If p>0 and limx→px−px3+px2+qx=12, then the value of p−q is....
Explanation
Let x=p for the numerator, then
x3+px2+qx=0
p3+p⋅p2+q⋅p=0
2p3+pq=0
2p2+q=0...(1)
Use L'Hôpital's rule
x→plim13x2+2px+q=12
3p2+2p⋅p+q=12
3p2+2p2+q=12
5p2+q=12...(2)
Substitute equation (1) into equation (2)
5p2+q=12
5p2+(−2p2)=12
3p2=12
p2=4
p=±2
Because the condition is p>0, then p=2. Substitute into
2p2+q=0
2⋅22+q=0
q=−8
Thus p−q=2−(−8)=10.