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JEE Main 20206 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2020 · 6 September · Evening Shift · Mathematics · Functions

For a suitably chosen real constant a , let a function, f: R - \ -...

For a suitably chosen real constant aa, let a function, f:R−{−a}→Rf: R - \left\{- a\right\} \to R be defined by f(x)=a−xa+xf\left(x\right) = \dfrac{a - x}{a + x}. Further supposed that for any real number x≠−a,x \ne - a,and f(x)≠−a,(fof)(x)=xf ( x ) \ne - a, \left(fof\right)\left(x\right) = x. Then f(−12)f\left(- \dfrac{1}{2}\right) is equal to :

Options

  1. A13\dfrac{1}{3}
  2. B−13- \dfrac{1}{3}
  3. C−3- 3
  4. D33

Correct answer

D. 33

Step-by-step solution

fof(x)=a−f(x)a+f(x)=xfof ( x ) = \dfrac{a - f\left(x\right)}{a + f ( x )} = x

⇒a−ax1+x=f(x)⇒ \dfrac{a - ax}{1 + x} = f ( x )

⇒a(1−x)1+x=a−xa+x⇒ \dfrac{a ( 1 - x )}{1 + x} = \dfrac{a - x}{a + x} (∵a=1)( ∵ a = 1 )

so f(x)=1−x1+xf ( x ) = \dfrac{1 - x}{1 + x}

f(−12)=3f\left(- \dfrac{1}{2}\right) = 3

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