A certain biased coin is tossed with probability of showing head being $p$. Let $Q$ denote the probability that when the coin is tossed four times, the number of heads obtained is even. Then, which of the following options are correct?
(A) there is no value of $p$ if $Q=\frac{1}{4}$
(B) there is exactly one value of $p$ if $Q=\frac{3}{4}$
(C) there are exactly two values of $p$ if $Q=\frac{3}{5}$
(D) there are exactly four values of $p$ if $Q=\frac{4}{5}$
What I attempted:- It seems like that we have to do calculation for each different value of $Q$ in order to check whether the claims raised in the question are true or not. Proceeding in the usual way, we have
$Q=\binom{4}{0}(1-p)^4+\binom{4}{2}p^2(1-p)^2+\binom{4}{4}p^4$
I don't know if the answerer is supposed to use some other tricks, which may consume less time. However, recently I came across a similar type of problem where an expression was provided to compute the probability of even number of heads in $n$ trials of a coin.It is not difficult to derive that if $X \sim Bin(n,p)$ then $P(X=\mbox{even})=\frac{1}{2}\left(1+(q-p)^n\right)=\frac{1}{2}\left(1+(1-2p)^n\right)$
In our case $Q=\frac{1}{2}\left(1+(1-2p)^4\right)$. Thus it easily follows that
If $Q= \frac{1}{4}, \left(1-2p\right)^4=-\frac{1}{2}$ [For any $0\le p \le 1 $ this is impossible. So option $(A)$ is obviously correct ]
If $Q= \frac{3}{4}, \left(1-2p\right)^4=\frac{1}{2}$
If $Q= \frac{3}{5}, \left(1-2p\right)^4=\frac{1}{6}$
If $Q= \frac{4}{5}, \left(1-2p\right)^4=\frac{3}{5}$
Consider the function $f(p)=\left(1-2p\right)^4$. I made an attempt to use tools from calculus in order to study the nature of the function.
We have $f'(p)=-8\left(1-2p\right)^3$.
$f'(p)=0$ gives $p=\frac{1}{2}$. It is easy to see that $f'(p)<0$ for $0\le p < \frac{1}{2}$ and $f'(p)>0$ for $\frac{1}{2}<p\le 1$. Thus $f(p)$ is a $U$ shaped curve for $0\le p \le 1$ as shown below. . 
It attains its maxima (which is $1$) at $p=0$ and $p=1$ and becomes zero when $p=\frac{1}{2}$. For every $0 < f(p) \le 1$, there are exactly two arguments where the function attains the same value.
For the values of $Q$ given in $(B), (C)$ and $(D)$, the corresponding values of $f(p)$ are obtained above, and it is seen that they lies between $0$ and $1$. Thus, we can conclude that for all these values of $Q$, we can have exactly two values of $p$. Consequently, the correct options are $(A)$ and $(C)$.
Am I correct? Is there any other way to reach the same?
As a fourth degree equation must have four roots (over $\mathbb{C}$) what happens with the other two values of $p$? I think they are not real.
(I am not sure if I have used correct tags. Hope someone would edit in case of any discrepancy.)
My answer would be (C) because of the symmetry in your first equation. That is, swapping $p$ and $(1-p)$ yields the same $Q$. So, for any possible $Q$, there are always $2$ values of $p$ where $p \neq (1-p)$ for a biased coin.