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27 Apr 2026, 02:43
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If \(2f(x) + f(x^2 - 1) = 1\), for all \(x\), then what is the value of \(f(-\sqrt{2})\) ?
A. \(-\frac{1}{3}\)
B. \(0\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{3}\)
E. \(1\)
A. \(-\frac{1}{3}\)
B. \(0\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{3}\)
E. \(1\)
27 Apr 2026, 02:43
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Official Solution:
If \(2f(x) + f(x^2 - 1) = 1\), for all \(x\), then what is the value of \(f(-\sqrt{2})\) ?
A. \(-\frac{1}{3}\)
B. \(0\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{3}\)
E. \(1\)
We are asked to find \(f(-\sqrt{2})\).
Substitute \(x = -\sqrt{2}\) in the given equation \(2f(x) + f(x^2 - 1) = 1\):
\(2f(-\sqrt{2}) + f(1) = 1\)
So, to get \(f(-\sqrt{2})\), we now need \(f(1)\).
Since the equation holds for all \(x\), it must also hold for \(x = 1\). Substitute \(x = 1\):
\(2f(1) + f(0) = 1\)
We now need \(f(0)\). Substitute \(x = 0\):
\(2f(0) + f(-1) = 1\)
We now need \(f(-1)\). Substitute \(x = -1\):
\(2f(-1) + f(0) = 1\)
From the last two equations:
\(2f(0) + f(-1) = 1\)
\(2f(-1) + f(0) = 1\)
Subtracting gives \(f(0) = f(-1)\).
Substitute this into either equation: \(3f(0) = 1\), so \(f(0) = f(-1) = \frac{1}{3}\).
Substitute \(f(0) = \frac{1}{3}\) into \(2f(1) + f(0) = 1\) to get \(f(1) = \frac{1}{3}\).
Finally, from the first equation:
\(2f(-\sqrt{2}) + \frac{1}{3} = 1\), so \(f(-\sqrt{2}) = \frac{1}{3}\).
Answer: C
If \(2f(x) + f(x^2 - 1) = 1\), for all \(x\), then what is the value of \(f(-\sqrt{2})\) ?
A. \(-\frac{1}{3}\)
B. \(0\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{3}\)
E. \(1\)
We are asked to find \(f(-\sqrt{2})\).
Substitute \(x = -\sqrt{2}\) in the given equation \(2f(x) + f(x^2 - 1) = 1\):
\(2f(-\sqrt{2}) + f(1) = 1\)
So, to get \(f(-\sqrt{2})\), we now need \(f(1)\).
Since the equation holds for all \(x\), it must also hold for \(x = 1\). Substitute \(x = 1\):
\(2f(1) + f(0) = 1\)
We now need \(f(0)\). Substitute \(x = 0\):
\(2f(0) + f(-1) = 1\)
We now need \(f(-1)\). Substitute \(x = -1\):
\(2f(-1) + f(0) = 1\)
From the last two equations:
\(2f(0) + f(-1) = 1\)
\(2f(-1) + f(0) = 1\)
Subtracting gives \(f(0) = f(-1)\).
Substitute this into either equation: \(3f(0) = 1\), so \(f(0) = f(-1) = \frac{1}{3}\).
Substitute \(f(0) = \frac{1}{3}\) into \(2f(1) + f(0) = 1\) to get \(f(1) = \frac{1}{3}\).
Finally, from the first equation:
\(2f(-\sqrt{2}) + \frac{1}{3} = 1\), so \(f(-\sqrt{2}) = \frac{1}{3}\).
Answer: C
30 Apr 2026, 10:10
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This is a great question that’s helpful for learning.
