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30 Sep 2021, 05:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
63% (02:29) correct
37%
(02:34)
wrong
based on 187
sessions
History
Date
Time
Result
Not Attempted Yet
If \(m - n = 1\), where \(m \neq -n \neq 0\), which of the following must be true ?
I. \(m(m-1)=n(n+1)\)
II. \(\frac{1}{mn^2}-\frac{1}{m^2n}+1=\frac{1+m^2n^2}{m^2n^2}\)
III. \(\frac{1}{mn^2}-\frac{1}{m^2n}+1=\frac{1+mn}{mn}\)
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
I. \(m(m-1)=n(n+1)\)
II. \(\frac{1}{mn^2}-\frac{1}{m^2n}+1=\frac{1+m^2n^2}{m^2n^2}\)
III. \(\frac{1}{mn^2}-\frac{1}{m^2n}+1=\frac{1+mn}{mn}\)
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
yashikaaggarwal
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Joined: 19 Jan 2020
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30 Sep 2021, 22:42
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M = N+1
Substituting value in LHS of options.
1) m(m-1) = (n+1)(n+1-1) = n(n+1) = RHS (( Correct ))
2) \(\frac{1}{m*n^2} - \frac{1}{m^2*n} + 1\)
=> \(\frac{m^2*n - m*n^2 + m^3*n^3 }{ m^3*n^3}\)
Taking mn common from numerator and denominator
=> \(\frac{m-n+m^2*n^2}{m^2*n^2}\)
Put m = n+1
=> \(\frac{n+1-n+m^2*n^2}{m^2*n^2}\)
=> \(\frac{1+m^2*n^2}{m^2*n^2}\)
(Option B is correct)
C's RHS is not equivalent of \(\frac{1}{m*n^2}\) - \(\frac{1}{m^2*n}\) + 1
Answer is D
Posted from my mobile device
Substituting value in LHS of options.
1) m(m-1) = (n+1)(n+1-1) = n(n+1) = RHS (( Correct ))
2) \(\frac{1}{m*n^2} - \frac{1}{m^2*n} + 1\)
=> \(\frac{m^2*n - m*n^2 + m^3*n^3 }{ m^3*n^3}\)
Taking mn common from numerator and denominator
=> \(\frac{m-n+m^2*n^2}{m^2*n^2}\)
Put m = n+1
=> \(\frac{n+1-n+m^2*n^2}{m^2*n^2}\)
=> \(\frac{1+m^2*n^2}{m^2*n^2}\)
(Option B is correct)
C's RHS is not equivalent of \(\frac{1}{m*n^2}\) - \(\frac{1}{m^2*n}\) + 1
Answer is D
Posted from my mobile device
12 Oct 2021, 07:49
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Bunuel
Number plugging :
let m \( = 2 \) and n \( = 1 \)
I. \(m(m-1)=n(n+1) -> 2= 2 \) True.
II. \(\frac{1}{mn^2}-\frac{1}{m^2n}+1=\frac{1+m^2n^2}{m^2n^2}\)
= \(\frac{1}{2} - \frac{1}{4} + 1 = \frac{1 + 4*1}{4}\)
->\( \frac{5}{4} = \frac{5}{4} \) True.
III. \(\frac{1}{mn^2}-\frac{1}{m^2n}+1=\frac{1+mn}{mn}\)
- >\( \frac{5}{4} \neq \frac{3}{2} \) Hence Not True
Only I and II must be true .
Ans D
Hope it's clear.
