50 AMC Lectures Problem Book 2 (36) Substitution Method

Transcription

1 0 AMC Letures Prolem Book Sustitution Metho PROBLEMS Prolem : Solve for rel : = Prolem : Solve for rel : Prolem : Show tht if 8 Prolem : Show tht + + if rel numers,, n stisf + + = Prolem : If, show tht = = log log log,, re positive numers not equl to Prolem : Solving 0, 0 Prolem 7: Show tht one of,, must e less thn or equl to / if 0 < <, 0 < <, 0 < < Prolem 8: Fin the vlue of if Prolem 9: Fin if 8

2 0 AMC Letures Prolem Book Sustitution Metho 9 Prolem 0: Compute Prolem : Ftor + + Prolem : Ftor Prolem : Solve for rel : 0 9 Prolem : Solve for rel n : Prolem : Fin if + = 9 Prolem : Solve for rel : log + log = Prolem 7: Solve for rel : Prolem 8: Solve for positive integers solutions of Prolem 9: Solve for : 7 Prolem 0: Show tht 9 for rel,, n

3 0 AMC Letures Prolem Book Sustitution Metho Prolem : Fin the omin of log [log log ] Prolem : Fin the smllest n gretest vlues of = + if Prolem : Fin the mimum n minimum vlues of Prolem : Fin the minimum vlue of + + if Prolem : Fin the mimum n minimum vlues of if + 9 = n re rel Prolem : Fin the mimum n minimum vlues of + + if + + = n + + = n re rel Prolem 7: As shown in the figure, the sie length of the squre is Cirles of rius re rwn with the enters t four verties of the squre Fin the she re Prolem 8: If, fin 0

4 0 AMC Letures Prolem Book Sustitution Metho SOLUTIONS Prolem : Solution: Let 9 =, 9 = The given eqution eomes + + = + or + + = + + So = 0 Tht is, 99 = 0 The solutions re,,, Prolem : Solution: Let The originl eqution eomes: = 0 Solving we get n So we hve = 0 Solving we get, So we hve = 0 Solving we get =, = Prolem : Solution: 8 Let Then 8 n 8 Sustituting the vlue of into the first term of the eqution, we hve: Similrl we hve 8

5 0 AMC Letures Prolem Book Sustitution Metho Therefore the left hn sie = Prolem : Solution: Sine + + =, the verge of,, n is Let = +, = +, = = 0 So + + = = = Prolem : Solution: Let log log log = k Then log = k +, log = k + log = k + So k, k, k, ] [ ] [ k k k Similrl k, k Therefore Prolem : Solution: Let m = +, n = The given sstem of equtions eomes: 0 0 m m n n Prolem 7: Solution: Sine 0 < <, let = sin Similrl we let = sin n = sin, where,, re ll ute ngles

6 0 AMC Letures Prolem Book Sustitution Metho The prolem hs een onverte to show tht one of os sin, os sin, os sin must e less thn or equl to / or one of os sin, os sin, os sin must e less thn or equl to / We ssume tht os sin >, os sin >, os sin > Then we hve os sin + os sin + os sin >, We know tht os sin os + sin os sin os + sin os sin os + sin Aing them we get: os sin + os sin + os sin, We see tht ontrit to Therefore one of must e less thn or equl to /,, Prolem 8: Solution: Let = k 8 Then = 8k, = k, = k 8k k k k Prolem 9: Solution: Let Then = + The originl epression 9 9 Prolem 0: Solution:

7 0 AMC Letures Prolem Book Sustitution Metho Let 0000 = = = Prolem : Solution: Let + = t + + = tt = t t = t t + = + + Prolem : Solution: Let + =, + = Then + = = + + = = + + Prolem : Solution: We see tht The given eqution n e written s Let eomes 9 0 = 0 Solving we get 0, 0 0 When, 0 + = 0 =, When, n this eqution hs no rel solutions We heke n =, re the solutions Prolem : Solution: Let u, v The given equtions eome:

8 0 AMC Letures Prolem Book Sustitution Metho, v u v u From, we get: u = v Sustituting into : v + v = Or v v + = 0 Solving we get v =, v = Sustituting v =, v = into we get u =, u = So ;, v u v u ;, Solving ove sstems of equtions: ;,, We heke n ll re the solutions Prolem : Solution: Divie oth sie : Let = The given eqution eomes = 0 Solving we get n etrneous

9 0 AMC Letures Prolem Book Sustitution Metho So Prolem : Solution: We know tht log log Let log The originl eqution n e written s + = 0 Solving we get, = From log we get = From log we get = Prolem 7: Solution: Let Then The given eqution eomes: Or 0 + = 0 Solving we get, = For, we hve For, we hve We heke n oth n re the solutions Prolem 8: Solution: Sine + + =, + = The verge vlue of n is

10 0 AMC Letures Prolem Book Sustitution Metho 7 Let, Where + = = eomes + + = Or = 0 Therefore Solving this inequlit, we get: Therefore n onl e,, Similrl n n lso onl e,, The solutions re:,,,,, Prolem 9: Solution: Therefore The originl eqution eomes + = 7 Or = 0 Sine + 8 0, 9 = 0, = = Solving we get = 0 or = We heke n the re the solutions of the eqution Prolem 0: Solution: Let + =, + =, + = Then + + = + + The given inequlit is equivlent to 9

11 0 AMC Letures Prolem Book Sustitution Metho = + = = 9 Note tht when > 0, > 0, we hve 9 Sine + + > 0, Prolem : Solution: Let log log u Then log [log log ] log u B the efinition of the logrithm funtion, we hve u > 0 Tht is, log log 0 From log log 0, we get log > Therefore the omin is > Prolem : Solution: Let = t = = t = = t Therefore = t t + t = t + t Or t Sine, t So t When, t m When =, t min = Therefore when t, m ; When t =, min = 0 8

12 0 AMC Letures Prolem Book Sustitution Metho Prolem : Solution: Let t Then = t + When t = 0, or 0, =, min = Prolem : Solution: Let k Then = k +, = k, n = k + So P = + + = k When k, P min = Prolem : Solution: Let = k Eliminting we get + k + k 9 = 0 From 0, we get k When,, the mimum vlue is When,, the minimum vlue is Prolem : Solution: k Solving we get k k, Sustituting them into the mile eqution, we hve + k k k = 0 = k k k 0 Mimum vlue is n the minimum vlue is Prolem 7: Solution: We lel eh region s follows: 9

13 0 AMC Letures Prolem Book Sustitution Metho 0,, + Prolem 8: Solution: Let f then f, n f 8 f