Applications Of Derivatives

Question

For $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$, where $a > 0$. Let $p$ and $q$ be the points of local maximum and local minimum respectively. If $p^2 = q$, then $a$ equals:

RankGuru · Accepted Answer

1. Find the derivative $f'(x)$:
$f'(x) = 6x^2 - 18ax + 12a^2$
$f'(x) = 6(x^2 - 3ax + 2a^2) = 6(x - a)(x - 2a)$.

2. Identify critical points:
The critical points are $x = a$ and $x = 2a$.
Since $a > 0$, we have $a < 2a$.

3. Determine Max/Min:
The leading coefficient is positive, so the curve goes Up-Down-Up.
Thus, $x = a$ is the local Maximum ($p$).
$x = 2a$ is the local Minimum ($q$).

4. Apply condition $p^2 = q$:
$(a)^2 = 2a$
$a^2 - 2a = 0 \implies a(a - 2) = 0$.
Since $a > 0$, we must have $a = 2$.

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Applications Of Derivatives

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More applications of derivatives Practice Questions

The set of all values of 'a' for which f(x)=(a+2)x3−3ax2+9ax−1 decreases for all real ...

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The set of all values of ' $a$ ' for which $f (x) = (a + 2) x^{3} - 3 a x^{2} + 9 a x - 1$ decreases for all real ...

The minimum value of $f (x) = ∣ x - 1∣ + ∣ x - 2∣ + ... + ∣ x - 2025∣$ occurs at:

The acute angle between the curves $y = ∣ x^{2} - 1∣$ and $y = 1$ at their points of intersection in th...