Let 'a' be a real number such that the function f(x) = ax² + 6x - 15, x ∈ R is increasing in ( - ∞ ,3/4 ) and decreasing in ( 3/4,∞ ). Then the function g(x) = ax² - 6x + 15, x∈R has a :

Solution

Maths sir · 4 steps

The turning point of f gives you a. Then find g's own turning point, which sits on the other side of zero.

1. Find a. f changes from increasing to decreasing at x = 3/4, so f'(3/4) = 0. With f'(x) = 2ax + 6, this gives 3a/2 + 6 = 0, so a = -4.
2. Check the sign. With a = -4, f'(x) = -8x + 6. It is positive for x < 3/4 and negative for x > 3/4, which matches the given behaviour.

   g(x) = -4x² - 6x + 15, g'(x) = -8x - 6

3. Turning point of g. g'(x) = 0 gives x = -3/4. Since g''(x) = -8 < 0, this point is a local maximum.
4. So. g has a local maximum at x = -3/4, which is option (a).

The idea to keep. For ax² + bx + c the turning point is at x = -b/2a. It is a maximum when a < 0 and a minimum when a > 0. Changing the sign of b only reflects the vertex across the y-axis.

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