JEE Main 2024 · 8 April · morning · Single correct · +4 / −1
Let f(x) be a positive function such that the area bounded by y=f(x), y=0 from x=0 to x=a>0 is e⁻ᵃ+4 a²+a-1. Then the differential equation, whose general solution is y=c₁ f(x)+c₂, where c₁ and c₂ are arbitrary constants, is
- (a) (8 eˣ+1) (d² y)/(d x²)-(d y)/(d x)=0
- (b) (8 eˣ+1) (d² y)/(d x²)+(d y)/(d x)=0
- (c) (8 eˣ-1) (d² y)/(d x²)-(d y)/(d x)=0
- (d) (8 eˣ-1) (d² y)/(d x²)+(d y)/(d x)=0
Answer and solution
Answer (b) (8 eˣ+1) (d² y)/(d x²)+(d y)/(d x)=0
Solution
Maths sir · 4 steps
Differentiate the area to get f(x), then differentiate y twice so both constants disappear.
- Find f(x). The area from 0 to a is ∫₀ᵃ f(x) dx, so f(a) is the derivative of the given expression. Differentiating e⁻ᵃ+4a²+a-1 gives f(x) = -e⁻ˣ+8x+1.
- Remove c2 and c1. From y = c₁ f(x) + c₂, one derivative removes c₂: y' = c₁ f'(x) = c₁(e⁻ˣ+8). A second derivative gives y'' = c₁ f''(x) = -c₁ e⁻ˣ.
- Divide to cancel c1. Dividing y' by y'' removes c₁, leaving an equation with no arbitrary constants.
y'/y'' = (e⁻ˣ+8)/-e⁻ˣ = -(1+8eˣ)
- So. y' = -(8eˣ+1)y'', that is, (8eˣ+1)d²y/dx²+dy/dx=0. The answer is (b).
The idea to keep. Area up to a given as an expression? Differentiate it to get f. Then differentiate the general solution once per arbitrary constant and eliminate them all.
Answer from the NTA final answer key. Solution by PrepWiser.