Let f(x)=(x-1/x+1), x ∈ R- 0,-1,1). If fn+1(x)=f(fn(x)) for all n ∈ N, then f6(6)+f7(7) is equal to:

Question

Let f(x)=(x-1/x+1), x ∈ R- 0,-1,1). If fn+1(x)=f(fn(x)) for all n ∈ N, then f6(6)+f7(7) is equal to: (A) (7/6) (B) -(3/2) (C) (7/12) (D) -(11/12)

Tardigrade Admin · Accepted Answer

f ( x )=( x -1/ x +1) ⇒ f 2( x )= f ( f ( x ))=(( x -1/ x +1)-1/( x -1) x +1+1)=-(1/ x ) f 3( x )= f ( f 2( x ))= f (-(1/ x ))=( x +1/1- x ) ⇒ f 4( x )= f (( x +1/1- x ))=-(1/ x ) ⇒ f 6( x )=-(1/ x ) ⇒ f 6(6)=-(1/8) f 7( x )=(-(1/ x ))=( x +1/1- x ) ⇒ f 7(7)=(8/-6)=-(4/3) ∴-(1/6)+-(4/3)=-(3/2)

Q. Let $f (x) = \frac{x - 1}{x + 1}, x \in R - {0, - 1, 1)$ . If $f^{n + 1} (x) = f (f^{n} (x))$ for all $n \in N$ , then $f^{6} (6) + f^{7} (7)$ is equal to:

$\frac{7}{6}$

$- \frac{3}{2}$

$\frac{7}{12}$

$- \frac{11}{12}$

Q. Let f(x)=x+1x−1​,x∈R−{0,−1,1). If fn+1(x)=f(fn(x)) for all n∈N, then f6(6)+f7(7) is equal to:

67​

−23​

127​

−1211​

f(x)=x+1x−1​ ⇒f2(x)=f(f(x))=x+1x−1​+1x+1x−1​−1​=−x1​ f3(x)=f(f2(x))=f(−x1​)=1−xx+1​ ⇒f4(x)=f(1−xx+1​)=−x1​ ⇒f6(x)=−x1​⇒f6(6)=−81​ f7(x)=(−x1​)=1−xx+1​ ⇒f7(7)=−68​=−34​ ∴−61​+−34​=−23​

Q. Let $f (x) = \frac{x - 1}{x + 1}, x \in R - {0, - 1, 1)$ . If $f^{n + 1} (x) = f (f^{n} (x))$ for all $n \in N$ , then $f^{6} (6) + f^{7} (7)$ is equal to:

$\frac{7}{6}$

$- \frac{3}{2}$

$\frac{7}{12}$

$- \frac{11}{12}$