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#1
Report Thread starter 14 years ago
#1
Just two questions:

1) f(x) = ax + b and g(x) = bx + a , a and b are constant and unequal, find the relationship between a and b for which fg(x) = gf(x).

Answer:a+b = 1 HOW?


2)

p:x --> ax + b
q:x -->cx + d

Where a b c d are constant, given that pq(x) = qp(x),
show that:

d(a-1) = b(c-1)


Thank you all!
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dvs
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#2
Report 14 years ago
#2
fg(x)=f(bx+a)=a(bx+a)+b=abx+a²+ b
gf(x)=g(ax+b)=b(ax+b)+a=abx+b²+ a
fg(x)=gf(x)
abx+a²+b=abx+b²+a
a²-b²=a-b
(a-b)(a+b)=(a-b)
Since a & b are unequal, a-b doesn't equal zero.. So divide by a-b to get:
a+b=1

2.
p(x)=ax+b q(x)=cx+d
pq(x)=p(cx+d)=a(cx+d)+b=acx+ad+b
qp(x)=q(ax+b)=c(ax+b)+d=acx+cb+d
pq(x)=qp(x)
acx+ad+b=acx+cb+d
ad+b=cb+d
ad-d=cb-b
d(a-1)=b(c-1)
QED
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