If x is a real, then minimum value of \(x^2 - 8x + 17\) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1
Completing the square:
\(x^2 - 8x + 17 = (x^2 - 8x + 16) + 1\)
\(= (x - 4)^2 + 1\)
Since \((x - 4)^2 \ge 0\) for every real x, with equality at x = 4, the least value is 0 + 1 = 1.
NCERT Exemplar:
Application of Derivatives
