IPMAT Indore 2025 · Quantitative Ability · Question 34

This one is won on the domain, not the algebra — check the base first.

Step 1 — the base must be valid. A logarithm’s base must be positive and \ne 1. The base here is x+\frac{1}{x}.

If x>0: by AM–GM, x+\frac{1}{x}\ge 2. So the base is greater than 1. ✓

If x<0: x+\frac{1}{x}\le -2, which is negative — not a legal base. ✗

So we are restricted to x>0 from the outset.

Step 2 — solve the inequality on that domain. Since the base exceeds 1, {\log}_{\text{base}}T>0 means T>1:

{\log}_{2}\left( \frac{x-1}{x+2}\right)>1 \Rightarrow \frac{x-1}{x+2}>2

Step 3 — solve the rational inequality.

\frac{x-1}{x+2}-2>0 \Rightarrow \frac{x-1-2x-4}{x+2}>0 \Rightarrow \frac{-\left( x+5 \right)}{x+2}>0

\Rightarrow \frac{x+5}{x+2}<0 \Rightarrow -5<x<-2

Step 4 — compare with the domain. Step 3 demands x be negative; Step 1 demands x>0. The two can never hold together.

\text{Solution set}=⌀

Answer: Option 4 — Null set