The correct option is D a^2 a^43÷ a^ 23 =a^43a^ 23 =a^43 ( 23 #160; ) #160; #160; #160; #160; #160; ( x^mx^n =x^m n ...

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Standard IX

Mathematics

Fractional Exponent

Evaluate a4...

Question

Evaluate $a^{\frac{4}{3}} \div a^{- \frac{2}{3}}$

\frac{1}{a^{2}}

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\frac{1}{a^{- 3}}

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a^{2}

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a

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Solution

The correct option is D $a^{2}$
$a^{\frac{4}{3}} \div a^{- \frac{2}{3}}$

$= \frac{a^{\frac{4}{3}}}{a^{\frac{- 2}{3}}}$

$= a^{\frac{4}{3} - (\frac{- 2}{3})}$ $(\frac{x^{m}}{x^{n}} = x^{m - n})$

$= a^{\frac{4}{3} + \frac{2}{3}}$

$= a^{\frac{6}{3}}$

$= a^{2}$

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e^{2} - 1 = \frac{1 + \frac{a^{2}}{2!} + \frac{a^{4}}{3!} + \frac{a^{6}}{4!} + . . .}{1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^{2}}{4!} + . . .}

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Q. Evaluate the following:
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{(\sqrt{x + 1} + \sqrt{x - 1})}^{6} + {(\sqrt{x + 1} - \sqrt{x - 1})}^{6}

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{(x + \sqrt{x^{2} - 1})}^{6} + {(x - \sqrt{x^{2} - 1})}^{6}

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{(\sqrt{3} + 1)}^{5} - {(\sqrt{3} - 1)}^{5}

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{\{a^{2} + \sqrt{a^{2} - 1}\}}^{4} + {\{a^{2} - \sqrt{a^{2} - 1}\}}^{4}

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$(i i) (x + \sqrt{x^{2} - 1})^{6} + (x - \sqrt{x^{2} - 1})^{6}$

$(i i i) (1 + 2 \sqrt{x})^{5} + (1 - 2 \sqrt{x})^{5}$

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$(v i) (2 + \sqrt{3})^{7} + (2 - \sqrt{3})^{7}$

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$(v i i i) (0.99)^{5} + (1.01)^{5}$

$(i x) (\sqrt{3} + \sqrt{2})^{6} - (\sqrt{3} - \sqrt{2})^{6}$

$(x) {a^{2} + \sqrt{a^{2} - 1}}^{4} + {a^{2} - \sqrt{a^{2} - 1}}^{4}$

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Evaluate a43÷a−23

The correct option is D a2a43÷a−23=a43a−23=a43−(−23) (xmxn=xm−n)=a43+23=a63=a2

Evaluate $a^{\frac{4}{3}} \div a^{- \frac{2}{3}}$

The correct option is D $a^{2}$
$a^{\frac{4}{3}} \div a^{- \frac{2}{3}}$

$= \frac{a^{\frac{4}{3}}}{a^{\frac{- 2}{3}}}$

$= a^{\frac{4}{3} - (\frac{- 2}{3})}$ $(\frac{x^{m}}{x^{n}} = x^{m - n})$

$= a^{\frac{4}{3} + \frac{2}{3}}$

$= a^{\frac{6}{3}}$

$= a^{2}$