JEE Main 2020 · 5 September · evening · Single correct · +4 / −1

Which of the following points lies on the tangent to the curve
x⁴eʸ + 2√(y + 1) = 3 at the point (1, 0)?

  1. (a) (2, 2)
  2. (b) (–2, 4)
  3. (c) (2, 6)
  4. (d) (–2, 6)
Answer and solution

Answer (d) (–2, 6)

Solution

Maths sir · 3 steps

Differentiate implicitly, put in (1, 0) to get the slope, then test each option in the tangent line.

  1. Differentiate implicitly. Differentiate both sides with respect to x, treating y as a function of x. Use the product rule on x⁴eʸ and the chain rule on 2√(y+1).

    4x³eʸ + x⁴eʸ dy/dx + 1/√(y+1) dy/dx = 0

  2. Slope and tangent. At (1, 0) this becomes 4 + dy/dx + dy/dx = 0, so the slope is -2. The tangent is y - 0 = -2(x - 1), that is, 2x + y = 2.
    OptionPoint2x + yOn the tangent?
    (a)(2, 2)4 + 2 = 6No
    (b)(-2, 4)-4 + 4 = 0No
    (c)(2, 6)4 + 6 = 10No
    (d)(-2, 6)-4 + 6 = 2Yes
  3. So. Only (-2, 6) satisfies 2x + y = 2, so the answer is (d).

The idea to keep. For a curve given implicitly, differentiate term by term, keep every dy/dx, and substitute the point before solving. Then use point-slope form for the tangent.

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

Study Applications of Derivatives

Asked about 160 times in JEE Main since 2020. This question is from 2020.