An enemy DRONE is flying along the curve y = x2 + 7. A soldier is placed at (3, 7). The nearest distance of the DRONE from soldier’s position is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sqrt{5}\)
A point of the curve is \((x,\ x^2+7)\). Its distance from (3, 7) is least where the square of the distance is least:
\(D(x) = (x-3)^2 + (x^2+7-7)^2 = (x-3)^2 + x^4\)
\(D'(x) = 2(x-3) + 4x^3 = 4x^3 + 2x - 6\)
Setting \(D'(x) = 0\):
\(2x^3 + x - 3 = 0\)
\((x-1)(2x^2+2x+3) = 0\)
The quadratic \(2x^2+2x+3\) has discriminant \(4 - 24 \lt 0\), so x = 1 is the only critical point. Also \(D''(x) = 12x^2 + 2 \gt 0\), so it gives the minimum.
At x = 1 the drone is at (1, 8):
\(D(1) = (1-3)^2 + 1^4 = 5\)
So the nearest distance is \(\sqrt{5}\).
NCERT (2026-27 edition):
Maximum and Minimum Values of a Function in a Closed Interval
