1(1+a )+(a2-1)x=0
so we have (a+1){1+(a-1)x}=0
so a=-1 or x= 1/(1-a)
what do we get from the second one .Do we have to ignore it??
if f(x) =1 + ax (a≠0) is the inverse of itself then what is the value of a?
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7 Answers
62f(x) = 1+ax
f(f(x))=1
because f(f-1x) = x
but f-1x=fx
1+a(1+ax) = x
1+a+(a2-1)x =0 for all x
hence a+1=0 ... so a=-1
also (a2-1) = 0 so a=±1
but common to both is a=-1
1sir i am not getting the solution
how did you get f(f(x))=1 f(f-1(x))=x not 1
162yes we ignore it becasue we have to find an a for which x=1/(1-a) is true for all x which is not possible
1let fx=ax+b/cx=d, x is not = -d/c . if d=-a show tht f(fx)=x is an identity .?? im nt gettin wht approach shud be applied