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Let f:R- α 6→ R be defined by f(x)=( 5x+3 6x-α ). Then the value of α for which (fof)(x)=x, for all x∈ R- α 6, is

JEE Main 2021 — Mathematics Algebra

2021mcqeasy

Let f:R−α6→Rf:R-{\frac{\alpha }{6}}\rightarrow Rf:R−6α​→R be defined by f(x)=(5x+36x−α)f(x)=(\frac{5x+3}{6x-\alpha })f(x)=(6x−α5x+3​). Then the value of α\alphaα for which (fof)(x)=x(fof)(x)=x(fof)(x)=x, for all x∈R−α6,x\in R-{\frac{\alpha }{6}},x∈R−6α​, is

Official previous-year question

Held on 20 Jul 2021 · Verified 6 Jul 2026.

Options

  1. A

    No such α\alphaα exists

  2. B

    555

  3. C

    888

  4. D

    666

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