Question

Let $f:R\to R$ be a polynomial function such that $f\left(x^2\right)=x^{3}f\left(x\right)+x^3-1$ for every $x\in R$, and $f\left(3\right)=26.$ Then

• A. $f\left(x\right)$ is an invertible function
• B. Fundamental period of $f\left(\sin x\right)$ is $2\pi$
• C. $f\left(x\right)$ is an odd function
• D. $2f\left(3\right)-3f\left(2\right)=31$

Answer 1

Answer: (A), (B), (D)

$2f\left(3\right)-3f\left(2\right)=31$

$f\left(x^2\right)=x^{3}f\left(x\right)+x^3-1\dots \left(1\right)$
Here is degree of $f\left(x\right)$ is $3$.
Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
From $\left(1\right),$
$a{x}^6+b{x}^4+c{x}^2+d=a{x}^6+b{x}^5+c{x}^4+d{x}^3+x^3-1$
$\therefore b=c=0,d=-1$
$f\left(x\right)=a{x}^3-1$
$\because f\left(3\right)=26$ then $26=a\cdot 27-1$
$\Rightarrow a=1$
$\therefore f\left(x\right)=x^3-1$
$2f\left(3\right)-3f\left(2\right)=2\left(26\right)-3\left(7\right)=52-21=31$
$\because f\left(x\right)=x^3-1$ is one-one and onto function
$\Rightarrow f\left(x\right)$ is invertible function
$f\left(\sin x\right)={\sin }^{3}x-1$ has period $2\pi$
[$\because$ Fundamental period of ${\sin }^{n}x$ is $2\pi$ if $n$ is odd]
$f(-x)=-x^3-1\ne -f\left(x\right)$
$\Rightarrow f\left(x\right)$ is not an odd function