The center of hyperbola C is at the origin, the right focus is f (2 √ 3 / 3,0), and the asymptotic equation is y = ± √ 3x (1) find the equation of hyperbola C (2) let the line L: y = KX + 1 intersect the hyperbola C at two points a and B, and ask: when the value of K is, the circle with diameter AB crosses the focus

The center of hyperbola C is at the origin, the right focus is f (2 √ 3 / 3,0), and the asymptotic equation is y = ± √ 3x (1) find the equation of hyperbola C (2) let the line L: y = KX + 1 intersect the hyperbola C at two points a and B, and ask: when the value of K is, the circle with diameter AB crosses the focus

(1) Xsquare-ysquare / 3 = 1
(2) Let's use Weida's theorem to calculate x1x2 =? X1 + x2 =?
Then because the circle passes through the origin, the vector OA * the vector ob = 0
Substitute the previous calculation of x1x2 X1 + x2 into k to calculate it

If f (x) = 1 / 2Sin (2x) + sin (x), then f '(x) is ()
The answer is "even functions with both maximum and minimum"

F '(x) = 1 / 2 * 2cos (2x) + cosx = cos2x + cosx = 2 (cosx) ^ 2 + cosx-1. First, because cosx is an even function, f' (x) is an even function. Then we can find that t = cosx, the range of T [- 1,1] = 2T ^ 2 + T-1 = 2 (T ^ 2 + 1 / 2T + 1 / 16) - 9 / 8 = 2 (t-1 / 4) ^ 2-9 / 8t = 1 / 4 is the minimum, and T = - 1 is the maximum

If a + C > b + C, then
If √ a > b, then a > B
Which is right?
Why?

If √ a > b, then a > B
That's right
Look at their range
The first one doesn't hold if a and B are negative