JEE Advanced · Mathematics · 25. AOD
Paragraph:
Consider the polynomial \(f(x)=1+2 x+3 x^2+4 x^3\). Let \(s\) be the sum of all distinct real roots of \(f(x)\) and let \(t=|s|\).Question:
The real number \(s\) lies in the interval
- A
\(\left(-\frac{1}{4}, 0\right)\)
- B
\(\left(-11,-\frac{3}{4}\right)\)
- C
\(\left(-\frac{3}{4},-\frac{1}{2}\right)\)
- D
\(\left(0, \frac{1}{4}\right)\)
Answer & Solution
Correct Answer
(C)
\(\left(-\frac{3}{4},-\frac{1}{2}\right)\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=4 x^3+3 x^2+2 x+1\)
\[
\begin{aligned}
& f^{\prime}(x)=2\left(6 x^2+3 x+1\right) \\
& D=9-24 < 0
\end{aligned}
\]
Hence, \(f(x)=0\) has only one real root.
\[
\begin{aligned}
f\left(-\frac{1}{2}\right) & =1-1+\frac{3}{4}-\frac{4}{8}>0 \\
f\left(-\frac{3}{4}\right) & =1-\frac{6}{4}+\frac{27}{16}-\frac{108}{64} \\
& =\frac{64-96+108-108}{64} < 0
\end{aligned}
\]
\(f(x)\) changes its sign in \(\left(-\frac{3}{4}, \frac{-1}{2}\right)\), hence \(f(x)=0\) has a root in \(\left(\frac{-3}{4}, \frac{-1}{2}\right)\).
\[
\begin{aligned}
& f^{\prime}(x)=2\left(6 x^2+3 x+1\right) \\
& D=9-24 < 0
\end{aligned}
\]
Hence, \(f(x)=0\) has only one real root.
\[
\begin{aligned}
f\left(-\frac{1}{2}\right) & =1-1+\frac{3}{4}-\frac{4}{8}>0 \\
f\left(-\frac{3}{4}\right) & =1-\frac{6}{4}+\frac{27}{16}-\frac{108}{64} \\
& =\frac{64-96+108-108}{64} < 0
\end{aligned}
\]
\(f(x)\) changes its sign in \(\left(-\frac{3}{4}, \frac{-1}{2}\right)\), hence \(f(x)=0\) has a root in \(\left(\frac{-3}{4}, \frac{-1}{2}\right)\).
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