Maths M.C.Q

3. $2^{\frac{1}{6}}$

Solution:

$\sqrt[4]{\sqrt[3]{2^2}}=\sqrt[4]{(2^2)^{\frac{1}{3}}}=\Big(2^{\frac{2}{3}}\Big)^{\frac{1}{4}}=2^{\frac{2}{3}\times\frac{1}{4}}=2^{\frac{1}{6}}$

Hence, (c) is the correct answer.

2. $5\sqrt{6}$

Solution:

$\sqrt{10}, \sqrt{15}=\sqrt{2.5}\sqrt{3.5}=\sqrt{2}\sqrt{5}\sqrt{3}\sqrt{5}=5\sqrt{6}$

3. $3\sqrt{3}$

Solution:

Given $2\sqrt{3}+\sqrt{3}=(2+1)\sqrt{3}=3\sqrt{3}$

Hence, (c) is the correct answer.

2. $2$

Solution:

LCM of 3, 4 and 12 = 12

$\sqrt[3]{2}=\sqrt[12]{2^4}\ [\because\sqrt[\text{m}]{\text{a}}=\sqrt[\text{mn}]{\text{a}^\text{n}}]$

$\sqrt[4]{2}=\sqrt[12]{2^3}$

and $\sqrt[12]{32}=\sqrt[12]{2^5}$

$\therefore\text{product of }\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{2^3}.\sqrt[12]{2^5}=\sqrt[12]{2^4.2^3.2^5}$

$=12\sqrt{2^{4+3+5}}=\sqrt[12]{2^{12}}=2^{12\times\frac{1}{12}}=2\ [\because\sqrt[\text{m}]{\text{a}^\text{n}}=\text{a}^{\frac{\text{n}}{\text{m}}}]$

4. $3+2\sqrt{2}$

Solution:

$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{3-2\sqrt{2}}=\frac{1}{3-2\sqrt{2}}\cdot\frac{3+2\sqrt{2}}{3+2\sqrt{2}}$ $[\because\sqrt{8}=\sqrt{2\times2\times2}=2\sqrt{2}]$

$[$multiplying numerator and denominator by $3+2\sqrt{2}]$

$\frac{3+2\sqrt{2}}{9-(2-\sqrt{2})^2}$$[\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$

$=\frac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}$

4. Sometimes rational, sometimes irrational.

Solution:

Sometimes rational, sometimes irrational The product of any two irrational numbers is sometimes rational and sometimes irrational.

Hence, (D) is the correct answer.

For example:

rational

$(2+\sqrt{3})(2-\sqrt{3})$

$(2)^2-(\sqrt{3})^2$

$4-3=1$

irrational

$(2+\sqrt{3})(1-\sqrt{3})$

$2(1-\sqrt{3})+\sqrt{3}(1-\sqrt{3})$

$2-2\sqrt{3}+\sqrt{3}-3$

$-1-\sqrt{3}$

4.  $0.4014001400014...$

Solution:

A number is irrational if and only of its decimal representation is non - terminating and nonrecurring.

1. $0.14$ is a terminating decimal and therefore cannot be an irrational number.

2. $0.14\overline{16}$ is a non - terminating and recurring decimal and therefore cannot be irrational.

3. $0.\overline{1416}$ is a non - terminating and recurring decimal and therefore cannot be irrational.

4. $0.4014001400014...$ is a non - terminating and non - recurring decimal and therefore is an irrational number.

3. $2$

Solution:

Let x = 1.999...

Now, 10x = 19.999...

On subtracting Eq. (i) from Eq. (ii), we get

10x - x = (19.999...) - (1.9999...)

⇒ 9x = 18

$\therefore\text{x}=\frac{18}{9}=2$

2. 19

Solution:

$\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}$

$[$multiplying numerator and denominator by $3\sqrt{3}+2\sqrt{2}]$

$=\frac{7(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2}$ $[\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$

$=\frac{7(3\sqrt{3}+2\sqrt{2})}{27-8}$ $\big[\because\text{fraction}=\frac{\text{numerator}}{\text{denominator}}\big]$

$=\frac{7(3\sqrt{3}+2\sqrt{2})}{19}$

Hence, after rationalising the denominator of $=\frac{7}{(3\sqrt{3}-2\sqrt{2})}$ we get the denominator as 19.

4. Non - terminating non - recurring.

Solution:

The decimal expansion of the number $\sqrt{2}$ is non-terminating non - recurring. Because $\sqrt{2}$ is an irrational number.

Also, we know that an irrational number is non - terminating non - recurring.

3. $\sqrt{7}$

Solution:

$\because\sqrt{\frac{4}{9}}=\frac{2}{3}\text{(rational)},$

$\frac{\sqrt{12}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3}}=2\text{(rational)},$

$\sqrt{81}=9\text{(rational)}$

but $\sqrt{7}$ is an irrational number.

Hence, $\sqrt{7}$ is an irrational number.

1. $\frac{\sqrt{7}+2}{3}$

Solution:

$\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times\frac{\sqrt{7}+2}{\sqrt{7}+2}$

$\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}$

Hence, (a) is the correct answer.

3.  $1.5$

Solution:

We know that

$\sqrt{2}.=1.4142135...$ and $\sqrt{3}.=1.732050807...$

We see that 1.5 is a rational number which lies between 1.4142135... and 1.732050807...

Hence, (c) is tthe correct answer.

1. $\frac{1}{9}$

Solution:

$\sqrt[4]{(81)^{-2}}=\sqrt[4]{\Big(\frac{1}{81}\Big)^2}=\sqrt[4]{\Big\{\Big(\frac{1}{9}\Big)^2\Big\}^2}=\sqrt[4]{\Big(\frac{1}{9}\Big)^4}=\Big(\frac{1}{9}\Big)^{4\times\frac{1}{4}}=\frac{1}{9}$

Hence, (a) is the correct answer.

3. $(\sqrt{\text{x}^3})^{\frac{2}{3}}$

Solution:

1. $\text{x}^{\frac{12}{7}}-\text{x}^{\frac{5}{7}}\neq\text{x}$

2. $\sqrt[12]{\text{(x}^4})^{\frac{1}{3}}=\sqrt[12]{\text{x}^{4\times\frac{1}{3}}}=\Big(\text{x}^\frac{4}{3}\Big)^\frac{1}{12}=\text{x}^{\frac{4}{3}\times\frac{1}{12}}=\text{x}^\frac{1}{9}\neq\text{x}$

3. $\Big((\text{x}^3)^\frac{1}{2}\Big)^\frac{2}{3}=\text{(x)}^{\frac{3}{2}\times\frac{2}{3}}=\text{x}^1=\text{x}$

4. $\text{x}^{\frac{12}{7}}\times\text{x}^\frac{7}{12}=\text{x}^{\frac{12}{7}+\frac{7}{12}}=\text{x}^\frac{193}{84}\neq\text{x}$

Hence, (c) is the correct answer.

2. 2

Solution:

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16\times2}+\sqrt{16\times3}}{\sqrt{4\times2}+\sqrt{4\times3}}$

$=\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}$ 

$=\frac{4}{2}=2$

Hence, (b) is correct answer.

3. 0.4142

Solution:

$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\sqrt{\frac{\sqrt{2}-1}{{\sqrt{2}+1}}\cdot\frac{\sqrt{2}-1}{\sqrt{2}-1}}$

$[$Inside the root, multiplying numerator and denominator by $(\sqrt{2}-1)]$

$=\sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2-(1)^2}}$ $ [\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$ 

$=\sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}=\sqrt{2}-1=(1.4142...)-1=0.4142...$

1. 4

Solution:

$(256)^{0.16}\times(256)^{0.09}=(256)^{\frac{16}{100}}\times(256)^{\frac{9}{100}}$

$=(256)^{\frac{16}{100}+\frac{9}{100}}\ [\because\text{x}^\text{a}\cdot\text{x}^{\text{b}}=\text{x}^{\text{a+b}}]$

$(256)^{\frac{25}{100}}=(256)^{\frac{1}{4}}$

$=(4^4)^{\frac{1}{4}}=4\ [\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$