Question

If $f\left(x\right)=(x-1)(x-3)(x-4)(x-6)+10$, then which of the following statements(s) is/are correct?

• A. $f\left(x\right)=0$ has $4$ distinct real roots.
• B. $f\left(x\right)=0$ has no real roots.
• C. $f\left(x\right)$ is always positive for all $x\in R.$
• D. $f\left(x\right)$ has negative values for some real values of $x.$

Answer 1

Answer: (B), (C)

$f\left(x\right)$ is always positive for all $x\in R.$

$f\left(x\right)=(x-1)(x-3)(x-4)(x-6)+10\Rightarrow f\left(x\right)=(x-1)(x-6)(x-3)(x-4)+10\Rightarrow f\left(x\right)=(x^2-7x+6)(x^2-7x+12)+10$
Assuming $x^2-7x+6=t$, so
$f\left(x\right)=t(t+6)+10\Rightarrow f\left(x\right)=t^2+6t+10\Rightarrow f\left(x\right)=(t+3{)}^2+1\Rightarrow f(x)=(x^2-7x+9{)}^2+1$

As $f\left(x\right)\ge 1$, so no real roots of $f\left(x\right)=0$.