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# If $a,b,c \in R$ and satisfy $3a + 5b + 15c = 0$, the equation $a{x^4} + b{x^2} + c = 0$ hasA.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$\left( { - 1,1} \right)$

Last updated date: 13th Jun 2024
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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.

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.