Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 0510:00 AM PDT
-11:00 AM PDT
GMAT Inequalities is a high-frequency topic in GMAT Quant, but many students struggle because the concepts behave differently from standard algebra. Understanding the right rules, patterns, and edge cases can significantly improve both speed and accuracy.- May 06
08:30 AM PDT
-09:30 AM PDT
In Episode 3 of our GMAT Ninja Critical Reasoning series, we tackle Discrepancy, Paradox, and Explain an Oddity questions. You know the feeling: the passage gives you two facts that seem completely contradictory.... - May 08
08:00 AM PDT
-08:30 AM PDT
Join the special YouTube live-stream for selecting the winners of GMAT Club MBA Scholarships sponsored by Juno live. Watch who gets these coveted MBA scholarships offered by GMAT Club and Juno.
27 Apr 2026, 02:43
Kudos
Bookmarks
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
Kudos
Bookmarks
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
Kudos
Bookmarks
This is a great question that’s helpful for learning.
