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 $\mathrm{f}(\mathrm{x})$ and let $\mathrm{t}=|\mathrm{s}|$. The real number s lies in the interval

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

(A) $\left(-\frac{1}{4}, 0\right)$

(B) $\left(-11,-\frac{3}{4}\right)$

(D) $\left(0, \frac{1}{4}\right)$

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Quadratic Equations and Inequalities

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 ∣

.

The real number s lies in the interval

Answer: (C)

Consider

f (x) = 1 + 2 x + 3 x^{2} + 4 x^{3}

.

Evaluating at test points:

f (- \frac{1}{2}) = 1 + 2 (- \frac{1}{2}) + 3 (\frac{1}{4}) + 4 (- \frac{1}{8}) = 1 - 1 + \frac{3}{4} - \frac{1}{2} = \frac{1}{4} > 0

f (- \frac{3}{4}) = 1 + 2 (- \frac{3}{4}) + 3 (\frac{9}{16}) + 4 (- \frac{27}{64})

= 1 - \frac{3}{2} + \frac{27}{16} - \frac{27}{16} = 1 - \frac{3}{2} = - \frac{1}{2} < 0

Since

f (- \frac{3}{4}) < 0

and

f (- \frac{1}{2}) > 0

, by the Intermediate Value Theorem, there is a root in

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

.

Since

f^{'} (x) = 2 + 6 x + 12 x^{2}

has discriminant

36 - 96 = - 60 < 0

f^{'} (x) > 0

for all

x

, so

f

is strictly increasing and has exactly one real root.

Therefore,

s

(the sum of all distinct real roots, which is just the single root) lies in

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

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option