Read the statements carefully and select the correct option.Statement-1: The value of
$ \frac{\sqrt[6]{2}\left[(625)^{3 / 5} \times(1024)^{-6 / 5} \p(25)^{3 / 5}\right]^{1 / 2}}{\left[(\sqrt[3]{128})^{-5 / 2}\right] \times(125)^{1 / 5}} $ is 1
Statement-2: The expression $ \left(x^{-2 p} y^{3 q}\right)^{6} \p\left(x^{3} y^{-1}\right)^{-4 p} $, after simplification becomes independent of both $ x $ and $ y $.A. Both Statement- 1 and Statement-2 are true.
B. Statement- 1 is true but Statement- 2 is false.
C. Statement- 1 is false but Statement- 2 is true.
D. Both Statement- 1 and Statement- 2 are false.

1.

$\frac{\sqrt[6]{2}[(625)^{3 / 5} \times(1024)^{-6 / 5} \div(25)^{3 / 5}]^{1 / 2}}{[(\sqrt[3]{128})^{-5 / 2}] \times(125)^{1 / 5}}$

$625=5^4, 1024=2^{10}, 25=5^2, 128=2^7, 125=5^3$

Therefore,

$\frac{\sqrt[6]{2}[(625)^{3 / 5} \times(1024)^{-6 / 5} \div(25)^{3 / 5}]^{1 / 2}}{[(\sqrt[3]{128})^{-5 / 2}] \times(125)^{1 / 5}}=\frac{(2)^{\frac{1}{6}}[(5^4)^{3 / 5} \times(2^{10})^{-6 / 5} \div(5^2)^{3 / 5}]^{1 / 2}}{[(2^{\frac{7}{3}})^{-5 / 2}] \times(5^3)^{1 / 5}}$

$=\frac{(2)^{\frac{1}{6}}[(5)^{4\times3 / 5} \times(2)^{10\times -6 / 5} \div(5)^{2\times3 / 5}]^{1 / 2}}{[(2)^{\frac{7}{3}\times-5 / 2}] \times(5)^{3\times1 / 5}}$

$ \begin{array}{l}
=\frac{( 2)^{\frac{1}{6}}\left[( 5)^{\frac{6}{5}} \times ( 2)^{-12}\right]^{\frac{1}{2}}}{( 2)^{\frac{-35}{6}} \times ( 5)^{\frac{3}{5}}}\\
=\frac{( 2)^{\frac{1}{6}}\left[( 5)^{\frac{6}{5} \times \frac{1}{2}} \times ( 2)^{-12\times \frac{1}{2}}\right]}{( 2)^{\frac{-35}{6}} \times ( 5)^{\frac{3}{5}}}\\
=\frac{( 2)^{\frac{1}{6}}\left[( 5)^{\frac{3}{5}} \times ( 2)^{-6}\right]}{( 2)^{\frac{-35}{6}} \times ( 5)^{\frac{3}{5}}}\\
=( 2)^{\frac{1}{6} -6-\left(\frac{-35}{6}\right)} \times ( 5)^{\frac{3}{5} -\frac{3}{5}}\\
=( 2)^{\frac{1}{6} -6+\frac{35}{6}} \times ( 5)^{0}\\
=( 2)^{\frac{1-36+35}{6}} \times 1\\
=( 2)^{\frac{36-36}{6}}\\
=( 2)^{0}\\
=1
\end{array}$

Therefore, statement 1 is true.

2.

$(x^{-2 p} y^{3 q})^{6} \div (x^{3} y^{-1})^{-4 p}=[(x)^{-2p \times 6} (y)^{3q \times 6}] \div [(x)^{3\times -4p} (y)^{-1 \times -4p}]$

$=[(x)^{-12p} (y)^{18q}] \div [(x)^{-12p} (y)^{4p}]$

$=(x)^{-12p-(-12p)} (y)^{18q-4p}$

$=(x)^{-12p+12p} (y)^{18q-4p}$

$=(x)^{0} (y)^{18q-4p}$

$=1 \times (y)^{18q-4p}$

$=(y)^{18q-4p}$

Therefore, statement 2 is false.

Option B is correct.

Tutorialspoint

Updated on: 10-Oct-2022

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