If š(š„)=1š„, show that š(š)āš(š)=š(šš/šāš).
Solution:
Given f(x)=1xf(x)=\dfrac1xf(x)=x1ā
LHS=f(a)āf(b)=1aā1b=bāaabLHS=f(a)-f(b) \\=\dfrac1a-\dfrac1b \\=\dfrac{b-a}{ab}LHS=f(a)āf(b)=a1āāb1ā=abbāaā
RHS=f(abbāa)=1abbāa=bāaabRHS=f(\dfrac{ab}{b-a}) \\=\dfrac{1}{\dfrac{ab}{b-a}} \\=\dfrac{b-a}{ab}RHS=f(bāaabā)=bāaabā1ā=abbāaā
So, LHS=RHS
Hence, proved.
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