JEE Mains Model Paper 1

f(x) = x ln x - 2x f'(x) = ln x + 1 - 2 = ln x - 1 For maximum or minimum, f'(x) = 0 ⇒ ln x = 1 ⇒ x = e Now evaluate f at the endpoints and at the critical point: f(1) = 1(ln 1 - 2) = -2 f(e) = e(ln e - 2) = e(1 - 2) = -e ≈ -2.718 f($e^2$) = $e^2$(ln $e^2$ - 2) = $e^2$(2 - 2) = 0 ∴ minimum value = -e (at x = e) and maximum value = 0 (at x = $e^2$) Difference = 0 - (-e) = e Hence the correct answer is e.