Question:

Step 1: Set up the differential inequality. We are given $P'(x) + (x-1)^3 \geq P(x) + 1$, i.e., $P'(x) - P(x) \geq 1 - (x-1)^3$. Also $P(x) \leq x^3 + 3x + 1$ and $P(0) = 1$. Step 2: Check if $P(x) = x^3 + 3x + 1$ satisfies the equality. Let $Q(x) = x^3 + 3x + 1$. Then $Q'(x) = 3x^2 + 3$, $Q(0) = 1$. Check: $Q'(x) + (x-1)^3 = 3x^2 + 3 + x^3 - 3x^2 + 3x - 1 = x^3 + 3x + 2$. And $Q(x) + 1 = x^3 + 3x + 2$. So equality holds: $Q'(x) + (x-1)^3 = Q(x) + 1$. Since $P(0) = Q(0) = 1$ and $P(x) \leq Q(x)$, by the differential inequality and equality condition, $P(x) = x^3 + 3x + 1$. Step 3: Verify monotonicity. $P'(x) = 3x^2 + 3 > 0$ for all $x \in [0,\infty)$. So $y = P(x)$ is strictly increasing (monotonic). Option (a) is correct. Step 4: Compute the area bounded by $y = P(x)$, $x$-axis, $x=0$ and $x=1$. Since $P(x) = x^3 + 3x + 1 > 0$ on $[0,1]$: $$\text{Area} = \int_0^1 (x^3 + 3x + 1)\,dx = \left[\frac{x^4}{4} + \frac{3x^2}{2} + x\right]_0^1 = \frac{1}{4} + \frac{3}{2} + 1 = \frac{1}{4} + \frac{6}{4} + \frac{4}{4} = \frac{11}{4}$$ Option (b) is correct. Step 5: Check option (c). $P(x) = x^3 + 3x + 1$ is defined on $[0,\infty)$, so $\int_{-1}^{1} P(x)\,dx$ requires $P$ on $[-1,0)$ which is not given. Option (c) cannot be confirmed. Step 6: Check bijectivity. $P(x) = x^3 + 3x + 1$ on $[0,\infty)$ is strictly increasing (injective) and maps $[0,\infty) \to [1,\infty)$, not onto $\mathbb{R}$, so it is not bijective onto $\mathbb{R}$. Option (d) is incorrect.