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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2011 · Mathematics · Application of Derivatives

The function ( f(x)= 2 x^ 3-3 x^ 2-12 x+ 4), has

The function (f(x)=\mathrm{f}(\mathrm{x})=2 \mathrm{x}^3-3 \mathrm{x}^2-12 x+\mathrm{x}+4), has

Options

  1. Atwo points of local maximum
  2. Btwo points of local minimum
  3. Cone maxima and one minima
  4. Dnomaxima or minima

Correct answer

C. one maxima and one minima

Step-by-step solution

f(x)=2x3−3x2−12x+4⇒f′(x)=6x2−6x−12=6(x2−x−2)=6(x−2)(x+1)\begin{aligned}& f(x)=2 x³-3 x²-12 x+4 \\ & \Rightarrow \quad f^{\prime}(x)=6 x²-6 x-12=6 \left(x^2-x-2\right) \\ & =6(x-2)(x+1) \end{aligned})ForFormaxima and minima (\mathrmf^{\prime}(\mathrm{x})=0) (\therefore \quad 6(\mathrm{x}-2)()(\mathrm{x}+1)=0⇒1)=0 \Rightarrow\mathrm{x}=2,-1) Now, (\mathrmf^{\prime \prime}(\mathrm{x})=1212\mathrm{x}-6)  At x=2;\mathrmf′′(x)=24−6=18>0∴x=2, local min. point  At x=−1;\mathrmf′′(x)=12(−1)−6=−18<0∴x=−1 local max. point \begin{aligned}& \text{ At } \mathrm{x}=2 ; \mathrmf^{\prime \prime}(\mathrm{x})=24-6=18 \gt 0 \\ & \therefore \quad \mathrm{x}=2, \text{ local min. point }\\ & \text{ At }\mathrm{x}=-1 ; \mathrmf^{\prime \prime}(\mathrm{x})=12(-1)-6=-18 \lt 0 \\ & \therefore \mathrm{x}=-1 \text{ local max. point } \end{aligned}
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