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JEE Main 2019 · 10 Jan · Shift II

JEE Main 2019 Mathematics question on Application of Derivatives

The shortest distance between the point (32,0)\left(\dfrac{3}{2}, 0\right) and the curve y=xy = \sqrt{x}, (x>0)(x > 0), is
  1. A.
    32\dfrac{3}{2}
  2. B.
    54\dfrac{5}{4}
  3. C.
    32\dfrac{\sqrt{3}}{2}
  4. D.
    52\dfrac{\sqrt{5}}{2}
Show answer and solution

Correct option: D

Solution

Let point on curve be (t², t) for t > 0. Distance² from (3/2, 0) is D = (t² - 3/2)² + t².

dD/dt = 2(t² - 3/2)(2t) + 2t = 2t[2t² - 3 + 1] = 2t[2t² - 2] = 4t(t² - 1).

Setting dD/dt = 0 gives t = 1, so the closest point is (1, 1).

Distance = √((1 - 3/2)² + 1²) = √(1/4 + 1) = √(5/4) = √5/2.

[JEE Main 2019 · 10 Jan · Shift II]

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