Question:

Step 1: For $f(x)$ to be injective (one-one) on $\mathbb{R}$, it must be strictly monotonic, so $f'(x) \geq 0$ or $f'(x) \leq 0$ for all $x \in \mathbb{R}$. $$f'(x) = 3x^2 + 6x + (4-k)$$ Step 2: Since the leading coefficient of $f'(x)$ is positive, we need $f'(x) \geq 0$ for all $x$. This requires the discriminant to be $\leq 0$. $$\Delta = 36 - 12(4-k) \leq 0$$ $$36 - 48 + 12k \leq 0$$ $$12k \leq 12$$ $$k \leq 1$$ Step 3: Identify the positive integral values of $k$. Since $k \leq 1$, the only positive integer satisfying this is $k = 1$. So the number of positive integral values is 1, and the maximum positive integral value is 1. Step 4: Identify non-negative integral values of $k$. Non-negative integers satisfying $k \leq 1$ are $k = 0$ and $k = 1$, giving 2 values. So option (d) is incorrect. Step 5: Check each option: - (a) Maximum positive integral value of $k$ is 1. ✓ (True) - (b) Minimum positive integral value of $k$ is 1. ✓ (True, since $k=1$ is the only positive integer) - (c) Number of positive integral values of $k$ is 1. ✓ (True) - (d) Number of non-negative integral values of $k$ is 1. ✗ (False, there are 2: $k=0,1$) Thus correct options are (a), (b), (c).