If loga ^ 2 (3-ax) (a is not equal to 0 and a is not equal to plus or minus 1) is a decreasing function on [0,2], then the range of a

If loga ^ 2 (3-ax) (a is not equal to 0 and a is not equal to plus or minus 1) is a decreasing function on [0,2], then the range of a

Logarithmic base a > 0
So the coefficient of X in 2 (3-ax) - 2a0
The true number is greater than 0
True and decreasing
So when x = 2, the minimum true number is 2 (3-2a) > 0
A
The discontinuous point of function f (x) = (X-2) / (LN | X-1 |) is
1,|x-1|x =1;
2, | X-1 | = 1 = > x = 2 or x = 0;
So the discontinuity point is x = 0,1,2
X = 1 and x = 2
The discontinuous point of function f (x) = (X-2) / (LN | X-1 |) is x = 1
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, and the sum of the first n terms of the arithmetic sequence {BN} be TN. the common ratio of the known sequence {BN} is Q (Q > 0), A1 = B1 = 1, S5 = 45, T3 = a3-b2
(I) to find the general term formula of sequence {an}, {BN};
(II) seeking
Q
a 1 a 2 +
Q
a 2 a 3 +… +
Q
a n a n+1
One
S5=5a1+5×4d/2=5+10d=45 10d=40
D=4
an=a1+(n-1)d=1+4(n-1)=4n-3
When n = 1, A1 = 4-3 = 1, which is also satisfied
The general formula of sequence {an} is an = 4n-3
T3=b1+b2+b3=b1(1+q+q^2)=1×(1+q+q^2)=1+q+q^2
a3-b2=4×3-3-b1q=9-q
T3=a3-b2
1 + Q + Q ^ 2 = 9-q
q^2+2q-8=0
(q+4)(q-2)=0
q=-4(
S5=45=5a3
a3=9
a1=1
2d=a3-a1=8
D=4
an=4n-3
T3=a3-b2
b1+b2+b3=a3-b2
1+2b2+b3=9
1+2q+q^2=9
q^2+2q-8=0
(q-2)(q+4)=0
Q=2
bn=2^(n-1)
The second question is not very clear
If we know that (1,0,1,0) t is a basic solution system of AX = 0, and a = (A1, A2, A3, A4) is a matrix of order 4, we know that the rank of a is 3?
Isn't it a basic solution system? This is the sixth multiple choice question of the 2011 postgraduate entrance examination. I don't understand it
There is only one basic solution system, doesn't it mean that the rank is 3?
If there are two basic solution systems, the rank is 2
Number of solution systems + rank = order 4
Since it has been said that it is a basic solution system, it means that there is only one column vector in all the basic solution systems of the equations, so r of a is 4-1 = 3
Given that the function f (x) = - x * 3 + ax is an increasing function on [0,1], we can find the value range of real number a
Method 1: let 1 > x2 > X1 > 0f (x2) = - x2 ^ 3 + A * X2F (x1) = - X1 ^ 3 + A * x1f (x2) - f (x1) = (x2-x1) (a - (x1 ^ 2 + x2 ^ 2 + X1 * x2)) make f (x) = - x ^ 3 + ax an increasing function on (0,1), then f (x2) - f (x1) = (x2-x1) [a - (x1 ^ 2 + x2 ^ 2 + X1 * x2)] > = 0A > = X1 ^ 2 + x2 ^ 2 + X1 * x2 because 1 > x2 > X1 > 0
f'(x)=-3x^2+a
x∈[0，1]，f'(x)≥0
And f '(x) is a decreasing function on [0, + ∞),
So only f '(1) = - 3 + a ≥ 0
A ≥ 3
The number of discontinuities of the function f (x) = 1 / SiNx in the interval (- 2 π, 2 π) is
sinx=0；
x=-π,0,π；
So there are three short cuts
Let {an} be an equal ratio sequence with common ratio Q, and Sn be the sum of its first n terms. If {Sn} is an equal difference sequence, then Q=______ .
Let the first term be A1, then S1 = A1, S2 = a1 + a1qs3 = a1 + a1q + a1q2. Because {Sn} is an arithmetic sequence, so 2 (a1 + a1q) = a1 + A1 + a1q + a1q2q2q2-q = 0, the solution is q = 1. So the answer is: 1
Let a be a 5 * 4 matrix of rank 3, A1, A2, A3 be a system of non-homogeneous linear equations, ax = B has three different solutions, if (A1) + (A2) + 2 (A3) = (2,0,0,0) ^ t, 3A1 + A2 = (2,4,6,8) ^ t, then the general solution of AX = B is?
Solution: because R (a) = 3, the basic solution of AX = 0 contains 4-R (a) = 1 solution vector
So (3A1 + A2) - (a1 + A2 + 2A3) = (0,4,6,8) ^ t ≠ 0 is the basic solution system of AX = 0
(1 / 4) (a1 + A2 + 2A3) = (1 / 2,0,0,0) ^ t is a special solution of AX = B
So the general solution of AX = B is (1 / 2,0,0) ^ t + C (0,4,6,8) ^ t
Given that the function f (x) = - x2-ax-5, (x ≤ 1) ax, (x > 1) is an increasing function on R, then the value range of a is ()
A. -3≤a＜0B. -3≤a≤-2C. a≤-2D. a＜0
∵ function f (x) = - x2-ax-5, (x ≤ 1) ax, (x ＞ 1) is an increasing function on R. let g (x) = - x2-ax-5 (x ≤ 1), H (x) = ax (x ＞ 1) from the properties of piecewise function, the function g (x) = - x2-ax-5 monotonically increases in (- ∞, 1], the function H (x) = ax monotonically increases in (1, + ∞), and G (1) ≤ H (1) ∵ - A2 ≥ 1a ∵ 0-a-6 ≤ a ∵ a ≤ - 2A ∵ 0; 0; a ≥ - 3 solution can be obtained, - 3 ≤ a ≤ - 2, so B is selected
X = 0 is the function f (x) = 1 / 3 + 2 / X * SiNx / 3, which is the breakpoint of x0 or jump? Why?
Teach you a little trick to distinguish between the reachable infinity and the reachable discontinuity
For example, the following formula of X-1 / (x ^ 2 + 2x-3) can be decomposed into (x + 3) (x-1). At this time, we observe that x = 1 is removable. Why? Because the denominator of the molecule tends to 0. If - 3 is infinite discontinuity, because the molecule tends to - 4 and the denominator tends to 0, we can distinguish between removable and infinite discontinuity
As for the jumping discontinuity, the left and right limits are not equal, that is, the jumping is very good, and 0 is the jumping discontinuity
The left limit is 1, the right limit is 0, and the function is a jump breakpoint at 0..