If the function \( f(x)=\ x^2-2kx+7 \) is increasing on [1, 5] then \( k \) belongs to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( (\ -\infty,1] \)
\( \displaystyle f'(x)=2x-2k=2(x-k) \)
f is increasing on [1, 5] when \( f'(x)\ge 0 \) there, i.e. \( x\ge k \) for every x in [1, 5]. The smallest such x is 1, so \( k\le 1 \).
Hence \( k\in(-\infty,1] \).