Let f(x)=xsinax+1,x≠-1,f≠1,f(x)≠0. Then for what value of sinais f(f(x))=x?
3
2
1
-1
Explanation for the correct answer
Given that: f(x)=xsinax+1,x≠-1,f≠1,f(x)≠0.
Now, calculatingf(fx)):
⇒f(fx)=f(x)sinaf(x)+1⇒f(fx)=xsinax+1.sinaxsinax+1+1⇒f(fx)=xsin2axsina+x+1⇒x=xsin2axsina+x+1Giventhat:f(fx)=x⇒xsina+x+1=sin2a⇒sin2a-xsina-x-1=0⇒sin2a+sina-xsina-x-1-sina=0⇒sina(sina-1-x)+1(sina-x-1)=0⇒(sina+1)(sina-x-1)=0⇒sina=-1or(x+1)
Hence, the correct answer is Option (D).
Let f(x)=axx+1,x≠-1. Then for what values of a is ffx=x.
If af(x)+bf(1x)=x-1,x≠0, and a≠b, then f(2)=