Question

How many real solutions does the equation $x^7+14x^5+16x^3+30x-560=0$ have?

• A. $5$
• B. $7$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

Let $f\left(x\right)=x^7+14x^5+16x^3+30x-560$
$\therefore$ $f^{\prime }\left(x\right)=7x^6+70x^4+48x^2+30$
$\Rightarrow$ $f^{\prime }\left(x\right)>0$ $\forall xϵR$
i.e.$f\left(x\right)$ is an strictly increasing function.
So it can have at the most one solution. It can be shown that it has exactly one solution.