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Let f(x)=ax+b and g(x)=cx+d, a≠0, c≠0. Assume a=1,b=2. If fog(x)=gof(x) for all x, what can you say about c and d

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a

c and d both arbitrary

b

c = 1, d arbitrary

c

c arbitrary, d = 1

d

c = 1, d = 1

answer is B.

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Detailed Solution

fog(x)=fgx=acx+d+band gof(x)=gfx=cax+b+dfog(x)=gof(x) and at a=1, b=2⇒cx+d+2=cx+2c+d⇒c=1 and d is arbitrary

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Let f(x)=ax+b and g(x)=cx+d, a≠0, c≠0. Assume a=1,b=2. If fog(x)=gof(x) for all x, what can you say about c and d