Applications of Derivatives Question - None

Question

If the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$, where $a > 0$, attains its maximum and minimum at $p$ and $q$ respectively, such that $p^2 = q$, then $a$ is equal to

Accepted Answer

Step 1: Given $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$

Step 2: Find critical points by setting $f'(x) = 0$:

$f'(x) = 6x^2 - 18ax + 12a^2 = 0$

$6(x^2 - 3ax + 2a^2) = 0$

Step 3: Factorize:

$x^2 - 3ax + 2a^2 = 0$

$x^2 - 2ax - ax + 2a^2 = 0$

$x(x - 2a) - a(x - 2a) = 0$

$(x - a)(x - 2a) = 0$

$x = a ext{ or }x = 2a$

Step 4: Check second derivative:

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

At $x = a$: $f''(a) = 12a - 18a = -6a < 0$ (maximum)

At $x = 2a$: $f''(2a) = 24a - 18a = 6a > 0$ (minimum)

Step 5: So $p = a$ and $q = 2a$

Step 6: Use the condition $p^2 = q$:

$a^2 = 2a$

$a^2 - 2a = 0$

$a(a - 2) = 0$

Since $a > 0$, we have $a = 2$

∴ Answer is C.

Question:

Step-by-Step Solution

Master Applications of Derivatives with Mathbee