JEE Mains · Maths · STD 12 - 6. Application of derivatives
If \(S_1\) and \(S_2\) are respectively the sets of local minimum and local maximum points of the function. \(f(x) = 9{x^4} + 12{x^3} - 36{x^2} + 25,x \in R\), then
- 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 \({S_1} = \left\{ { - 1} \right\};{S_2} = \left\{ {0,2} \right\}\)
Answer & Solution
Correct Answer
(A) \({S_1} = \left\{ { - 2,1} \right\};{S_2} = \left\{ 0 \right\}\)
Step-by-step Solution
Detailed explanation
\(f(x)=9 x^{4}+12 x^{3}-36 x^{2}+25\) \(f(x)=36 x^{3}+36 x^{2}-72 x\) \(=36 x\left(x^{2}+x-2\right)\) \(=36 x(x-1)(x+2)\) Point of minima \(=\{-2,1\}=\mathrm{S}_{1}\) Point of maxima \(=\{0\}=\mathrm{S}_{2}\)
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