Three problems in mathematics
1. Fill in the blanks (6 small questions in this question, 4 points for each small question, out of 24 points. Fill in the answers on the lines in the questions)
(1) If its derivative function is continuous at x=0, the value range of is _ _ _ _.
(2) The given curve is tangent to the X-axis, which can be expressed by a as _ _ _ _ _ _ _ _.
(3) Let a>0 and D represent the whole plane, then = _ _ _ _.
(4) setting an n-dimensional vector; E is identity matrix of order n, matrix.
, ,
Where the inverse matrix of A is B, then A = _ _ _ _ _
(5) Let the correlation coefficient of random variables X and Y be 0.9, and if it is, the correlation coefficient of Y and Z is _ _ _ _ _ _ _ _.
(6) Let the population X obey the exponential distribution with parameter 2 and be a simple random sample of the population X, then when convergence in probability is in _ _ _ _ _.
Second, the multiple-choice question (this question ***6 small questions, each small question 4 points, out of 24 points. Only one of the four options given in each small question meets the requirements of the topic, and the letter before the selected option is filled in brackets after the topic)
(1) Let f(x) be a odd function that is not equal to zero and it exists, then the function.
(a) The left limit does not exist when x=0. (b) There is jump discontinuity x=0.
(c) The right limit does not exist when x=0. (d) It is enough for Rolle theorem to be applied to point x=0. The condition f(0)+f( 1)+f(2)=3 is equivalent to 1 between the maximum value of f(x). Finally, the intermediate value theorem can be used.
Because f(x) is continuous in the world and derivable in (c, 3), it must exist from Rolle's theorem, which makes.
Comment mean value theorem, differential mean value theorem and integral mean value theorem are all common knowledge points, which are generally tested in pairs. This problem is a typical case of the combination of mean value theorem and differential mean value theorem.
Nine, (full mark for this question 13)
Known homogeneous linear equation
When trying to discuss the relationship between B and B,
(1) equations have only zero solutions;
(2) The equations have nonzero solutions. When there is a non-zero solution, find the basic solution system of the equations.
The number of analytical equations is the same as the number of unknowns. The problem is transformed into whether the determinant of the coefficient matrix is zero, and the calculation of the coefficient determinant has obvious characteristics: all the corresponding elements in the column are equal after adding up. You can first add all the corresponding elements in a column, then put forward the common factor, and then multiply the (-1) of the first row by other rows to calculate the value of the determinant.
Detailed solution of coefficient determinant of equation
=
(1) If and when rank (A)=n, the system of equations has only zero solution.
(2) When b=0, the homosolution equation of the original equation is
It can be seen that it is not all zero. It can be assumed that the basic solution system of the original equation is
, ,
The coefficient matrix of the original equation can be simplified as
(Multiply-1 in line1by other lines, and then multiply from line 2 by line n)
(Multiply the number of times from the nth line to the second line by 1 line, and then move 1 line to the last line)
Thus, the same solution equation of the original equation is obtained as follows.
, , .
The basic solution system of the original equation is
In fact, the difficulty in commenting on this topic can also be analyzed as follows: at this time, the rank of the coefficient matrix is n- 1 (there is a sub-formula of order n- 1 that is not zero), which is obviously a non-zero solution of the equations and can be used as the basic solution system.
Ten, (full mark for this question 13)
Set quadratic form
,
The sum of eigenvalues of quadratic matrix A is 1, and the product of eigenvalues is-12.
(3) find the values of a and b;
(4) Transform the quadratic form into the standard form by orthogonal transformation, and write the orthogonal transformation and the corresponding orthogonal matrix.
The sum of eigenvalues is the sum of elements on the main diagonal of A, and the product of eigenvalues is the determinant of A, from which the values of A and B can be obtained. Further find the eigenvalues and eigenvectors of A, orthogonalize the eigenvectors with the same eigenvalues (if necessary), and then unitize the eigenvectors and take them as columns to construct the matrix, which is the obtained orthogonal matrix.
The matrix of (1) quadratic form f is detailed as follows.
Let the eigenvalue of a be a topic, and there are
,
The solution is a = 1 and b =-2.
(2) Through the characteristic polynomial of matrix A
,
Get the eigenvalue of a
For solving homogeneous linear equations, the basic solution system is obtained.
,
For solving homogeneous linear equations, the basic solution system is obtained.
Because it is already an orthogonal vector group, in order to get a normal orthogonal vector group, we only need to unitize it, thus obtaining
, ,
Order matrix
,
Then q is an orthogonal matrix. Under the orthogonal transformation X=QY, there are
,
The standard form of quadratic form is
Comment on finding a and b in this question, you can also find the characteristic polynomial first, and then use the relationship between root and coefficient to determine:
The characteristic polynomial of matrix A corresponding to quadratic form F is
Let the eigenvalue of A be, then it is set by the question.
,
The solution is a = 1 and b = 2.
Eleven, (full mark for this question 13)
Let the probability density of random variable x be
F(x) is the distribution function of X. Find the distribution function of random variable Y=F(X).
The concrete form of the distribution function F(x) can be obtained first, so that Y=F(X) can be determined, and then the distribution function of y can be obtained according to the definition. Note that the value range of Y=F(X) should be determined first, and then y should be discussed in sections.
A detailed explanation is easy to see when X.
Yes, there is.
Let G(y) be the distribution function of random variable Y=F(X). G(y)= 1。
Yes, there is.
=
=
Therefore, the distribution function of Y=F(X) is
Explain that in fact, this problem X can be any continuous random variable, and Y=F(X) still obeys uniform distribution:
When y < 0, g (y) = 0;
g(y)= 1；
When 0,
=
=
Twelve, (full mark for this question 13)
Let the random variables X and Y be independent, where the probability distribution of X is
,
And the probability density of y is f(y). Find the probability density g(u) of the random variable u = x+y 。
Using the distribution function method to analyze and find the distribution of two-dimensional random variable function is generally converted into finding the corresponding probability. Note that there are only two possible values of x, and the probability can be calculated by the total probability formula.
Let F(y) be the detailed distribution function of y, then the distribution function of U=X+Y can be known from the total probability formula as follows.
=
= .
Since x and y are independent, it can be seen that
g(u)= 1
=
In this way, the probability density of u is obtained.
=
Explanatory question is a new question type, and it is difficult and comprehensive to find the distribution of the sum of two random variables, one of which is continuous and the other is discrete, which needs to be calculated by full probability formula.