If P is on the image of curve y = LNX, q is on the line y = x, and PQ ‖ Y axis, then the minimum distance from P to the line y = x is? And the minimum value of PQ is?

When the distance from P to the line y = x is the smallest, the tangent of y = LNX at P is parallel to y = x, and the slopes of both are 1
y = lnx
y' = 1/x = 1,x = 1
P(1,0)
The minimum distance from P to the line X - y = 0 d = | 1 - 0 | / √ (1 + 1) = √ 2 / 2
Let g (x) = x - LNX
g'(x) = 1 - 1/x = 0
x = 1
When x = 1, the distance of PQ is the smallest; P (1,0), q (1,1), the distance is 1

It is known that rational number x satisfies 1 / 2 (3x-1) - 7 / 3 ≥ X-1 / 3 (5 + 2x). If the maximum value of x-3 - x + 2 is p and the minimum value is Q, find PQ

1/2(3x-1)-7/3≥x-1/3(5+2x)
Multiply both sides by 6
3(3x-1)-14≥6x-2(5+2x)
9x-3-14≥6x-10-4x
7x≥7
x≥1
Y = x-3 - x + 2
(1) x≥3 y=(x-3)-(x+2)=-5
(2)1≤x

It is known that 3x-1 / 2-7 / 3 > = - 5 + 2x / 3, if the maximum value of | x-3 | - | x + 2 | is p and the minimum value is Q, then PQ=

3x-17/6>=-5+(2/3)x
(7/3)x>=-13/6
x>=-(14/13)
According to the situation (1)

If the points P (- 2, a), q (B, 3) are parallel to the Y-axis and the length of PQ is 2, then a =? B =?

Because of PQ / / Y axis, the abscissa of P and Q is equal, so B = - 2, so p (- 2, a), q (- 2.3) because the length of PQ is 2, so / A-3 / = 2, a = 1 or a = 5

If P is on the image of curve y = LNX, q is on the line y = x, and PQ ‖ Y axis, then the minimum distance from P to the line y = x is? And the minimum value of PQ is?