
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
465k+ views
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.
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.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
