Question

Consider the equation $x^6-x-1=0$, then the number of real roots of the equation is

• A. 6
• B. 4
• C. 2
• D. 0

Answer 1

Answer: (C)

$f(x)=x^6-x-1\Rightarrow f^{\prime }(x)=6x^5-1=0\Rightarrow x={\left(\frac{1}{6}\right)}^{1/5}$
is a point of local minimum also $f\left({\left(\frac{1}{6}\right)}^{1/5}\right)<0$
$\Rightarrow f(x)=0$ has exactly 2 real roots.