Step-by-step Solution

1. Analyze the function f(x) and its derivative.

The given polynomial is $f(x) = 1 + 2x + 3x^2 + 4x^3.$ To understand the behavior of the function, we find its derivative: $f'(x) = {d \over dx}(1 + 2x + 3x^2 + 4x^3) = 2 + 6x + 12x^2.$ To determine the sign of f'(x), we can check the discriminant of the quadratic expression $12x^2 + 6x + 2$. The discriminant is $Δ = b^2 - 4ac = 6^2 - 4(12)(2) = 36 - 96 = -60.$ Since the leading coefficient (12) is positive and the discriminant is negative, the quadratic f'(x) is always positive for all real values of x. This implies that f(x) is a strictly increasing function.

2. Find the real roots of f(x).

Since f(x) is a strictly increasing function, it can cross the x-axis at most once. Therefore, f(x) has exactly one real root. Let's call this root α. We can locate this root by evaluating f(x) at some points:

• $f(0) = 1 + 2(0) + 3(0)^2 + 4(0)^3 = 1$
• $f(-1) = 1 + 2(-1) + 3(-1)^2 + 4(-1)^3 = 1 - 2 + 3 - 4 = -2$ Since f(-1) < 0 and f(0) > 0, and f(x) is continuous, by the Intermediate Value Theorem, the real root α must lie in the interval (-1, 0).

3. Determine s and t.

s is the sum of all distinct real roots of f(x). Since there is only one real root α, we have s = α. t is defined as t = |s| = |α|. Since -1 < α < 0, α is negative. Therefore, t = -α. This means 0 < t < 1.

4. Calculate the required area.

The area is bounded by y = f(x), x = 0, y = 0, and x = t. Since 0 < t < 1, we are considering the area in the first quadrant. For x in the interval [0, t], we have x > α, and since f(x) is increasing, f(x) > f(α) = 0. So, the curve is above the x-axis. The area A is given by the definite integral: $A = \int_0^t f(x) \, dx = \int_0^t (1 + 2x + 3x^2 + 4x^3) \, dx$ Evaluating the integral: $A = \left[ x + x^2 + x^3 + x^4 \right]_0^t = (t + t^2 + t^3 + t^4) - 0 = t + t^2 + t^3 + t^4.$

5. Find the value of t.

Since α is a root of f(x), we have f(α) = 0. Substituting α = -t: $f(-t) = 1 + 2(-t) + 3(-t)^2 + 4(-t)^3 = 0$ $1 - 2t + 3t^2 - 4t^3 = 0$ Let $g(t) = 1 - 2t + 3t^2 - 4t^3$. We need to find the root of g(t)=0 which we know is in (0, 1). Let's estimate the value of t by testing some values:

• g(1/2) = 1 - 2(1/2) + 3(1/4) - 4(1/8) = 1 - 1 + 3/4 - 1/2 = 1/4 > 0
• g(3/4) = 1 - 2(3/4) + 3(9/16) - 4(27/64) = 1 - 3/2 + 27/16 - 27/16 = -1/2 < 0 Since g(1/2) > 0 and g(3/4) < 0, the root t must lie between 1/2 and 3/4. So, $1/2 < t < 3/4.$

6. Find the interval for the area A.

The area is $A(t) = t + t^2 + t^3 + t^4$. The derivative of the area function is $A'(t) = 1 + 2t + 3t^2 + 4t^3$. For t > 0, A'(t) is clearly positive, so A(t) is an increasing function. Since 1/2 < t < 3/4, we can find the bounds for A: $A(1/2) < A(t) < A(3/4)$

• Lower bound: $A(1/2) = {1 \over 2} + \left({1 \over 2}\right)^2 + \left({1 \over 2}\right)^3 + \left({1 \over 2}\right)^4 = {1 \over 2} + {1 \over 4} + {1 \over 8} + {1 \over 16} = {8+4+2+1 \over 16} = {15 \over 16}$
• Upper bound: $A(3/4) = {3 \over 4} + \left({3 \over 4}\right)^2 + \left({3 \over 4}\right)^3 + \left({3 \over 4}\right)^4 = {3 \over 4} + {9 \over 16} + {27 \over 64} + {81 \over 256}$ $A(3/4) = {3 \cdot 64 + 9 \cdot 16 + 27 \cdot 4 + 81 \over 256} = {192 + 144 + 108 + 81 \over 256} = {525 \over 256}$ So, the area A lies in the interval $\left({15 \over 16}, {525 \over 256}\right).$

7. Compare with the given options.

Let's convert the bounds to decimals: 15/16 = 0.9375 and 525/256 ≈ 2.05. So, the area A is in the interval (0.9375, 2.05). Now let's examine the options:

• A: (3/4, 3) = (0.75, 3)
• B: (21/64, 11/16) = (0.328125, 0.6875)
• C: (9, 10)
• D: (0, 21/64) = (0, 0.328125)

The interval we calculated, (0.9375, 2.05), is completely contained within the interval (0.75, 3). None of the other option intervals contain our calculated interval. Therefore, the correct option is A.