Let α ,β and γ be three positive real numbers. Let f(x)=α x^5+β x^3+γ x,x∈ R and g:R→ R be such that g(f(x))=x for all x∈ R. If a_1,a_2,a_3,… ,a_n be…

Given $\alpha ,\beta$α,β and $\gamma$γ be three positive real numbers. Let $f(x)=\alpha {x}^{5}+\beta {x}^{3}+\gamma x,x\in R$f(x)=αx5+βx3+γx,x∈R and $g:R\rightarrow R$g:R→R be such that $g(f(x))=x$g(f(x))=x for all $x\in R$x∈R. If ${a}_{1},{a}_{2},{a}_{3},\ldots ,{a}_{n}$a1​,a2​,a3​,…,an​ be in arithmetic progression with mean zero, then the value of $\displaystyle f(g(\frac{1}{n}\sum _{i=1}^{n}f({a}_{i})))$f(g(n1​i=1∑n​f(ai​))) is equal to

Given,

Mean is zero, so $\frac{{a}_{1}+{a}_{2}+{a}_{3}+\ldots ..+{a}_{n}}{n}=0$na1​+a2​+a3​+…..+an​​=0

Now this means that first and last term, second and second last and so on are equal in magnitude but opposite in sign.

Given, $f(x)=\alpha {x}^{5}+\beta {x}^{3}+\gamma x$f(x)=αx5+βx3+γx

Now finding $\displaystyle \sum _{i=1}^{n}f({a}_{i})=\alpha ({a}_{1}^{5}+{a}_{2}^{5}+{a}_{3}^{5}+\ldots .+{a}_{n}^{5})+\beta ({a}_{1}^{3}+{a}_{2}^{3}+\ldots .+{a}_{n}^{3})+\gamma ({a}_{1}+{a}_{2}+\ldots .+{a}_{n})$i=1∑n​f(ai​)=α(a15​+a25​+a35​+….+an5​)+β(a13​+a23​+….+an3​)+γ(a1​+a2​+….+an​)

$=0\alpha +0\beta +0\gamma =0$=0α+0β+0γ=0 ......equation$(1)$(1) {as first & last, second and second last and so on are equal in magnitude and opposite in sign}

Now given $g(f(x))=x$g(f(x))=x

$$\Rightarrow f(g(f(x)))=f(x)$$⇒f(g(f(x)))=f(x)

So, $\displaystyle f(g(\frac{1}{n}\sum _{i=1}^{n}f({a}_{i}))]=\frac{1}{n}\sum _{i=1}^{n}f({a}_{i})=0$f(g(n1​i=1∑n​f(ai​))]=n1​i=1∑n​f(ai​)=0 {from equation$(1)$(1) }