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14 May 2024, 01:30
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Difficulty:
95%
(hard)
Question Stats:
53% (02:22) correct
47%
(02:31)
wrong
based on 151
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If \(x\) is a prime number and \(\sqrt{(a + b + c - 1 + x)^2} + |\frac{1}{1-x}- a - b - c| + \sqrt{c^2}=0\), what is the value of \(x + c\)?
A. -2
B. 0
C. 1
D. 2
E. Cannot be determined from the given information
A. -2
B. 0
C. 1
D. 2
E. Cannot be determined from the given information
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01 Jun 2024, 11:15
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Official Solution:
If \(x\) is a prime number and \(\sqrt{(a + b + c - 1 + x)^2} + |\frac{1}{1-x}- a - b - c| + \sqrt{c^2}=0\), what is the value of \(x + c\)?
A. -2
B. 0
C. 1
D. 2
E. Cannot be determined from the given information
We have that the sum of three non-negative values (the square root of a square, the absolute value of some number, and the square root of a square again) is 0. For this to be true, each of the three must be 0.
So:
Equate (i) and (ii): \(1 - x = \frac{1}{1-x}\). This results in \((1-x)^2 = 1\).
Take the square root: \(|1-x| = 1\). This yields \(x = 0\) or \(x = 2\). It's given that \(x\) is a prime number, so \(x = 2\).
Answer: D
If \(x\) is a prime number and \(\sqrt{(a + b + c - 1 + x)^2} + |\frac{1}{1-x}- a - b - c| + \sqrt{c^2}=0\), what is the value of \(x + c\)?
A. -2
B. 0
C. 1
D. 2
E. Cannot be determined from the given information
We have that the sum of three non-negative values (the square root of a square, the absolute value of some number, and the square root of a square again) is 0. For this to be true, each of the three must be 0.
So:
- (i) \(a + b + c - 1 + x=0\), which gives \(a + b + c = 1 - x\)
(ii) \(\frac{1}{1-x}- a - b - c=0\), which gives \(a + b + c =\frac{1}{1-x}\)
(iii) \(c^2=0\), which gives \(c=0\).
Equate (i) and (ii): \(1 - x = \frac{1}{1-x}\). This results in \((1-x)^2 = 1\).
Take the square root: \(|1-x| = 1\). This yields \(x = 0\) or \(x = 2\). It's given that \(x\) is a prime number, so \(x = 2\).
- \(x + c=2+0=2\).
Answer: D
General Discussion
14 May 2024, 02:05
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Bunuel
\(\sqrt{(a + b + c - 1 + x)^2} + |\frac{1}{1-x}- a - b - c| + \sqrt{c^2}=0\)
\(|a + b + c - 1 + x| + |\frac{1}{1-x}- a - b - c| + |c|=0\)
As each term in the modulus is non-negative, each of the term equals 0
\(|a + b + c - 1 + x| = 0\)
\(a + b + c = 1 - x\) → (1)
\(\frac{1}{1-x}- a - b - c = 0\)
\(\frac{1}{1-x} = a + b + c\) → (2)
\(|c| = 0\)
\(c = 0\) → (3)
From (1) & (2)
\(1 - x = \frac{1}{1-x}\)
\((1 - x)^2 = 1\)
\(1 - x = \pm 1\)
\(1 - x = -1\) ⇒ x = 2
\(1 - x = 1\) ⇒ x = 0
As x is a prime number, the value of \(x\) cannot be \(0\). It has to be \(2\).
\(x + c = 2 + 0 = 2 \)
Option D
