The correct option is C (−3,−2)
1log4(x+1x+2)>1log4(x+3) ⋯(1)
log is defined when
x+3>0⇒x∈(−3,∞) ⋯(i)
And x+1x+2>0
⇒x∈(−∞,−2)∪(−1,∞) ⋯(ii)
From (i) and (ii)
∴x∈(−3,−2)∪(−1,∞) ⋯(2)
Case 1: x∈(−1,∞) ...(3)
⇒x+1x+2∈(0,1) and x+3∈(2,∞)
⇒log4(x+1x+2)<0 and log4(x+3)>0
which does not satisfies inequation (1)
Case 2: x∈(−3,−2) ...(4)
⇒x+1x+2∈(2,∞) and x+3∈(0,1)
⇒log4(x+1x+2)>0 and log4(x+3)<0
⇒log4(x+1x+2)>log4(x+3)
⇒x+1x+2>x+3
⇒x+1x+2−x+3>0
⇒−(x2+4x+5)x+2>0
⇒(x+2)2+1x+2<0
⇒x<−2 ...(5)
Now, from (2),(4) and (5)
x∈(−3,−2)