question_answer

If \[{{S}_{1}}\]and \[{{S}_{2}}\] are respectively the sets of local minimum and local maximum points of the function, \[(x)=9{{x}^{4}}+12{{x}^{3}}36{{x}^{2}}+25,x\in R,\] then : [JEE Main 8-4-2019 Morning]

A) \[{{S}_{1}}=\left\{ 2,1 \right\};{{S}_{2}}=\left\{ 0 \right\}\]

B) \[{{S}_{1}}=\left\{ 2,0 \right\};{{S}_{2}}=\left\{ 1 \right\}\]

C) \[{{S}_{1}}=\left\{ 2 \right\};{{S}_{2}}=\left\{ 0,1 \right\}\]

D) \[{{S}_{1}}=\left\{ 1 \right\};{{S}_{2}}=\left\{ 0,2 \right\}\]

Correct Answer: A

Solution :

\[\left( x \right)=9{{x}^{4}}+12{{x}^{3}}36{{x}^{2}}+25\] \['\left( x \right)=36{{x}^{3}}+36{{x}^{2}}72x\] \[=36x\left( {{x}^{2}}+x2 \right)\] \[=36x\left( x1 \right)\left( x+2 \right)\] Points of minima \[=\left\{ 2,1 \right\}={{S}_{1}}\] Point of maxima \[=\left\{ 0 \right\}={{S}_{2}}\]