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06 Jul 2022, 19:15
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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:
59% (02:12) correct
41%
(02:27)
wrong
based on 213
sessions
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Not Attempted Yet
If \(3x-1 > 0\) and y is a real number, which of the following is NOT always true?
A. \(x(3x-1) > x-\frac{1}{3}\)
B. \(y(3x-1) < y^2(3x-1)\)
C. \((y+1)(1-3x) < y(1-3x)\)
D. \((y+1)(3x-1) > y(3x-1)\)
E. \((2-y)(1-3x) < (1-y)(1-3x)\)
A. \(x(3x-1) > x-\frac{1}{3}\)
B. \(y(3x-1) < y^2(3x-1)\)
C. \((y+1)(1-3x) < y(1-3x)\)
D. \((y+1)(3x-1) > y(3x-1)\)
E. \((2-y)(1-3x) < (1-y)(1-3x)\)
06 Jul 2022, 20:10
Kudos
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If \(3x-1 > 0\) and y is a real number, which of the following is NOT always true?
\(3x-1>0………x>\frac{1}{3}\)
A. \(x(3x-1) > x-\frac{1}{3}\)
\(x*3(x-\frac{1}{3}) > x-\frac{1}{3}\)
Divide by x-1/3 without changing the sign as x>1/3.
3x>0…..x>0….As x>1/3>0, x>0 is always true.
B. \(y(3x-1) < y^2(3x-1)\)
\(y(3x-1) < y^2(3x-1)……….y<y^2\)
Divide by 3x-1 without changing the sign as 3x-1>0
We know nothing about y. If 0<y<1, then not true. Otherwise true.
Our ANSWER
C. \((y+1)(1-3x) < y(1-3x)\)
Divide by 1-3x and change the sign as 1-3x<0
\(y+1>y……..1>0\)
Always true
D. \((y+1)(3x-1) > y(3x-1)\)
Divide by 3x-1 without changing the sign as 3x-1>0
\(y+1>y……1>0\)
Always true.
E. \((2-y)(1-3x) < (1-y)(1-3x)\)
Divide by 1-3x and change the sign as 1-3x<0
2-y>1-y or 2>1…..Always true.
B
\(3x-1>0………x>\frac{1}{3}\)
A. \(x(3x-1) > x-\frac{1}{3}\)
\(x*3(x-\frac{1}{3}) > x-\frac{1}{3}\)
Divide by x-1/3 without changing the sign as x>1/3.
3x>0…..x>0….As x>1/3>0, x>0 is always true.
B. \(y(3x-1) < y^2(3x-1)\)
\(y(3x-1) < y^2(3x-1)……….y<y^2\)
Divide by 3x-1 without changing the sign as 3x-1>0
We know nothing about y. If 0<y<1, then not true. Otherwise true.
Our ANSWER
C. \((y+1)(1-3x) < y(1-3x)\)
Divide by 1-3x and change the sign as 1-3x<0
\(y+1>y……..1>0\)
Always true
D. \((y+1)(3x-1) > y(3x-1)\)
Divide by 3x-1 without changing the sign as 3x-1>0
\(y+1>y……1>0\)
Always true.
E. \((2-y)(1-3x) < (1-y)(1-3x)\)
Divide by 1-3x and change the sign as 1-3x<0
2-y>1-y or 2>1…..Always true.
B
03 Mar 2023, 01:48
Kudos
Bookmarks
Bunuel
If \(3x-1 > 0\) and y is a real number, which of the following is NOT always true?
A. \(x(3x-1) > x-\frac{1}{3}\)
B. \(y(3x-1) < y^2(3x-1)\)
C. \((y+1)(1-3x) < y(1-3x)\)
D. \((y+1)(3x-1) > y(3x-1)\)
E. \((2-y)(1-3x) < (1-y)(1-3x)\)
Solution: A. \(x(3x-1) > x-\frac{1}{3}\)
B. \(y(3x-1) < y^2(3x-1)\)
C. \((y+1)(1-3x) < y(1-3x)\)
D. \((y+1)(3x-1) > y(3x-1)\)
E. \((2-y)(1-3x) < (1-y)(1-3x)\)
- We are given \(3x-1 > 0\) or \(x>\frac{1}{3}\) and x is positive
- y is a real number
- Let us check each option one by one
- A. \(x(3x-1) > x-\frac{1}{3}\)
- We already know this is true and thus this inequality is always true
\(⇒3x(3x-1)>3x-1\)
\(⇒3x>1\) (we could cancel 3x - 1 from both sides without changing inequality sign because we know 3x - 1 > 0)
\(⇒x>\frac{1}{3}\)
- B. \(y(3x-1) < y^2(3x-1)\)
- \(y<y^2\) is not going to be true for all real values of y
- Thus, this inequality is not always true
\(⇒y<y^2\) (cancelled 3x - 1)
Hence the right answer is Option B
- C. \((y+1)(1-3x) < y(1-3x)\)
- Always true
\(⇒(y+1)(3x-1)>y(3x-1)\)
\(⇒y+1>y\)
\(⇒1>0\)
- D. \((y+1)(3x-1) > y(3x-1)\)
- Same as option C. Always true
\(⇒y+1>y\)
- E. \((2-y)(1-3x) < (1-y)(1-3x)\)
- Always true
\(⇒(2-y)(3x-1)>(1-y)(3x-1)\)
\(⇒2-y>1-y\)
\(⇒2>1\)
