Question

If sum of maximum and minimum value of $y=\log _2\left(x^4+x^2+1\right)-\log _2\left(x^4+x^3+2 x^2+x+1\right)$ can be expressed in form $\left(\left(\log _2 m\right)-n\right)$, where $m$ and $n$ are coprime then compute $(m+n)$.

Answer 1

We have $y =\log _2\left(\frac{\left( x ^2+ x +1\right)\left( x ^2- x +1\right)}{ x ^4+ x ^3+ x ^2+ x +1}\right)=\log _2\left(\frac{ x ^2- x +1}{ x ^2+1}\right)$
Assume $z=\frac{x^2-x+1}{x^2+1}$
$(z-1) x^2+x+(z-1)=0$. As $x \in R$, so
$D \geq 0 \Rightarrow 1-4( z -1)^2 \geq 0 \Rightarrow( z -1)^2 \leq \frac{1}{4} \Rightarrow \frac{1}{2} \leq z \leq \frac{3}{2}$
So, $y _{\min }=\log _2\left(\frac{1}{2}\right)=-1$ and $y _{\max }=\log _2\left(\frac{3}{2}\right)=\left(\log _2 3-1\right)$
Hence $y _{\min .}+ y _{\max .}=\log _2 3-2 \equiv\left(\left(\log _2 m \right)- n \right)$
$\therefore m =3, n =2$
Hence $( m + n )=3+2=5$