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

$1$

Explanation for the correct option:

Find the real solution of the given equation:

Let $f\left(x\right)=x^7+14x^5+16x^3+30x–560=0$

$\Rightarrow$$f\text{'}\left(x\right)=7x^6+70x^4+48x^2+30$

$f\text{'}\left(x\right)>0\forall x\in Ri.e.f\left(x\right)$ is a strictly increasing function.

The given function cuts $X$- axis and $Y$- axis at only one point.

So the number of real solution of the given function is $\mathbf{1}$.

Hence, option (C) is correct answer.