Let f(x) = { x³ - x² + 10x - 7,, x ≤ 1; - 2x + log ₂(b² - 4),, x > 1 .
Then the set of all values of b, for which f(x) has maximum value at x = 1, is :

Solution

Maths sir · 4 steps

The left piece climbs to f(1) = 3, so the right piece must start at or below 3.

1. Left piece rises. For x ≤ 1, f'(x) = 3x² - 2x + 10. Its discriminant is 4 - 120 < 0, so f'(x) > 0 always and f increases right up to x = 1, where f(1) = 1 - 1 + 10 - 7 = 3.
2. Right piece falls. For x > 1, f(x) = -2x + log₂(b² - 4) is decreasing. Its values come down from the limit -2 + log₂(b² - 4) as x → 1⁺. For x = 1 to be the maximum, this starting height must not exceed 3.

   -2 + log₂(b² - 4) ≤ 3 ⇒ b² - 4 ≤ 2⁵ = 32 ⇒ b² ≤ 36

3. Log must exist. The logarithm needs b² - 4 > 0, so b² > 4. Together this gives 4 < b² ≤ 36, which means 2 < |b| ≤ 6.
4. So. b ∈ [-6, -2) ∪ (2, 6], which is option (c). At b = ± 6 the right piece is -2x + 5 < 3 for every x > 1, so the endpoints are included.

The idea to keep. For a maximum at the joining point of a piecewise function, the left piece must rise into it and the right piece must start no higher than f there. Do not forget the domain of any log.

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