Considering that both P & Q are integers, lets see how can 4 be written as (base)^exponent
4 = 2^2 Or (-2)^2 Or 4^1 Or (-4)^1
So we have cases:
P = 2, Q = 2
P = -2, Q = 2 (in both these cases, |P|=2)
P = 4, Q = 1
P = -4, Q = 1 (in both these cases, |P|=4)
as we can see, P can be both positive and negative, but Q can only be positive
Statement 1. |P| + |Q| = |P+Q|
If P is negative and Q is positive, then above statement
cannot be trueBut if P and Q are both positive, then above statement
is definitely true
So we get that P is positive, thus it answers NO to the question asked.
SufficientStatement 2. P^2 - 5|P| + 4 = 0
Note that P^2 can also be written as |P|^2
So we have |P|^2 - 5|P| + 4 = 0
Factorise this quadratic equation to get
(|P| - 1)(|P| - 4) = 0
So |P| = 1 OR |P| = 4
Now |P| = 1 is not possible if |P|^Q = 4
But for |P| = 4, we have |P| =4, Q = 1 so that 4^1 = 4
But still here we can have P = 4 or P = -4
Cant say whether P is positive or negative.
InsufficientHence answer is
A