If the equation is continuous between -1,1 then it should be continuous on x=0 also
lim x>0 [_/(1+px) - _/(1-px)]/x
using L'hospital rule
1*p/2_/(1+px) + 1*p/2_/(1-px)
at x=0
p/2 + p/2 = p
This should be equal to value at 0 given by other function so that it gets continuous
F(0) for (2x+1)/(x-2) = -1/2
so p = -1/2
pic
This is reply by AI
@abhishek.exl this one looks interesting
If the equation is continuous between -1,1 then it should be continuous on x=0 also
lim x>0 [_/(1+px) - _/(1-px)]/x
using L'hospital rule
1*p/2_/(1+px) + 1*p/2_/(1-px)
at x=0
p/2 + p/2 = p
This should be equal to value at 0 given by other function so that it gets continuous
F(0) for (2x+1)/(x-2) = -1/2
so p = -1/2
mention anyone using @username while adding comments