Q130 / 225 · JEE Main 2003 · Mathematics

If the function $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2-9 a \mathrm{x}^2+12 \mathrm{a}^2 \mathrm{x}+1$, where $\mathrm{a}>0$, attains its maximum and minimum at $\mathrm{p}$ and $\mathrm{q}$ respectively such that $\mathrm{p}^2=\mathrm{q}$, then a equals

  • A
    $\frac{1}{2}$
  • B
    3
  • C
    1
  • Correct answer
    D
    2

Solution

$f(x)=2 x^3-9 a x^2+12 a^2 x+1$ $f^{\prime}(x)=6 x^2-18 a x+12 a^2 ; f^{\prime \prime}(x)=12 x-18 a$ For max. or min. $6 x^2-18 a x+12 a^2=0 \Rightarrow x^2-3 a x+2 a^2=0$ $\mathrm{x}=\mathrm{a}$ or $\mathrm{x}=2 \mathrm{a}$, at $\mathrm{x}=\mathrm{a}$ max.and at $\mathrm{x}=2 \mathrm{a} \min$. $p^2=q$ $a^2=2 a \Rightarrow a=2$ or $a=0$ but $\mathrm{a}>0$, therefore, $\mathrm{a}=2$.

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

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