The value of z such that z > 1, satisfying int_1^z t * ln(t) dt = 1/4 is A. sqrt(e) B. e C. e^(1/4) D. e^(-1)

GATE 2025 · Engineering Mathematics · Definite Integral · medium

Answer: sqrt(e) (option A)

1. Integration by parts: int t*ln(t) dt: int t*ln(t) dt = (t^2/2)*ln(t) - int (t^2/2)*(1/t) dt = (t^2/2)*ln(t) - int (t/2) dt = (t^2/2)*ln(t) - t^2/4 + C.
2. Evaluate the definite integral: = [(z^2/2)*ln(z) - z^2/4] - [(1/2)*ln(1) - 1/4] = (z^2/2)*ln(z) - z^2/4 - (0 - 1/4) = (z^2/2)*ln(z) - z^2/4 + 1/4.
3. Set equal to 1/4 and solve: (z^2/2)*ln(z) - z^2/4 = 0 => z^2*(ln(z)/2 - 1/4) = 0. Since z > 1, z != 0, so ln(z)/2 = 1/4 => ln(z) = 1/2 => z = e^(1/2) = sqrt(e).