If the function f(x)=2 x³-9 a x²+12 a² x+1, where a>0, attains its local maximum and local minimum values at p and q , respectively, such that p²=q, then f(3) is equal to :

Solution

Maths sir · 4 steps

Factor f'(x), see which root is the maximum, and the condition p² = q pins down a.

1. Critical points. f'(x) = 6x² - 18ax + 12a² = 6(x² - 3ax + 2a²) = 6(x - a)(x - 2a). So the critical points are x = a and x = 2a.
2. Which is which. Since a > 0, we have a < 2a. For a cubic with a positive leading coefficient, f' is positive before a, negative between a and 2a, and positive after 2a. So the local maximum is at p = a and the local minimum is at q = 2a.
3. Find a. p² = q gives a² = 2a, so a(a - 2) = 0. Since a > 0, a = 2.

   f(x) = 2x³ - 18x² + 48x + 1

4. So. f(3) = 54 - 162 + 144 + 1 = 37, which is option (b).

The idea to keep. For a cubic with a positive leading coefficient, the smaller critical point is the local maximum and the larger one is the local minimum. Factor f'(x) first and the roots tell you everything.

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