Question

find
$\frac{1}{1 + {x}^{a - b} } + \frac{1 +}{1 + {x}^{b - a} }$
​

Answer 1

Answer:-

To find:

$\sf \dfrac{1}{1 + {x}^{a - b} } + \dfrac{1}{1 + {x}^{b - a} }$

Using $\sf \dfrac{a^m}{a^n} = a^{m - n}$ we get,

$\sf \implies \dfrac{1}{1 + \dfrac{ {x}^{a} }{ {x}^{b} } } + \dfrac{1}{1 + \dfrac{ {x}^{b} }{ {x}^{a} } }$

Taking LCM in denominators we get,

$\sf \implies \: \dfrac{1}{ \dfrac{ {x}^{b} + {x}^{a} }{ {x}^{b} } } + \dfrac{1}{ \dfrac{ {x}^{a} + {x}^{b} }{ {x}^{a} } } \\ \\ \sf \implies \dfrac{ {x}^{b} }{ {x}^{a} + {x}^{b} } + \dfrac{ {x}^{a} }{ {x}^{a} + {x}^{b} }\\ \\ \sf \implies \: \dfrac{ {x}^{a} + {x}^{b} }{ {x}^{a} + {x}^{b} } \\ \\ \implies \: \sf \large{ \red{1}}$

Some Important Formulae:

• $\sf a^m \times a^n = a^{m + n}$

• $\sf a^m / a^n = a^{m - n}$

• $\sf {({a}^{m})}^{n} = a^{mn}$

• $\sf { \bigg(\dfrac{a}{b}\bigg) }^{-m} = { \bigg(\dfrac{b}{a}\bigg) }^{m}$

• a⁰ = 1

• a¹ = a