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Q)

In the set of real numbers $R,$ an operation $^\ast $is defined by $a^\ast b=\sqrt{a^{2}+b^{2}}.$ Then the value of $(3^\ast4)^\ast 5$ is

\[\begin{array}{1 1}(1)5&(2)5\sqrt{2}\\(3)25&(4)50\end{array}\]

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A)
$a*b =\sqrt {a^2+b^2}$
$(3 * 4)*5=\sqrt {3^2+4^2} *5$
$\qquad =\sqrt {9 +16}* 5$
$\qquad= 5 *5 =\sqrt {5^2 +5^2}$
$\qquad= 5 \sqrt 2$
Hence 2 is the correct answer.
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