Let the function f(x) = 2x³ + (2p - 7)x² + 3(2p - 9)x - 6 have a maxima for some value of x < 0 and a minima for some value of x > 0. Then, the set of all values of p is

Solution

Maths sir · 4 steps

The cubic's maximum and minimum sit at the two roots of f'(x), so make those roots straddle zero.

1. Differentiate. f'(x) = 6x² + 2(2p-7)x + 3(2p-9). The maximum and minimum of f occur where this quadratic is zero.
2. Which root is which. The leading coefficient of f is positive, so f' goes from + to - at the smaller root (maximum) and from - to + at the larger root (minimum). We need the smaller root negative and the larger root positive.
3. Roots of opposite sign. An upward-opening parabola has one root on each side of 0 exactly when it is negative at 0. Then two real roots are guaranteed, so no separate discriminant check is needed.

   f'(0) = 3(2p-9) < 0 ⇒ p < 9/2

4. So. Every p in (-∞, 9/2) works, with no lower limit. The answer is (d).

The idea to keep. For a quadratic ax²+bx+c with a>0, roots on opposite sides of a point k means just one condition: the quadratic is negative at k.

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