JEE Main 2020MathematicsFunctionsHardMCQ

JEE Main 2020 — Functions Question with Solution

JEE Main 2020 (06 Sep Shift 2)

Question

For a suitably chosen real constant a, let a function, f : R--a→R be defined by fx=a-xa+x. Further supposed that for any real number x≠-a,and f(x)≠-a, fofx=x. Then f-12 is equal to : 

Choose an option

Show full solutionCorrect option: D
Correct answer
D3

Step-by-step explanation

fof(x)=a-fxa+f(x)=x

⇒a-ax1+x=f(x)

⇒a(1-x)1+x=a-xa+x        (∵a=1)

so f(x)=1-x1+x

f-12=3

 

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About this question

This is a previous-year question from JEE Main 2020, covering the Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.