CAT 2025 Slot 1 QA Question & Solution

AlgebraMedium

Question

Let $3\leq x\leq6$ and $\left[x^{2}\right] =\left[x\right]^{2}$ , where [x] is the greatest integer not exceeding x . If set S represents all feasible values of x, then a possible subset of Sis

Options

$\left(3,\sqrt{10}\right)\cup \left[5,\sqrt{26}\right)\cup \left\{6\right\}$

$\left[3,\sqrt{10}\right] \cup \left[5,\sqrt{26}\right]$

$\left[3,\sqrt{10}\right]\cup \left[4,\sqrt{17}\right]\cup \left\{6\right\}$

$\left[4,\sqrt{18}\right]\cup \left[5,\sqrt{27}\right]\cup \left\{6\right\}$

Solution

For n=3,4,5 and $$x\in[n,n+1)$$ we have $$\lfloor x\rfloor=n$$, so the equation

$$\lfloor x^2\rfloor=\lfloor x\rfloor^2=n^2$$

$$x^2\in[n^2,n^2+1)$$, i.e. $$x\in[n,\sqrt{n^2+1})$$

Thus for $$3\le x\le6$$

$$S=[3,\sqrt{10})\ \cup\ [4,\sqrt{17})\ \cup\ [5,\sqrt{26})\ \cup{6}$$

Option B and C have $$\sqrt{10}$$ included, which is not part of the original set. And Option D has $$\sqrt{18}$$. So, it is not possible.

Option A is the answer.