If sum of maximum and minimum value of y= log 2(x4+x2+1)- log 2(x4+x3+2 x2+x+1) can be expressed in form (( log 2 m)-n), where m and n are coprime then compute (m+n).

Question

If sum of maximum and minimum value of y= log 2(x4+x2+1)- log 2(x4+x3+2 x2+x+1) can be expressed in form (( log 2 m)-n), where m and n are coprime then compute (m+n). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

We have y = log 2((( x 2+ x +1)( x 2- x +1)/ x 4+ x 3+ x 2+ x +1))= log 2(( x 2- x +1/ x 2+1)) Assume z=(x2-x+1/x2+1) (z-1) x2+x+(z-1)=0. As x ∈ R, so D ≥ 0 ⇒ 1-4( z -1)2 ≥ 0 ⇒( z -1)2 ≤ (1/4) ⇒ (1/2) ≤ z ≤ (3/2) So, y min = log 2((1/2))=-1 and y max = log 2((3/2))=( log 2 3-1) Hence y min .+ y max .= log 2 3-2 ≡(( log 2 m )- n ) ∴ m =3, n =2 Hence ( m + n )=3+2=5

Q. If sum of maximum and minimum value of y=log2​(x4+x2+1)−log2​(x4+x3+2x2+x+1) can be expressed in form ((log2​m)−n), where m and n are coprime then compute (m+n).

Q. If sum of maximum and minimum value of $y = lo g_{2} (x^{4} + x^{2} + 1) - lo g_{2} (x^{4} + x^{3} + 2 x^{2} + x + 1)$ can be expressed in form $((lo g_{2} m) - n)$ , where $m$ and $n$ are coprime then compute $(m + n)$ .