Given a, B, x, y ∈ R, and a ^ 2 + B ^ 2 = 1, x ^ 2 + y ^ 2 = 4, then the maximum value of AX + by is Such as the title

Given a, B, x, y ∈ R, and a ^ 2 + B ^ 2 = 1, x ^ 2 + y ^ 2 = 4, then the maximum value of AX + by is Such as the title

Let a = sinm, then B ^ 2 = 1 - (sinm) ^ 2 = (COSM) ^ 2
The range of COSM is symmetric with respect to the origin
So let B = COSM
Let x = 2cosn, then y ^ 2 = 4 (Sinn) ^ 2
So y = 2sinn
ax+by=2sinmcosn+2cosmsinn=2sin(m+n)
So max = 2
The maximum value of F (x) = the square of X + ax + 3 on 0 less than or equal to x less than or equal to 1
If the axis of symmetry is on the left side of x = 0, f (x) increases in the interval, min = f (0), max = f (1) 2. If the axis of symmetry is on the right side of x = 1, f (x) decreases in the interval, min = f (1), min = f (0) 3. If the axis of symmetry is on (0,1), min = f (...)
If the square of x plus ax minus 15 equals [x plus 3] times [x plus b], then the value of a is
(x+3)(x+b)
=x²+3x+bx+3b
=x²+(3+b)x+3b
=x²+ax-15
According to the equivalence of the corresponding terms
3b=-15 3+b=a
b=-5
a=-2
x²+ax-15=(x+3)(x-b)=x²+(3-b)x-3b
So we have - 3B = - 15, 3-B = a
b=5,a=-2
x^2+ax-15=(x+3)*(x+b)=x^2+(3+b)x+3b
So 3B = - 15
b=-5
a=3+b=3-5=-2
x^2+ax-15=(x+3)(x+b)
===> x^2+ax-15=x^2+(3+b)x+3b
===> ax-15=(3+b)x+3b
===> [a-(3+b)]x=3b+15
If the above formula holds for any x, then:
a-(3+b)=0；3b+15=0
So, B = - 5, a = 3 + (- 5) = - 2
Given that the image of power function y = x & nbsp; m2-2m-3 (m ∈ n *) is symmetric about y axis and is a decreasing function on (0, + ∞), the range of a satisfying inequality (2A2 + 1) - M ＜ (4-A) - M is obtained
∵ power function y = x & nbsp; m2-2m-3 (m ∈ n *) is a decreasing function on (0, + ∞), and ∵ m2-2m-3 ＜ 0. The solution is - 1 ＜ m ＜ 3, ∵ m ∈ n *, ∵ M = 1 or 2. When m = 1, y = x-4 is an even function satisfying the condition, and when m = 2, y = x-3 is an odd function not satisfying the condition
If the function f (x) = a-2x + 1 is an odd function on R, then the real number a=
F (x) = 1 / (a-2x + 1) is an odd function,
Then f (- x) = - f (x), in particular,
There is f (- 1) = - f (1)
That is 1 / (a + 3) = - 1 / (A-1)
So a + 3 = 1-A,
The solution is a = - 1
Given that the image of power function y = xm-3 (m ∈ n *) is symmetric about y axis and monotonically decreasing on (0, + ∞), then M=______ .
The image of power function y = xm-3 is symmetric about y axis, and decreases at (0, + ∞), M-3 ＜ 0, and M-3 is even number. From M-3 ＜ 0 to m ＜ 3, let m be a positive integer, so the value of M may be 1 or 2. Only when m = 1 is verified, can M-3 be even number, so m = 1 is the solution
Find the range of function y = x & # 178; + 2ax-3 on [- 2,4]
Syllogism, axis of symmetry and three paragraph discussion of region (- 2,4)
Axis of symmetry B / (- 2A) = - A
① - A is in the (- 2,4) region, i.e. - 2
(a) = ^ 2, - A-3
There are three cases according to the vertex
-A2 is an increasing function on [- 2,4], the minimum is 4-4a-3 = 1-4a, and the maximum is 16 + 8a-3 = 13 + 8A
-a> 4, i.e. A4, i.e. a
F (x) = (m ^ 2 + m-1) x ^ m is a power function to find the value of real number M
∵ f (x) = (M & # 178; + m-1) x ^ m is a power function
　　∴m²+m-1=1
　　∴m²+m-2=0
　　(m+2)(m-1)=0
‖ M = - 2 or M = 1
It is known that the range of the function f (x) = ax2-2ax + 2 + B (a > 0) in the interval [2,3] is [2,5] (I) find the value of a and B; (II) if the function g (x) = f (x) - (M + 1) x in the interval [2,4] is monotone, find the range of the real number M
(I) ∵ a ＞ 0, so the opening of parabola is upward and the axis of symmetry is x = 1. The function f (x) increases monotonically on [2,3]. From the condition, f (2) = 2F (3) = 5, that is, 2 + B = 23a + 2 + B = 5, the solution is a = 1, B = 0. So a = 1, B = 0
Given that the image of power function f (x) = xm2-m-3 has no intersection with y axis and is symmetric with respect to y axis, m belongs to Z to find M
Y axis has no intersection point and is symmetrical about y axis. When x = 0, y = 0 is passing through the origin
So 0-m-3 = 0 m = - 3