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Functions question

2022 · 26 Jun · Shift 1 · Q21
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  5. /2022 · 26 Jun · Shift 1 · Q21

Functions question

2022 · 26 Jun · Shift 1 · Q21

JEE MainMathematicsFunctionsMCQ+4 / −1
Let f(x)=x−1x+1, x∈R−{0,−1,1}f(x) = {{x - 1} \over {x + 1}},\,x \in R - \{ 0, - 1,1\}f(x)=x+1x−1​,x∈R−{0,−1,1}. If fn+1(x)=f(fn(x)){f^{n + 1}}(x) = f({f^n}(x))fn+1(x)=f(fn(x)) for all n ∈\in∈ N, then f6(6)+f7(7){f^6}(6) + {f^7}(7)f6(6)+f7(7) is equal to :
  1. A
    76{7 \over 6}67​
  2. B
    −32- {3 \over 2}−23​
  3. C
    712{7 \over {12}}127​
  4. D
    −1112- {{11} \over {12}}−1211​
View written solutionFree

Correct answer: B

  1. Given function and iterates

We have f(x)=x−1x+1f(x)=\frac{x-1}{x+1}f(x)=x+1x−1​ and fn+1(x)=f(fn(x))f^{n+1}(x)=f(f^n(x))fn+1(x)=f(fn(x)) with the usual meaning of repeated composition.

We need to find f6(6)+f7(7).f^6(6)+f^7(7).f6(6)+f7(7).

  1. Find a useful pattern by composing fff with itself

Let y=f(x)=x−1x+1.y=f(x)=\frac{x-1}{x+1}.y=f(x)=x+1x−1​. Then f2(x)=f(f(x))=y−1y+1.f^2(x)=f(f(x))=\frac{y-1}{y+1}.f2(x)=f(f(x))=y+1y−1​. Now compute:

y−1=x−1x+1−1=x−1−(x+1)x+1=−2x+1,y-1=\frac{x-1}{x+1}-1=\frac{x-1-(x+1)}{x+1}=\frac{-2}{x+1},y−1=x+1x−1​−1=x+1x−1−(x+1)​=x+1−2​,

y+1=x−1x+1+1=x−1+x+1x+1=2xx+1.y+1=\frac{x-1}{x+1}+1=\frac{x-1+x+1}{x+1}=\frac{2x}{x+1}.y+1=x+1x−1​+1=x+1x−1+x+1​=x+12x​.

Therefore, f2(x)=−2x+12xx+1=−1x.f^2(x)=\frac{\frac{-2}{x+1}}{\frac{2x}{x+1}}=-\frac{1}{x}.f2(x)=x+12x​x+1−2​​=−x1​.

So, f2(x)=−1x.f^2(x)=-\frac{1}{x}.f2(x)=−x1​.

  1. Compute higher iterates

Now, f4(x)=f2(f2(x))=f2(−1x)=−1−1/x=x.f^4(x)=f^2(f^2(x))=f^2\left(-\frac{1}{x}\right)=-\frac{1}{-1/x}=x.f4(x)=f2(f2(x))=f2(−x1​)=−−1/x1​=x.

Hence, f4(x)=x.f^4(x)=x.f4(x)=x.

So the iterates repeat with period 444: fn+4(x)=fn(x).f^{n+4}(x)=f^n(x).fn+4(x)=fn(x).

  1. Compute f6(6)f^6(6)f6(6)

Since 6≡2(mod4)6\equiv 2 \pmod 46≡2(mod4), f6(6)=f2(6)=−16.f^6(6)=f^2(6)=-\frac{1}{6}.f6(6)=f2(6)=−61​.

  1. Compute f7(7)f^7(7)f7(7)

Since 7≡3(mod4)7\equiv 3 \pmod 47≡3(mod4), f7(7)=f3(7)=f(f2(7))=f(−17).f^7(7)=f^3(7)=f(f^2(7))=f\left(-\frac{1}{7}\right).f7(7)=f3(7)=f(f2(7))=f(−71​).

Now,

=\frac{-\frac{8}{7}}{\frac{6}{7}}=-\frac{8}{6}=-\frac{4}{3}.$$ So, $$f^7(7)=-\frac{4}{3}.$$ --- 6. **Add the two values** $$f^6(6)+f^7(7)=-\frac{1}{6}-\frac{4}{3} =-\frac{1}{6}-\frac{8}{6} =-\frac{9}{6}=-\frac{3}{2}.$$ --- 7. **Match with the options** $$-\frac{3}{2}$$ corresponds to **Option B**. --- 8. **Comparison with stored correct answer** Stored correct answer: **B** Our derived answer: **B** They agree.
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