The polynomial \(p(x)\) of degree 98 satisfies the condition \(p(x) = \frac{1}{x}\) for \(x = 1, 2, \ldots, 98\). To determine \(p(100)\), the discussion suggests that if \(p(x)\) were monic, then \(p(100) = 99!\). However, a revised approach using the polynomial \(q(x) = xp(x) - 1\) reveals that \(q(x)\) is of degree 99 and has roots at \(x = 1, 2, \ldots, 99\). Consequently, \(p(100) = \frac{1}{50}\) after calculating \(q(100)\).
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jacks said:If $p(x)$ be a polynomial of degree $98$ such that $\displaystyle p(x) = \frac{1}{x}$ for $x=1,2,3,...,98$
Then value of $p(100)=$
In that case, let $q(x) = xp(x)-1$. Then $q(x)$ is a polynomial of degree 99, and $q(x)=0$ for $x=1,2,3,\ldots,99$. By the factor theorem, $q(x) = k(x-1)(x-2)\cdots(x-99)$ for some constant $k$. Also, $q(0)=-1$, and therefore $k=1/99!$. It follows that $q(100) = 1$, from which $p(100) = 1/50$.If $p(x)$ be a polynomial of degree $98$ such that $\displaystyle p(x) = \frac{1}{x}$ for $x=1,2,3,...,\color{red}{99}$
Then value of $p(100)=$