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STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 6. Application of derivatives4 Marks
Question
If the function $f(x) = 2{x^3} - 9a{x^2}$ $ + 12{a^2}x + 1,$ where $a > 0$ attains its maximum and minimum at $ p$  and $ q$ respectively such that ${p^2} = q$, then $a$ equals
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Answer

c
(c) $f(x) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$

$f'(x) = 6{x^2} - 18ax + 12{a^2}$

$f''(x) = 12x - 18a$

For maximum and minimum ,

$6{x^2} - 18ax + 12{a^2} = 0 \Rightarrow {x^2} - 3ax + 2{a^2} = 0$

$x = a$ or $x = 2a$, at $x = a$ maximum and at $x = 2a$ minimum

$\because$ ${p^2} = q$,

$\therefore $ ${a^2} = 2a \Rightarrow a = 2$ or $a = 0$

But $a > 0$, therefore $a = 2.$

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