fof(x)=a+f(x)a−f(x)=x
⇒1+xa−ax=f(x)
⇒1+xa(1−x)=a+xa−x (∵a=1)
so f(x)=1+x1−x
f(−21)=3
JEE Main 2020 — Mathematics Algebra
For a suitably chosen real constant a, let a function, f:R−−a→R be defined by f(x)=a+xa−x. Further supposed that for any real number x=−a,and f(x)=−a,(fof)(x)=x. Then f(−21) is equal to :
Held on 6 Sept 2020 · Verified 6 Jul 2026.
31
−31
−3
3
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