JEE Main 2020 · 7 January · morning · Single correct · +4 / −1

Let xᵏ + yᵏ = aᵏ, (a, k > 0 ) and dy/dx + ( y/x )^(1/3) = 0, then k is:

  1. (a) 1/3
  2. (b) 2/3
  3. (c) 4/3
  4. (d) 3/2
Answer and solution

Answer (b) 2/3

Solution

Maths sir · 3 steps

Differentiate both sides, write dy/dx as a power of y/x, then match the exponent with 1/3.

  1. Differentiate implicitly. Differentiate xᵏ + yᵏ = aᵏ with respect to x. The right side is a constant, so its derivative is 0, and yᵏ gives k yᵏ⁻¹dy/dx by the chain rule.

    k xᵏ⁻¹ + k yᵏ⁻¹dy/dx = 0 ⇒ dy/dx = -xᵏ⁻¹/yᵏ⁻¹ = -(y/x)¹⁻ᵏ

  2. Flip the fraction. The given condition says dy/dx = -(y/x)^(1/3), with y on top. Since (x/y)ᵏ⁻¹ = (y/x)¹⁻ᵏ, compare the exponents in that form: 1 - k = 1/3.
  3. So. k = 1 - 1/3 = 2/3, which is option (b).

The idea to keep. Before you match exponents, bring both expressions to the same base. Here, turn (x/y)ᵏ⁻¹ into (y/x)¹⁻ᵏ first, and only then compare the powers.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Differentiation

Asked about 40 times in JEE Main since 2020. This question is from 2020.