Question

Let f be a function defined on R (the set of all real numbers) such that,\[f'\left( x \right) = 2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4}\], for all $x \in R$. If g is a function defined on R with values in the interval (0, $\infty $) such that f (x) = ln (g(x)), for all $x \in R$, then the number of points in R at which g has a local maximum is
$\left( a \right)$ 0
$\left( b \right)$ 1
$\left( c \right)$ 2
$\left( d \right)$ 3

Answer 1

Hint: In this particular question to find the maxima and minima differentiate the given function w.r.t x and equate to zero and solve for x, then again differentiate the given function and calculate its value on previous calculated x value if we got positive than it is a minima and if we got negative than it is a maxima so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
Let f be a function defined on R (the set of all real numbers) such that, \[f'\left( x \right) = 2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4}\]................... (1), for all $x \in R$.
Now the given function is
f (x) = ln (g(x)), for all $x \in R$.
Now take antilog on both sides we have,
$ \Rightarrow {e^{f\left( x \right)}} = g\left( x \right)$
Now differentiate the above equation w.r.t x and equate to zero we have,
$ \Rightarrow \dfrac{d}{{dx}}{e^{f\left( x \right)}} = \dfrac{d}{{dx}}g\left( x \right) = 0$
Now as we know that $\dfrac{d}{{dx}}{e^{f\left( x \right)}} = {e^{f\left( x \right)}}\dfrac{d}{{dx}}f\left( x \right)$ so we have,
$ \Rightarrow {e^{f\left( x \right)}}\dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}g\left( x \right) = 0$
$ \Rightarrow {e^{f\left( x \right)}}f'\left( x \right) = g'\left( x \right) = 0$............... (2)
Now substitute the value of f’ (x) from equation (1) in the above equation we have,
$ \Rightarrow {e^{f\left( x \right)}}2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} = 0$
So from the above equation we can say that $2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} = 0$ and ${e^{f\left( x \right)}} \ne 0$.
$ \Rightarrow 2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} = 0$
So from this the values of x are,
$ \Rightarrow x = 2009, 2010, 2011, 2012$
Now again differentiate the equation (2) we have
$ \Rightarrow \dfrac{d}{{dx}}g'\left( x \right) = \dfrac{d}{{dx}}\left[ {{e^{f\left( x \right)}}f'\left( x \right)} \right]$
$ \Rightarrow \dfrac{d}{{dx}}g'\left( x \right) = g''\left( x \right) = \left[ {{e^{f\left( x \right)}}\dfrac{d}{{dx}}f'\left( x \right) + f'\left( x \right)\dfrac{d}{{dx}}{e^{f\left( x \right)}}} \right]$
$ \Rightarrow \dfrac{d}{{dx}}g'\left( x \right) = g''\left( x \right) = \left[ {{e^{f\left( x \right)}}f''\left( x \right) + f'\left( x \right){e^{f\left( x \right)}}f'\left( x \right)} \right]$
$ \Rightarrow g''\left( x \right) = {e^{f\left( x \right)}}f''\left( x \right) + {\left[ {f'\left( x \right)} \right]^2}{e^{f\left( x \right)}}$............... (3)
Now differentiate equation (1) w.r.t x we have,
\[ \Rightarrow \dfrac{d}{{dx}}f'\left( x \right) = \dfrac{d}{{dx}}\left[ {2010\left( {x-2009} \right){{\left( {x - 2010} \right)}^2}{{\left( {x - 2011} \right)}^3}{{\left( {x - 2012} \right)}^4}} \right]\]
Now as we know that $\dfrac{d}{{dx}}abcd = a\dfrac{d}{{dx}}bcd + bcd\dfrac{d}{{dx}}a$, $\dfrac{d}{{dx}}bcd = b\dfrac{d}{{dx}}cd + cd\dfrac{d}{{dx}}b$, $\dfrac{d}{{dx}}cd = c\dfrac{d}{{dx}}d + d\dfrac{d}{{dx}}c$, $\dfrac{d}{{dx}}{\left( {f\left( x \right)} \right)^n} = n{\left( {f\left( x \right)} \right)^{n - 1}}\dfrac{d}{{dx}}f\left( x \right)$ so use this property in the above equation we have,
\[
   \Rightarrow \dfrac{d}{{dx}}f'\left( x \right) = 2010\left( {x-2009} \right)\dfrac{d}{{dx}}\left[ {{{\left( {x - 2010} \right)}^2}{{\left( {x - 2011} \right)}^3}{{\left( {x - 2012} \right)}^4}} \right] \\
   + 2010\left[ {{{\left( {x - 2010} \right)}^2}{{\left( {x - 2011} \right)}^3}{{\left( {x - 2012} \right)}^4}} \right]\dfrac{d}{{dx}}\left( {x - 2009} \right) \\
\]
\[ \Rightarrow f''\left( x \right) = 2010\left\{ {\left( {x-2009} \right)\dfrac{d}{{dx}}\left[ {{{\left( {x - 2010} \right)}^2}{{\left( {x - 2011} \right)}^3}{{\left( {x - 2012} \right)}^4}} \right] + \left[ {{{\left( {x - 2010} \right)}^2}{{\left( {x - 2011} \right)}^3}{{\left( {x - 2012} \right)}^4}} \right]} \right\}\]
\[
   \Rightarrow f''\left( x \right) = 2010{\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} + 2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}\dfrac{d}{{dx}}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} \\
   + 2010\left( {x-2009} \right){\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4}\dfrac{d}{{dx}}{\left( {x - 2010} \right)^2} \\
\]
\[
   \Rightarrow f''\left( x \right) = 2010{\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} + 2010\left( {x-2009} \right){\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4}2\left( {x - 2010} \right) \\
   + 2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}\dfrac{d}{{dx}}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} \\
\]
\[
   \Rightarrow f''\left( x \right) = 2010{\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} + 2010\left( {x-2009} \right){\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4}2\left( {x - 2010} \right) \\
   + 2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}\dfrac{d}{{dx}}{\left( {x - 2012} \right)^4} + 2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2012} \right)^4}\dfrac{d}{{dx}}{\left( {x - 2011} \right)^3} \\
\]
\[
   \Rightarrow f''\left( x \right) = 2010{\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} + 2010\left( {x-2009} \right){\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4}2\left( {x - 2010} \right) \\
   + 2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}4{\left( {x - 2012} \right)^3} + 2010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2012} \right)^4}3{\left( {x - 2011} \right)^2} \\
\]
Now substitute the values of f’(x) and f’’(x) in equation (3) we have,
$ \Rightarrow g''\left( x \right) = {e^{f\left( x \right)}}f''\left( x \right) + {\left[ {f'\left( x \right)} \right]^2}{e^{f\left( x \right)}}$
Now when, x = 2009
\[ \Rightarrow f''\left( {2009} \right) = 2010{\left( {2009 - 2010} \right)^2}{\left( {2009 - 2011} \right)^3}{\left( {2009 - 2012} \right)^4} + 0 + 0 + 0\]
\[ \Rightarrow f''\left( {2009} \right) = 2010{\left( { - 1} \right)^2}{\left( { - 2} \right)^3}{\left( { - 3} \right)^4}\]
\[ \Rightarrow f''\left( {2009} \right) = 2010\left[ {\left( 1 \right)\left( { - 8} \right)\left( {81} \right)} \right]\]
\[ \Rightarrow f''\left( {2009} \right) = - 2010\left( 8 \right)\left( {81} \right)\]
Now when, x = 2010, 2011, 2012
$ \Rightarrow f''\left( {2010} \right) = f''\left( {2011} \right) = f''\left( {2012} \right) = 0$
And when, x = 2009, 2010, 2011, 2012 the value of f’ (x) is
\[f'\left( {2009} \right) = f'\left( {2010} \right) = f'\left( {2011} \right) = f'\left( {2012} \right) = 0\]
Now substitute the values of f’(x) and f’’(x) in equation (3) we have,
$ \Rightarrow g''\left( x \right) = {e^{f\left( x \right)}}f''\left( x \right) + {\left[ {f'\left( x \right)} \right]^2}{e^{f\left( x \right)}}$
$ \Rightarrow g''\left( {2009} \right) = {e^{f\left( {2009} \right)}}f''\left( {2009} \right) + {\left[ {f'\left( {2009} \right)} \right]^2}{e^{f\left( {2009} \right)}} = {e^{f\left( {2009} \right)}}\left( { - 2010\left( {81} \right)\left( 8 \right)} \right)$ = negative so it is a local maxima.
$ \Rightarrow g''\left( {2009} \right) = g''\left( {2010} \right) = g''\left( {2011} \right) = g''\left( {2012} \right) = 0$, So at these values function is neither maximum nor minimum.
So there is only 1 point in R at which g has a local maximum which is 2009.
So this is the required answer.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation property which is given as $\dfrac{d}{{dx}}abcd = a\dfrac{d}{{dx}}bcd + bcd\dfrac{d}{{dx}}a$, $\dfrac{d}{{dx}}bcd = b\dfrac{d}{{dx}}cd + cd\dfrac{d}{{dx}}b$, $\dfrac{d}{{dx}}cd = c\dfrac{d}{{dx}}d + d\dfrac{d}{{dx}}c$, $\dfrac{d}{{dx}}{\left( {f\left( x \right)} \right)^n} = n{\left( {f\left( x \right)} \right)^{n - 1}}\dfrac{d}{{dx}}f\left( x \right)$, $\dfrac{d}{{dx}}\left( {x - a} \right) = 1$.