Kerala KEAM 5 June 2024 Solved Paper

We are asked to evaluate the limit: $$L = \lim_{x \to 0^-} \frac{x \left( \left\lfloor x \right\rfloor + | x | \right)}{ | x | }$$Step 1: Consider the behavior of $\left\lfloor x \right\rfloor$ and $| x |$ as $x \to 0^-$:- As $x$ approaches 0 from the left, $x$ is negative, so $| x | = -x$.- For $x \in (-1, 0)$, $\left\lfloor x \right\rfloor = -1$, since $\left\lfloor x \right\rfloor$ is the greatest integer less than or equal to $x$.Step 2: Substitute these values into the expression: $$L = \lim_{x \to 0^-} \frac{x \left( -1 + (-x) \right)}{-x}$$Simplify the expression: $$L = \lim_{x \to 0^-} \frac{x \left( -1 - x \right)}{-x} = \lim_{x \to 0^-} \frac{x(-1 - x)}{-x}$$ $$L = \lim_{x \to 0^-} \left( 1 + x \right)$$Step 3: Now, take the limit as $x \to 0^-$: $$L = 1 + 0 = 1$$Thus, the value of the limit is $1$.Therefore, the correct answer is option (D).Quick Tip: When dealing with floor functions in limits, carefully examine the behavior of $\left\lfloor x \right\rfloor$ as $x$ approaches the desired point. For negative values of $x$, $\left\lfloor x \right\rfloor$ is the greatest integer less than or equal to $x$, and $| x | = -x$.