result for $p=2$! Finally for fun:
**Corollary 3** Let $G$ be a finite group of *odd* order and $H$ a subgroup with $|G:H|=65537$. Then $H$ is normal.
Yes. If $p$ is the smallest prime dividing the order of $G$ and $[G : H] = p$, then $H \trianglelefteq G$. This is a corollary of what is commonly referred to as the [Strong Cayley Theorem](http://www.mathreference.com/grp-act,cayley.html). See the final corollary of [this document](https://math.la.asu.edu/~kawski/classes/mat444/handouts/strongCayley.pdf) for a proof.