If f(x) = x^3 - x^2 + 100x + 1001,then: | vedclass

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(D) Given $f(x) = x^3 - x^2 + 100x + 1001$. To determine the nature of the function,we find its derivative: $f'(x) = 3x^2 - 2x + 100$. The discriminan

Vedclass · Accepted Answer

$f\left(\frac{1}{1999}\right) > f\left(\frac{1}{2000}\right)$

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(D) Given $f(x) = x^3 - x^2 + 100x + 1001$. To determine the nature of the function,we find its derivative: $f'(x) = 3x^2 - 2x + 100$. The discriminant of this quadratic is $D = (-2)^2 - 4(3)(100) = 4 - 1200 = -1196$. Since $D  0$ for all $x \in \mathbb{R}$. Thus,$f(x)$ is a strictly increasing function. For a strictly increasing function,if $a > b$,then $f(a) > f(b)$. Since $\frac{1}{1999} > \frac{1}{2000}$,it follows that $f\left(\frac{1}{1999}\right) > f\left(\frac{1}{2000}\right)$. Therefore,option $D$ is correct.

If $f(x) = x^3 - x^2 + 100x + 1001$,then:

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