If a_ is the greatest term in the sequence _n=n^3n^4+147,n=1,2,3,, then is equal to ______…

Question

If ${a}_{\alpha }$ is the greatest term in the sequence ${\alpha }_{n}=\frac{{n}^{3}}{{n}^{4}+147},n=1,2,3,\ldots$, then $\alpha $ is equal to \_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

$\begin{gathered} y=\frac{{x}^{3}}{{x}^{4}+147} \ \Rightarrow \frac{dy}{dx}=\frac{({x}^{4}+147)	imes 3{x}^{2}-{x}^{3}(4{x}^{3})}{{({x}^{4}+147)}^{2}} \ =\frac{3{x}^{6}+441{x}^{2}-4{x}^{6}}{{({x}^{4}+147)}^{2}}=\frac{441{x}^{2}-{x}^{6}}{{({x}^{4}+147)}^{2}} \end{gathered}$

For maxima/minima, put $\frac{dy}{dx}=0$

$\begin{gathered} \Rightarrow 441{x}^{2}-{x}^{6}=0\Rightarrow {x}^{4}=441 \ \Rightarrow x=\pm \sqrt{21},\pm \sqrt{21}i \end{gathered}$

Now, by descrates rule on number line we have

Since sign changes from negative to positive at 0 .

$	herefore $ Maximum value of is at $x=\sqrt{21}=4.58$

Now, $4<4.5<5$

$\begin{gathered} 	herefore x=4=\frac{64}{403}=0.159 \ yx=5=\frac{125}{772}=0.162 \end{gathered}$

So, $y$ is maximum at $x=5$

$	herefore \alpha =5$

Solutions · 1