Question

Consider the polynomial, $f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}$. Let s be the sum of all the distinct real roots of f (x), the real number of s lies in the interval.
$\left( a \right)\left( {\dfrac{{ - 1}}{4},0} \right)$
$\left( b \right)\left( { - 11,\dfrac{{ - 3}}{4}} \right)$
$\left( c \right)\left( {\dfrac{{ - 3}}{4},\dfrac{{ - 1}}{2}} \right)$
$\left( d \right)\left( {0,\dfrac{1}{4}} \right)$

Answer 1

In this particular question use the concept that after differentiating the given function if the function is always positive for every value of x then it is a strictly increasing function, so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given data:
$f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}$
Now differentiate the above equation w.r.t x we have,
$\Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {1 + 2x + 3{x^2} + 4{x^3}} \right)$
Now as we know that $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ so we have,
$\Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \left( {0 + 2 + 6x + 12{x^2}} \right)$
$\Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = 2\left( {1 + 3x + 6{x^2}} \right)$
Now as we know that $6{x^2} > 3x$
$\Rightarrow 2\left( {1 + 3x + 6{x^2}} \right) > 0$
$\Rightarrow f'\left( x \right) > 0$
So, f (x) is a strictly increasing function.
Now as we see that f’(x) is a quadratic equation so the discriminant (D) of the quadratic equation is given as,
$D = \sqrt {{b^2} - 4ac}$, where a = 6, b = 3, c = 1.
$\Rightarrow D = \sqrt {{3^2} - 4\left( 6 \right)\left( 1 \right)} = \sqrt {9 - 24} = \sqrt { - 15}$ , so as we see that the discriminant is complex so two of the roots of the cubic equation is complex and one of the root is real.
So the given cubic equation has only one real root.
Now as we know that if the root of the given equation is lies in the interval say (a, b), then the value of f (x) is greater than zero at one of the extremities of the interval and less than zero at the other extremities of the interval so that the given equation changes its sign than there is a root of the equation in that interval otherwise not.
Now check option (a)
$\Rightarrow f\left( {\dfrac{{ - 1}}{4}} \right) = 1 + 2\left( {\dfrac{{ - 1}}{4}} \right) + 3{\left( {\dfrac{{ - 1}}{4}} \right)^2} + 4{\left( {\dfrac{{ - 1}}{4}} \right)^3} = 1 - \dfrac{1}{2} + \dfrac{3}{{16}} - \dfrac{1}{{16}} = \dfrac{5}{8} > 0$
$\Rightarrow f\left( 0 \right) = 1 + 2\left( 0 \right) + 3{\left( 0 \right)^2} + 4{\left( 0 \right)^3} = 1 > 0$
So at the both value function is greater than zero so the function does not have any root in this interval.
Now check option (b)
$\Rightarrow f\left( { - 11} \right) = 1 + 2\left( { - 11} \right) + 3{\left( { - 11} \right)^2} + 4{\left( { - 11} \right)^3} = 1 - 22 + 363 - 5324 = - 4982 < 0$
$\Rightarrow f\left( {\dfrac{{ - 3}}{4}} \right) = 1 + 2\left( {\dfrac{{ - 3}}{4}} \right) + 3{\left( {\dfrac{{ - 3}}{4}} \right)^2} + 4{\left( {\dfrac{{ - 3}}{4}} \right)^3} = 1 - \dfrac{3}{2} + \dfrac{{27}}{{16}} - \dfrac{{27}}{{16}} = - \dfrac{1}{2} < 0$
So at both values the function is less than zero so the function does not have any root in this interval.
Now check option (c)
$\Rightarrow f\left( {\dfrac{{ - 3}}{4}} \right) = 1 + 2\left( {\dfrac{{ - 3}}{4}} \right) + 3{\left( {\dfrac{{ - 3}}{4}} \right)^2} + 4{\left( {\dfrac{{ - 3}}{4}} \right)^3} = 1 - \dfrac{3}{2} + \dfrac{{27}}{{16}} - \dfrac{{27}}{{16}} = - \dfrac{1}{2} < 0$
$\Rightarrow f\left( {\dfrac{{ - 1}}{2}} \right) = 1 + 2\left( {\dfrac{{ - 1}}{2}} \right) + 3{\left( {\dfrac{{ - 1}}{2}} \right)^2} + 4{\left( {\dfrac{{ - 1}}{2}} \right)^3} = 1 - 1 + \dfrac{3}{4} - \dfrac{1}{2} = \dfrac{1}{4} > 0$
So as we see that at one value function is less than zero and at other value function is greater than zero so there is a root of the given equation in that interval.
Now check option (d)
$\Rightarrow f\left( 0 \right) = 1 + 2\left( 0 \right) + 3{\left( 0 \right)^2} + 4{\left( 0 \right)^3} = 1 > 0$
$\Rightarrow f\left( {\dfrac{1}{4}} \right) = 1 + 2\left( {\dfrac{1}{4}} \right) + 3{\left( {\dfrac{1}{4}} \right)^2} + 4{\left( {\dfrac{1}{4}} \right)^3} = 1 + \dfrac{1}{2} + \dfrac{3}{{16}} + \dfrac{1}{{16}} = \dfrac{7}{4} > 0$
So at the both value function is greater than zero so the function does not have any root in this interval.
As the given equation has only one real root, so the sum of all the distinct real roots of f (x) lies in the interval, $\left( {\dfrac{{ - 3}}{4},\dfrac{{ - 1}}{2}} \right)$,
So this is the required answer.
Hence option (c) is the correct answer.

Note : Whenever we face such types of questions the key concept we have to remember is that if the root of the given equation is lies in the interval say (a, b), then the value of f (x) is greater than zero at one of the extremities of the interval and less than zero at the other extremities of the interval so that the given equation changes its sign than there is a root of the equation in that interval otherwise not.