Question

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

• A. 1
• B. 3
• C. 5
• D. 7

Answer 1

The correct option is A 1

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

Diiferentiating both sides w.r.t $x$ we get

${f\left(x\right)}^{\prime }=7x^6+70x^4+48x^2+30$ for all $x\in R$

and the function has its first derivative whose sign is not changed.

$\therefore$ it is monotonic.

and $f\left(0\right)$ is negative$=-560$

$\Rightarrow$ it will continuously increase till infinity and hence will cross the $x-$axis only once.

Thus, it has only one real root.