[5 marks sum]

Question 15 Marks

Prove that : $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1 A$

Answer

$\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$
$\text{L.H.S}. =\frac{1}{1+ x ^{ a - b }}+\frac{1}{1+ x ^{ b - a }}$
$=\frac{1}{ x ^{ a - a }+ x ^{ a - b }}+\frac{1}{ x ^{ b - b }+ x ^{ b - a }}$
$=\frac{1}{ x ^{ a } \cdot x ^{- a }+ x ^{ a } \cdot x ^{- b }}+\frac{1}{ x ^{ b } \cdot x ^{- b }+ x ^{ b } \cdot x ^{- a }}$
$=\frac{1}{ x ^{ a }\left( x ^{- a }+ x ^{- b }\right)}+\frac{1}{ x ^{ b }\left( x ^{- b }+ x ^{- a }\right)}$
$=\frac{1}{\left( x ^{- a }+ x ^{- b }\right)}\left[\frac{1}{ x ^{ a }}+\frac{1}{ x ^{ b }}\right]$
$=\frac{1}{ x ^{- a }+ x ^{- b }}\left[ x ^{- a }+ x ^{- b }\right]=1=\text { R. H.S. }$

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Question 25 Marks

Prove that : $\left(\frac{x^a}{x^b}\right)^{\frac{1}{a b}}\left(\frac{x^b}{x^c}\right)^{\frac{1}{b c}}\left(\frac{x^c}{x^a}\right)^{\frac{1}{c a}}=1$

Answer

$\left(\frac{x^a}{x^b}\right)^{\frac{1}{a b}}\left(\frac{x^b}{x^c}\right)^{\frac{1}{b c}}\left(\frac{x^c}{x^a}\right)^{\frac{1}{c a}}=1$
$\text{L. H.S}.=\left(\frac{x^a}{x^b}\right)^{\frac{1}{a b}}\left(\frac{x^b}{x^c}\right)^{\frac{1}{b c}}\left(\frac{x^c}{x^a}\right)^{\frac{1}{c a}}$
$\left(x^{a-b}\right)^{\frac{1}{a b}}\left(x^{b-c}\right)^{\frac{1}{b c}}\left(x^{c-a}\right)^{\frac{1}{c a}}$
$ =x^{\frac{a-b}{a b}} x^{\frac{b-c}{b c}} x^{\frac{c-a}{c a}} \ldots \ldots \ldots \ldots\left\{\left(x^a\right)^b=x^{a b}\right\}$
$=x^{\frac{ a - b }{ ab }+\frac{ b - c }{ bc }+\frac{ c - a }{ ca }}$
$=x^{\frac{a c-b c+a b-a c+b c-a b}{a b c}}$
$=x^0=1= \text{R. H.S}.\left(\because x^0=1\right)$

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MATHS [5 marks sum]

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