If the function \(f(x) = 2x^2 - kx +7\), is increasing on [1, 2], then \(k\) lies in the interval
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \((-\infty, 4)\)
f is increasing on [1, 2] when \(f'(x) \ge 0\) there:
\(f'(x) = 4x - k\)
\(4x - k\) increases with x, so its least value on [1, 2] is at x = 1:
\(4(1) - k \ge 0\)
\(k \le 4\)
Every k in \((-\infty, 4)\) gives \(f'(x) \gt 0\) on [1, 2], so f is increasing. (The end value k = 4 also works, since then \(f'(x) = 4(x - 1)\) is zero only at x = 1; but it is not needed to choose the option.) Option (2) is the only option all of whose values of k work.
Why the other options are incorrect
Option (1) \((-\infty, 8)\) — Incorrect. This comes from testing x = 2 instead of x = 1. With k = 6, \(f'(1) = 4 - 6 = -2 \lt 0\), so f decreases near x = 1.
Option (3) \((4, \infty)\) — Incorrect. For every k > 4, \(f'(1) = 4 - k \lt 0\).
Option (4) \((8, \infty)\) — Incorrect. For k > 8, \(f'(x) = 4x - k \lt 0\) on the whole of [1, 2], so f is decreasing there.