Question

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. $\left( {0,\dfrac{{21}}{{64}}} \right)$

Answer 1

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.