Question

Let $f\left(x\right)=x^3-6x^2+15x+3.$ Then

• A. $f\left(x\right)>0$ for all $xϵR$
• B. $f\left(x\right)>f(x+1)$ does not hold for any real x
• C. $f\left(x\right)$ is invertible
• D. $f\left(x\right)$ is a one -one function

Answer 1

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

The correct options are
B $f\left(x\right)>f(x+1)$ does not hold for any real x
C $f\left(x\right)$ is invertible
D $f\left(x\right)$ is a one -one function

We have,
$f\left(x\right)=x^3-6x^2+15x+3$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-12x+15=3(x^2-4x+5)$
Clearly discriminant of $x^2-4x+5=0$ is $=16-20=-4<0$
Hence $f^{\prime }\left(x\right)>0$ for all $x\in R$
$\Rightarrow f\left(x\right)<f(x+1)\forall x\in R$
Thus $f$ is strictly increasing function.
Also $f$ is cubic polynomial so it's range is R $=$ co-domain
Therefore f is bijective function.