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JEE Main 202226 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
26 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2022 · 26 June · Morning Shift · Mathematics · Functions

Let f(x) = (x - 1)/(x + 1), x ∈ R - 0, - 1, 1 ) . If f ^ n + 1 (x)...

Let f\left(x\right) = \dfrac{x - 1}{x + 1}, x \in R - { 0, - 1, 1 ) . If fn+1(x)=f(fn(x)){f}^{n + 1}\left(x\right) = f\left({f}^{n}\left(x\right)\right) for all n∈Nn \in N, then f6{f}^{6}\left(6\right) + {f}^{7}(7)\left(7\right) is equal to

Options

  1. A76\dfrac{7}{6}
  2. B−32- \dfrac{3}{2}
  3. C712\dfrac{7}{12}
  4. D−1112- \dfrac{11}{12}

Correct answer

B. −32- \dfrac{3}{2}

Step-by-step solution

Given f(x)=x−1x+1=f1(x)f\left(x\right) = \dfrac{x - 1}{x + 1} = {f}^{1}\left(x\right)

⇒f2(x)=f(f(x))=x−1x+1−1x−1x+1+1=−1x⇒ {f}^{2}\left(x\right) = f\left(f\left(x\right)\right) = \dfrac{\dfrac{x - 1}{x + 1} - 1}{\dfrac{x - 1}{x + 1} + 1} = - \dfrac{1}{x}

Now f3(x)=f(f2(x))=f(−1x)=x+11−x{f}^{3}\left(x\right) = f\left({f}^{2}\left(x\right)\right) = f\left(- \dfrac{1}{x}\right) = \dfrac{x + 1}{1 - x}

⇒f4(x)=f(x+11−x)=−1x⇒ {f}^{4}\left(x\right) = f\left(\dfrac{x + 1}{1 - x}\right) = - \dfrac{1}{x}

⇒f6(x)=−1x⇒f6(6)=−16⇒ {f}^{6}\left(x\right) = - \dfrac{1}{x} ⇒ {f}^{6}\left(6\right) = - \dfrac{1}{6}

So f7(x)=(−1x)=x+11−x{f}^{7}\left(x\right) = \left(- \dfrac{1}{x}\right) = \dfrac{x + 1}{1 - x}

⇒f7(7)=8−6=−43⇒ {f}^{7}\left(7\right) = \dfrac{8}{- 6} = - \dfrac{4}{3}

∴(−16)+(−43)=−32∴ \left(- \dfrac{1}{6}\right) + \left(- \dfrac{4}{3}\right) = - \dfrac{3}{2}

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