1x = 0
Maximum of LHS is minimum of RHS
Show that there is exactly one value of x satisfying
2cos^{2}(x^{3}+x)=2^{x}+2^{-x}
-
UP 0 DOWN 0 0 12
12 Answers
13show that the eqn
x3+7x-14(n2+1)=0
has no integral root for any integer n
1x3+7x-14(n2+1)=0
let c be an integral root.
=> c3+7c = 14k [k=n2+1]
=> c3+7c is a multiple of 14.
=> c2+7 = 14k/c is an integer.
=> c2 = 7 [2k/c - 1]
for RHS to be a perfect square,
2k/c -1 = 7m2
=> 2k/c = 7 m2+1
=> n2 = [7cm2+c]/2 -1 has to b a perfect square.
put values ... n u will find it is impossible to have a c for which [7cm2+c]/2 -1 is a perfect square....
aur achha kuchh aya dimag me toh bataungi ...
1@skygal,
=> c2 = 7 [2k/c - 1]
for RHS to be a perfect square,
2k/c-1 = 7m2
rite,
1[12]
am sry :( i didn get you...
that is rite only... if am not wrong :O
1arey oho!
sry sry...
so stupid of me... cant find the typo even after gordo pointed it ! huh!
9proof of skys statement
let
n2+1 =7λ
let n=7β+δ where δ is 1,2,3,4,5,6
δ2+1 =7λ
now u see that for diff values of δ ,
LHS =2,5,10,17,26,37 all not multiples of 7
1yes celes :) my statement was based on this observation only... that is putting values n finding...
but how can we know for sure that after some 10000000 no.s there wont be a multiple of seven which is of the kind n2+1 ....
9for that only ive gvn proof abv
if it doesnt satisfy for 1 - 6 itll never satisfy