JEE Main 2020 · 3 September · morning · Single correct · +4 / −1

If y² + logₑ (cos²x) = y,
x ∈ ( - π/2,π/2 ), then :

  1. (a) |y''(0)| = 2
  2. (b) |y'(0)| + |y''(0)| = 3
  3. (c) y''(0) = 0
  4. (d) |y'(0)| + |y"(0)| = 1
Answer and solution

Answer (a) |y''(0)| = 2

Solution

Maths sir · 4 steps

First find y at x = 0, then differentiate twice and put x = 0 each time.

  1. Value at zero. At x = 0, cos² 0 = 1 and ln 1 = 0. So y² = y, which gives y(0) = 0 or y(0) = 1. In both cases 2y(0) - 1 = ± 1, which is never zero.
  2. Differentiate once. The derivative of ln(cos² x) is (2 cos x(-sin x))/(cos² x) = -2 tan x. So 2yy' - 2 tan x = y', which gives y'(2y - 1) = 2 tan x. At x = 0 the right side is 0, so y'(0) = 0.

    2(y')² + 2y y'' - 2 sec² x = y''

  3. Differentiate again. Put x = 0 in the second derivative above, using y'(0) = 0 and sec² 0 = 1. This gives y''(0) (2y(0) - 1) = 2. If y(0) = 0, then y''(0) = -2. If y(0) = 1, then y''(0) = 2.
  4. So. On either branch, |y''(0)| = 2 and |y'(0)| + |y''(0)| = 0 + 2 = 2. Only option (a) is true.

The idea to keep. In implicit differentiation, find the value of y at the point first. Then differentiate step by step and substitute at each stage, instead of solving for y' in general.

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.