Question

If \[a,b,c \in R\] and satisfy \[3a + 5b + 15c = 0\], the equation \[a{x^4} + b{x^2} + c = 0\] has
A.At least one root in\[\left( { - 1,0} \right)\]
B.At least one root in\[\left( {0,1} \right)\]
C.At least two roots in\[\left( { - 1,1} \right)\]
D.No root in\[\left( { - 1,1} \right)\]

Answer 1

Hint: Here we will first find a function by integrating the equation. Then we will find the value of the function at different point of \[x\]. We will find the value of the differentiation of the function. Then we will apply the Rolle’s Theorem to find the roots in those intervals.

Complete step-by-step answer:
Given equation is \[a{x^4} + b{x^2} + c = 0\] ……………….\[\left( 1 \right)\].
It satisfies the equation \[3a + 5b + 15c = 0\]………………\[\left( 2 \right)\].
Firstly, we will integrate the equation \[\left( 1 \right)\] to get the value of the function \[f\left( x \right)\]. So, we get
\[f\left( x \right) = \int {\left( {a{x^4} + b{x^2} + c} \right)} dx\]
After integrating the terms, we get
\[ \Rightarrow f\left( x \right) = \dfrac{{a{x^5}}}{5} + \dfrac{{b{x^3}}}{3} + cx\]
On taking LCM, we get
\[ \Rightarrow f\left( x \right) = \dfrac{{3a{x^5} + 5b{x^3} + 15cx}}{{15}}\]
Since the function \[f\left( x \right)\] is the polynomial function. So, it must be continuous and differentiable at every point. Now we will find the value of the function at \[x = 0,x = 1,x = - 1\]. Therefore, we get
\[f\left( 0 \right) = \dfrac{{0 + 0 + 0}}{{15}} = 0\]……………..\[\left( 3 \right)\]
\[f\left( 1 \right) = \dfrac{{3a + 5b + 15c}}{{15}}\]
As from the equation \[\left( 2 \right)\] we know that \[3a + 5b + 15c = 0\]. So, we get
\[f\left( 1 \right) = \dfrac{{3a + 5b + 15c}}{{15}} = 0\]……………..\[\left( 4 \right)\]
\[f\left( { - 1} \right) = \dfrac{{ - 3a - 5b - 15c}}{{15}} = \dfrac{{ - \left( {3a + 5b + 15c} \right)}}{{15}} = 0\]……………..\[\left( 5 \right)\]
Now differentiating this function \[f\left( x \right)\], we get
\[\begin{array}{l}f'\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{{a{x^5}}}{5} + \dfrac{{b{x^3}}}{3} + cx} \right)\\ \Rightarrow f'\left( x \right) = a{x^4} + b{x^2} + c\end{array}\]
From the equation \[\left( 1 \right)\] we know that \[a{x^4} + b{x^2} + c = 0\]. Therefore,
\[ \Rightarrow f'\left( x \right) = 0\]……………..\[\left( 6 \right)\]
So according to Rolle ’s Theorem for the equation \[\left( 3 \right)\], equation \[\left( 4 \right)\]and equation \[\left( 6 \right)\]. We can say that there exists \[x \in \left( {0,1} \right)\].
Similarly according to Rolle ’s Theorem for the equation \[\left( 3 \right)\], equation \[\left( 4 \right)\] and equation \[\left( 6 \right)\]. We can say that there exists \[x \in \left( { - 1,0} \right)\].
Therefore, the equation will have at least two roots in \[\left( { - 1,1} \right)\].
So, option C is the correct option.

Note: Here we should know the Rolle’s Theorem to find the intervals of the root.
So Rolle’s Theorem states that if a function is continuous in the interval \[\left[ {a,b} \right]\] and differentiable in the interval \[\left( {a,b} \right)\] such that \[f\left( a \right) = f\left( b \right)\] then \[f'\left( x \right) = 0\] for \[a \le x \le b\].
We need to remember that if the differentiation of the function is zero, then there is a root of that function in that interval.