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16 Sep 2019, 21:47
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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If function f(x) satisfies \(f(1 - x) = f(2 + x)\) for all x, which of the following can be true?
A. \(f(x) = 3-x\)
B. \(f(x) = 1-x^2\)
C. \(f(x) = 2 - x - x^2\)
D. \(f(x) = 2x - (1-x)^2\)
E. \(f(x) = \frac{1-x}{2 + x}\)
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16 Sep 2019, 22:35
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A. f(x)=3−xf(x)=3−x ....
B. f(x)=1−x2f(x)=1−x2
C. f(x)=2−x−x2f(x)=2−x−x2
D. f(x)=2x−(1−x)2f(x)=2x−(1−x)2.....after substitution.....OA
E. f(x)=1−x2+x
B. f(x)=1−x2f(x)=1−x2
C. f(x)=2−x−x2f(x)=2−x−x2
D. f(x)=2x−(1−x)2f(x)=2x−(1−x)2.....after substitution.....OA
E. f(x)=1−x2+x
Updated on: 17 Sep 2019, 03:00
Originally posted by freedom128 on 16 Sep 2019, 22:39.
Last edited by freedom128 on 17 Sep 2019, 03:00, edited 4 times in total.
Last edited by freedom128 on 17 Sep 2019, 03:00, edited 4 times in total.
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f(1-x)=f(2+x) means that
x=0 --> f(1)=f(2)
x=1 --> f(-1)=f(),
x=2 --> f(-1)=f(4), etc
It is clear that f(x) has to be symmetrical at x=1.5.
A. f(x)=3−x
NOT symmetrical at x=1.5
B. f(x)=1−x^2
NOT symmetrical at x=1.5. In fact, f(x) is symmetrical at x=0.
C. f(x)=2−x−x^2=(2+x)(1-x)
NOT symmetrical at x=1.5. In fact, f(x) is symmetrical at x=-0.5.
D. f(x)=2x−(1−x)^2= 2x-(x^2-2x+1)=-(x^2+1)
NOT symmetrical at x=1.5. In fact, f(x) is symmetrical at x=0.
E. f(x)=(1−x)^2/(2+x)
Symmetrical at x=1.5
Correct answer is (E)
x=0 --> f(1)=f(2)
x=1 --> f(-1)=f(),
x=2 --> f(-1)=f(4), etc
It is clear that f(x) has to be symmetrical at x=1.5.
A. f(x)=3−x
NOT symmetrical at x=1.5
B. f(x)=1−x^2
NOT symmetrical at x=1.5. In fact, f(x) is symmetrical at x=0.
C. f(x)=2−x−x^2=(2+x)(1-x)
NOT symmetrical at x=1.5. In fact, f(x) is symmetrical at x=-0.5.
D. f(x)=2x−(1−x)^2= 2x-(x^2-2x+1)=-(x^2+1)
NOT symmetrical at x=1.5. In fact, f(x) is symmetrical at x=0.
E. f(x)=(1−x)^2/(2+x)
Symmetrical at x=1.5
Correct answer is (E)
