If a function f(x) = x^2 + bx + 1 is increasing in the interval [1, 2…

The correct answer is Option (3) → −2

f is increasing on [1, 2] when \(f'(x) \ge 0\) for every x in [1, 2]:

    \(f'(x) = 2x + b\)

2x + b grows with x, so it is smallest at x = 1. The condition at x = 1 is therefore enough:

    \(2(1) + b \ge 0\)
    \(b \ge -2\)

So the least value is b = −2. Then f’(x) = 2x − 2, which is 0 only at the end point x = 1 and positive on (1, 2], so f is increasing on [1, 2].

Why the other options are incorrect

• Option (1) 5 — Incorrect. f’(x) = 2x + 5 > 0, so f is increasing, but −2 is a smaller value that also works.

• Option (2) 0 — Incorrect. f’(x) = 2x > 0 on [1, 2], but again −2 is smaller.

• Option (4) −4 — Incorrect. It satisfies f’(2) ≥ 0 only. With b = −4, f’(x) = 2x − 4 < 0 for 1 ≤ x < 2, so f is decreasing there.