Q.Solve , when
The inequality simplifies to . For integer , the solution set is ; for real , it is the open interval .
The core idea here is simple: an inequality like is solved almost exactly like an equation — you isolate by doing the same operation on both sides. The only twist is that if you ever multiply or divide by a negative number, the inequality sign flips. That doesn’t happen here, so we’re in straightforward territory.
Once we have , the answer depends entirely on what kind of numbers is allowed to be. That’s the whole point of the two parts: the same inequality, but different number systems give different solution sets.
Let’s work through it.
- Simplify the inequality. Start with . Add to both sides:
Then divide both sides by (positive, so no sign flip):
That’s the core result. Every solution must satisfy .
- Part (i): is an integer. Integers are whole numbers (…, -3, -2, -1, 0, 1, 2, 3, …). We need all integers strictly less than 2. That means can be — all integers down to negative infinity. There is no smallest integer here (since you can always go more negative), so we list it as:
Often in exams, they write it as or simply say “all integers less than 2”.
A common mistake is to include 2 in the solution. The inequality is strict (), so 2 is not allowed. For integers, the largest value is 1.
- Part (ii): is a real number. Real numbers include everything on the number line — fractions, decimals, irrationals, everything. The condition now describes an entire interval: all real numbers from negative infinity up to, but not including, 2. In interval notation, this is:
The round bracket at 2 means “open” — 2 is excluded. Negative infinity always gets a round bracket because it’s not a number you can reach.
If the inequality had been , the interval would be , with a square bracket at 2. The bracket shape tells you whether the endpoint is included.
So the two answers are different in form, but come from the same simple inequality.
- For integer , the solution set is ;
- for real , the solution set is .
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