JEE Main 2020 · 7 January · evening · Single correct · +4 / −1
The value of α for which
4α ∫₋₁² e^(- α | x |)dx = 5, is:
- (a) log ₑ2
- (b) log ₑ√2
- (c) log ₑ( 4/3 )
- (d) log ₑ( 3/2 )
Answer and solution
Answer (a) log ₑ2
Solution
Maths sir · 4 steps
The modulus changes sign at 0, so split the integral there and you get a quadratic in e^(-α).
- Split at zero. For x < 0, |x| = -x, so the integrand is e^(α x). For x ≥ 0, |x| = x, so it is e^(-α x).
∫₋₁² e^(-α|x|) dx = ∫₋₁⁰ e^(α x) dx + ∫₀² e^(-α x) dx = (1 - e^(-α))/α + (1 - e^(-2α))/α
- Use the condition. Multiply by 4α and set the result equal to 5: 4(2 - e^(-α) - e^(-2α)) = 5. Put t = e^(-α) to get 4t² + 4t - 3 = 0.
- Solve for t. Factorise: (2t - 1)(2t + 3) = 0, so t = 1/2 or t = -3/2. Since e^(-α) is always positive, only t = 1/2 is allowed.
- So. e^(-α) = 1/2 gives e^α = 2, so α = logₑ 2, which is option (a).
The idea to keep. Whenever you see |x| inside an integral, break the limits where x changes sign. Then look for a substitution like t = e^(-α) that turns the equation into a quadratic.
Answer from the NTA final answer key. Solution by PrepWiser.