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If f:R→R and g: R→R be defined as f(x) = x+1 and g(x) = x – 1 . Then for all x ∈ R
- a)f o g = g o f
- b)g o f does not exist
- c)f o g does not exist
- d)f o g = g o g
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If f:R→R and g: R→R be defined as f(x) = x+1 and g(x) = x ...
Given f:R→R and g: R→R
=> fog : R→R and gof : R→R
We know that I(R) : R→R
So the domains of gof, fog I(R) are the same.
fog(x) = f(g(x)) = f(x-1) = x-1+1 = I(R)x…..(1)
gof(x) = g(f(x)) = g(x-1) = x-1+1 = I(R)x……(2)
From (1) and (2), we get
fog(x) = gof(x) = I(R) for all x belongs to R
Hence fog = gof
=> fog : R→R and gof : R→R
We know that I(R) : R→R
So the domains of gof, fog I(R) are the same.
fog(x) = f(g(x)) = f(x-1) = x-1+1 = I(R)x…..(1)
gof(x) = g(f(x)) = g(x-1) = x-1+1 = I(R)x……(2)
From (1) and (2), we get
fog(x) = gof(x) = I(R) for all x belongs to R
Hence fog = gof
Most Upvoted Answer
If f:R→R and g: R→R be defined as f(x) = x+1 and g(x) = x ...
Fog(x) = f(g(x)) = f(x-1) = (x-1) + 1 = xgof(x) =
g(f(x)) = g(x+1) = (x+1) - 1 = x∵ fog = gof.....so option A is correct.clearly, fog &
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Community Answer
If f:R→R and g: R→R be defined as f(x) = x+1 and g(x) = x ...
Given f:R→R and g: R→R
=> fog : R→R and gof : R→R
We know that I(R) : R→R
So the domains of gof, fog I(R) are the same.
fog(x) = f(g(x)) = f(x-1) = x-1+1 = I(R)x…..(1)
gof(x) = g(f(x)) = g(x-1) = x-1+1 = I(R)x……(2)
From (1) and (2), we get
fog(x) = gof(x) = I(R) for all x belongs to R
Hence fog = gof
=> fog : R→R and gof : R→R
We know that I(R) : R→R
So the domains of gof, fog I(R) are the same.
fog(x) = f(g(x)) = f(x-1) = x-1+1 = I(R)x…..(1)
gof(x) = g(f(x)) = g(x-1) = x-1+1 = I(R)x……(2)
From (1) and (2), we get
fog(x) = gof(x) = I(R) for all x belongs to R
Hence fog = gof
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If f:R→R and g: R→R be defined as f(x) = x+1 and g(x) = x – 1 . Then for all x ∈ Ra)f o g = g o fb)g o f does not existc)f o g does not existd)f o g = g o gCorrect answer is option 'A'. Can you explain this answer?
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