University of Science and Technology in 2014 for Love science experiment class selection of questions - Mathematics | Math173
Source: Lan Qi page
A fill in the blank (the title of 5 smaller problems, each of 8 points, 40 points total)
1, set \ (f (x) = \ dfrac {x} {\ sqrt {1 + x ^ 2}} \), then \ (n-\) weight composite function \ (f_n (x) = f (f (\ cdots f (x) \ cdots)) = \) _______.
2, provided polynomial \ (p (x) \) satisfies \ (p \ left (x ^ 2 + 1 \ right) = \ left (p (x) \ right) ^ 2 + 1 \) and \ (p (0 ) = 0 \), then \ (p (x) = \) _______.
3, set \ (S_n = \ sum \ limits_ {k = 1} ^ n \ dfrac {6 ^ k} {\ left (3 ^ {k + 1} -2 ^ {k + 1} \ right) \ left ( 3 ^ k-2 ^ k \ right)} \), then the limit \ (\ lim \ limits_ {n \ to \ infty} S_n = \) _______.
4, \ (x> 0 \), the function \ (f (x) = \ dfrac {\ left (x + \ dfrac1x \ right) ^ 6- \ left (x ^ 6 + \ dfrac1 {x ^ 6} \ right minimum) -2} {\ left (x + \ dfrac1x \ right) ^ 3 + \ left (x ^ 3 + \ dfrac1 {x ^ 3} \ right)} \) is _______.
5, assuming that \ (20 \) students in every student can learn from one to six selected six courses offered, one can not choose. Try to determine the following proposition is correct: there is \ (5 \) students and two courses, making this \ (5 \) students have selected these two courses, or did not choose, optional "right" or "no "_______.
II (the title of 14 minutes)
1, if \ (a \) is a positive integer and \ (\ sqrt a \) is not an integer, prove that: \ (\ sqrt a \) is an irrational number.
2, the test card: In addition to \ (0,0,0 \), there is no other integer \ (m, n, p \) such that \ [m + n \ sqrt2 + p \ sqrt3 = 0 \.] III (the title a total of 16 points) Let \ (a, b, c \) is the length of the sides of a triangle, \ (p = \ dfrac {a + b + c} 2 \), \ (r \) is the radius of the inscribed circle, proved : \ [\ dfrac1 {(pa) ^ 2} + \ dfrac1 {(pb) ^ 2} + \ dfrac1 {(pc) ^ 2} \ geqslant \ dfrac1 {r ^ 2} \.]
IV (the title of 12 points) demonstrate: Let \ (m \) is a positive integer, the \ (a_m = \ dfrac12 + \ dfrac13 + \ dfrac14 + \ dfrac15 + \ cdots + \ dfrac1 {2 ^ m} \) is not an integer.
V. (the title a total of 18 points) The figure is 2013 Evergrande Football Club, home of the planning and FC Seoul football team finals of the AFC Champions posters, constant brigade left, the right is the Seoul team, what is the moral of the poster? Requiring a simple derivation of the results of two mathematical formulas poster. A mathematical formula is \ (\ sqrt {1 + 2 \ sqrt {1 + 3 \ sqrt {1 + 4 \ sqrt {1+ \ cdots}}}} \) (Ramanujan formula), the other is \ (\ mathrm e ^ {\ pi \ mathrm i} +1 \) (known Euler formula \ (\ mathrm e ^ {\ pi \ mathrm i} = \ cos \ alpha + \ mathrm i \ sin \ alpha \) ).
Answers
First, fill in the blank
1, \ (\ dfrac {x} {\ sqrt {1 + nx ^ 2}} \).
2, \ (x \) prompt equation \ (p (x) -x = 0 \) there are numerous zero, then \ (p (x) = x \).
3, \ (2 \) prompt split entry for \ (\ dfrac {2 ^ k } {3 ^ k-2 ^ k} - \ dfrac {2 ^ {k + 1}} {3 ^ {k + 1} - 2 ^ {k + 1}} \).
4, \ (6 \) prompt function \ (f (x) = 3 \ left (x + \ dfrac 1x \ right) \).
5, no prompt \ (6 \) selected courses \ (3 \) Total door \ ({\ rm C} _6 ^ 3 = 20 \) different combinations, so that each student are selected from a composition, then any option number two courses at the same time and at the same time the students are not selected \ (4 \).
Second, a little
Tips are by contradiction.
Third, slightly
Tip \ (pr = S \), and \ (S = \ sqrt {p (pa) (pb) (pc)} \).
Fourth, slightly
Tip arranged to the right of the common denominator \ [\ left [2,3,4, \ cdots, 2 ^ m \ right] = 2 ^ m \ cdot p, \] where \ (P \) is an odd number, while on both sides by to the common denominator, it is even on the left, while the right side is odd.
Note an advantage of this method may prove \ (\ sum \ limits_ {i = 1} ^ n {\ dfrac 1i} \), where \ (n \ in \ mathcal N ^ * \) and \ (n \ geqslant 2 \) They are not an integer. In addition, this method is not necessary to start from the \ (2 \) a power of.
Note II may also on both sides simultaneously multiplying \ (\ dfrac {[2,3, \ cdots, 2 ^ m]} p \) after which the \ (P \) decomposition of the quality factor is the denominator of the right side of each of the maximum number of odd prime factor, the Bertrand - Chebyshev Theorem containing \ (P \) is the only item, to obtain further.
Fifth, \ (3: 0 \)
Tip Ramanujan identities noted \ [n = \ sqrt {1+ (n-1) (n + 1)}, \] so \ [\ begin {split} 3 & = \ sqrt {1 + 2 \ cdot 4} \\ & = \ sqrt { 1 + 2 \ sqrt {1 + 3 \ cdot 5}} \\ & = \ sqrt {1 + 2 \ sqrt {1 + 3 \ sqrt {1 + 4 \ cdot 6}} } \\ & = \ cdots \ end {split}. \]
University of Science and Technology in 2015 for Love science experiment class selection of questions - Mathematics | Math173
First, fill in the blank
1, parabolic \ (y ^ 2 = 2 \ sqrt 2x \), assuming that its focal point is \ (F \), \ (y \) on the axle axis positive point is \ (N \). If there is a unique point in the line of \ (P \) such that \ (\ angle NPF = 90 ^ \ circ \), then \ (N \) ordinate points is _______.
2, \ (\ dfrac {1} {\ sqrt 1+ \ sqrt 2} + \ dfrac {1} {\ sqrt 2+ \ sqrt 3} + \ cdots + \ dfrac {1} {\ sqrt {255} + \ sqrt {256}} = \) ______.
3, if it is known \ (\ lim \ limits_ {n \ to + \ infty} \ left (\ sum \ limits_ {i = 1} ^ n \ dfrac {1} {i} - \ ln n \ right) \) exists, \ (\ sum \ limits_ {i = 0} ^ {+ \ infty} {\ dfrac {(- 1) ^ {i + 2}} {i + 1}} = \) _______.
4, the side length \ (1 \) in a square (including the boundary) taken \ (9 \) points, which must be \ (3 \) points, which constitute the triangle area is not more than _______.
5, a person shooting hit the ring 8, the ring 9, the ring 10, respectively, the probability \ (0.15 \) \ (.25 \) \ (0.2 \), he is now three shots, not less than \ (28 \ ) probability ring is _______.
Second, answer questions
6, if for any real number \ (x, y \), there are \ (f \ left ((xy) ^ 2 \ right) = \ left (f (x) \ right) ^ 2-2x \ cdot f (y) + y ^ 2 \), seeking \ (f (x) \).
7, find all \ (a, b \), so \ (\ left | \ sqrt {1-x ^ 2} -ax-b \ right | \ leqslant \ dfrac {\ sqrt 2-1} 2 \) established, where \ (x \ in [0,1] \).
8, if the complex \ (Z \) satisfies \ (| z | = 1 \), seeking \ (\ left | z ^ 3-z + 2 \ right | ^ 2 \) minimum.
9, it is known cubic equation \ (x ^ 3 + ax ^ 2 + bx + c = 0 \) has three real roots.
(1) If the three real roots \ (x_1, x_2, x_3 \), and \ (x_1 \ leqslant x_2 \ leqslant x_3 \), \ (a, b \) is a constant demand \ (C \) Change \ (x_3-x_1 \) ranges;
(2) If the three real roots \ (a, b, c \), find \ (a, b, c \).
Answers
First, fill in the blank
1, \ (2 \) Tip: hypotenuse \ midpoint (of NF \) a \ (M \) on the parabola, coordinates \ (\ left (\ dfrac {\ sqrt {2}} 4,1 \ right) \).
2、\(15\)
3、\(\ln 2\)
4, \ (\ dfrac 18 \)
Tip: As shown.
Second, answer questions
6, \ (f (x) = x \ lor f (x) = x + 1 \) Note: Order \ (x = y \) to give \ [f (0) = \ left (f (x) -x \ right) ^ 2, \] then make \ (x = 0 \) available \ (f (0) = 0 \ lor f (0) = 1 \).
7, \ (a = -1 \ land b = \ dfrac {\ sqrt 2 + 1} 2 \) Tip: triangle by substitution, \ (x = \ cos \ theta \), where \ (\ theta \ in \ left [0, \ dfrac {\ pi} 2 \ right] \), then the original formula is deformed into a \ [\ left | \ sqrt {1 + a ^ 2} \ sin \ left (\ theta + \ varphi \ right) -b \ right | \ leqslant \ dfrac {\ sqrt 2-1} 2, \] noted algebraic \ (\ sqrt {1 + a ^ 2} \ sin \ left (\ theta + \ varphi \ right) \) of the interval length range not exceed \ (\ sqrt 2-1 \), then \ (a = -1 \), and further \ (b = \ dfrac {\ sqrt 2 + 1} 2 \).
8, \ (\ dfrac {8} {27} \) Tip: using a complex conjugate, and so \ (x = z + \ bar z \), then there is the original formula is equal to \ (2x ^ 3-x ^ 2-8x +8 \), where \ (x \ in [-2,2] \).
9、（1）\(\left[\sqrt{a^2-3b},2\sqrt{\dfrac{a^2}3-b}\right]\)；
(2) is understood to \ ((a, b, c) = (0,0,0), (1, -1, -1), (1, -2,0) \), understood as non \ ( \ left (- \ dfrac 1b, b, \ dfrac 2b-b \ right) \), where \ (b = t + \ dfrac 2 {3t} \), and \ (t = \ sqrt [3] {- 1+ \ sqrt {\ dfrac {19} {27}}} \).
Solution [29] general cubic equation of a problem Daily | Math173
Today's question is from 2011, the second World Championship Mathematics youth group relay a second round of questions began.
Of Equation \ [(x + 1) (x ^ 2 + 1) (x ^ 3 + 1) = 30x ^ 3 \] of real roots and all.
Powerful, our teacher is a change in the inverse triangle triangles _ (• ω • "∠) _
With this problem today is not the problem, just the tips of it. Today's issue is competition title of the 16th century (the time, mathematicians often find their secret, but challenged peers, so that they solve the same problem. This is surely a very intellectual temper, and attractive contest ):
Solution of \ (X \) equation \ [x ^ 3 + px + q = 0. \]
The key is how to properly change yuan.
Noting \ [\ left (t + \ dfrac 1t \ right) ^ 3 = t ^ 3 + \ dfrac 1 {t ^ 3} +3 \ left (t + \ dfrac 1t \ right). \] That is, if the \ (p = -3 \), then we do transducer element \ (x = t + \ dfrac 1t \), the equation is converted to \ [t ^ 3 + \ dfrac 1 {t ^ 3} + q = 0, \] i.e. \ [(t ^ 3) ^ 2 + q \ cdot t ^ 3 + 1 = 0, \] can be obtained by using the quadratic root formula \ (t ^ 3 \), then find \ (T \) and then on behalf of the back \ (x = t + \ dfrac 1t \), root-finding process is complete.
How to deal with the difficulties now facing \ (p \), you need to change a little transformation primitives.
Because of \ [\ left (t + \ dfrac ut \ right) ^ 3 = t ^ 3 + \ dfrac {u ^ 3} {t ^ 3} + 3u \ left (t + \ dfrac ut \ right), \] Accordingly Order \ (x = t + \ dfrac ut \), where \ (U \) coefficients to be determined, then the original equation becomes \ [t ^ 3 + \ dfrac {u ^ 3} {t ^ 3} + (3u + p) \ . cdot \ left (t + \ dfrac ut \ right) + q = 0 \] in this equation, so that \ (u = - \ dfrac p3 \), and will become about the same as before \ (t ^ 3 \ ) of the quadratic equation, the following slightly.
Indeed, any of a cubic equation \ [ax ^ 3 + bx ^ 2 + cx + d = 0, a \ neq 0 \] may utilize full cube formula \ [\ left (x + \ dfrac b {3a} \ right) ^ 3 = x ^ 3 + \ dfrac bax ^ 2 + \ dfrac {b ^ 2} {3a ^ 2} x + \ dfrac {b ^ 3} {27a ^ 3} = 0 \] by formulation into \ [x ^ 3 + px + q = 0 \] form. Therefore mastered this method, it is equivalent to master the general solution of cubic equations.
In general solution of cubic equations, we used to change yuan \ (x = t + \ dfrac ut \) solution is also an important high-order equations by substitution. To note that, in every step of the solution process, the first figure out the real root of the equation is still seeking all the roots.
The last remaining one exercises.
On demand \ (X \) equation \ [x ^ 5 + 10x ^ 3 + 20x-4 = 0 \] of all the root.
The answer is \ [x = \ left (2 ^ {\ frac 35} -2 ^ {\ frac 25} \ right) \ cos \ dfrac {2k \ pi} {5} + \ left (2 ^ {\ frac 35} +2 ^ {\ frac 25} \ right) \ mathcal {i} \ sin \ dfrac {2k \ pi} {5}, k = 0,1,2,3,4. \] where substitution is used \ (x = t- \ dfrac 2t \).