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QJEEM22Ma51d8: JEE Main 2022 Mathematics - Class 11 JEEM22Ma51d8 Solution: Let and be three positive real numbers. Let and be such that…

Let α,β\alpha, \beta and γ\gamma be three positive real numbers. Let f(x)=αx5+βx3+γx,x∈Rf(x)=\alpha x^{5}+\beta x^{3}+\gamma x, x \in \mathbf{R} and g:R→Rg: \mathbf{R} \rightarrow \mathbf{R} be such that g(f(x))=xg(f(x))=x for all x∈Rx \in \mathbf{R} . If a1,a2,a3,…,an\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}} be in arithmetic progression with mean zero, then the value of f(g(1n∑i=1nf(ai)))f\left(g\left(\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right) is equal to :

Multiple Choice Options

  • A 0
  • B 3
  • C 9
  • D 27

Step-by-Step Verified Solution

f(g(1n∑i=1nf(ai)))f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right)

a1+a2+a3 + ...... + ann=0{{{a_1} + {a_2} + {a_3}\, + \,......\, + \,{a_n}} \over n} = 0

∴\therefore First and last term, second and second last and so on are equal in magnitude but opposite in sign.

f(x)=αx5+βx3+γxf(x) = \alpha {x^5} + \beta {x^3} + \gamma x

∑i=1nf(ai)=α(a15+a25+a35 + ..... + an5)+β(a13+a23 + .... + an3)+γ(a1+a2 + .... + an)\sum\limits_{i = 1}^n {f({a_i}) = \alpha \left( {a_1^5 + a_2^5 + a_3^5\, + \,.....\, + \,a_n^5} \right) + \beta \left( {a_1^3 + a_2^3\, + \,....\, + \,a_n^3} \right) + \gamma \left( {{a_1} + {a_2}\, + \,....\, + \,{a_n}} \right)}

=0α+0β+0γ= 0\alpha + 0\beta + 0\gamma

=0= 0

∴\therefore f(g(1n∑i=1nf(ai)))=1n∑i=1nf(ai)=0f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right) = {1 \over n}\sum\limits_{i = 1}^n {f({a_i}) = 0}

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