Question

Q.21 Let f be a function on R ( the set of all real numbers ) such that f'(x) = 2010(x-2009)(x-2010)² (x-2011)³ (x-2012)² , 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 local maximum is

Answer 1

Dear Student,

$$\begin{gathered} f\text{'}\left(x\right)=2010\left(x-2009\right){\left(x-2010\right)}^2{\left(x-2011\right)}^3{\left(x-2012\right)}^4 \\ Andf\left(x\right)=\ln g\left(x\right) \\ Thus,g\left(x\right)=e^{f\left(x\right)} \\ Thus,g\text{'}\left(x\right)=e^{f\left(x\right)}f\text{'}\left(x\right) \\ Tofindmaxima,letg\text{'}\left(x\right)=0 \\ \Rightarrow e^{f\left(x\right)}f\text{'}\left(x\right)=0 \\ \Rightarrow f\text{'}\left(x\right)=0\left(\because e^x\ne 0\forall x\in \mathrm{ℝ}\right) \\ \Rightarrow 2010\left(x-2009\right){\left(x-2010\right)}^2{\left(x-2011\right)}^3{\left(x-2012\right)}^4=0 \\ \Rightarrow x=2009,2010,2011,2012 \\ Nowforx<2009wegetf\text{'}\left(x\right)>0 \\ andforx>2009wegetf\text{'}\left(x\right)<0 \\ Thusx=2009isapointoflocalmaxima. \\ Nowforx<2010wegetf\text{'}\left(x\right)<0 \\ andforx>2010wegetf\text{'}\left(x\right)<0 \\ Thusx=2010isapointofinflexion. \\ Nowforx<2011wegetf\text{'}\left(x\right)<0 \\ andforx>2011wegetf\text{'}\left(x\right)>0 \\ Thusx=2011isapointoflocalminima. \\ Nowforx<2012wegetf\text{'}\left(x\right)>0 \\ andforx>2012wegetf\text{'}\left(x\right)>0 \\ Thusx=2012isapointofinflexion. \\ Thusthereisonlyonepointofmaximaofg\left(x\right). \end{gathered}$$

Regards