Power, Indices and Surds MCQ Question with Answer

Solution: $\frac{\sqrt{24}+\sqrt{6}}{\sqrt{24}-\sqrt{6}}=\frac{2\sqrt{6}+\sqrt{6}}{2\sqrt{6}-\sqrt{6}}=\frac{3\sqrt{6}}{\sqrt{6}}=3$

Solution: $\sqrt{\left( \sqrt{8}+\sqrt{5} \right) ^2}=\sqrt{13+2\sqrt{40}}$

$\sqrt{\left( \sqrt{7}+\sqrt{6} \right) ^2}=\sqrt{13+2\sqrt{42}}$

$\sqrt{\left( \sqrt{10}+\sqrt{3} \right) ^2}=\sqrt{13+2\sqrt{30}}$

$\sqrt{\left( \sqrt{11}+\sqrt{2} \right) ^2}=\sqrt{13+2\sqrt{22}}$

Since $2\sqrt{22}$ is smallest then $\sqrt{11}+\sqrt{2}$ is smallest

Solution: $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}=2^{\frac{2^5-1}{2^5}}=2^{\frac{31}{32}}$

Rules: $\sqrt{\mathrm{X}\sqrt{\mathrm{X}\sqrt{\mathrm{X}\sqrt{\mathrm{X}\sqrt{\mathrm{X}}}}}}=\mathrm{X}^{\frac{\mathrm{X}^{\mathrm{n}}-1}{\mathrm{X}^{\mathrm{n}}}}$

where n is no. of times X repeated.

Solution: $\sqrt{5\sqrt{5\sqrt{5…\mathrm{\alpha}}}}=5$

Rules: $\sqrt{\mathrm{X}\sqrt{\mathrm{X}\sqrt{\mathrm{X}\sqrt{\mathrm{X}\sqrt{\mathrm{X}}}}}}=\mathrm{X}$

Solution:

$\sqrt{12+\sqrt{12+\sqrt{12….}}}=\mathrm{x}\left( \mathrm{let} \right)$
$\Rightarrow 12 +\mathrm{x} =\,\,\mathrm{x}^2$ ( square both side)
$\Rightarrow \mathrm{x}^2-\mathrm{x}-12=0$
$\Rightarrow \mathrm{x}^2-4\mathrm{x}+3\mathrm{x}-12=0$
$\Rightarrow \mathrm{x}\left( \mathrm{x}-4 \right) +3\left( \mathrm{x}-12 \right) =0$
$\Rightarrow \left( \mathrm{x}-4 \right) \left( \mathrm{x}+3 \right) =0$

∴ x = 4 or -3

Here –3 is in option.

Rules: $\sqrt{\mathrm{n}\left( \mathrm{n}+1 \right) +\sqrt{\mathrm{n}\left( \mathrm{n}+1 \right) +\sqrt{\mathrm{n}\left( \mathrm{n}+1 \right) ….}}}=-\mathrm{n},\left( \mathrm{n}+1 \right) $

Solution: $\sqrt{30-\sqrt{30-\sqrt{30….}}}=\mathrm{x}\left( \mathrm{let} \right)$
$\Rightarrow 30 -\mathrm{x} =\,\,\mathrm{x}^2$ ( square both side)
$\Rightarrow \mathrm{x}^2+\mathrm{x}-30=0$
$\Rightarrow \mathrm{x}^2+6\mathrm{x}-5\mathrm{x}-30=0$
$\Rightarrow \mathrm{x}\left( \mathrm{x}+6 \right) -5\left( \mathrm{x}+6 \right) =0$
$\Rightarrow \left( \mathrm{x}+6 \right) \left( \mathrm{x}-5 \right) =0$

∴ x = -6 or 5

Here 5 is in option.

Rules: $\sqrt{\mathrm{n}\left( \mathrm{n}+1 \right) -\sqrt{\mathrm{n}\left( \mathrm{n}+1 \right) -\sqrt{\mathrm{n}\left( \mathrm{n}+1 \right) ….}}}=\mathrm{n},-\left( \mathrm{n}+1 \right) $

Solution: $\sqrt{105\frac{4}{64}\,\,}=\,\,\sqrt{\frac{6724}{64}}=\frac{82}{8}=10\frac{1}{4}$

Solution: $\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{\sqrt{9}+\sqrt{8}}{\left( \sqrt{9}-\sqrt{8} \right) \left( \sqrt{9}+\sqrt{8} \right)}=\sqrt{9}+\sqrt{8}$

Similarly $\frac{1}{\sqrt{8}-\sqrt{7}}=\sqrt{8}+\sqrt{7}$

$\frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{7}+\sqrt{6}$

$\frac{1}{\sqrt{6}-\sqrt{5}}=\sqrt{6}+\sqrt{5}$

$\frac{1}{\sqrt{5}-\sqrt{4}}=\sqrt{5}+\sqrt{4}$

$\therefore \sqrt{9}+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}-\sqrt{6}-\sqrt{5}+\sqrt{5}+\sqrt{4}$

= 3 + 2 = 5

Solution:

$\left( \frac{1024}{243} \right) ^{-\frac{4}{5}}\,\,=\,\,\left( \frac{\left( 2 \right) ^{10}}{\left( 3 \right) ^5} \right) ^{^{-\frac{4}{5}}}=\,\,\frac{2^{10\times \left( -\frac{4}{5} \right)}}{3^{5\times \left( -\frac{4}{5} \right)}}\,\,=\,\,\frac{2^{-8}}{3^{-4}}=\left( \frac{3}{4} \right) ^4=\frac{81}{256}$

Solution: $\left( \sqrt{125} \right) ^{\frac{1}{3}}\,\,=\,\,\left( \sqrt{5^3} \right) ^{\frac{1}{3}}\,\,=\,\,\left( 5^{\frac{3}{2}} \right) ^{\frac{1}{3}}=\sqrt{5}$