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…

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
  1. $\left(-\frac{1}{4}, 0\right)$
  2. $\left(-11,-\frac{3}{4}\right)$
  3. $\left(-\frac{3}{4},-\frac{1}{2}\right)$
  4. $\left(0, \frac{1}{4}\right)$

Solution

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 $\left(\frac{-3}{4}, \frac{-1}{2}\right)$

Asked in: JEE Advanced 2010 (Paper 2)

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