If a_α is the greatest term in the sequence a_n= n^3 n^4+147,n=1,2,3...., then α is equal to ______

Let,

$$y=\frac{{x}^{3}}{{x}^{4}+147}=f(x)$$y=x4+147x3​=f(x)

We know that, for increasing function $\frac{dy}{dx}>0$dxdy​>0

So, differentiating the given function $f(x)=\frac{{x}^{3}}{{x}^{4}+147}$f(x)=x4+147x3​ we get,

$$\frac{dy}{dx}=\frac{-{x}^{2}({x}^{4}-441)}{({x}^{4}+147{)}^{2}}$$dxdy​=(x4+147)2−x2(x4−441)​

$$\Rightarrow \frac{-{x}^{2}({x}^{4}-441)}{({x}^{4}+147{)}^{2}}>0$$⇒(x4+147)2−x2(x4−441)​>0

$$\Rightarrow \frac{{x}^{2}({x}^{4}-441)}{({x}^{4}+147{)}^{2}}<0$$⇒(x4+147)2x2(x4−441)​<0

$$\Rightarrow {x}^{4}<441$$⇒x4<441

Now for maxima/minima $\frac{dy}{dx}=0$dxdy​=0

$$\Rightarrow {x}^{4}=441$$⇒x4=441

$$\Rightarrow x=\alpha ,4<\alpha <5$$⇒x=α,4<α<5

$\Rightarrow$⇒Maximum value of $f(x)$f(x) is at $x=4$x=4 or $x=5$x=5

$$f(4)=\frac{64}{403},f(5)=\frac{125}{772}$$f(4)=40364​,f(5)=772125​

$$\because f(5)>f(4)$$∵f(5)>f(4)

$$\Rightarrow \alpha =5$$⇒α=5