Re: 1/0.001<(1/10)^n-4
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26 May 2020, 15:14
we are given,
1/0.001 < (1/10)^(n-4) , this can be written as
10^3 < (10^-1)^(n-4) , upon further simplification
10^3 < (10) ^(4-n) , applying inequality to the exponents , we get
3< 4 - n , subtracting both side by 4
-1< -n , multiplying both sides by -1 and flipping the inequality, we get
1> n
take different values of n ,
case 1)
n = 0.99
A = I n-1I = I 0.99 - 1 I = I - 0.01I = 0.01
B = 1-n = 1 - 0.99 = 0.01
thus A = B for case 1
case 2 )
n= 0
A = I n - 1 I = I -1 I = 1
B= 1-n = 1
for this case also A=B
now take an extreme case like n = -1000
case 3 ) n = -1000
A= I n-1 I = I -1000 - 1 I = I -1001 I = 1001
B = 1- n = 1 - (-1000) = 1 + 1000 = 1001
thus for this case too A = B
hence we can conclude that A = B ( option c)