Medium MCQ +4 / -1 PYQ · JEE Mains 2022

Let $\alpha, \beta$ and $\gamma$ be three positive real numbers. Let $f(x)=\alpha x^{5}+\beta x^{3}+\gamma x, x \in \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be such that $g(f(x))=x$ for all $x \in \mathbf{R}$. 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 $$f\left(g\left(\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right)$$ is equal to :

  1. A 0 Correct answer
  2. B 3
  3. C 9
  4. D 27

Solution

<p>$f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right)$</p> <p>${{{a_1} + {a_2} + {a_3}\, + \,......\, + \,{a_n}} \over n} = 0$</p> <p>$\therefore$ First and last term, second and second last and so on are equal in magnitude but opposite in sign.</p> <p>$f(x) = \alpha {x^5} + \beta {x^3} + \gamma x$</p> <p>$$\sum\limits_{i = 1}^n {f({a_i}) = \alpha \left( {a_1^5 + a_2^5 + a_3^5\, + \,.....\, + \,a_n^5} \right) + \beta \left( {a_1^3 + a_2^3\, + \,....\, + \,a_n^3} \right) + \gamma \left( {{a_1} + {a_2}\, + \,....\, + \,{a_n}} \right)} $$</p> <p>$= 0\alpha + 0\beta + 0\gamma$</p> <p>$= 0$</p> <p>$\therefore$ $$f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right) = {1 \over n}\sum\limits_{i = 1}^n {f({a_i}) = 0} $$</p>

About this question

Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations

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