If vector a and vector B are not collinear, what conditions will be satisfied?

If vector a and vector B are not collinear, what conditions will be satisfied?

If the vector a coordinates are (x, y) and the vector b coordinates are (F, g), then XG FY is not equal to 0

f(x)=2sin^2x+cos^2x+sinxcosx,x∈R,f(π/12)
f(x)=2sin^2x+cos^2x+sinxcosx,x∈R
Ask for:
1) The value of F (π / 12)
2) The minimum value of F (x) and the corresponding value of X
3) The increasing interval of F (x)

f(x) = 2sin^2x+cos^2x+sinxcosx
= 1 + sin^2x + 1/2sin2x
= 1 + (1 - cos2x)/2 + 1/2sin2x
= 3/2 + 1/2(sin2x - cos2x)
= 3/2 + 1/sqrt(2) (sin(2x - π/4))
So f (π / 12) = 3 / 2 + 1 / sqrt (2) * sin (- π / 12) = (7 - sqrt (3)) / 4
Min (f (x)) = 3 / 2 - 1 / sqrt (2) = (3 - sqrt (2)) / 2, then 2x - π / 4 = - π / 2, x = - π / 8
F (x) increases on [- π / 8 + K π, 3 π / 8 + K π]

lim(x->0)(tan3x+2x)/(sin2x+3x)

The final result is 1

Given that the equation x ^ 2-2x-m + 1 = 0 of X has no real root, try to judge the case of the root of x ^ 2 - (M + 2) x - (2m + 1) = 0

When - 6-2 root 7

If the equation x2-2x-m + 1 = 0 has no real roots, we prove that the equation X2 - (2m-1) x + M2-2 = 0 has two unequal real roots

It is proved that: ∵ equation x2-2x-m + 1 = 0 has no real root, ∵ (- 2) 2-4 × 1 × (- M + 1) ＜ 0, the discriminant of the root of M ＜ 0, X2 - (2m-1) x + M2-2 = 0 △ = (2m-1) 2-4 (M2-2) = - 4m + 8, ∵ m ＜ 0, ∵ - 4m + 8 ＜ 0 is obtained, that is, ∵ equation X2 - (2m-1) x + M2-2 = 0 has two unequal real roots

Given the quadratic function y = (1-m) x square of the image opening down, the value range of M

Quadratic coefficient 1-m1

If the line y = m intersects with the image of quadratic function y = 1-x square, the range of m value is obtained

∵ the line y = m intersects the image of quadratic function y = 1-x ^ 2
The equations of y = m and y = 1-x ^ 2 have solutions
∴m=1-x^2,
That is, x ^ 2 = 1-m has a solution
∴1-m≥0
∴m≤1

If the square of the function y = (m-2) x is a quadratic function with respect to X and its image opening is downward, then the value range of M is

m-2

In the univariate quadratic equation (k-1) x ^ 2 - (K + 2) X-1 = 0, when k is a value, the equation has only one real root
It's to be finished by seven tomorrow morning, thanks

(k-1)x^2-(k+2)x-1=0
There is only one real root
k-1≠0
Discriminant
△=(k+2)²+4(k-1)=0
k²+8k=0
k(k+8)=0
We get k = 0 or K = - 8
To sum up
K = 0 or K = - 8

On the quadratic equation of X with one variable X-2 (2-k) x + K + 12 = 0, there are real roots, and the range of K is obtained

According to the meaning of the title
Discriminant = 4 (2-k) &# 178; - 4 (k) &# 178; + 12) ≥ 0
k²-4k+4-k²-12≥0
4k≤-8
k≤-2