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  • The product 3sqroot2,4sqroot2,12sqroot32 equals: (A)sqroot2(B)2 (C)12 sqroot 2 (D)12 sqroot 32

The product \sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32} equals:

(A)\sqrt{2}
(B)2
(C)12\sqrt{2}
(D)12\sqrt{32}

Answers (1)

Answer.        [B]
Solution.        
We have, \sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}

We know that \sqrt[n]{a}= \left ( a \right )^{\frac{1}{n}}

 \sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}=\left ( 2 \right )^{\frac{1}{3}}\left ( 2 \right )^{\frac{1}{4}}\left ( 2\cdot 2\cdot 2\cdot 2\cdot 2 \right )^{\frac{1}{12}}

= \left ( 2 \right )^{\frac{1}{3}}\left ( 2 \right )^{\frac{1}{4}}\left ( 2^{5} \right )^{\frac{1}{12}}                                                                             

= \left ( 2 \right )^{\frac{1}{3}}\left ( 2 \right )^{\frac{1}{4}}\left ( 2 \right )^{\frac{5}{12}}     \because \left ( a^{m} \right )^{n}= a^{m\times n}
= \left ( 2 \right )^{\frac{1}{3}+ \frac{1}{4}+\frac{5}{12}}
\because a^{m}\times a^{n}= a^{m+n}               

Now,  \frac{1}{3}+\frac{1}{4}+\frac{5}{12}= \frac{4+3+5}{12}= \frac{12}{12}= 1

So, \left ( 2 \right )^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}=2^{1}= 2

Hence option B is correct.

Posted by

infoexpert27

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