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Question: Let f be a function defined on R (the set of all real numbers) such that,\[f'\left( x \right) = 2010...

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)4f'\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∈Rx \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∈Rx \in R, then the number of points in R at which g has a local maximum is
(a)\left( a \right) 0
(b)\left( b \right) 1
(c)\left( c \right) 2
(d)\left( d \right) 3

Explanation

Solution

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′(x)=2010(x−2009)(x−2010)2(x−2011)3(x−2012)4f'\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∈Rx \in R.
Now the given function is
f (x) = ln (g(x)), for all x∈Rx \in R.
Now take antilog on both sides we have,
⇒ef(x)=g(x)\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,
⇒ddxef(x)=ddxg(x)=0\Rightarrow \dfrac{d}{{dx}}{e^{f\left( x \right)}} = \dfrac{d}{{dx}}g\left( x \right) = 0
Now as we know that ddxef(x)=ef(x)ddxf(x)\dfrac{d}{{dx}}{e^{f\left( x \right)}} = {e^{f\left( x \right)}}\dfrac{d}{{dx}}f\left( x \right) so we have,
⇒ef(x)ddxf(x)=ddxg(x)=0\Rightarrow {e^{f\left( x \right)}}\dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}g\left( x \right) = 0
⇒ef(x)f′(x)=g′(x)=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,
⇒ef(x)2010(x−2009)(x−2010)2(x−2011)3(x−2012)4=0\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(x−2009)(x−2010)2(x−2011)3(x−2012)4=02010\left( {x-2009} \right){\left( {x - 2010} \right)^2}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} = 0 and ef(x)≠0{e^{f\left( x \right)}} \ne 0.
⇒2010(x−2009)(x−2010)2(x−2011)3(x−2012)4=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,
⇒x=2009,2010,2011,2012\Rightarrow x = 2009, 2010, 2011, 2012
Now again differentiate the equation (2) we have
⇒ddxg′(x)=ddx[ef(x)f′(x)]\Rightarrow \dfrac{d}{{dx}}g'\left( x \right) = \dfrac{d}{{dx}}\left[ {{e^{f\left( x \right)}}f'\left( x \right)} \right]
⇒ddxg′(x)=g′′(x)=[ef(x)ddxf′(x)+f′(x)ddxef(x)]\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]
⇒ddxg′(x)=g′′(x)=[ef(x)f′′(x)+f′(x)ef(x)f′(x)]\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]
⇒g′′(x)=ef(x)f′′(x)+[f′(x)]2ef(x)\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,
⇒ddxf′(x)=ddx[2010(x−2009)(x−2010)2(x−2011)3(x−2012)4]\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 ddxabcd=addxbcd+bcdddxa\dfrac{d}{{dx}}abcd = a\dfrac{d}{{dx}}bcd + bcd\dfrac{d}{{dx}}a, ddxbcd=bddxcd+cdddxb\dfrac{d}{{dx}}bcd = b\dfrac{d}{{dx}}cd + cd\dfrac{d}{{dx}}b, ddxcd=cddxd+dddxc\dfrac{d}{{dx}}cd = c\dfrac{d}{{dx}}d + d\dfrac{d}{{dx}}c, ddx(f(x))n=n(f(x))n−1ddxf(x)\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,

⇒ddxf′(x)=2010(x−2009)ddx[(x−2010)2(x−2011)3(x−2012)4] \+2010[(x−2010)2(x−2011)3(x−2012)4]ddx(x−2009)  \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\\}

⇒f′′(x)=2010(x−2010)2(x−2011)3(x−2012)4+2010(x−2009)(x−2010)2ddx(x−2011)3(x−2012)4 \+2010(x−2009)(x−2011)3(x−2012)4ddx(x−2010)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 - 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} \\\ ⇒f′′(x)=2010(x−2010)2(x−2011)3(x−2012)4+2010(x−2009)(x−2011)3(x−2012)42(x−2010) \+2010(x−2009)(x−2010)2ddx(x−2011)3(x−2012)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}\dfrac{d}{{dx}}{\left( {x - 2011} \right)^3}{\left( {x - 2012} \right)^4} \\\ ⇒f′′(x)=2010(x−2010)2(x−2011)3(x−2012)4+2010(x−2009)(x−2011)3(x−2012)42(x−2010) \+2010(x−2009)(x−2010)2(x−2011)3ddx(x−2012)4+2010(x−2009)(x−2010)2(x−2012)4ddx(x−2011)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}\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} \\\ ⇒f′′(x)=2010(x−2010)2(x−2011)3(x−2012)4+2010(x−2009)(x−2011)3(x−2012)42(x−2010) \+2010(x−2009)(x−2010)2(x−2011)34(x−2012)3+2010(x−2009)(x−2010)2(x−2012)43(x−2011)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}{\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,
⇒g′′(x)=ef(x)f′′(x)+[f′(x)]2ef(x)\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
⇒f′′(2009)=2010(2009−2010)2(2009−2011)3(2009−2012)4+0+0+0\Rightarrow f''\left( {2009} \right) = 2010{\left( {2009 - 2010} \right)^2}{\left( {2009 - 2011} \right)^3}{\left( {2009 - 2012} \right)^4} + 0 + 0 + 0
⇒f′′(2009)=2010(−1)2(−2)3(−3)4\Rightarrow f''\left( {2009} \right) = 2010{\left( { - 1} \right)^2}{\left( { - 2} \right)^3}{\left( { - 3} \right)^4}
⇒f′′(2009)=2010[(1)(−8)(81)]\Rightarrow f''\left( {2009} \right) = 2010\left[ {\left( 1 \right)\left( { - 8} \right)\left( {81} \right)} \right]
⇒f′′(2009)=−2010(8)(81)\Rightarrow f''\left( {2009} \right) = - 2010\left( 8 \right)\left( {81} \right)
Now when, x = 2010, 2011, 2012
⇒f′′(2010)=f′′(2011)=f′′(2012)=0\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′(2009)=f′(2010)=f′(2011)=f′(2012)=0f'\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,
⇒g′′(x)=ef(x)f′′(x)+[f′(x)]2ef(x)\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)}}
⇒g′′(2009)=ef(2009)f′′(2009)+[f′(2009)]2ef(2009)=ef(2009)(−2010(81)(8))\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.
⇒g′′(2009)=g′′(2010)=g′′(2011)=g′′(2012)=0\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 ddxabcd=addxbcd+bcdddxa\dfrac{d}{{dx}}abcd = a\dfrac{d}{{dx}}bcd + bcd\dfrac{d}{{dx}}a, ddxbcd=bddxcd+cdddxb\dfrac{d}{{dx}}bcd = b\dfrac{d}{{dx}}cd + cd\dfrac{d}{{dx}}b, ddxcd=cddxd+dddxc\dfrac{d}{{dx}}cd = c\dfrac{d}{{dx}}d + d\dfrac{d}{{dx}}c, ddx(f(x))n=n(f(x))n−1ddxf(x)\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), ddx(x−a)=1\dfrac{d}{{dx}}\left( {x - a} \right) = 1.