Question

$f:R\to R$ ; $f\left(x^2+x+3\right)+2f\left(x^2-2x+5\right)=8x^2-10x+17$; $\forall x\in R$ then

• A. $f$ is strictly decreasing
• B. $f\left(x\right)=0$ has a root in $(0,2)$
• C. $f\left(x\right)$ is an odd function
• D. $f\left(x\right)$ is invertible

Answer 1

Answer: (B), (D)

$f\left(x\right)=0$ has a root in $(0,2)$
$f\left(x\right)$ is invertible

$f:R\to R;f(x^2+x+3)+2f(x^2-2x+5)=8x^2-10x+17;\forall xϵR$

Let $x^2+x+3=k$

$f\left(k\right)+2f(k-3x+2)=8(k-x-3)-10x+17$

$f\left(k\right)+2f(k-3x+2)=8k-18x-7$

at $k=0,x=0$

$f\left(0\right)+2f\left(2\right)=-7$

$f\left(0\right)=-f\left(2\right)-7$

Therefore, $f\left(0\right)$ and $f\left(2\right)$ are of opposite signs

And thus $f\left(x\right)=0$ has a root in $(0,2)$

And also

$f^{\prime }(x^2+x+3)\times (2x+1)+2f^{\prime }(x^2-2x+5)(2x-2)=16x-10$

Function is differentiable and therefore invertible.