Let $\alpha, \beta$ and $\gamma$ be three positive real numbers. Let $\mathrm{f}(\mathrm{x})=\alpha \mathrm{x}^{5}+\beta \mathrm{x}^{3}+\gamma \mathrm{x}, \mathrm{x} \in \mathrm{R}$ and $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ be such that $\mathrm{g}(\mathrm{f}(\mathrm{x}))=\mathrm{x}$ for all $\mathrm{x} \in \mathrm{R}$. If $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}$ be in arithmetic progression with mean zero, then the value of $\mathrm{f}\left(\mathrm{g}\left(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{a}_{\mathrm{i}}\right)\right)\right)$ is equal to :

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Sets Relations and Functions

Let

α, β

and

γ

be three positive real numbers. Let

f (x) = α x^{5} + β x^{3} + γ x, x \in R

and

g : R \to R

be such that

g (f (x)) = 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 (g (\frac{1}{n} \sum_{i = 1}^{n} f (a_{i})))

is equal to :

Answer: (A)

Since

f (x) = α x^{5} + β x^{3} + γ x

is an odd function and

g

is its inverse,

g

is also odd.

The sequence

a_{1}, a_{2}, \dots, a_{n}

is in AP with mean zero, so

\sum_{i = 1}^{n} a_{i} = 0

.

Since

f

is odd:

\sum_{i = 1}^{n} f (a_{i}) = f (\sum_{i = 1}^{n} a_{i}) = f (0) = 0

(by properties of odd functions on symmetric sets).

Therefore:

\frac{1}{n} \sum_{i = 1}^{n} f (a_{i}) = 0

.

Thus:

f (g (0)) = f (0) = 0

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option