Kerala KEAM 5 June 2024 Solved Paper

We are given the integral: $$I = \int_0^{\sqrt{2}} \left\lfloor x^2 \right\rfloor \, dx$$where $\left\lfloor x^2 \right\rfloor$ represents the greatest integer less than or equal to $x^2$.Step 1: Analyze the function $\left\lfloor x^2 \right\rfloor$ for $x \in [0, \sqrt{2}]$:- For $x \in [0, 1)$, $x^2 \in [0, 1)$, so $\left\lfloor x^2 \right\rfloor = 0$.- For $x \in [1, \sqrt{2})$, $x^2 \in [1, 2)$, so $\left\lfloor x^2 \right\rfloor = 1$.- At $x = \sqrt{2}$, $x^2 = 2$, and $\left\lfloor 2 \right\rfloor = 2$, but the integral does not consider this single point.Step 2: Break the integral into two parts: $$I = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx$$Step 3: Evaluate the integrals: $$I = 0 + \int_1^{\sqrt{2}} 1 \, dx = \left[ x \right]_1^{\sqrt{2}} = \sqrt{2} - 1$$Thus, the value of the integral is $\sqrt{2} - 1$.Thus, the correct answer is option (C).Quick Tip: When dealing with integrals involving the greatest integer function, divide the integral based on the intervals where the greatest integer function remains constant. In this case, it was split at $x = 1$, where the value of $\left\lfloor x^2 \right\rfloor$ changed.