Question

Let A =R-{3},B=R-{1}, Let f: A+ B be defined by f(x) = * - 3
then
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Answer 1

A=R−{3}

B=R−{1}

f:A→B

f(x)=x−3x−2

f(x1)=f(x2)

x1−3x1−2=x2−3x2−2

(x2−3)(x1−2)=(x2−2)(x1−3)

x1x2−3x1−2x2+6=x1x2−3x2−2x1+6

−3x1−2x2=−3x2−2x1

−x1=−x2

x1=x2

So, f(x) is one-one

f(x)=x−3x−2

y=x−3x−2

y(x−3)=x−2

yx−3y=x−2

yx−x=3y−2

x(y−1)=3y−2

x=(y−1)3y−2

f(x)=x−3x−2

=y−13y−2−3y−13y−2−2

=y−13y−2−3(y−1)y−13y−2−2(y−1)

=3y−2−3y+33y−2−2y+2

=−2+33y−2y

=y

f(x)=y

f(x) is onto.

So f(x) is bijective and invertible

f(x)=x−3x−2

y=x−3x−2

x=y−3y−2

x(y−3)=y−2

xy−3x=y−2

xy−y=3x−2

y(x−1)=3x−2

y=x−13x−2

f−1(x)=x−13x−2\

hope it helps you