If we know that the vector a = (root 3 * cosx, 0), B = (0, SiNx), the square of the function f (x) = (a + b) + root 3 * sin2x, then if the image of the function is translated according to the vector D, The obtained image is centrosymmetric with respect to the origin, and increases monotonically on [0, π / 4]. Find the vector D with the smallest length,

If we know that the vector a = (root 3 * cosx, 0), B = (0, SiNx), the square of the function f (x) = (a + b) + root 3 * sin2x, then if the image of the function is translated according to the vector D, The obtained image is centrosymmetric with respect to the origin, and increases monotonically on [0, π / 4]. Find the vector D with the smallest length,


The function is reduced to f (x) = 2Sin (2x + π / 6) + 2
According to the vector (π / 12, - 2), the moving distance is the smallest, d = under the root sign (π & # 178 / 144 + 4)
(the result is so awkward ~)



Given that an acute angle of a right triangle is 60 ° and the length of the hypotenuse is 1, then the perimeter of the right triangle is ()
A. 3B. 3C. 3+2D. 3+32


As shown in the figure, in RT △ ABC, ∠ B = 60 °, ab = 1, then ∠ a = 90 ° - 60 ° = 30 °, so BC = 12ab = 12 × 1 = 12, AC = AB2 − BC2 = 12 − (12) 2 = 32, so the perimeter of the triangle is 3 + 32



(with graph) points D and E are on the sides AB and BC of triangle ABC respectively. Please find a point P on AC to minimize the perimeter of triangle Dep
Please write down the practice
I've thought about this method on the first floor. Please give me the reasons


1. Make the symmetric point m of point e with respect to AC
2. Connect DM to AC at point P
Then point P is the point
In this case, the perimeter of △ dep is the shortest
The de length is fixed, as long as EP + DP is minimum
You can take any Q in AC, then DP + EP



As shown in the figure, it is known that △ ABC, points D and E are on AB and BC respectively. Please make a point P on AC to minimize the perimeter of △ dep. (keep the trace of drawing)


As shown in the figure: make the symmetric point F of point d with respect to AC, connect EF, and the intersection point with AC is the calculated point P. suppose that Q is the calculated point, does not coincide with point P, and connects QD, QE, QF, QE + QF > ef (that is: EP + PD), so point P is the calculated point