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← JEE Main 2022 · 26 Jun · Shift 1
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JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 1

Let f(x)=x−1x+1f(x)=\dfrac{x-1}{x+1}, x∈R−{0,−1,1}x\in\mathbb R-\{0,-1,1\}. If fn+1(x)=f(fn(x))f^{n+1}(x)=f\left(f^n(x)\right) for all n∈Nn\in\mathbb N, then f6(6)+f7(7)f^6(6)+f^7(7) is equal to :

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Answer and worked solution

Official answer: B

Iterate. f2(x)=x−1x+1−1x−1x+1+1=−1xf^2(x)=\dfrac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\dfrac1x, so f4(x)=xf^4(x)=x and the cycle has length 4.

Reduce. f6=f2f^6=f^2 and f7=f3f^7=f^3, with f3(x)=f(−1x)=−1+xx−1f^3(x)=f\left(-\frac1x\right)=-\dfrac{1+x}{x-1}.

Values. f6(6)=−16f^6(6)=-\frac16 and f7(7)=−86=−43f^7(7)=-\frac86=-\frac43; the sum is −32-\frac32.

(B)

Topics:Composition and inverse functions
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