Let f be a function defined on $\mathbf{R}$ (the set of all real numbers) such that $\mathrm{f}^{\prime}(\mathrm{x})=2010(\mathrm{x}-2009)(\mathrm{x}-2010)^{2}(\mathrm{x}-2011)^{3}(\mathrm{x}-2012)^{4}$, for all $\mathrm{x} \in \mathbf{R}$. If $g$ is a function defined on $\mathbf{R}$ with values in the interval $(0, \infty)$ such that \[ \mathrm{f}(\mathrm{x})=\ell n(\mathrm{~g}(\mathrm{x})), \text { for all } \mathrm{x} \in \mathbf{R}, \] then the number of points in $\mathbf{R}$ at which $g$ has a local maximum is

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

Let f be a function defined on

R

(the set of all real numbers) such that

f^{'} (x) = 2010 (x - 2009) (x - 2010)^{2} (x - 2011)^{3} (x - 2012)^{4}

, for all

x \in R

. If

g

is a function defined on

R

with values in the interval

(0, \infty)

such that

f (x) = ℓ n (g (x)), for all x \in R,

then the number of points in

R

at which

g

has a local maximum is

Answer: 1

Given

f (x) = ln (g (x))

, we have:

g (x) = e^{f (x)}

Differentiating:

g^{'} (x) = e^{f (x)} \cdot f^{'} (x)

Since

e^{f (x)} > 0

for all

x

, the sign of

g^{'} (x)

is determined by the sign of

f^{'} (x)

.

The critical points of

g

occur where

f^{'} (x) = 0

2010 (x - 2009) (x - 2010)^{2} (x - 2011)^{3} (x - 2012)^{4} = 0

This gives

x = 2009, 2010, 2011, 2012

.

To determine local maxima, we check the sign change of

f^{'} (x)

(and hence

g^{'} (x)

) around each critical point:

At $x = 2009$ : The factor $(x - 2009)$ changes sign (power 1), so $f^{'} (x)$ changes from negative to positive → local minimum of $g$ .
At $x = 2010$ : The factor $(x - 2010)^{2}$ does not change sign (even power) → not an extremum.
At $x = 2011$ : The factor $(x - 2011)^{3}$ changes sign (odd power), and considering all factors, $f^{'} (x)$ changes from positive to negative → local maximum of $g$ .
At $x = 2012$ : The factor $(x - 2012)^{4}$ does not change sign (even power) → not an extremum.

Therefore,

g

has exactly 1 local maximum.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Integer answer