JEE Main 2021 · 31 August · morning · Single correct · +4 / −1
If dy/dx = (2ˣ⁺ʸ - 2ˣ)/2ʸ, y(0) = 1, then y(1) is equal to :
- (a) log₂(2 + e)
- (b) log₂(1 + e)
- (c) log₂(2e)
- (d) log₂(1 + e²)
Answer and solution
Answer (b) log₂(1 + e)
Solution
Maths sir · 4 steps
Pull 2ˣ out of the top, and the x and y parts separate cleanly.
- Factor the numerator. 2ˣ⁺ʸ - 2ˣ = 2ˣ(2ʸ - 1), so dy/dx = 2ˣ(2ʸ - 1)/2ʸ. Move all the y terms to the left: 2ʸ/(2ʸ - 1) dy = 2ˣ dx.
- Integrate both sides. Put u = 2ʸ - 1, so du = 2ʸ ln 2 dy and the left side becomes 1/(ln 2)ln(2ʸ - 1). The right side is 2ˣ/(ln 2). Multiply through by ln 2.
ln(2ʸ - 1) = 2ˣ + C
- Find the constant. At x = 0, y = 1: ln(2 - 1) = 1 + C, so 0 = 1 + C and C = -1.
- So. At x = 1: ln(2ʸ - 1) = 2 - 1 = 1, so 2ʸ - 1 = e and y = log₂(1 + e), which is option (b).
The idea to keep. When the numerator has a common factor like 2ˣ, take it out first. The equation often turns variable separable right away. And always use the initial condition to fix C before you plug in the new point.
Answer from the NTA final answer key. Solution by PrepWiser.