Mathematics - Functions Question with Solution | TestHub

Given $\alpha ,\beta$ and $\gamma$ be three positive real numbers. Let $f\left(x\right)=\alpha {\mathrm{x}}^5+\beta {\mathrm{x}}^3+\gamma \mathrm{x},\mathrm{x}\in \mathrm{R}$ and $g:R\to R$ be such that $g\left(f\left(x\right)\right)=x$ for all $x\in R$. If $a_1,a_2,a_3,\dots ,a_n$ be in arithmetic progression with mean zero, then the value of $f\left(g\left(\frac{1}{n}\sum _{i=1}^{n}f\left(a_i\right)\right)\right)$ is equal to

Given,

Mean is zero, so $\frac{a_1+a_2+a_3+\dots ..+a_n}{n}=0$

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

Given, $f\left(x\right)=\alpha x^5+\beta x^3+\gamma x$

Now finding $\sum _{i=1}^{n}f\left(a_i\right)=\alpha \left(a_1^5+a_2^5+a_3^5+\dots .+a_n^5\right)+\beta \left(a_1^3+a_2^3+\dots .+a_n^3\right)+\gamma \left(a_1+a_2+\dots .+a_n\right)$

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

Now given $g\left(f\left(x\right)\right)=x$

$\Rightarrow f\left(g\left(f\left(x\right)\right)\right)=f\left(x\right)$

So, $f\left(g\left(\frac{1}{n}\sum _{i=1}^{n}f\left(a_i\right)\right)\right]=\frac{1}{n}\sum _{i=1}^{n}f\left(a_i\right)=0$ {from equation$\left(1\right)$ }