Given the function f (x) = 4x + 1 X. (1) find the zeros of the function y = f (x) - 4; (2) prove that the function f (x) is an increasing function in the interval (12, + ∞)

Given the function f (x) = 4x + 1 X. (1) find the zeros of the function y = f (x) - 4; (2) prove that the function f (x) is an increasing function in the interval (12, + ∞)

(1) because f (x) - 4 = 4x + 4x-4, Let f (x) - 4 = 0, and get 4x + 1-4 = 0, that is, 4x2-4x2-4x + 1 = 0, and the solution is x = 12. So the zero point of the function y = f (x) - 4 is 12. Let (2) let x1, X2 is any two real numbers on the interval (12, + ∞) (12, + ∞) and x1 \x X2, then f (x1) - f (x1) - f (x1) - f (x1) - f (x (x1) - f (x2) = 4x1 + 4x1 + 1 (x (x (x (x1) - 4 (x1) - f (x1) - f (x1) - f (x1) - f (x1) - f (x1) - f (x1) - f (x1) - f (x (x1) - f (x1) - f (x (x (x (x2) - 4 (x1) - 4) is (x1) - 4 (x1) - 4) 4 (x1-x2) 4 (x1-x2) X1X 2-14 x 1 x 2 ＞ 0, then f (x 1) ＞ f (x 2), so the function f (x) is an increasing function in the interval (12, + ∞)

If f (x) = X-1 / x, then the zero of F (4x) = x is zero

Plus or minus six root sign three

The function f (x) = (radical 1 + A / 2) sin (2x - π / 4) - 1 (a belongs to R, a is a constant), and π / 4 is the zero point of function y = f (x)
1. Find the value of a and the minimum positive period
2. When x belongs to [0, π / 2], find the maximum value of the function and the corresponding x value when f (x) gets the maximum value
To process, urgent!

(π / 4) = (1 / 4) = (1 / 4) = (1 / 2) = (1 / 2) = (1 / 2) = (1 A / 2) = (1 A / 2) = (1 A / 2) = root 2, if you here (root 1 A / 2) refers to the root ((1 a) / 2) then a = 3; if you here (root 1 A / 2) refers to the root ((1 a) / 2) then a = 3; if you here (root 1 A / 2) refers to the root ((1 (1 (A / 2) (π / 2)) (2) then a = 3; make 2x π π / 4 = π / 2 2K / 2, 2K π π / 2K π = > x = 3 π / 8 2K / 2K π π / 8 2K π / 8 2K π = 3 π / 8 K π π / 8, because x belongs to [0, π [0, π [0, π= root 2

Two tangent lines of a circle are drawn from a point m (x0, Y0) outside the circle. The line connecting the tangent points is called the tangent chord of point m about the circle,
If the equation of a circle is x ^ 2 + y ^ 2 = R ^ 2 and the point m (x0, Y0) is outside the circle, prove that the linear equation of the tangent chord of the circle is x0 * x + Y0 * y = R ^ 2

Let two tangent points be a (x1, Y1), B (X2, Y2)
Then the tangent passing through point a is X1X + y1y = R ^ 2
The tangent through point B is X2X + Y2Y = R ^ 2
∵ both tangents pass through point m (x0, Y0)
∴ x1x0+y1y0=r^2
x2x0+y2y0=r^2
The points a (x1, Y1) and B (X2, Y2) satisfy the equation x0x + y0y = R ^ 2
The equation of line AB is x0x + y0y = R ^ 2

For point P (x0, Y0) and circle C: x2 + y2 = R2, write a continuous program to judge the position relationship between P and circle C, and use pseudo code to express it

read Xo
read Yo
read r
M=Xo^2+Yo^2
If (m ＞ R ^ 2) then print "out of circle"
Else if (M = R ^ 2) then print "on a circle"
Else print "inside the circle"

If M (x0, Y0) is a point in circle x2 + y2 = A2 (a > 0) which is different from the center of the circle, then the positional relationship between the line x0x + y0y = A2 and the circle is ()
A. Tangency B. intersection C. separation D. tangency or intersection

According to the equation of circle, the center coordinate of circle is (0,0), radius r = a, and M is a point in the circle. If X02 + Y02 < A, then the distance from the center of circle to the known straight line d = |− A2 | X02 + Y02 > A2A = a = R, so the position relationship between the straight line and the circle is: phase separation. So select C

Given that P (x0, Y0) is a point in the circle x ^ 2 + y ^ 2 = a ^ 2 which is different from the center of the circle, then the position relationship between the line X * x0 + y * Y0 = a ^ 2 and the circle is ()?

P (x0, Y0) is the point in the circle x ^ 2 + y ^ 2 = a ^ 2 which is different from the center of the circle
So x0 ^ 2 + Y0 ^ 2

When passing through point P (2,3) and making two tangent lines PA, Pb of circle x2 + y2 = 1, a and B are tangent points, then the equation of straight line AB is___ ．

As shown in the figure, point P connects the origin o of coordinates, then OP = 9 + 4 = 13oa and ob are perpendicular to PA and Pb respectively, the angle between OP and OA is α, then cos α = the distance from the center of 113 circle to the straight line AB: D = Oh = aocos α = 113, the slope of straight line OP K '= 32, then the slope of straight line AB k = - 23, let the linear equation be y = - 23x + B, that is, 2x + 3Y -

The tangent point PA, Pb, a, B of the leading circle x ^ 2 + y ^ 2 = 4 passing through point P (1,2) is used to solve the linear equation passing through the tangent point a, B_________
Such as the title
It is the tangent point of PA, Pb, a and B passing through the point P (3,6) leading circle x ^ 2 + y ^ 2 = 4. Find the linear equation passing through the tangent point a and B_________

Obviously, we can get the position of a tangent point (0,2)
And the slope of OP is 2
Then the slope of AB is - 1 / 2
So the equation of AB is y = - X / 2 + 2
Very speechless
In that case,
OP=3√5
Draw a circle with P as the center and PA as the radius
Then PA ^ 2 = OP ^ 2-oa ^ 2 = 45-4 = 41
So the circle is
(x-3)^2+(y-6)^2=41
The equation of subtracting circle
We get - 6x + 9-12y + 36 = 41-4
3x+6y-4=0

If the tangent of the curve f (x) = x · SiNx + 1 at x = π 2 is perpendicular to the straight line ax + 2Y + 1 = 0, then the real number a is equal to______ ．

F '(x) = SiNx + xcosx, f' (π 2) = 1, that is, the slope of the tangent of the function f (x) = xsinx + 1 at the point x = π 2 is 1, and the slope of the straight line ax + 2Y + 1 = 0 is − A2, so (− A2) × 1 = − 1, the solution is a = 2