JEE MAIN - Mathematics (2003 - No. 14)
If the function $$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1,$$ where $$a>0,$$ attains its maximum and minimum at $$p$$ and $$q$$ respectively such that $${p^2} = q$$ , then $$a$$ equals
$${1 \over 2}$$
$$3$$
$$1$$
$$2$$
Explanation
$$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$
$$f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};$$
$$\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a$$
For max. or min.
$$6{x^2} - 18ax + 12{a^2} = 0$$
$$ \Rightarrow {x^2} - 3ax + 2{a^2} = 0$$
$$ \Rightarrow x = a$$ or $$x=2a.$$
At $$x=a$$
$$f''(a) = 12a - 18a = -6a < 0$$
At $$x=a$$ maximum As f''($$a$$) < 0.
At $$x=2a$$
$$f''(a) = 24a - 18a = 6a > 0$$
At $$x=2a$$ minimum As f''(2$$a$$) > 0.
$$\therefore$$ $$p=a$$ and $$q=2a$$
As per question $${p^2} = q$$
$$\therefore$$ $${a^2} = 2a $$
$$ \Rightarrow $$ $$a(a - 2) = 0$$
$$ \therefore $$ $$a$$ = 0, 2
but $$a > 0,$$ therefore, $$a=2.$$
$$f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};$$
$$\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a$$
For max. or min.
$$6{x^2} - 18ax + 12{a^2} = 0$$
$$ \Rightarrow {x^2} - 3ax + 2{a^2} = 0$$
$$ \Rightarrow x = a$$ or $$x=2a.$$
At $$x=a$$
$$f''(a) = 12a - 18a = -6a < 0$$
At $$x=a$$ maximum As f''($$a$$) < 0.
At $$x=2a$$
$$f''(a) = 24a - 18a = 6a > 0$$
At $$x=2a$$ minimum As f''(2$$a$$) > 0.
$$\therefore$$ $$p=a$$ and $$q=2a$$
As per question $${p^2} = q$$
$$\therefore$$ $${a^2} = 2a $$
$$ \Rightarrow $$ $$a(a - 2) = 0$$
$$ \therefore $$ $$a$$ = 0, 2
but $$a > 0,$$ therefore, $$a=2.$$
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