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If \[{{a}_{0}},{{a}_{1}},{{a}_{2}},{{a}_{3}}\] are positive, then $4{{a}_{0}}{{x}^{3}}+3{{a}_{1}}{{x}^{2}}+2{{a}_{2}}x+{{a}_{3}}=0$ has at least one root in (-1,0) if:
A. ${{a}_{0}}+{{a}_{2}}={{a}_{1}}+{{a}_{3}}\text{ and 4}{{a}_{0}}+2{{a}_{2}}>3{{a}_{1}}+{{a}_{3}}$
B. $4{{a}_{0}}+2{{a}_{2}}<3{{a}_{1}}+{{a}_{3}}$
C. $4{{a}_{0}}+2{{a}_{2}}=3{{a}_{1}}+{{a}_{3}}\text{ and }{{a}_{0}}+{{a}_{2}}<{{a}_{1}}+{{a}_{3}}$
D. None of these

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Last updated date: 20th Jun 2024
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Answer
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Hint: We have been given that \[{{a}_{0}},{{a}_{1}},{{a}_{2}},{{a}_{3}}\] are positive numbers. For $4{{a}_{0}}{{x}^{3}}+3{{a}_{1}}{{x}^{2}}+2{{a}_{2}}x+{{a}_{3}}=0$ have at least one root in $\left( -1,0 \right)$, it should follow Rolle’s theorem according to which if $f\left( a \right)=f\left( b \right)$ where $f\left( x \right)$ is a continuous and differentiable function in $\left[ a,b \right]$ then there exists one value of ‘c’ in $\left[ a,b \right]$ such that $f'\left( c \right)=0$ . Here, we will take the given equation as $f'\left( x \right)$. Thus, to solve this question, we will keep the integral of the given equation equal for -1 and 0. Also, for any polynomial $p\left( x \right)$ to have at least one root in $\left( a,b \right)$, the value of $p\left( a \right)p\left( b \right)$ should be less than 0. Hence, the product of the values of the integral of the given equation at -1 and 0 should be less than 0. We will use both these properties and find our required condition. Hence, we will have our answer.

Complete step-by-step solution:
Now, we have been given the equation $4{{a}_{0}}{{x}^{3}}+3{{a}_{1}}{{x}^{2}}+2{{a}_{2}}x+{{a}_{3}}=0$
Let this equation be $f'\left( x \right)$.
Therefore, $f\left( x \right)$ is given by:
$f\left( x \right)=\int{f'\left( x \right)}$
Putting the value of $f'\left( x \right)$in the above equation, we will get the value of $f\left( x \right)$
Thus, $f\left( x \right)$ is given as:
$\begin{align}
  & f\left( x \right)=\int{\left( 4{{a}_{0}}{{x}^{3}}+3{{a}_{1}}{{x}^{2}}+2{{a}_{2}}x+{{a}_{3}}=0 \right).dx} \\
 & \Rightarrow f\left( x \right)=4{{a}_{0}}\left[ \dfrac{{{x}^{4}}}{4} \right]+3{{a}_{1}}\left[ \dfrac{{{x}^{3}}}{3} \right]+2{{a}_{2}}\left[ \dfrac{{{x}^{2}}}{2} \right]+{{a}_{3}}\left[ x \right]+C \\
 & \Rightarrow f\left( x \right)={{a}_{0}}{{x}^{4}}+{{a}_{1}}{{x}^{3}}+{{a}_{2}}{{x}^{2}}+{{a}_{3}}x+C \\
\end{align}$
Here, ‘C’ is the constant of integration.
 Now, $f\left( x \right)$ is a polynomial function hence it is both continuous and differentiable in $\left[ -1,0 \right]$
For $f'\left( x \right)=0$ in $\left[ -1,0 \right]$, $f\left( -1 \right)=f\left( 0 \right)$
Hence, putting after putting in the values, we get:
$\begin{align}
  & {{a}_{0}}{{\left( -1 \right)}^{4}}+{{a}_{1}}{{\left( -1 \right)}^{3}}+{{a}_{2}}{{\left( -1 \right)}^{2}}+{{a}_{3}}\left( -1 \right)+C={{a}_{0}}{{\left( 0 \right)}^{4}}+{{a}_{1}}{{\left( 0 \right)}^{3}}+{{a}_{2}}{{\left( 0 \right)}^{2}}+{{a}_{3}}\left( 0 \right)+C \\
 & \Rightarrow {{a}_{0}}-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}+C=0+0+0+0+C \\
 & \Rightarrow {{a}_{0}}-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}=0 \\
 & \Rightarrow {{a}_{0}}+{{a}_{2}}={{a}_{1}}+{{a}_{3}} \\
\end{align}$
Thus, we have got out first required condition.
Now, for $f'\left( x \right)$ to have one root in $\left( -1,0 \right)$ , $f'\left( -1 \right)f'\left( 0 \right)<0$
Applying putting in the values in the above condition, we get:
$\begin{align}
  & f'\left( -1 \right)f\left( 0 \right)<0 \\
 & \Rightarrow \left( 4{{a}_{0}}{{\left( -1 \right)}^{3}}+3{{a}_{1}}{{\left( -1 \right)}^{2}}+2{{a}_{2}}\left( -1 \right)+{{a}_{3}} \right)\left( 4{{a}_{0}}{{\left( 0 \right)}^{3}}+3{{a}_{1}}{{\left( 0 \right)}^{3}}+2{{a}_{2}}\left( 0 \right)+{{a}_{3}} \right)<0 \\
 & \Rightarrow \left( -4{{a}_{0}}+3{{a}_{1}}-2{{a}_{2}}+{{a}_{3}} \right)\left( 0+0+0+{{a}_{3}} \right)<0 \\
 & \Rightarrow \left( -4{{a}_{0}}+3{{a}_{1}}-2{{a}_{2}}+{{a}_{3}} \right){{a}_{3}}<0 \\
\end{align}$
Now, we have been given in the question that \[{{a}_{0}},{{a}_{1}},{{a}_{2}},{{a}_{3}}\] are positive numbers. Thus we can say that:
$\begin{align}
  & -4{{a}_{0}}+3{{a}_{1}}-2{{a}_{2}}+{{a}_{3}}<0 \\
 & \Rightarrow 3{{a}_{1}}+{{a}_{3}}<4{{a}_{1}}+2{{a}_{2}} \\
\end{align}$
Thus, we have our second required condition.
Thus, option (A) is the correct option.

Note: Be careful while integrating the equation and don’t forget to include the constant of integration ‘C’. Also, Rolle’s theorem is only applicable if the function is both continuous and differentiable in the given range. Since we have been given a polynomial function, its integral will also be a polynomial and polynomial functions are continuously differentiable everywhere.