Statement-I: The sequence a_n = frac{n^2}{n^3 | vedclass

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(C) To check the behavior of $f(x) = \frac{x^2}{x^3 + 200}$,we find its derivative $f'(x)$.Using the quotient rule: $f'(x) = \frac{(x^3 + 200)(2x) - (

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Statement-$I$ is true,Statement-$II$ is false.

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(C) To check the behavior of $f(x) = \frac{x^2}{x^3 + 200}$,we find its derivative $f'(x)$.Using the quotient rule: $f'(x) = \frac{(x^3 + 200)(2x) - (x^2)(3x^2)}{(x^3 + 200)^2} = \frac{2x^4 + 400x - 3x^4}{(x^3 + 200)^2} = \frac{400x - x^4}{(x^3 + 200)^2} = \frac{x(400 - x^3)}{(x^3 + 200)^2}$.For local extrema,set $f'(x) = 0$,which gives $x = 0$ or $x^3 = 400$,so $x = \sqrt[3]{400} \approx 7.368$.Since $f'(x) > 0$ for $x  \sqrt[3]{400}$,the function has a local maximum at $x = \sqrt[3]{400}$.For the sequence $a_n$,we compare $a_7$ and $a_8$.$a_7 = \frac{49}{343 + 200} = \frac{49}{543} \approx 0.0902$.$a_8 = \frac{64}{512 + 200} = \frac{64}{712} = \frac{8}{89} \approx 0.0898$.Since $a_7 > a_8$ and $a_7 > a_6$ (as $f(x)$ is increasing until $x \approx 7.368$),$a_7$ is indeed the largest term.However,Statement-$II$ claims the maximum is at $x = 7$,which is incorrect as the maximum is at $x = \sqrt[3]{400} \approx 7.368$.

Statement-$I$: The sequence $a_n = \frac{n^2}{n^3 + 200}, n \in N$ has its $7^{th}$ term as the largest term.
Statement-$II$: The function $f(x) = \frac{x^2}{x^3 + 200}$ attains a local maximum at $x = 7$.

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Statement-$I$: The sequence $a_n = \frac{n^2}{n^3 + 200}, n \in N$ has its $7^{th}$ term as the largest term.Statement-$II$: The function $f(x) = \frac{x^2}{x^3 + 200}$ attains a local maximum at $x = 7$.