Let $3 \leq x \leq 6$ and $[x^2] = [x]^2$ , where $[x]$ is the greatest integer not exceeding $x$. If set $S$ represents all feasible values of $x$ , then a possible subset of $S$ is

A) $[3,\sqrt{10}]\cup[5,\sqrt{26}]$

B) $[3,\sqrt{10}]\cup[4,\sqrt{17}]\cup\{6\}$

D) $(4,\sqrt{18})\cup[5,\sqrt{27})\cup\{6\}$

CAT 2025 Slot 1 QA Question 3

Multiple choice (+3 / −1) · Algebra · Functions · Try it, then check the answer and solution below.

CAT 2025 Slot 1QAAlgebra • FunctionsHard

Let

3 \leq x \leq 6

and

[x^2] = [x]^2 , where

[x]

is the greatest integer not exceeding

. If set

represents all feasible values of

x , then a possible subset of

S

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Official Convenor Answer Key: C

Official Correct Answer: C. The question says: Let 3 ≤ x ≤ 6 and [x²] = [x]², where [x] is the greatest integer not exceeding x. Set S represents all feasible values of x. We need to find a possible subset of S from the given options. First, We analyze what [x] means. It's the floor function, right? So [x] is the greatest integer less than or equal to x. For example, [3.2] is 3, [5] is 5, and [6] is 6. The equation given is [x²] = ([x])². So, the floor of x squared equals the square of the floor of x. We find all x in [3,6] that satisfy this condition. Let [x] as n, where n is an integer. Since x is between 3 and 6, n can be 3, 4, 5, or 6. Wait, but x can be up to 6, so n can be 3, 4, 5, or 6. But if x is exactly 6, then [x] is 6, and [x²] is [36] which is 36, and ([x])² is 36, so that works. So x=6 is included. Now, for each possible n, I can find the range of x where [x] = n, and then check when [x²] = n². Let's break it down for each n: 1. **Case 1: n = 3** - Here, x is in [3,4). So, 3 ≤ x < 4. - Then, [x²] should be equal to 9, since [x] = 3, so ([x])² = 9. - So, [x²] = 9. That means 9 ≤ x² < 10. - Taking square roots, we get 3 ≤ x < √10. - But since x is already in [3,4), the overlap is [3, √10). - So, for n=3, x is in [3, √10). 2. **Case 2: n = 4** - Here, x is in [4,5). So, 4 ≤ x < 5. - Then, [x²] should be equal to 16, since [x] = 4, so ([x])² = 16. - So, [x²] = 16. That means 16 ≤ x² < 17. - Taking square roots, we get 4 ≤ x < √17. - But x is in [4,5), so the overlap is [4, √17). - So, for n=4, x is in [4, √17). 3. **Case 3: n = 5** - Here, x is in [5,6). So, 5 ≤ x < 6. - Then, [x²] should be equal to 25, since [x] = 5, so ([x])² = 25. - So, [x²] = 25. That means 25 ≤ x² < 26. - Taking square roots, we get 5 ≤ x < √26. - But x is in [5,6), so the overlap is [5, √26). - So, for n=5, x is in [5, √26). 4. **Case 4: n = 6** - Here, x is exactly 6. - Then, [x²] = [36] = 36, and ([x])² = 6² = 36. - So, x=6 satisfies the condition. Now, let's compile all the intervals where the condition holds: - From n=3: [3, √10) - From n=4: [4, √17) - From n=5: [5, √26) - And the point x=6. But For n=4, x is in [4,5), and [x²] must be 16. So x² must be in [16,17), which means x is in [4, √17). Since √17 is approximately 4.123, which is less than 5, so the interval is [4, √17). Similarly, for n=5, x is in [5,6), and x² is in [25,26), so x is in [5, √26), which is approximately [5,5.099). Wait, but √26 is about 5.099, which is less than 6, so the interval is [5, √26). So, putting it all together, the feasible set S is: S = [3, √10) ∪ [4, √17) ∪ [5, √26) ∪ {6} But looking at the options, Option A: [3,√10] ∪ [5,√26] - This includes √10 and √26, but from our analysis, these points are excluded because at x=√10, [x] would be 3, but x²=10, so [x²]=10, which is not equal to 9. Similarly for √26, [x²]=26, which is not 25. So A is incorrect. Option B: [3,√10] ∪ [4,√17] ∪ {6} - Again, includes √10 and √17, which are excluded, so B is incorrect. Option C: [3,√10) ∪ [5,√26) ∪ {6} - This includes [3,√10), [5,√26), and {6}. But wait, what about the interval [4,√17)? It's missing here. So does that mean that the feasible set S includes [4,√17) as well? But looking at the options, none of them include [4,√17). Hmm, maybe I made a mistake. The question says "a possible subset of S". So S is the union of all feasible intervals, which includes [3,√10), [4,√17), [5,√26),

Answer and solution

Answer: C) $[3,\sqrt{10})\cup[5,\sqrt{26})\cup\{6\}$

The question says: Let 3 ≤ x ≤ 6 and [x²] = [x]², where [x] is the greatest integer not exceeding x. Set S represents all feasible values of x. We need to find a possible subset of S from the given options.

First, We analyze what [x] means. It's the floor function, right? So [x] is the greatest integer less than or equal to x. For example, [3.2] is 3, [5] is 5, and [6] is 6.

The equation given is [x²] = ([x])². So, the floor of x squared equals the square of the floor of x. We find all x in [3,6] that satisfy this condition.

Let [x] as n, where n is an integer. Since x is between 3 and 6, n can be 3, 4, 5, or 6. Wait, but x can be up to 6, so n can be 3, 4, 5, or 6. But if x is exactly 6, then [x] is 6, and [x²] is [36] which is 36, and ([x])² is 36, so that works. So x=6 is included.

Now, for each possible n, I can find the range of x where [x] = n, and then check when [x²] = n².

Let's break it down for each n:

Case 1: n = 3

Here, x is in [3,4). So, 3 ≤ x < 4.

Then, [x²] should be equal to 9, since [x] = 3, so ([x])² = 9.

So, [x²] = 9. That means 9 ≤ x² < 10.

Taking square roots, we get 3 ≤ x < √10.

But since x is already in [3,4), the overlap is [3, √10).

So, for n=3, x is in [3, √10).

Case 2: n = 4

Here, x is in [4,5). So, 4 ≤ x < 5.

Then, [x²] should be equal to 16, since [x] = 4, so ([x])² = 16.

So, [x²] = 16. That means 16 ≤ x² < 17.

Taking square roots, we get 4 ≤ x < √17.

But x is in [4,5), so the overlap is [4, √17).

So, for n=4, x is in [4, √17).

Case 3: n = 5

Here, x is in [5,6). So, 5 ≤ x < 6.

Then, [x²] should be equal to 25, since [x] = 5, so ([x])² = 25.

So, [x²] = 25. That means 25 ≤ x² < 26.

Taking square roots, we get 5 ≤ x < √26.

But x is in [5,6), so the overlap is [5, √26).

So, for n=5, x is in [5, √26).

Case 4: n = 6

Here, x is exactly 6.

Then, [x²] = [36] = 36, and ([x])² = 6² = 36.

So, x=6 satisfies the condition.

Now, let's compile all the intervals where the condition holds:

From n=3: [3, √10)

From n=4: [4, √17)

From n=5: [5, √26)

And the point x=6.

But For n=4, x is in [4,5), and [x²] must be 16. So x² must be in [16,17), which means x is in [4, √17). Since √17 is approximately 4.123, which is less than 5, so the interval is [4, √17). Similarly, for n=5, x is in [5,6), and x² is in [25,26), so x is in [5, √26), which is approximately [5,5.099).

Wait, but √26 is about 5.099, which is less than 6, so the interval is [5, √26). So, putting it all together, the feasible set S is:

S = [3, √10) ∪ [4, √17) ∪ [5, √26) ∪ {6}

But looking at the options, Option A: [3,√10] ∪ [5,√26]

This includes √10 and √26, but from our analysis, these points are excluded because at x=√10, [x] would be 3, but x²=10, so [x²]=10, which is not equal to 9. Similarly for √26, [x²]=26, which is not 25. So A is incorrect.

Option B: [3,√10] ∪ [4,√17] ∪ {6}

Again, includes √10 and √17, which are excluded, so B is incorrect.

Option C: [3,√10) ∪ [5,√26) ∪ {6}

This includes [3,√10), [5,√26), and {6}. But wait, what about the interval [4,√17)? It's missing here. So does that mean that the feasible set S includes [4,√17) as well? But looking at the options, none of them include [4,√17). Hmm, maybe I made a mistake.

The question says "a possible subset of S". So S is the union of all feasible intervals, which includes [3,√10), [4,√17), [5,√26),

CAT 2025 Slot 1 QA Question 3

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