Mean Value Theorems

The Lagrange mean value theorem states that for a function continuous on [a,b] and differentiable on (a,b), there is a point c where f'(c) equals the average rate of change (f(b)-f(a))/(b-a). Here f(x) = \ln x is smooth on [1,e], with f(1) = 0 and f(e) = 1, so the secant slope is (1-0)/(e-1) = 1/(e-1). Since f'(x) = 1/x, we set 1/c = 1/(e-1), giving c = e - 1. Because e - 1 \approx 1.718 lies strictly between 1 and e \approx 2.718, the point is valid. Option 1 is the left endpoint where the slope 1 exceeds the secant slope. Option (e+1)/2 is the arithmetic midpoint, which would only be the answer for a quadratic, not a logarithm. Option \sqrt{e} is the geometric mean, relevant to different problems, not this secant condition. The governing JEE pattern is solving f'(c) = secant slope for the Lagrange point. Plausibility check: the tangent to \ln x is steeper near x = 1 and flatter near x = e, so the matching point sits left of centre, and e - 1 \approx 1.72 is indeed left of the midpoint 1.86, confirming consistency.