800_gal wrote:
If n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be.
A) less than -10
B) between -1 and - 1/10
C) between -1/10 and 0
D) between 0 and 1/10
E) greater than 10
kkalyan wrote:
so when n^2 < 1/100 then n<-1/10 It means when we take square root on both side we have to change the sign?
Can anyone explain the logic?
\(n^2<\frac{1}{100}\)
\(n^2<\frac{1}{(10)^2}\)
Taking square root both sides:
\(\sqrt{n^2}<\sqrt{\frac{1}{(10)^2}}\)
\(\hspace{3} \sqrt{n^2}=|n|\)
\(\therefore \hspace{3} |n|<\frac{1}{10}\)
\(|x|<y\) is equivalent to \(-y<x<+y\)
Similarly,
\(-\frac{1}{10}<n<\frac{1}{10}\)
From the stem we know, \(n<0\)
So,
\(-\frac{1}{10}<n<0\)
Write those separately:
\(-\frac{1}{10}<n\)
Flip RHS and LHS and change the symbol to ">"
\(n>-\frac{1}{10}\)
Multiply both sides by -10:
\(-10*n<1\)
Divide each side by n(n is -ve so signs flip again)
\(-10>\frac{1}{n}\)
Flip RHS and LHS and change the symbol to "<"
\(\frac{1}{n}<-10\)--------1
Also,
\(n<0\)-----------2
In 1 and 2; 1 is more limiting
\(\frac{1}{n}<-10\)
Ans: "A"
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Note: Few of these steps could be avoided, but I just wanted to make clear how the inequality symbols and polarity signs change.