If f (x) = (1 + TaNx) cosx, 0 ≤ X

If f (x) = (1 + TaNx) cosx, 0 ≤ X

f(x)=(1+tanx)cosx
=cosx+sinx
=√2((√2/2)cosx+(√2/2)sinx)
=√2sin(x+π/4)
F (x) increases in [0, π / 4] and decreases in [π / 4, π / 2]
So the maximum value is: when x = π / 4, f (π / 4) = √ 2
The minimum value is: when x = 0, f (0) = √ 2 / 2
F (x) = cosx + SiNx = radical 2Sin (x + π / 4)
The maximum and minimum values are positive and negative root sign 2
Using trigonometric identity transformation
Given vector M = (cosx, 1-asinx), n = (cosx, 2), where a ∈ R, X ∈ R, Let f (x) = Mn, and the maximum value of function f (x) is g (a)
1. Find the analytic expression of the function g (a)
2, let 0 ≤ x < 2 π, find the maximum and minimum of G (2cosx + 1) and the corresponding x value
(1) f(x)=mn=(cosx)^2+2-2asinx=1-(sinx)^2+2-2asinx=-(sinx+a)^2+a^2+3
When a ∈ [- 1,1], G (a) = a ^ 2 + 3. When A1, G (a) = (A-1) ^ 2 + A ^ 2 + 3
(2) G (2cosx + 1) = ① (2cosx + 1) ^ 2 + 3, when x ∈ [2,3 π / 2]
② [(2cosx + 1) - 1] ^ 2, when x ∈ [0, π / 2) ∪ (3 π / 2, π)
The rest is worth your own calculation
If the equation AX ^ 2 + 2x + 1 = 0 has no negative root, then the value set of a is
When a = 0, then the equation is a linear equation of one variable. We can directly calculate the root x = - 1 / 2 of the equation, which is obviously not satisfactory, so a cannot be equal to 0
When a is not equal to 0, then the equation is a quadratic equation of one variable. According to the meaning of the problem, if the equation has no negative root, then it can only be in the following three cases: 1. No real root; 2. There are two equal non negative roots (positive roots or 0); 3. There are two unequal non negative roots (positive roots or 0)
For 1, let △ = 4-4a1;
For 2, let △ = 4-4a = 0, and get a = 1, but at this time, the two roots of the equation are x = - 1, which is not suitable for the problem, so this case does not hold for the problem
For 3, let △ = 4-4a > 0, and get A1}
If the equation is a linear equation, then a = 0, x = - 1 / 2;
If the equation is quadratic, if the equation has no root, then △ = 4-4a1;
When the equation has one root, a = 1, the root of the equation is x = - 1, which is not suitable for the problem;
When the equation has two unequal roots, there are only two positive roots
... unfold
If the equation is a linear equation, then a = 0, x = - 1 / 2;
If the equation is quadratic, if the equation has no root, then △ = 4-4a1;
When the equation has one root, a = 1, the root of the equation is x = - 1, which is not suitable for the problem;
When the equation has two unequal roots, there are only two positive roots
If - 2 / a > 0, 1 / a > 0, a does not exist.
In conclusion, the value set of a is {a | a > 1}. Put it away
If f (x) is a function defined on R, f (0) = 1, and for any real number x, y always has f (x + Y / 2) = f (x) + y (2x + y + 1), find the analytic expression of F (x)
In F (x + Y / 2) = f (x) + y (2x + y + 1), let x = 0
f(y/2)=f(0)+y(y+1)=y²+y+1
Let Y / 2 = x, then we get
f(x)=4x²+2x+1
Note: if the equation f (x + Y / 2) = f (x) + y (2x + y + 1) is strictly considered, it is impossible to hold for any X and Y. the reasons are as follows
Let y = - 4x in the formula, then the above formula becomes
f(-x)=f(x)-4x(-2x+1) (1)
That is, formula (1) holds for all x ∈ R
In formula (1), X is replaced by - x, and
f(x)=f(-x) +4x(2x+1) (2)
So (2) holds for all x ∈ R
(1) + (2) get
f(-x)+f(x)=f(x)+f(-x)+16x² (3)
That is to say, formula (3) holds for any x ∈ R, but it is obviously impossible,
Because (3) is 16x & # 178; = 0, it holds only when x = 0
f(x+y/2)=f(x)+y(2x+y+1)
Let y = - 2x
f(0)=f(x)+(-2x)(2x-2x+1)
=f(x)+(-2x)(2x+(-2x)+1)
=f(x)-2x=1
So f (x) = 2x + 1, thank you. Your answer is exactly the same as that in the book. Here's what I do: in F (x + Y / 2) = f (x) + y (2x + y + 1), let x = 0, then f (Y / 2) = f (0) + y (y + 1) =... Expand
f(x+y/2)=f(x)+y(2x+y+1)
Let y = - 2x
f(0)=f(x)+(-2x)(2x-2x+1)
=f(x)+(-2x)(2x+(-2x)+1)
=f(x)-2x=1
So f (x) = 2x + 1 question: Thank you, your answer is exactly the same as that in the book. Here's what I do: in F (x + Y / 2) = f (x) + y (2x + y + 1), let x = 0, get f (Y / 2) = f (0) + y (y + 1) = y & sup2; + y + 1, and then let Y / 2 = x, get f (x) = 4x & sup2; + 2x + 1. What I do is reasonable. What's wrong with me?
Can the "root of equation M2 + 1 = 0" form a set? Can the number of elements in the set be 0?
It can be a set, which is an empty set
Let f (x) be a function defined on R and satisfy f (0) = 1, and for any real number x, y, f (X-Y) = f (x) - Y (2x-y + 1), find the analytic expression of F (x)
Because for any real number x, y, f (X-Y) = f (x) - Y (2x-y + 1)
So let x = y and substitute it into the above formula
f(0)=f(x)-x(2x-x+1)
f(x)=x^2+x+f(0)
Because f (0) = 1
So f (x) = x ^ 2 + X + 1
Given that a, B, X are positive numbers, and LG (BX) is multiplied by LG (AX) + 1 = 0, the range of a / B is obtained
LG (BX) LG (AX) + 1 = 0, and a, B, X are positive numbers, then (LGA + lgx) (LGB + lgx) + 1 = 0 (lgx) ^ 2 + (LGA + LGB) lgx + 1 + lgalgb = 0 this equation has a solution, so (LGA + LGB) ^ 2-4lgalgb-4 ≥ 0 (LGA) ^ 2 + 2lgalhb + (LGB) ^ 2-4lgalgb-4 ≥ 0 (LGA LGB) ^ 2 ≥ 4 LGA LGB ≥ 2 or LGA -
Let f (x) be a function on R and satisfy f (0) = 1, and for any real number x, y, f (X-Y) = f (x) - Y (2x-y + 1),
Find the expression of F (x)
Let y = x be:
f(x - x) = f(x) - x(2x - x + 1)
f(0) = f(x) - x(x + 1)
1 = f(x) - x(x + 1)
f(x) = x(x + 1) + 1
f(x) = x² + x + 1
If a, B, X are positive numbers and LG (AX) LG (BX) + 1 = 0, the value range of a / B is obtained
Expand the original formula with a formula
(lga+lgx)(lgb+lgx)+1=0
(lgx)^2+(lga+lgb)lgx+lgalgb+1=0
If LG (AX) LG (BX) + 1 = 0, there is a solution
Then there must be lgx satisfying Formula 1
Δ≥0
(lga+lgb)^2-4(lgalgb+1)≥0
(lga-lgb)^2≥4
[lg(a/b)]^2≥4
LG (A / b) ≥ 2, LG (10 ^ 2) = 2, so a / b ≥ 100 or LG (A / b) ≤ - 2, LG (1 / 100) = - 2, so
Zero
A / b > 100, or 0
Let f (x) be a function defined on R. for any x, y ∈ R, f (x + y) = f (x) + F (y),
1. Find the value of F (0)
2. Prove that f (x) is an odd function
3. If the function f (x) is an increasing function on R, we know that f (1) = 1 and f (2a) > F (A-1) + 2, then we can find the value range of A
For any x, y ∈ R, f (x + y) = f (x) + F (y) 1. Find the value of F (0), f (x + y) = f (x) + F (y) let x = 0, y = 0; f (x + y) = f (0) = f (0) + F (0) f (0) = 02. Prove that f (x) is an odd function, f (x + y) = f (x) + F (y) let y = - x, then f (x-x) = f (x) + F (- x) = f (0) = 0, so f (- x) = 0
This is a problem of abstract function. We just need to assign a value. Let x = y = 0 get f (0) = F
(0) + F (0) the value of F (0) is 0
Let y = - x, then f (0) = f (x) + F (- x), which can be proved by knowing f (x) + F (- x) = 0
3 is also simple
(1) Let x = y = 0, f (0) = f (0) + F (0) f (0) = 0
(2) Let - x = y, f (0) = f (0) + F (0) f (- x) + F (x) = 0 be an odd function
(3) If f (x) is an increasing function on R, we know that f (1) = 1, f (2) = 2
f(2a)>f(a-1)+2 f(2a)>f(a-1)+f（2）=f(a+1)
If f (x) is an increasing function of R, 2A > A + 1, a > 1
1. Let x = y = 0, f (0) = 2F (0),
∴f(0)=0.
2.f(x)+f(-x)=f(x-x)=f(0)=0,
∴f(-x)=-f(x),
F (x) is an odd function.
3.f(2)=2f(1)=2,
∴f(a-1)+2=f(a+1),
∵ f (x) is an increasing function on R, f (2a) > F (A-1) + 2 = f (a + 1),
2 a > A + 1, a > 1.