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Q.
Consider the equation $x^6-x-1=0$, then the number of real roots of the equation is
Application of Derivatives
Solution:
$f(x)=x^6-x-1 \Rightarrow f^{\prime}(x)=6 x^5-1=0 \Rightarrow x=\left(\frac{1}{6}\right)^{1 / 5}$
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.