Exams › JEE Advanced › Maths
Correct answer: 2
In I2, substitute x² = t, so 2x dx = dt. When x = 0, t = 0; when x = 1/sqrt(2), t = 1/2. Then I2 = (1/2)*integral from 0 to 1/2 of e^t * ln(1-t) dt. But we need full limits; note that I1 = integral from 0 to 1 of e^x*ln(1-x) dx. Actually substitution x² = u gives I2 = (1/2)*I1, so I1/I2 = 2.