JEE Main 2022 — Functions Question with Solution

Given,

$$f(x)=\frac{x-1}{x+1}$$

Also given,

$$f^{n+1}(x)=f(f^n(x))$$ ..... (1)

$$\therefore$$ For $$n=1$$

$$f^{1+1}(x)=f(f^1(x))$$

$$\Rightarrow f^2(x)=f(f(x))$$

$$=f\left(\frac{x-1}{x+1}\right)$$

$$=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}$$

$$=\frac{\frac{x-1-x-1}{x+1}}{\frac{x-1+x+1}{x+1}}$$

$$=\frac{-2}{2x}$$

$$=-\frac{1}{x}$$

From equation (1), when n = 2

$$f^{2+1}(x)=f(f^2(x))$$

$$\Rightarrow f^3(x)=f(f^2(x))$$

$$=f\left(-\frac{1}{x}\right)$$

$$=\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}$$

$$=\frac{\frac{-1-x}{x}}{\frac{-1+x}{x}}$$

$$=\frac{-1-x}{-1+x}=\frac{-(x+1)}{x-1}$$

Similarly,

$$f^4(x)=f(f^3(x))$$

$$=f\left(\frac{-x+1}{x-1}\right)$$

$$=\frac{\frac{-(x+1)}{x-1}-1}{\frac{-(x+1)}{x-1}+1}$$

$$=\frac{\frac{-x-1-x+1}{x-1}}{\frac{-x-1+x-1}{x-1}}$$

$$=\frac{-2x}{-2}=x$$

$$\therefore$$ $$f^5(x)=f(f^4(x))$$

$$=f(x)$$

$$=\frac{x-1}{x+1}$$

$$f^6(x)=f(f^5(x))$$

$$=f\left(\frac{x-1}{x+1}\right)$$

$$=-\frac{1}{x}$$ (Already calculated earlier)

$$f^7(x)=f(f^6(x))$$

$$=f\left(-\frac{1}{x}\right)$$

$$=\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}$$

$$=\frac{-(x+1)}{x-1}$$

$$\therefore$$ $$f^6(6)=-\frac{1}{6}$$

and $$f^7(7)=\frac{-(7+1)}{7-1}=-\frac{8}{6}$$

So, $$f^6(6)+f^7(7)$$

$$=-\frac{1}{6}-\frac{8}{6}$$

$$=-\frac{3}{2}$$