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The graph of the function y = xcosx + SiNx is approximately () A. B. C. D.
Because the function y = xcosx + SiNx is odd, option B is excluded. When x = π 2, y = π 2 × cos π 2 + sin π 2 = 1 > 0; when x = π, y = π × cos π + sin π = - π < 0. Thus option a and option C can be excluded. Therefore, the correct option is d. therefore, D is selected
Is x the analytic expression of Y function image y =, or x=
x=
If it's y, it's y about X=
It is proved that the function y = 1 / xsin1 / X is unbounded in the interval (0,1), but it is not infinite when x ~ 0
Let xn satisfy 1 / xn = 2n π + π / 2, when n -- > ∞ X -- > 0, when y = 2n π + π / 2 -- > infinity, most of them are unbounded
The de sequence yn satisfies 1 / yn = 2n π X -- > 0 when y = 0
So y is oscillatory, not infinite
It is known that a is a prime number, and a + 6, a + 8, a + 12 and a + 14 are prime numbers. What is such a prime number a
It's five
If a is divided by 5, then a + 14 can be divided by 5
If a is divided by 5 and there are more than 2, a + 8 can be divided by 5
If a is divided by 5 and 3 remains, a + 12 can be divided by 5
If a is divided by 5 and there are 4, a + 6 can be divided by 5
So a should be a multiple of 5, and because a is prime, a must be 5
Proving loga B = 1 / (logb a)
A is the base number, B is the true number
The proof process is required, not formula directly
Let loga B = X
b=a^x
Then logb a = loga ^ x a = 1 / X
So 1 / (logb a) = 1 / (1 / x) = X
So loga B = 1 / (logb a)
Let f (x) = LG (the square of MX - 2x + 1)
Given the function f (x) = LG (the square of MX - 2x + 1), find (1) the value range of real number m if the domain of function is R; (2) the value range of real number m if the domain of function is r
Detailed steps. Thank you~
Domain MX ^ 2-2x + 1 > 0
If M = 0, then - 2x + 1 > 0 is not constant
If M is not equal to 0, then the quadratic function is always greater than 0, so the opening is upward, M > 0
And the discriminant is less than 0
So 4-4m1
So m > 1
The range is r
Then the true number must take all positive numbers
If M = 0, then MX ^ 2-2x + 1 = - 2x + 1 can take all positive numbers
If M is not equal to 0, it is a quadratic function
At this time, the opening must be upward, M > 0, and the minimum value cannot be greater than 0, otherwise the positive number between 0 and the minimum value cannot be taken
So quadratic function and X axis have intersection, that is, the discriminant is greater than or equal to 0
So 4-4m > = 0
0
A 6 / 5 fractional unit is a fraction. Adding a few such fractional units is the smallest prime number
A 6 / 5 fractional unit is one fifth. Adding four such fractional units is the smallest prime number (2)
26. Let the probability density of two-dimensional random variables be: (1) the marginal probability density of X and Y; (2)
26. Let the probability density of two-dimensional random variables be: (1) the marginal probability density of X and Y; (2)
1
fx(x)=∫(0~2)1/6 dy=1/3 (x~(0,3))
fy(y)=∫(0~3)1/6 dx=1/2(y~(0,2))
two
∫(0~2)∫(0~2-y) 1/6 dxdy
=∫(0~2)(2-y)/6 dy
=y/3-y²/12|(0~2)
=2/3-4/12
=1/3
The function f (x) = cos2x - (cosx-1) cosx is known
Comprehensive application of double angle formula and substitution method
f(x)=2(cosx)^2-1-(cosx)^2+cosx=(cosx)^2+cosx-1
Let t = cosx be [- 1,1]
Then f (T) = T ^ 2 + T-1 = (T + 1 / 2) ^ 2-5 / 4
Because t belongs to [- 1,1]
So when t = - 1 / 2, there is a minimum of - 5 / 4
Three seventh of a is B, that is, what is a equal to
From the question, 3A = 7b, that is, a = 7 / 3 B
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