02nd sets of simulated exam for Mathematics Competition Training of Gannan Normal University

 

 

1. find $ \ DPS {\ int _ \ VGA y ^ 2 \ RD s} $, $ \ VGA $ consists of $ \ DPS {\ sedd {\ BA {RL} x ^ 2 + y ^ 2 + Z ^ 2 & = a ^ 2 \ x + Z & = A \ EA }}$ decision.

Answer: $ \ VGA $: $ \ Bex \ sedd {\ BA {RL} \ sex {X-\ cfrac {A} {2} ^ 2 + y ^ 2 + \ sex {z-\ cfrac {A} {2 }}^ 2 & =\ cfrac {A ^ 2} {2 }\\\ sex {X-\ cfrac {A} {2 }}+ \ sex {Y-\ cfrac {A} {2 }}& = 0 \ EA }. \ EEx $ transform $ \ Bex u = x-\ cfrac {A} {2}, \ quad v = Y, \ quad W = z-\ cfrac {A} {2 }, \ EEx $ then $ \ beex \ Bea \ int _ \ VGA y ^ 2 \ rd s & =\ int_l V ^ 2 \ RD s \ quad \ sex {L: \ sedd {\ BA {RL} U ^ 2 + V ^ 2 + w ^ 2 & =\ cfrac {A ^ 2} {2} \ U + W = 0 \ EA }}\\<\int_0 ^ {2 \ PI} \ cfrac {A ^ 2} {2} \ sin ^ 2 \ TT \ SQRT {\ sex {\ cfrac {\ rd u }{\ RD \ TT }}^ 2 + \ sex {\ cfrac {\ RD v }{\ rd T }}^ 2 + \ sex {\ cfrac {\ rd W }{\ rd T }}^ 2} \ RD \ TT \ & \ quad \ sex {L: \ sedd {\ BA {RL} u = \ cfrac {A} {2} \ cos \ TT \ v = \ cfrac {A} {\ SQRT {2 }}\ sin \ TT \ W =-\ cfrac {A} {2} \ cos \ TT \ EA }, 0 \ Leq \ TT \ Leq 2 \ PI }\\\&=\ int_0 ^ {2 \ PI} \ cfrac {A ^ 2} {2} \ sin ^ 2 \ TT \ cdot \ cfrac {A }{\ SQRT {2 }}\ RD \ TT \\\&=\ cfrac {A ^ 3 \ PI }{ 2 \ SQRT {2 }}. \ EEA \ eeex $

 

2. set $ A, B, and C $ to positive values, calculate curved points $ \ Bex \ iint_s x ^ 3 \ rd y \ rd z + y ^ 3 \ rd z \ RD x + Z ^ 3 \ RD x \ RD y, \ EEx $ where $ S $ is the upper half elliptical sphere $ \ Bex \ frac {x ^ 2} {A ^ 2} + \ frac {y ^ 2} {B ^ 2} + \ frac {z ^ 2} {C ^ 2} = 1, \ quad Z \ geq 0, \ EEx $ direction up.

Answer: $ \ Bex \ mbox {original second-Type Curved Surface points} & = & 3 \ iiint _ {\ frac {x ^ 2} {A ^ 2} + \ frac {y ^ 2} {B ^ 2} + \ frac {z ^ 2} {C ^ 2} \ Leq 1 \ atop Z \ geq 0} (x ^ 2 + y ^ 2 + Z ^ 2) \ RD x \ rd y \ rd z \ & = & 3abc \ iiint _ {x ^ 2 + y ^ 2 + Z ^ 2 \ Leq 1 \ atop Z \ geq 0} (a ^ 2x ^ 2 + B ^ 2y ^ 2 + C ^ 2Z ^ 2) \ RD x \ rd y \ rd z \ & = & 3abc (a ^ 2 + B ^ 2 + C ^ 2) \ iiint _ {x ^ 2 + y ^ 2 + Z ^ 2 \ Leq 1 \ atop Z \ geq 0} x ^ 2 \ RD x \ rd y \ rd z \\& = & ABC (a ^ 2 + B ^ 2 + C ^ 2) \ iiint _ {x ^ 2 + y ^ 2 + Z ^ 2 \ Leq 1 \ atop Z \ geq 0} (x ^ 2 + y ^ 2 + Z ^ 2) \ RD x \ rd y \ rd z \ & =& \ frac {1} {2} ABC (a ^ 2 + B ^ 2 + C ^ 2) \ iiint _ {x ^ 2 + y ^ 2 + Z ^ 2 \ Leq 1} (x ^ 2 + y ^ 2 + Z ^ 2) \ RD x \ rd y \ rd z \ & =& \ frac {1} {2} ABC (a ^ 2 + B ^ 2 + C ^ 2) \ int_0 ^ 1 R ^ 2 \ cdot 4 \ Pi R ^ 2 \ RD r \ & = & \ frac {2 \ PI} {5} ABC (a ^ 2 + B ^ 2 + C ^ 2 ). \ EEx $

 

3. calculate surface points $ \ DPS {\ iint_s \ frac {\ RD s} {\ SQRT {x ^ 2 + y ^ 2 + (Z + 2) ^ 2 }}$, $ S $ is the top part cut by $ Z =\sqrt {x ^ 2 + y ^ 2} $ on the Sphere centered at the origin and with the radius of $2 $.

Answer: $ \ beex \ Bea \ iint_s \ frac {\ RD s} {\ SQRT {x ^ 2 + y ^ 2 + (Z + 2) ^ 2 }}& =\ iint_s \ frac {\ RD s }{\ SQRT {8 + 2Z }}\\\& =\ iint _ {x ^ 2 + y ^ 2 \ leq 2} \ frac {1} {\ SQRT {8 + 2 \ SQRT {4-x ^ 2-y ^ 2 }}\ cdot \ frac {2} {\ SQRT {4-x ^ 2-y ^ 2 }}\ RD x \ RD y \\\&=\ int_0 ^{\ SQRT {2 }}\ frac {\ SQRT {2 }}{\ SQRT {4 + \ SQRT {4-r ^ 2 }}\ cdot 2 \ pi r \ rd r \\\&=\ SQRT {2 }\ pi \ int_0 ^ 2 \ frac {\ rd s }{ \ SQRT {4 + \ SQRT {4-s }}\ cdot \ SQRT {4-s }}\ quad \ sex {S = R ^ 2 }\\\&=\ SQRT {2 }\ pi \ int _ {\ SQRT {2 }}^ 2 \ frac {1 }{\ SQRT {4 + t} \ cdot 2t \ RD t \ quad \ sex {\ SQRT {4-s} = T }\\\& = 2 \ SQRT {2} \ pi \ int _ {\ SQRT {2 }}^ 2 \ frac {1 }{\ SQRT {4 + T }}\ rd T = 4 \ SQRT {2} \ pi \ sex {\ SQRT {6}-\ SQRT {4 + \ SQRT {2 }}}. \ EEA \ eeex $

 

4. set $ \ DPS {f (x) = \ sum _ {n = 0} ^ \ infty \ frac {1} {x + 2 ^ n }}$, $ x \ in [0, \ infty) $. proof:

(1) $ F $ on $ [0, \ infty) $;

(2) $ \ DPS {\ lim _ {x \ To \ infty} f (x) = 0 }$; (3) for everything $ x \ In (0, \ infty) $ \ Bex 0 <F (x)-\ frac {\ ln (1 + x )} {x \ ln 2} <\ frac {1} {1 + x }. \ EEx $

Proof:

(1) by $ \ Bex \ frac {1} {x + 2 ^ n} <\ frac {1} {2 ^ n} \ EEx $ Zhi $ \ DPS {\ sum _ {n = 0} ^ \ infty \ frac {1} {x + 2 ^ n }}$ about $ x $ consistent convergence. while $ f (x) $ is continuous as the sum function.

(2) by $ \ beex \ Bea f (x) & =\ sum _ {n = 0} ^ \ infty \ frac {1} {x + 2 ^ n }\\& =\ sum _ {n = 0} ^ n \ frac {1} {x + 2 ^ n} + \ sum _ {n = n + 1} ^ \ infty \ frac {1} {x + 2 ^ n }\\& <\ frac {n + 1} {x} + \ frac {1} {2 ^ n} \ & \ equiv I _1 + I _2 \ EEA \ eeex $ any fixed $ \ ve> 0 $, you can get $ N $ \ DPS {I _2 <\ frac {\ ve} {2 }}$ first, and then get $ N $, $ \ Bex \ exists \ x> 0, \ st x> X \ rA I _1 <\ frac {\ ve} {2 }. \ EEx $ (3) by $ \ Bex \ frac {1} {x + 2 ^ {n + 1 }}< \ frac {1} {x + 2 ^ t} <\ frac {1} {x + 2 ^ n} \ Quad (n <t <n + 1) \ EEx $ Zhi $ \ Bex \ sum _ {n = 1} ^ \ infty \ frac {1} {x + 2 ^ n} <\ int_0 ^ \ infty \ frac {1} {x + 2 ^ t} \ rd t <\ sum _ {n = 0} ^ \ infty \ frac {1} {x + 2 ^ n }. \ EEx $ and $ \ Bex \ int_0 ^ \ infty \ frac {1} {x + 2 ^ t} \ rd t <F (x) = \ sum _ {n = 0} ^ \ infty \ frac {1} {x + 2 ^ n} <\ int_0 ^ \ infty \ frac {1} {x + 2 ^ t} \ rd t + \ frac {1} {1 + x }. \ EEx $ $ \ beex \ Bea \ int_0 ^ \ infty \ frac {1} {x + 2 ^ t} \ rd t & =\ frac {1 }{\ ln 2} \ int _ {x + 1} ^ \ infty \ frac {1} {y} \ cdot \ frac {\ RD y} {Y-x} \ quad \ sex {x + 2 ^ 6 = y }\\&=\ frac {1} {x \ ln 2} \ cdot \ left. \ ln \ frac {Y-x} {y} \ right | _ {Y = x + 1} ^ {Y = \ infty} \\\\&=\ frac {\ Ln (1 + x )} {x \ ln 2}, \ EEA \ eeex $ we finally get $ \ Bex 0 <F (x)-\ frac {\ ln (1 + x )} {x \ ln 2} <\ frac {1} {1 + x }. \ EEx $

 

5. set $ F $ to a bounded continuous function on $ (0, \ infty) $, and set $ \ sed {R_N} $ to any given positive sequence. test Certificate: The Positive Sequence $ \ sed {x_n} $ exists, making $ \ Bex \ lim _ {n \ To \ infty} [F (x_n + R_N) -F (x_n)] = 0. \ EEx $

Proof: (1) first prove: For any fixed $ n \ In \ BBN $, $ \ bee \ label {113.8: EQ1} \ exists \ Y_n> 0, \ st F (Y_n + R_N)-f (Y_n) <\ frac {1} {n}; \ EEE $ \ bee \ label {113.8: eq2} \ exists \ Z_N> 0, \ st F (Z_N + R_N)-f (Z_N)>-\ frac {1} {n }. the proof of \ EEE $ asserted \ eqref {113.8: EQ1} is similar to that of \ eqref {113.8: eq2. we only use the reverse verification method to prove \ eqref {113.8: EQ1 }. if $ \ Bex y> 0 \ rA f (y + R_N)-f (y)> \ frac {1} {n }, \ EEx $ then $ \ beex \ Bea F (1 + kr_n) & =\ sum _ {J = 0} ^ {k-1} [F (1 + (J + 1) R_N)-f (1 + jr_n)] + F (1) \ &> \ frac {k} {n} + F (1 ). \ EEA \ eeex $ order $ k \ To \ infty $, we found that $ F $ has no upper bound. this is a conflict.

(2) further proof: $ \ Bex \ exists \ x_n> 0, \ ST \ sev {f (x_n + R_N)-f (x_n )} <\ frac {1} {n }. \ EEx $

(A) If $ \ DPS {f (Y_n + R_N)-f (Y_n)>-\ frac {1} {n }}$, you can use $ x_n = Y_n $;

(B) If $ \ DPS {f (Z_N + R_N)-f (Z_N) <\ frac {1} {n }}$, you can use $ x_n = Z_N $;

(C) If $ \ Bex F (Y_n + R_N)-f (Y_n) \ Leq-\ frac {1} {n}, \ quad F (Z_N + R_N) -F (Z_N) \ geq \ frac {1} {n}, \ EEx $ by $ \ Bex \ SEZ {-\ frac {1} {n }, \ frac {1} {n }}\ subset [F (Y_n + R_N)-f (Y_n), F (Z_N + R_N)-f (Z_N)] \ EEx $ and the parameter theorem of continuous functions $ \ Bex \ exists \ x_n \ mbox {between} Y_n, Z_N \ mbox }, \ st F (x_n + R_N)-f (x_n) = 0 \,! \ EEx $

 

6. set $ F: (0, + \ infty) \ To (0, + \ infty) $ is a monotonically increasing function. if $ \ Bex \ lim _ {x \ to + \ infty} \ frac {f (2x)} {f (x)} = 1, \ EEx $ proof: for any $ m> 0 $, $ \ Bex \ lim _ {x \ to + \ infty} \ frac {f (MX)} {f (x )} = 1. \ EEx $

Proof: (1) the question must be verified only for $ m \ geq 1 $. in fact, for $0 <m <1 $ \ Bex \ lim _ {x \ to + \ infty} \ cfrac {f (MX)} {f (x )} = \ lim _ {x \ to + \ infty} \ cfrac {1} {\ cfrac {f (x)} {f (MX )}} =\ cfrac {1} {\ DPS {\ lim _ {x \ to + \ infty} \ cfrac {f \ sex {\ frac {1} {m} s }}{ F (s )}}}. \ EEx $

(2) The question must be verified only for $ x = 2 ^ k \ (k \ In \ BBN) $. in fact, for $ m \ geq 1 $, $ \ Bex \ exists \ K \ In \ BBN, \ ST 2 ^ k \ Leq m <2 ^ {k + 1 }. \ EEx $ and $ \ Bex \ cfrac {F (2 ^ kx)} {f (x)} \ Leq \ cfrac {f (MX )} {f (x)} \ Leq \ cfrac {F (2 ^ {k + 1} x)} {f (x )}. \ EEx $ (3) when $ x = 2 ^ K $, $ \ Bex \ lim _ {x \ to + \ infty} \ cfrac {F (2 ^ kx)} {f (x )} = \ lim _ {x \ to + \ infty} \ prod _ {I = 1} ^ k \ cfrac {F (2 ^ IX )} {F (2 ^ {I-1} x )} = \ prod _ {I = 1} ^ k \ lim _ {s \ to + \ infty} \ cfrac {F (2 S)} {f (s)} = 1. \ EEx $

 

7. For $ \ DPS {x \ In \ sex {0, \ cfrac {\ PI} {2 }}$, test certificate:

(1) $ \ DPS {\ cfrac {1} {3} \ Tan x + \ cfrac {2} {3} \ SiN x> X} $;

(2) $ \ DPS {\ cfrac {1} {\ sin ^ 2 x}-\ cfrac {1} {x ^ 2} \ Leq 1-\ cfrac {4} {\ PI ^ 2 }}$.

Proof: (1) When $ \ DPS {x \ In \ sex {0, \ frac {\ PI} {2 }}$, by $ \ beex \ Bea \ Tan x & =\ int_0 ^ x \ sec ^ 2t \ RD t \ & =\ int_0 ^ X (1 + \ tan ^ 2 T) \ RD t \\\&> \ int_0 ^ x (1 + t ^ 2) \ RD t \ quad \ sex {\ mbox {drawing on the unit circle }\\& = x + \ frac {x ^ 3} {3 }; \ EEA \ eeex $ \ beex \ Bea \ SiN x & =\ int_0 ^ x \ cos t \ RD t \ & =\ int_0 ^ x \ SEZ {1- \ int_0 ^ t \ sin s \ RD s} \ RD t \ &> \ int_0 ^ x \ SEZ {1-\ int_0 ^ t s \ RD s} \ RD t \ \ & = x-\ frac {x ^ 3} {6} \ EEA \ eeex $ \ Bex \ frac {1} {3} \ Tan x + \ frac {2} {3} \ SiN x> X. \ EEx $

(2) $ \ beex \ Bea \ cfrac {1} {x ^ 2}-\ cfrac {4} {\ PI ^ 2 }{\ cfrac {1 }{\ sin ^ 2x}-1} & =\ cfrac {-\ cfrac {2} {y ^ 3 }}{-2 \ cfrac {\ cos y }{\ sin ^ 2y}} \ quad \ sex {x <Y <\ cfrac {\ PI} {2 }}\\\&=\ cfrac {2 \ sin Z \ Cos Z \ Tan Z + \ sin ^ 2Z \ sec ^ 2Z} {3z ^ 2} \ quad \ sex {0 <z <y }\\&=\ cfrac {2 \ sin ^ 2z + \ tan ^ 2 z} {3z ^ 2 }\\\&=\ cfrac {\ SEZ {\ sex {\ SQRT {2 }}^ 2 + 1 ^ 2 }\ SEZ {\ sex {\ SQRT {2} \ sin z} ^ 2 + \ sex {\ Tan Z} ^ 2} {9z ^ 2} \ & \ geq \ cfrac {(\ SQRT {2} \ cdot \ SQRT {2} Z + 1 \ cdot \ Tan Z) ^ 2} {9z ^ 2 }\\\&=\ sex {\ cfrac {2 \ sin Z + \ Tan Z} {3z }}^ 2 \\&> 1. \ EEA \ eeex $

 

8. Set $ F $ to $ [0, \ infty) $ to the monotonic function:

(1) for any $ a> 0 $, $ \ Bex \ lim _ {n \ To \ infty} \ int_0 ^ A f (x) \ sin NX \ RD x = 0; \ EEx $

(2) If you set $ \ DPS {\ vlm {x} f (x) = 0 }$, verify that: $ \ Bex \ vlm {n} \ int_0 ^ \ infty f (x) \ sin NX \ RD x = 0. \ EEx $

Proof: (1) by the second point of the integral value theorem, $ \ beex \ Bea \ sev {\ int_0 ^ A f (x) \ sin NX \ RD x} & =\ sev {f (0) \ int_0 ^ \ Xi \ sin NX \ RD x + f () \ int _ \ Xi ^ A \ sin NX \ RD x }\\&=\ sev {f (0) \ cfrac {1-\ cos n \ Xi} {n} + f () \ cfrac {\ cos n \ Xi-\ cos Na} {n }}\\ & \ Leq \ cfrac {2 \ SEZ {| f (0) | + | f (a) |}}{ n }\\\\ to 0 \ quad \ sex {n \ To \ infty }. \ EEA \ eeex $

(2) Write $ \ beex \ Bea \ int_0 ^ \ infty f (x) \ sin NX \ rd x & =\ frac {1} {n} \ int_0 ^ \ infty f \ sex {\ frac {t} {n} \ sin t \ RD t \ & =\ frac {1} {n} \ sum _ {k = 0} ^ \ infty \ SEZ {\ int _ {2 k \ PI} ^ {(2 k + 1) \ PI} f \ sex {\ frac {t} {n} \ sin t \ rd t + \ int _ {(2 k + 1) \ PI} ^ {(2 k + 2) \ PI} f \ sex {\ frac {t} {n} \ sin t \ RD t }. \ EEA \ eeex $ we have $ \ beex \ Bea & \ quad \ int_0 ^ \ infty f (x) \ sin NX \ RD x \ & \ geq \ frac {1} {n} \ sum _ {k = 0} ^ \ infty \ SEZ {f \ sex {\ frac {(2 k + 1) \ PI} {n }}\ int _ {2 k \ PI} ^ {(2 k + 1) \ PI} \ sin t \ rd t + f \ sex {\ frac {(2 k + 1) \ PI} {n }}\ int _ {(2 k + 1) \ PI} ^ {(2 k + 2) \ PI} \ sin t \ RD t} \ & = 0; \ EEA \ eeex $ \ beex \ Bea & \ quad \ int_0 ^ \ infty f (x) \ sin NX \ RD x \\\\=\ frac {1} {n} \ lim _ {M \ To \ infty} \ sum _ {k = 0} ^ m \ Sez {\ int _ {2 k \ PI} ^ {(2 k + 1) \ PI} f \ sex {\ frac {t} {n} \ sin t \ rd t + \ int _ {(2 k + 1) \ PI} ^ {(2 k + 2) \ PI} f \ sex {\ frac {t} {n} \ sin t \ RD t} \ & \ Leq \ frac {1} {n} \ lim _{ m \ To \ infty} \ sum _ {k = 0} ^ m \ SEZ {f \ sex {\ frac {2 k \ PI} {n }}\ int _ {2 k \ pi} ^ {(2 k + 1) \ PI} \ sin t \ rd t + f \ sex {\ frac {(2 k + 2) \ PI} {n }}\ int _ {(2 k + 1) \ PI} ^ {(2 k + 2) \ PI} \ sin t \ RD t} \ & =\ frac {2} {n} \ lim _ {M \ To \ infty} \ sum _ {k = 0} ^ m \ SEZ {f \ sex {\ frac {2 k \ PI} {n}-f \ sex {\ frac {2 (k + 1) \ PI} {n }}\\\&=\ frac {2} {n} \ lim _ {M \ To \ infty} \ SEZ {f (0) -F \ sex {\ frac {2 (m + 1) \ PI} {n }}\\\&=\ frac {2} {n} f (0) \ EEA \ eeex $ \ Bex 0 \ Leq \ int_0 ^ \ infty f (x) \ SiN x \ RD x \ Leq \ frac {2} {n} f (0 ). \ EEx $ there is a conclusion.

 

9. set $ F $ to continuous on $ [0, C] $, $ F (0) = 0 $, and when $ x \ In (0, c) $, $ f'' (x) <0 $. test Certificate: When $0 <A <B <a + B <C $, $ \ Bex f (a + B) <F (a) + F (B ). \ EEx $

Proof: For fixed $ B> 0 $, make $ \ Bex f (x) = f (x + B)-f (x)-f (B ), \ EEx $ then $ F (0) = 0 $; and by $ f'' (x) <0 $ $ \ Bex f' (X) = f' (x + B)-f' (x) <0. \ EEx $ then $ \ Bex F (a) <F (0) = 0. \ EEx $