A student Sumit is running on a playground along the curve given by \(y = x^2 + 7\). Another student Amit standing at a point with co-ordinates (3,7) on the playground wants to hit Sumit by paper ball, when Sumit is nearest to him.
Based on the above information answer the following.
Minimum value of D is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{5}\)
The distance between Sumit \((x, x^2 + 7)\) and Amit (3, 7) satisfies
\(D^2 = f(x) = (x - 3)^2 + x^4\)
\(f'(x) = 4x^3 + 2x - 6 = 2(x - 1)(2x^2 + 2x + 3)\)
Since \(2x^2 + 2x + 3 \gt 0\) always, f′(x) = 0 only at x = 1, and \(f''(x) = 12x^2 + 2 \gt 0\), so the minimum is at x = 1, the point (1, 8).
\( \displaystyle D = \sqrt{(1 - 3)^2 + 1^4} = \sqrt{4 + 1} \)
\( = \sqrt{5} \)
Option (1), 5, is the minimum of \(D^2\), not of D.
NCERT (2026-27 edition):
Maximum and Minimum Values of a Function in a Closed Interval
