Question

If $f:R\to R,f(x^2+x+3)+2f(x^2-3x+5)=6x^2-10x+17\forall x\in R$
then which among the following options are correct

• A. $f\left(x\right)$ is an odd function
• B. $f^{\prime }\left(x\right)$ is an even function
• C. $f\left(x\right)=0$ has a root in $(0,2)$
• D. $f\left(x\right)$ is invertible function.

Answer 1

Answer: (B), (C), (D)

$f\left(x\right)$ is invertible function.

Given ,$f(x^2+x+3)+2f(x^2-3x+5)=6x^2-10x+17$
By observation $f\left(x\right)$ should be linear function,
let $f\left(x\right)=ax+b$
$\therefore [a(x^2+x+3)+b]+2[a(x^2-3x+5)+b]=6x^2-10x+17$
on comparing $x^2$ coefficients $a+2a=6\Rightarrow a=2$
On comparing constants $3a+b+10a+2b=17$
$13a+3b=17\Rightarrow b=-3$
$\therefore f\left(x\right)=2x-3$
Clearly, $f\left(x\right)$ is neither even nor odd function. But $f^{\prime }\left(x\right)=2$ is an even function.
Also we can see, $f\left(x\right)=0\Rightarrow x=\frac{3}{2}$, which lies in $(0,2).$
$f:R\to R,f\left(x\right)=2x-3$ is bijective function. Hence it is invertible function.