Indices and Surd Questions and answers

Given that, \( \Large  a = 2 + \sqrt{3} \)

Then, \( \Large  \frac{1}{a} = 2 - \sqrt{3} \) [by conjugate property]

Now, we have, \( \Large  a^{2} + a^{-2} = \left(a + \frac{1}{a}\right)^{2} - 2 \)

= \( \Large  \left(2 + \sqrt{3} + 2 - \sqrt{3}\right)^{2} - 2 \)

= \( \Large  \left(4\right)^{2} - 2 = 16 - 2 = 14 \)

Given expression = \( \Large  \left[ \left(\sqrt{2}\right)^{\sqrt{2}} \right]^{\sqrt{2}} \)

= \( \Large  \left(\sqrt{2}\right)^{ \left(2\right)^{\frac{\sqrt{2}}{2}} } = \left(\sqrt{2}\right)^{ \left(2\right)^{\frac{1}{\sqrt{2}}} } \)

= \( \Large  \left(2\right)^{\frac{1}{2} \times 2^{\frac{1}{\sqrt{2}}}} = 2^{ \left(\frac{2}{2}^{\frac{1}{\sqrt{2}}}\right) } = \left(2\right)^{ \left(2\right)^{ \left(\frac{1}{\sqrt{2}}-1\right) } } \)

which denotes a real number but not a rational number.

Given that, \( \Large 16 \times 8^{n+2} = 2^{m} \)

=> \( \Large  \left(2\right)^{4} \times 2^{3 \left(n+2\right) } = 2^{m} \)

=> \( \Large  \left(2\right)^{ \left(4+3n+6\right) } = 2^{m} \)

=> \( \Large  2^{ \left(3n+10\right) }= 2^{m} \)

On comparing, we get

\( \Large  3n + 10 = m \)

=> m = 3n + 10

Given, \( \Large  \sqrt{10 + \sqrt[3]{x}} = 4 \)

On squaring both sides, we get

\( \Large  10 + \sqrt[3]{x} = 16 \)

=> \( \Large  \sqrt[3]{x} = 6 \)

Now, cubic on both sides, we get

\(  \Large x = \left(6\right)^{3} = 216 \)

If m and n are natural numbers, then \(  \Large  \sqrt[m]{n} \) is irrational unless n is mth power of an integer.

\( \Large  \sqrt{2}, \sqrt[3]{3}, \sqrt[6]{6} \)

Taking LCM of 2, 3 and 6 = 12

Now, \(  \Large \sqrt{2} = \left(2\right)^{\frac{1}{2}} = \left(2\right)^{\frac{6}{12}} = \sqrt[12]{2^{6}} = \sqrt[12]{64} \)

\( \Large  \sqrt[3]{3} = \left(3\right)^{\frac{1}{3}} = \left(3\right)^{\frac{4}{12}} = \sqrt[12]{3^{4}} = \sqrt[12]{81} \)

\( \Large  \sqrt[6]{6} = \left(6\right)^{\frac{1}{6}} = \left(6\right)^{\frac{2}{12}} = \sqrt[12]{6^{2}} = \sqrt[12]{36} \)

So, neither I nor II are correct.

\(  \Large  \left(\sqrt{1+a} + \sqrt{1-a}\right)^{2} \)

= \(  \Large  \left(1 + a\right) + \left(1 - a\right) + 2\sqrt{1-a^{2}} \)

=\(  \Large  2 \left(1 + \sqrt{1 - \frac{3}{4}}\right) \)

= \( \Large  2 \left(1 + \frac{1}{2}\right) = 2 \times \frac{3}{2} = 3 \)

Therefore, \( \Large  \left(\sqrt{1+a} + \sqrt{1-a}\right) = \sqrt{3} \)

\( \Large  6\sqrt{ \left(27\right)^{-\frac{2}{3}} + \left(8\right)^{-\frac{2}{3}} } = 6\sqrt{ \left(\frac{1}{27}\right)^{\frac{2}{3}} + \left(\frac{1}{8}\right)^{\frac{2}{3}} } \)

\( \Large  6\sqrt{ \left(\frac{1}{3^{3}}\right)^{\frac{2}{3}} } + \left(\frac{1}{2^{3}}\right)^{\frac{2}{3}} = 6\sqrt{\frac{1}{3^{2}} + \frac{1}{2^{2}}} \)

\( \Large  6\sqrt{\frac{1}{9} + \frac{1}{4}} = 6\sqrt{\frac{4+9}{36}} = 6 \times \frac{\sqrt{13}}{6} = \sqrt{13} \)

Given that, \(  \Large 2x^{\frac{1}{3}} + 2x^{-\frac{1}{3}} = 5 \)

Let \( \Large  x^{\frac{1}{3}} = m \), then \( 2m+\frac{2}{m} = 5 \)

=> \( \Large  2m^{2} - 5m + 2 = 0 \)

=> \(  \Large \left(2m - 1\right) \left(m - 2\right) = 0 \)

Therefore, \( \Large  m = \frac{1}{2} \) or m = 2

=> \( \Large  x^{\frac{1}{3}} = 2 \ or \ \frac{1}{2} \)