Question

If $P(x)=51x^{101}-2323x^{100}-45x+1035$, then $P(x)=0$ has at least one root in

• A. $(45,46)$
• B. $\left(45^{1/100},46\right)$
• C. $\left(45^{1/50},46\right)$
• D. $\left(45,46^{1/100}\right)$

Answer 1

Answer: (B)

$\int P(x)dx=\frac{x^{102}}{2}-23x^{101}-45\frac{x^2}{2}+1035x+c$
Now consider a function
$f(x)=\frac{x^2}{2}\left(x^{100}-45\right)-23x\left(x^{100}-45\right)=\frac{x}{2}(x-46)\left(x^{100}-45\right)$
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 $\left(45^{1/100},46\right)$