13 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.? This isn't a functional equation tho
ugh just ignore the title of doubt
and and the ques
Is this the entire question?
yeahh
x=(x+1)/(x+2) which means x^2+x-1=0 is one case.
Another case is -x=(x+1)/(x+2)
how u getting these cases
f(x1)=f(x2) only when x1=x2 . this is for one-one func tho
f(x1)=f(x2) will be valid for the given function when x2=x1 or x2=-x1 (even)
So just make those cases and solve.
huh how does this arise
isnt an even func one wherein f(x) = f(-x)
We are given two expressions right? If these expressions in argument (x and (x+1)/(x+2)) give equal values, for instance say 5, f(5)=f(5)
Similarly if one expression gives the negative of other, say, let x=5 and other expression=-5 f(5)=f(-5), and condition holds.
not necessarily tho
for all we know f(x) might be periodic
f(x)= f(x+T) where T is period
x not equal x+T ofc
similalry
x not equal to (x+1)/x+2
But the question says we only need 4 values na?
So from here we can make 2 quadratics to get 4 values. If some root doesn't lie in (-5,5) we need to look for something else.