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If the function $f(x)=2x^{3}-9ax^{2}+12a^{2}x+1,where\; a> 0,$ attains its maximum and minumum at $p$ and $q$ respectively such that $p^{2}=q,then\; a\; equals$

Option 1)
1

Option 2)
2

Option 3)
1/2

Option 4)
3

Answers (1)

As we learnt in

Rate Measurement -

Rate of any of variable with respect to time is rate of measurement. Means according to small change in time how much other factors change is rate measurement:

$\Rightarrow \frac{dx}{dt},\:\frac{dy}{dt},\:\frac{dR}{dt},(linear),\:\frac{da}{dt}$

$\Rightarrow \frac{dS}{dt},\:\frac{dA}{dt}(Area)$

$\Rightarrow \frac{dV}{dt}(Volume)$

$\Rightarrow \frac{dV}{V}\times 100(percentage\:change\:in\:volume)$

- wherein

Where dR / dt means Rate of change of radius.

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

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

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

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

$\therefore 6p^{2}-18ap+12a^{2}-6q^{2}+18aq-12a^{2}=0$

$6 (p^{2}-q^{2}) + 18 a (q-p) = 0$

$6(p-q)\left [ (p+q)-3a \right ]=0$

$p=0,$ or $p+q = 3a$

$q=0$

$\therefore p = a & \:\:\:\: q = 2a$

But $p^{2} = q$

$a^{2}=2a$

$\therefore a = 0, a=2$ but $a \neq 0$

$\therefore a = 2$

Option 1)

1

This option is incorrect

Option 2)

2

This option is correct.

Option 3)

1/2

This option is incorrect

Option 4)

3

This option is incorrect

Posted by

Sabhrant Ambastha

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