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Functions question

2020 · 6 Sep · Shift 2 · Q18
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Functions question

2020 · 6 Sep · Shift 2 · Q18

JEE MainMathematicsFunctionsMCQ+4 / −1
For a suitably chosen real constant a, let a function, f:R−{−a}→Rf:R - \left\{ { - a} \right\} \to Rf:R−{−a}→R be defined by f(x)=a−xa+xf(x) = {{a - x} \over {a + x}}f(x)=a+xa−x​. Further suppose that for any real number xe−ax e - axe−a and f(x)e−af(x) e - af(x)e−a, (fof)(x) = x. Then f(−12)f\left( { - {1 \over 2}} \right)f(−21​) is equal to :
  1. A
    13{1 \over 3}31​
  2. B
    –3
  3. C
    −13- {1 \over 3}−31​
  4. D
    3
View written solutionFree

Correct answer: D

  1. We are given f(x)=a−xa+x,x∈R∖{−a}f(x)=\frac{a-x}{a+x}, \qquad x\in \mathbb R\setminus\{-a\}f(x)=a+xa−x​,x∈R∖{−a} and for all real xxx such that x≠−ax\neq -ax=−a and f(x)≠−af(x)\neq -af(x)=−a, f(f(x))=x.f(f(x))=x.f(f(x))=x.

  2. Compute f(f(x))f(f(x))f(f(x)).

    Let y=f(x)=a−xa+x.y=f(x)=\frac{a-x}{a+x}.y=f(x)=a+xa−x​. Then f(f(x))=f(y)=a−ya+y.f(f(x))=f(y)=\frac{a-y}{a+y}.f(f(x))=f(y)=a+ya−y​.

    Substitute y=a−xa+xy=\dfrac{a-x}{a+x}y=a+xa−x​: f(f(x))=a−a−xa+xa+a−xa+x.f(f(x))=\frac{a-\frac{a-x}{a+x}}{a+\frac{a-x}{a+x}}.f(f(x))=a+a+xa−x​a−a+xa−x​​.

  3. Simplify numerator and denominator separately.

    Numerator:

    =\frac{a^2+ax-a+x}{a+x}.$$ Denominator: $$a+\frac{a-x}{a+x}=\frac{a(a+x)+(a-x)}{a+x} =\frac{a^2+ax+a-x}{a+x}.$$ Hence, $$f(f(x))=\frac{a^2+ax-a+x}{a^2+ax+a-x}.$$

  4. Since f(f)(x)=xf(f)(x)=xf(f)(x)=x for all admissible xxx, we must have a2+ax−a+xa2+ax+a−x=x.\frac{a^2+ax-a+x}{a^2+ax+a-x}=x.a2+ax+a−xa2+ax−a+x​=x.

    Cross-multiplying: a2+ax−a+x=x(a2+ax+a−x).a^2+ax-a+x=x(a^2+ax+a-x).a2+ax−a+x=x(a2+ax+a−x).

    Expanding the right side: a2+ax−a+x=a2x+ax2+ax−x2.a^2+ax-a+x=a^2x+ax^2+ax-x^2.a2+ax−a+x=a2x+ax2+ax−x2.

    Bring all terms to one side: 0=a2x+ax2+ax−x2−a2−ax+a−x.0=a^2x+ax^2+ax-x^2-a^2-ax+a-x.0=a2x+ax2+ax−x2−a2−ax+a−x.

    Simplify: 0=ax2−x2+a2x−a2+a−x.0=ax^2-x^2+a^2x-a^2+a-x.0=ax2−x2+a2x−a2+a−x.

    Factor: 0=(a−1)x2+(a2−1)x+(a−a2).0=(a-1)x^2+(a^2-1)x+(a-a^2).0=(a−1)x2+(a2−1)x+(a−a2).

    Further factor: 0=(a−1)(x2+(a+1)x−a).0=(a-1)\big(x^2+(a+1)x-a\big).0=(a−1)(x2+(a+1)x−a).

  5. This must hold for all admissible xxx. A nonzero quadratic cannot vanish for all real xxx, so we must have a−1=0  ⟹  a=1.a-1=0 \implies a=1.a−1=0⟹a=1.

  6. Now compute

    =\frac{1+1/2}{1-1/2} =\frac{3/2}{1/2}=3.$$

  7. Therefore, f(−12)=3.f\left(-\frac12\right)=3.f(−21​)=3.

So the correct option is D.

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