If the function f(x)=2 x³-9 ax²+12 a² x+1, a> 0 has a local maximum at x=α and a local minimum at x=α², then α and α² are the roots of the equation :

Solution

Maths sir · 4 steps

Factor f'(x) to find the two critical points, then match the max to α and the min to α².

1. Critical points. f'(x) = 6x² - 18ax + 12a² = 6(x-a)(x-2a), so the critical points are x = a and x = 2a.
2. Which is which. f''(x) = 12x - 18a. At x = a it is -6a < 0, a local maximum. At x = 2a it is 6a > 0, a local minimum (using a > 0).
3. Match the points. So α = a and α² = 2a. That gives a² = 2a, and since a > 0, a = 2. Hence α = 2 and α² = 4.

   (x-2)(x-4) = x² - 6x + 8 = 0

4. So. α and α² are the roots of x² - 6x + 8 = 0, which is option (a).

The idea to keep. For a cubic with positive leading coefficient, the smaller critical point is the local maximum and the larger one is the local minimum. Fix that order before you match the points to the given conditions.

Answer from the NTA final answer key. Solution by PrepWiser.