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JEE Main 2019 · 10 Jan · Shift II

JEE Main 2019 Mathematics question on Application of Derivatives

A helicopter is flying along the curve given by y−x3/2=7y - x^{3/2} = 7, (x≥0)(x \geq 0). A soldier positioned at the point (12,7)\left(\frac{1}{2}, 7\right) wants to shoot down the helicopter when it is nearest to him. Then, this nearest distance is
  1. A.
    1376\frac{1}{3}\sqrt{\frac{7}{6}}
  2. B.
    56\frac{\sqrt{5}}{6}
  3. C.
    1673\frac{1}{6}\sqrt{\frac{7}{3}}
  4. D.
    12\frac{1}{2}
Show answer and solution

Correct option: C

Solution

Let a point on the curve be (x, x^(3/2) + 7). Distance squared from (1/2, 7) is:

D = (x - 1/2)² + (x^(3/2))²= (x - 1/2)² + x³

dD/dx = 2(x - 1/2) + 3x² = 3x² + 2x - 1 = (3x - 1)(x + 1) = 0

Since x ≥ 0, we get x = 1/3.

At x = 1/3: D = (1/3 - 1/2)² + (1/3)³ = (-1/6)² + 1/27 = 1/36 + 1/27 = 3/108 + 4/108 = 7/108

Distance = √(7/108) = √7/(6√3) = (1/6)√(7/3).

[JEE Main 2019 · 10 Jan · Shift II]

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