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1.Relation and Function
hard
ધારો કે,$f(x)=\frac{x-1}{x+1}, x \in R -\{0,-1,1\} .$ ને પ્રત્યેક $n \in N$ માટે $f^{ n +1}(x)=f\left(f^{ n }(x)\right)$ તો $f^{6}(6)+f^{7}(7)=$
A$\frac{7}{6}$
B$-\frac{3}{2}$
C$\frac{7}{12}$
D$-\frac{11}{12}$
(JEE MAIN-2022)
Solution
$f(x)=\frac{x-1}{x+1}$
$\Rightarrow f^{2}(x)=f(f(x))=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\frac{1}{x}$
$f^{3}(x)=f\left(f^{2}(x)\right)=f\left(-\frac{1}{x}\right)=\frac{x+1}{1-x}$
$\Rightarrow f^{4}(x)=f\left(\frac{x+1}{1-x}\right)=-\frac{1}{x}$
$\Rightarrow f^{6}(x)=-\frac{1}{x} \Rightarrow f^{6}(6)=-\frac{1}{8}$
$f^{7}(x)=\left(-\frac{1}{x}\right)=\frac{x+1}{1-x}$
$\Rightarrow f^{7}(7)=\frac{8}{-6}=-\frac{4}{3}$
$\therefore-\frac{1}{6}+-\frac{4}{3}=-\frac{3}{2}$
$\Rightarrow f^{2}(x)=f(f(x))=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\frac{1}{x}$
$f^{3}(x)=f\left(f^{2}(x)\right)=f\left(-\frac{1}{x}\right)=\frac{x+1}{1-x}$
$\Rightarrow f^{4}(x)=f\left(\frac{x+1}{1-x}\right)=-\frac{1}{x}$
$\Rightarrow f^{6}(x)=-\frac{1}{x} \Rightarrow f^{6}(6)=-\frac{1}{8}$
$f^{7}(x)=\left(-\frac{1}{x}\right)=\frac{x+1}{1-x}$
$\Rightarrow f^{7}(7)=\frac{8}{-6}=-\frac{4}{3}$
$\therefore-\frac{1}{6}+-\frac{4}{3}=-\frac{3}{2}$
Standard 12
Mathematics
