Consider the polynomial \[f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}\]. Let \[s\] be the sum of all distinct real roots of \[f\left( x \right)\] and let \[t = \left| s \right|\]. The area bounded by the curve \[y = f\left( x \right)\] and the lines \[x = 0,y = 0\] and \[x = t\], lies in the interval
A. \[\left( {\dfrac{3}{4},3} \right)\]
B. \[\left( {\dfrac{{15}}{{16}},\dfrac{{525}}{{256}}} \right)\]
C. \[\left( {9,10} \right)\]
D. \[\left( {0,\dfrac{{21}}{{64}}} \right)\]

Hint: First of all, find the first derivative of the given function to have the values of \[s\] and which will lead to get the values of \[t\]. From these intervals we can obtain the area bounded by the given curve lies in which interval.

Complete step-by-step answer:
Given polynomial is \[f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}\].
Now consider the first derivative of \[f\left( x \right)\]
\[
   \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {1 + 2x + 3{x^2} + 4{x^3}} \right) \\
   \Rightarrow f'\left( x \right) = 2 + 6x + 12{x^2} > 0{\text{ for all }}x{\text{ in }}R \\
\]
Thus, \[f\left( x \right)\] is an increasing function on \[R\]. So, \[f\left( x \right)\] can have at most one root. It is clear that \[f\left( x \right)\] cannot have a positive real root.
We have \[f\left( {\dfrac{{ - 3}}{4}} \right) = 1 - \dfrac{3}{2} + \dfrac{{27}}{{16}} - \dfrac{{27}}{{16}} = - \dfrac{1}{2} < 0\]
And also, we have \[f\left( {\dfrac{{ - 1}}{2}} \right) = 1 - 1 + \dfrac{3}{4} - \dfrac{1}{2} = \dfrac{1}{4} > 0\]
Since, \[s\] is the sum of all distinct real roots of \[f\left( x \right)\] we have \[ - \dfrac{3}{4} < s < - \dfrac{1}{2}\]
Given that \[t = \left| s \right|\]. So, we have \[\dfrac{1}{2} < t < \dfrac{3}{4}\]
Now we have to calculate the area bounded by the curve \[y = f\left( x \right)\] and the lines \[x = 0,y = 0\] and \[x = t\]. So, we have
\[
   \Rightarrow \int\limits_0^{\dfrac{1}{2}} {f\left( x \right) < area < } \int\limits_0^{\dfrac{3}{4}} {f\left( x \right)} \\
   \Rightarrow \int\limits_0^{\dfrac{1}{2}} {\left( {4{x^3} + 3{x^2} + 2x + 1} \right)dx < area < } \int\limits_0^{\dfrac{3}{4}} {\left( {4{x^3} + 3{x^2} + 2x + 1} \right)dx} \\
   \Rightarrow \left[ {{x^4} + {x^3} + {x^2} + x} \right]_0^{\dfrac{1}{2}} < area < \left[ {{x^4} + {x^3} + {x^2} + x} \right]_0^{\dfrac{3}{4}} \\
   \Rightarrow \dfrac{1}{{16}} + \dfrac{1}{8} + \dfrac{1}{4} + \dfrac{1}{2} < area < \dfrac{{81}}{{256}} + \dfrac{{27}}{{64}} + \dfrac{9}{{16}} + \dfrac{3}{4} \\
  \therefore \dfrac{{15}}{{16}} < area < \dfrac{{525}}{{256}} \\
\]
Thus, the correct option is B. \[\left( {\dfrac{{15}}{{16}},\dfrac{{525}}{{256}}} \right)\]

Note: If \[a < x < b\] then \[ - b < \left| x \right| < - a\]. A function is said to be increasing when y-value increases as the x-value increases and a function is said to be decreasing when y-value decreases as the x-value decreases.