1.

1 - {1 + (a2 - 1)-1}-1 = ?1). \(\frac{1}{{{a^2}}}\)2). a23). \(- \frac{1}{{{a^2}}}\)4). –a2

Answer»

Solve the given question, using FOLLOWING laws of indices,

Laws of Indices,

1-: am × an = a{m + n}

2-: am ÷ an = a{m - n}

3-: [(am)n] = amn

4-: (a)1/m = $(\SQRT[m]{a})$

5-: (a)-m = 1/am

6-: (a)(m/n) = $(\sqrt[n]{{{a^m}}})$

7-: a0 = 1

⇒ 1 - {1 + (a2 -1)-1}-1

$(\begin{array}{l} \Rightarrow 1 - {\left\{ {1 + \frac{1}{{{a^{2\;}} - 1}}} \right\}^{ - 1}}\\ \Rightarrow 1 - {\left\{ {\frac{{{a^2}}}{{{a^2} - 1}}} \right\}^{ - 1}}\\ \Rightarrow 1 - \left\{ {\frac{{{a^2}}}{{{a^2} - 1}}} \right\}^{ - 1}\\ \Rightarrow \frac{1}{{{a^2}}} \END{array})$


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