<s>[INST] Finding the range of $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$ 
If the range of the function $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$ is $\left(\frac pq,\infty\right)$, where $p, q$ are in their lowest form then find $(p+q), (p-q)$

Looks like the domain is $\mathbb R$.
If $x^2=0$, $(x-1)^2=1\implies f(0)=\log_2\left(1+4\right)=\log_25$
If $(x-1)^2=0, x^2=1\implies f(1)=\log_25$
So, I think $p=\log_25, q=1$
So, $p+q=\log_25+1, p-q=\log_25-1$
But the options are $1/3/4/5/10$.
I tried converting the base to $10$ and writing $5$ as $10/2$ but in vain.
 [/INST] You have
$$\begin{align}4^{x^2}+4^{{(x-1)}^2}
&≥2\sqrt{4^{x^2+(x-1)^2}}\\
&=2^{{x^2+(x-1)^2}+1}\\
&≥2^{2\left(x-\frac 12\right)^2+\frac 32}\\
&≥2^{\frac 32}\end{align}$$
$$\log_2\left(4^{x^2}+4^{(x-1)^2}\right)≥\log_2{2^{\frac 32}}=\frac 32.$$
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