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 function $f^{\prime}(x)$ is

(B) decreasing in $\left(-\mathrm{t},-\frac{1}{4}\right)$ and increasing in $\left(-\frac{1}{4}, \mathrm{t}\right)$

(A) increasing in $\left(-\mathrm{t},-\frac{1}{4}\right)$ and decreasing in $\left(-\frac{1}{4}, \mathrm{t}\right)$

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Application of Derivatives

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 function

f^{'} (x)

(A)increasing in $(- t, - \frac{1}{4})$ and decreasing in $(- \frac{1}{4}, t)$
(B)decreasing in $(- t, - \frac{1}{4})$ and increasing in $(- \frac{1}{4}, t)$ ✓ Correct
(C)increasing in (-t, t)
(D)decreasing in (-t, t)

Answer: (B)

We have

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

.

To determine where

f^{'} (x)

is increasing or decreasing, we compute

f^{''} (x)

f^{''} (x) = 6 + 24 x

Setting

f^{''} (x) = 0

6 + 24 x = 0 ⟹ x = - \frac{1}{4}

For

x < - \frac{1}{4}

f^{''} (x) < 0

, so

f^{'} (x)

is decreasing.

For

x > - \frac{1}{4}

f^{''} (x) > 0

, so

f^{'} (x)

is increasing.

Since

t \in (\frac{1}{2}, \frac{3}{4})

, the interval

(- t, t)

contains

- \frac{1}{4}

.

Therefore,

f^{'} (x)

is decreasing in

(- t, - \frac{1}{4})

and increasing in

(- \frac{1}{4}, t)

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