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Q.
If $P(x)=51 x^{101}-2323 x^{100}-45 x+1035$, then $P(x)=0$ has at least one root in
Application of Derivatives
Solution:
$\int P(x) d x=\frac{x^{102}}{2}-23 x^{101}-45 \frac{x^2}{2}+1035 x+c$
Now consider a function
$f(x)=\frac{x^2}{2}\left(x^{100}-45\right)-23 x\left(x^{100}-45\right)=\frac{x}{2}(x-46)\left(x^{100}-45\right)$
It satisfies all conditions of Rolle's theorem in $\left[45^{1 / 100}, 46\right]$. Hence there exist at least one c $\in\left(45^{1 / 100}, 46\right)$ for which $f^{\prime}(c)=0$
$\Rightarrow P(c)=0$ which proves that there exist a root of $P(x)=0$ in $\left(45^{1 / 100}, 46\right)$