Let p be on the curve y = 12ex + 1 and Q be on the curve y = ln (2x-2), then the minimum value of | PQ | is () A. 1-ln2B. 2（2-ln2）C. 1+ln2D. 2（1+ln2）

Let p be on the curve y = 12ex + 1 and Q be on the curve y = ln (2x-2), then the minimum value of | PQ | is () A. 1-ln2B. 2（2-ln2）C. 1+ln2D. 2（1+ln2）

∵ function y = 12ex + 1 and function y = ln (2x-2) are reciprocal functions. The image of ∵ function y = 12ex + 1 and function y = ln (2x-2) is symmetric with respect to line y = X. the minimum value of | PQ | is twice the shortest distance from point P to line y = X. let the tangent of slope 1 on curve y = 12ex + 1 be y = x + B, ∵ y ′ = 12ex

Let P move on the circle (x + 1) ^ 2 + (Y-1) ^ 2 and Q move on the curve xy = 1 (x > 0), then the minimum value of PQ is 0
Such as the title

We can find the minimum value from O to Q, then the minimum value of PQ is OQ minus radius

What is the minimum value of PQ when p is on the curve y = 1 / 2ex and Q is on the curve y = ln (2x)

∵ function y = 1 / 2E ^ X and function y = ln (2x) are reciprocal functions. The distance from point P (x, 1 / 2 e ^ x) on symmetric function y = 1 / 2E ^ x to straight line y = x is d = | 1 / 2E ^ x-x | / √ 2. Let g (x) = 1 / 2 e ^ x-x, (x > 0), then G '(x) = 1 / 2 e ^ X-1. From G' (x) = 1 / 2 e ^ X-1 ≥ 0, X ≥ LN2 can be obtained. From G '(x) = 12 e ^ X-1 < 0, 0 < x < LN2 can be obtained, +When x = LN2, the function g (x) min = 1-ln2 Dmin = (1-ln2) / √ 2 is obtained from the symmetry of the image with respect to y = x: the minimum value of | PQ | is 2dmin = √ 2 (1-ln2)

Let p be on the curve y = 12ex and Q be on the curve y = ln (2x), then the minimum value of | PQ | is ()
A. 1-ln2B. 2(1-ln2)C. 1+ln2D. 2(1+ln2)

∵ function y = 12ex and function y = ln (2x) are reciprocal functions. The image is symmetric with respect to y = X. the distance from point P (x, 12ex) on function y = 12ex to straight line y = x is d = | 12ex-x | 2. Let g (x) = 12ex-x (x > 0), then G ′ (x) = 12ex-1, X ≥ LN2 can be obtained from G ′ (x) = 12ex-1 ≥ 0, and 0 < x < LN2 can be obtained from G ′ (x) = 12ex-1 < 0 When x = LN2, the function g (x) min = 1-ln2, Dmin = 1-ln22. From the symmetry of the image with respect to y = x, the minimum value of | PQ | is 2dmin = 2 (1-ln2)