JEE Main 2020 · 2 September · morning · Single correct · +4 / −1

If the tangent to the curve y = x + sin y at a point
(a, b) is parallel to the line joining ( 0,3/2 ) and ( 1/2,2 ), then :

  1. (a) b = a
  2. (b) |b - a| = 1
  3. (c) b = π/2 + a
  4. (d) |a + b| = 1
Answer and solution

Answer (b) |b - a| = 1

Solution

Maths sir · 5 steps

The slope is 1, so cos b = 0. Then the curve itself says b - a = sin b = ± 1.

  1. Slope of the line. The line through (0, 3/2) and (1/2, 2) has slope (2 - 3/2)/(1/2 - 0) = (1/2)/(1/2) = 1. So the tangent at (a, b) must also have slope 1.
  2. Differentiate implicitly. Differentiate y = x + sin y with respect to x. You get y' = 1 + cos y · y', so y'(1 - cos y) = 1.

    dy/dx = 1/(1 - cos y)

  3. Set slope to 1. At (a, b), 1/(1 - cos b) = 1 gives 1 - cos b = 1, so cos b = 0. That forces sin b = ± 1.
  4. Use the curve. The point lies on the curve, so b = a + sin b. That is, b - a = sin b = ± 1.
  5. So. |b - a| = 1, which is option (b).

The idea to keep. In implicit-curve tangent questions, the slope condition fixes a trig value at the point. Then put the point back into the curve equation to link a and b.

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.