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KEAM 2017 Math Paper

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Question : 27 of 120

Marks: +1, -0

Let

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

where a > 0. The minimum of f is attained at a point q and the maximum is attained at a point p. If p3 = q, then a is equal to

1
3
2
$\sqrt{2}$
1/2
None of the above

Solution:

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

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

For maximum or minimum,

f'(x)=0

\Rightarrow \;\; 6 x^{2}-18 a x+12 a^{2}=0

x = \frac{18 a \pm \sqrt{324 a^{2}-288 a^{2}}}{2 \times 6}

= \frac{18 a \pm \sqrt{36 a^{2}}}{12} = \frac{18 a \pm 6 a}{12} = 2a, a

Now,

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

At

x=2 a

f''(x)=24 a-18 a

=6 a>0,

maxima

\therefore p=f(2 a)=2 \times 8 a^{3}-36 a^{3}+24 a^{3}+1

=4 a^{3}+1

At

x=a

f''(x)=12 \times a-18 a

=-6 a < a,

minima

\therefore q=f(a)

=2 a^{3}-9 a^{3}+12 a^{3}+1

=5 a^{3}+1

Also given

p^{3}=q

\therefore (4 a^{3}+1)^{3}=(5 a^{3}+1)

\Rightarrow a=0,

but

a > 0

(given)

Hence, none of the given option is correct.

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