Let n be a natural number and let a be a real | vedclass

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(C) Let $f(x) = x^{2n+1} - (2n+1)x + a$. To find the number of zeroes in the interval $[-1, 1]$,we examine the derivative of $f(x)$. $f'(x) = (2n+1)x^

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at most one for every value of $a$

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(C) Let $f(x) = x^{2n+1} - (2n+1)x + a$. To find the number of zeroes in the interval $[-1, 1]$,we examine the derivative of $f(x)$. $f'(x) = (2n+1)x^{2n} - (2n+1) = (2n+1)(x^{2n} - 1)$. For $x \in (-1, 1)$,we have $|x| Thus,$x^{2n} - 1 Since $f'(x) \leq 0$ for $x \in [-1, 1]$,the function $f(x)$ is strictly decreasing on the interval $[-1, 1]$. $A$ strictly monotonic function can intersect the $X$-axis at most once. Therefore,the equation $f(x) = 0$ has at most one root in the interval $[-1, 1]$ for any value of $a$.

Let $n$ be a natural number and let $a$ be a real number. The number of zeroes of $x^{2n+1} - (2n+1)x + a = 0$ in the interval $[-1, 1]$ is:

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