Q75 / 125 · BITSAT 2011 · Mathematics

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

  • A
    two points of local maximum
  • B
    two points of local minimum
  • Correct answer
    C
    one maxima and one minima
  • D
    nomaxima or minima

Solution

\(\begin{aligned} & f(x)=2 x^3-3 x^2-12 x+4 \\ & \Rightarrow \quad f^{\prime}(x)=6 x^2-6 x-12=6\left(x^2-x-2\right) \\ & =6(x-2)(x+1) \end{aligned}\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\therefore \quad 6(\mathrm{x}-2)(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}=2,-1\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-6\) \(\begin{aligned} & \text { At } \mathrm{x}=2 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=24-6=18 > 0 \\ & \therefore \quad \mathrm{x}=2, \text { local min. point } \\ & \text { At } \mathrm{x}=-1 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=12(-1)-6=-18 < 0 \\ & \therefore \mathrm{x}=-1 \text { local max. point } \end{aligned}\)

Hence, the correct answer is option: 'C'.

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