Q51 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

If $6 x-x^2+12$ attains its extreme value $\beta$ at $x=\alpha$ then $\beta=$

  • Correct answer
    A
    $7 \alpha$
  • B
    $5 \alpha$
  • C
    $3 \alpha$
  • D
    $\alpha$

Solution

We are given that $$ \begin{aligned} & \mathrm{f}(\mathrm{x})=-\mathrm{x}^2+6 \mathrm{x}+12 \\ & \mathrm{f}^{\prime}(\mathrm{x})=-2 \mathrm{x}+6 \\ & \mathrm{f}^{\prime}(\mathrm{x})=-2 \\ & \text { Now } \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow-2 \mathrm{x}+6=0 \Rightarrow \mathrm{x}=3 \\ & \therefore \mathrm{f}(\alpha)=\mathrm{f}(3)=-9+18+12=-9+30=21 \\ & \Rightarrow \mathrm{f}(\alpha)=\mathrm{f}(3)=7 \times 3=7 \alpha \end{aligned} $$

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

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