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QJEEM22Ma868d: JEE Main 2022 Mathematics - Class 11 JEEM22Ma868d Solution: Let . If for all n N, then is equal to :

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

Multiple Choice Options

  • A 76{7 \over 6}
  • B −32- {3 \over 2}
  • C 712{7 \over {12}}
  • D −1112- {{11} \over {12}}

Step-by-Step Verified Solution

Given,

f(x)=x−1x+1f(x) = {{x - 1} \over {x + 1}}

Also given,

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

∴\therefore For n=1n = 1

f1+1(x)=f(f1(x)){f^{1 + 1}}(x) = f({f^1}(x))

⇒f2(x)=f(f(x))\Rightarrow {f^2}(x) = f(f(x))

=f(x−1x+1)= f\left( {{{x - 1} \over {x + 1}}} \right)

=x−1x+1−1x−1x+1+1= {{{{x - 1} \over {x + 1}} - 1} \over {{{x - 1} \over {x + 1}} + 1}}

=x−1−x−1x+1x−1+x+1x+1= {{{{x - 1 - x - 1} \over {x + 1}}} \over {{{x - 1 + x + 1} \over {x + 1}}}}

=−22x= {{ - 2} \over {2x}}

=−1x= - {1 \over x}

From equation (1), when n = 2

f2+1(x)=f(f2(x)){f^{2 + 1}}(x) = f({f^2}(x))

⇒f3(x)=f(f2(x))\Rightarrow {f^3}(x) = f({f^2}(x))

=f(−1x)= f\left( { - {1 \over x}} \right)

=−1x−1−1x+1= {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}

=−1−xx−1+xx= {{{{ - 1 - x} \over x}} \over {{{ - 1 + x} \over x}}}

=−1−x−1+x=−(x+1)x−1= {{ - 1 - x} \over { - 1 + x}} = {{ - (x + 1)} \over {x - 1}}

Similarly,

f4(x)=f(f3(x)){f^4}(x) = f({f^3}(x))

=f(−x+1x−1)= f\left( {{{ - x + 1} \over {x - 1}}} \right)

=−(x+1)x−1−1−(x+1)x−1+1= {{{{ - (x + 1)} \over {x - 1}} - 1} \over {{{ - (x + 1)} \over {x - 1}} + 1}}

=−x−1−x+1x−1−x−1+x−1x−1= {{{{ - x - 1 - x + 1} \over {x - 1}}} \over {{{ - x - 1 + x - 1} \over {x - 1}}}}

=−2x−2=x= {{ - 2x} \over { - 2}} = x

∴\therefore f5(x)=f(f4(x)){f^5}(x) = f({f^4}(x))

=f(x)= f(x)

=x−1x+1= {{x - 1} \over {x + 1}}

f6(x)=f(f5(x)){f^6}(x) = f({f^5}(x))

=f(x−1x+1)= f\left( {{{x - 1} \over {x + 1}}} \right)

=−1x= - {1 \over x} (Already calculated earlier)

f7(x)=f(f6(x)){f^7}(x) = f({f^6}(x))

=f(−1x)= f\left( { - {1 \over x}} \right)

=−1x−1−1x+1= {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}

=−(x+1)x−1= {{ - (x + 1)} \over {x - 1}}

∴\therefore f6(6)=−16{f^6}(6) = - {1 \over 6}

and f7(7)=−(7+1)7−1=−86{f^7}(7) = {{ - (7 + 1)} \over {7 - 1}} = - {8 \over 6}

So, f6(6)+f7(7){f^6}(6) + {f^7}(7)

=−16−86= - {1 \over 6} - {8 \over 6}

=−32= - {3 \over 2}

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