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24 Oct 2022, 01:30
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C
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Difficulty:
95%
(hard)
Question Stats:
39% (02:55) correct
61%
(02:23)
wrong
based on 64
sessions
History
Date
Time
Result
Not Attempted Yet
If \(2f(x) + f(x^2 - 1) = 1\), for all x, then what is the value of \(f(-\sqrt{2})\) ?
A. -1/3
B. 0
C. 1/3
D. 2/3
E. 1
A. -1/3
B. 0
C. 1/3
D. 2/3
E. 1
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24 Oct 2022, 04:26
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We can substitute the value of x = -\(\sqrt{2}\) in the equation. Once we have obtained the function whose value is not known, that variable can be substituted in the base equation to check if we have sufficient information to solve the question.
Ex:\(2f(x) + f(x^2 - 1) = 1\)
at x = -\(\sqrt{2}\)
\(2f(-\sqrt{2}) + f((-\sqrt{2})^2 - 1) = 1\)
\(f(-\sqrt{2})\) = \(\frac{1-f(1)}{2}\)
Now, we can substitute x = 1 in the base equation and solve further.
x = 1
\(2f(1) + f((1)^2 - 1) = 1\)
\(f(1) = \frac{1-f(0)}{2}\)
This process needs to continue, till we have all known values (of course, we will soon see a pattern!)
A detailed solution is attached.
IMO C
Ex:\(2f(x) + f(x^2 - 1) = 1\)
at x = -\(\sqrt{2}\)
\(2f(-\sqrt{2}) + f((-\sqrt{2})^2 - 1) = 1\)
\(f(-\sqrt{2})\) = \(\frac{1-f(1)}{2}\)
Now, we can substitute x = 1 in the base equation and solve further.
x = 1
\(2f(1) + f((1)^2 - 1) = 1\)
\(f(1) = \frac{1-f(0)}{2}\)
This process needs to continue, till we have all known values (of course, we will soon see a pattern!)
A detailed solution is attached.
IMO C
Attachments
Screenshot 2022-10-24 152220.jpg [ 191.82 KiB | Viewed 2813 times ]
24 Oct 2022, 20:43
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Bunuel
If \(2f(x) + f(x^2 - 1) = 1\), for all x, then what is the value of \(f(-\sqrt{2})\) ?
A. -1/3
B. 0
C. 1/3
D. 2/3
E. 1
A. -1/3
B. 0
C. 1/3
D. 2/3
E. 1
Such questions require you to substitute values of x in successive steps till you find that you can get a value of some function.
\(2f(x) + f(x^2 - 1) = 1\)
(1) \(x=(-\sqrt{2})\)
\(2f(-\sqrt{2}) + f((-\sqrt{2)}^2 - 1) = 1\)
\(2f(-\sqrt{2}) + f(1) = 1\)
(2) \(x=(1)\)
\(2f(1) + f(1^2 - 1) = 1\)
\(2f(1) + f(0) = 1\)
(3) \(x=(0)\)
\(2f(0) + f(0^2 - 1) = 1\)
\(2f(0) + f(-1) = 1\)………(3)
(4) \(x=(-1)\)
\(2f(-1) + f((-1)^2 - 1) = 1\)
\(2f(-1) + f(0) = 1\)……..(4)
Multiply (3) with 2 and subtract 4 from it.
\(3f(0)=1……f(0)=\frac{1}{3}\)
Substitute this in (2)
\(2f(1)+\frac{1}{3}=1\)
\(f(1)=\frac{1}{3}\)
Substitute this in (1)
\(2f(-\sqrt{2})+\frac{1}{3}=1\)
\(f(-\sqrt{2})=\frac{1}{3}\)
C
