Given that the line y = x + 1 is tangent to the curve y = ln (x + a), then the value of a is______ ．

Let the tangent point P (x0, Y0), then Y0 = x0 + 1, Y0 = ln (x0 + a), and the slope of the tangent equation y = x + 1 is 1, that is, y ′| x = x0 = 1x0 + a = 1, ∧ x0 + a = 1, ∧ Y0 = 0, x0 = - 1, ∧ a = 2

P is any point on the curve x-y-lnx = 0. Find the minimum distance from P to the straight line y = X-2
RT

Point P is any point on the curve y = x-lnx. When the tangent passing through point P is parallel to the straight line y = X-2, the distance from point P to the straight line y = X-2 is the smallest. The slope of the straight line y = X-2 is equal to 1, so that the derivative of y = x-lnx y ′ = 1-1 / x = K (slope) = 1, x = 1, or x = - 1 / 2 (rounding off), so the tangent parallel to the straight line y = X-2 on the curve y = x-lnx

If point P is any point on the curve y = x2-lnx, then the minimum distance from point P to the line y = X-2 is ()
A. 1B. 2C. 22D. 3

Let P (x0, x02-lnx0) have k = y ′| x = x0 = 2x0-1x0. | 2x0-1x0 = 1, | x0 = 1 or x0 = - 12 (rounding off). | P (1,1), | d = | 1 − 1 − 2 | 1 + 1 = 2

If point P is any point on the curve y = x2 LNX, then the minimum distance from point P to the straight line y = x + 2 is___ ．

Let P (x, y), then y ′ = 2x-1 x (x > 0), let 2x-1 x = 1, then (x-1) (2x + 1) = 0, ∵ x > 0, ∵ x = 1 ∵ y = 1, that is, the coordinate of the tangent point parallel to the straight line y = x + 2 and tangent to the curve y = x2 LNX is (1,1). From the distance formula from the point to the straight line, we can get d = | 1-1 + 2 | 2 = 2, so the answer is: 2

Point P is any point on the curve y = x ^ 2-lnx, then the minimum distance from point P to the straight line y = x + 2 is? Point P is the curve y=
If point P is any point on the curve y = x ^ 2-linx, then the minimum distance from point P to the line y = x + 2 is?
Point P is any point on the curve y = x ^ 2-lnx, then the minimum distance from point P to the straight line y = x + 2 is? And the coordinates of point P are obtained

Y '= 2x-1 / x, when y' = 1, x = 1, x = - 1 / 2 (rounding)
In this case, y = 1
The distance from point (1,1) to y = x + 2 is the root 2, which is the minimum distance from any point on the curve y = x ^ 2-linx to the straight line y = x + 2

Given that P is a point on the curve y = LNX, the minimum distance from point P to line y = x is ()
A. 1B. 22C. 2D. 2

Let P (x, LNX), x > 0, then the distance from point P to the straight line y = x D = | x-lnx | 2, let H (x) = x-lnx, then h ′ (x) = 1-1x, when 0 < x < 1, y ′ < 0.. H (x) min = H (1) = 1-ln1 = 1, and the minimum distance from point P to the straight line y = x is Dmin = | 1-ln1 | 2 = 22

P (x, y) is a moving point on the curve X ^ 2 / 4 + y ^ 2 = 1, then the maximum value of X + y is

P (x, y) is the moving point on the curve X ^ 2 / 4 + y ^ 2 = 1
Let P (2sina, COSA)
x+y= 2sina+cosa= √5sin（a+θ）
When sin (a + θ) = 1, the maximum value is √ 5

The curvature of any point on the curve (x-1) &# 178; + (Y-2) &# 178; = 9 is?
=I really can't do it,

A:
(x-1)²+(y-2)²=9
This is a circle with center (1,2) and radius r = 3
The curvature of a circle is always 1 / r = 1 / 3

Monotone increasing and decreasing interval of sine function and cosine function, for example, monotone increasing interval of y = SiNx is [2K Wu - Wu / 2,2k Wu + Wu / 2]

Y = SiNx increase [2K π - π / 2,2k π + π / 2]
Minus [2K π + π / 2,2k π + 3 π / 2]
Y = cosx increase [(2k-1) π, 2K π]
Minus [2K π, (2k + 1) π]
k∈Z

It is proved that the sine function y = SiNx is a continuous function on R
Delta y = sin (x + △ x) - SiNx = 2Sin (△ X / 2) cos (x + △ X / 2), which formula is used to get the equation? Explain in detail

Trigonometric function sum difference product formula sin α - sin β = 2cos [(α + β) / 2] · sin [(α - β) / 2]

Given that the line y = x + 1 is tangent to the curve y = ln (x + a), then the value of a is______ ．