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  • If  is the greatest term in the sequence <img alt="a_n=\frac{\mathrm{n}^3}{\mathrm{n}^4+147}, \mathrm{n}=1,2,3,

If  a_a is the greatest term in the sequence a_n=\frac{\mathrm{n}^3}{\mathrm{n}^4+147}, \mathrm{n}=1,2,3, \ldots, then a is equal to ___________.

Option: 1

0.158


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & f(x)=\frac{x^3}{x^4+147} \\ & f^{\prime}(x)=\frac{\left(x^4+147\right) 3 x^2-x^3\left(4 x^3\right)}{\left(x^4+147\right)^2} \\ & =\frac{3 x^6+147 \times 3 x^2-4 x^6}{+v e}=x^2\left(44-x^4\right) \end{aligned}

\begin{aligned} & f^{\prime}(x)=0 \text { at } x^6=147 \times 3 x^2 \\ & x^2=0, x^4=147 \times 3 \\ & x=0, x^2= \pm \sqrt{147 \times 3} \\ & x^2= \pm 21 \\ & x= \pm \sqrt{21} \end{aligned}

fmax at f(4) \text { or } f(5)

\begin{aligned} & f(4)=\frac{64}{403} \simeq 0.158 \quad f(5)=\frac{125}{772} \simeq 0.161 \\ & \therefore a=5 \end{aligned}

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Divya Prakash Singh

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