Q111 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Mathematics

In the interval \([-2,4]\), the absolute maximum of \(f(x)=2 x^3-3 x^2-12 x+5\) occurs \(x=\)

  • Correct answer
    A
    4
  • B
    -2
  • C
    -1
  • D
    2

Solution

Given, function \(f(x)=2 x^3-3 x^2-12 x+5\) So, \(f^{\prime}(x)=6 x^2-6 x-12=0\) [for maxima and minima \(f^{\prime}(x)=0\) ] \(\begin{aligned} & \Rightarrow \quad x^2-x-2=0 \Rightarrow \quad(x+1)(x-2)=0 \\ & \Rightarrow \quad x=-1,2 \in[-2,4] \\ & \because \quad f(-2)=-16-12+24+5=1 \\ & f(-1)=-2-3+12+5=12 \\ & f(2)=16-12-24+5=-15 \\ & \text {and } \quad f(4)=128-48-48+5=37 \end{aligned}\) The absolute maximum of \(f(x)\) occurs at \(x=4\). Hence, option (a) is correct.

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

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