1. Write each line in slope form

Given lines: $p(p^2+1)x-y+q=0$ and $(p^2+1)^2x+(p^2+1)y+2q=0$

We find their slopes.

2. Slope of the first line

From $p(p^2+1)x-y+q=0$ we get $-y=-p(p^2+1)x-q$ $y=p(p^2+1)x+q$

So slope of first line is $m_1=p(p^2+1)$

3. Slope of the second line

From $(p^2+1)^2x+(p^2+1)y+2q=0$ we get $(p^2+1)y=-(p^2+1)^2x-2q$ $y=-(p^2+1)x-\frac{2q}{p^2+1}$

So slope of second line is $m_2=-(p^2+1)$

4. Condition for both lines to be perpendicular to a common line

If two lines are both perpendicular to the same line, then they must be parallel to each other.

Hence we need: $m_1=m_2$

So, $p(p^2+1)=-(p^2+1)$

Since $p^2+1>0$ for all real $p$, we can divide by $(p^2+1)$: $p=-1$

Thus there is exactly one real value of $p$.

5. Check options

• A: exactly one value of $p$ ✔️
• B: exactly two values of $p$ ✘
• C: more than two values of $p$ ✘
• D: no value of $p$ ✘

Therefore, the correct option is: $\boxed{\text{A}}$