Maths M.C.Q

We have to find he seventh root of $x$ divided by the eighth root of $x, $so let it be $L.$ So,
$\text{L}=\frac{\sqrt[7]{\text{x}}}{\sqrt[8]{\text{x}}}$
$=\frac{\text{x}^{\frac{1}{7}}}{\text{x}^{\frac{1}{8}}}$
$=\text{x}^{\frac{1}{7}-\frac{1}{8}}$
$=\text{x}^{\frac{1\times8}{7\times8}-\frac{1\times7}{8\times7}}$
$=\text{L}=\text{x}^{\frac{8}{56}-\frac{7}{56}}$
$=\text{x}^{\frac{1}{56}}$
$=\sqrt[56]{\text{x}}$
The seventh root of $x$ divided by the eighth root of $x$ is $=\sqrt[56]{\text{x}}$
Hence the correct choice is $c.$

$\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$
$=\sqrt{\frac{\text{q}}{\text{p}}}.\sqrt{\frac{\text{r}}{\text{q}}}.\sqrt{\frac{\text{p}}{\text{r}}}$
$=\sqrt{\frac{\text{q}}{\text{p}}\times\frac{\text{r}}{\text{q}}\times\frac{\text{p}}{\text{r}}}$
$=\sqrt{1}$
$=1$
Hence, the correct option is $(c).$

A number is irrational if and only of its decimal representation is non $-$ terminating and nonrecurring.
$0.14$ is a terminating decimal and therefore cannot be an irrational number.
$0.14\overline{16}$ is a non $-$ terminating and recurring decimal and therefore cannot be irrational.
$0.\overline{1416}$ is a non $-$ terminating and recurring decimal and therefore cannot be irrational.
$0.4014001400014...$ is a non $-$ terminating and non $-$ recurring decimal and therefore is an irrational number.

$\frac{9^{\frac{1}{3}}\times27^\frac{1}{2}}{3^{\frac{-1}{6}}\times3^{\frac{1}{3}}}$
$\Rightarrow\frac{3^{\frac{2}{3}}\times3^{\frac{3}{2}}}{3^{\frac{-1}{6}}\times3^{\frac{1}{3}}}$
$\Rightarrow\frac{3^{\frac{2}{3}+\frac{3}{2}}}{3^{\frac{-1}{6}+\frac{1}{3}}}$
$\Rightarrow\frac{3\frac{4+9}{6}}{3^{\frac{-1+2}{6}}}$
$\Rightarrow\frac{3\frac{13}{6}}{3^{\frac{1}{6}}}=3^{\frac{13}{6}-\frac{1}{6}}$
$\Rightarrow3^{\frac{12}{6}}=9$
$=(\frac{1}{5})^{-1}=25$

$0.5030030003$
The decimal expansion that a rational number cannot have is non terminating, non repeating.

$\frac{-2}{3}<\frac{-4}{5}$
Taking $LCM$ of $3$ and $5,$
$LCM = 15,$
So, $\frac{-2\times5}{3\times5},\frac{-4\times3}{5\times3}$
$\Rightarrow\frac{-10}{15},\frac{-12}{15}$
Now, since both the denominator is equal so, we compare its numerator,
and since, $-10 < -12$
So, $\frac{-10}{15}>\frac{-12}{15}$
thus, $\frac{-2}{3}>\frac{-4}{5}$

$\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$
Now, put, $\text{x}=4-\sqrt{15}$
$\Rightarrow\frac{(-4\sqrt{15})^2+1}{4-\sqrt{15}}$
$\Rightarrow\frac{16+15-8\sqrt{15}+1}{4-\sqrt{15}}$
$\Rightarrow\frac{32-8\sqrt{15}}{4-\sqrt{15}}$
$\Rightarrow8$

Rationalisation factor of any number like $\sqrt{\text{a}}$ is $\frac{1}{\text{a}}$ or $\frac{1}{\text{a}}$ is $\sqrt{\text{a}}.$
So. Rationalisation factor of $\sqrt3$ is $\frac{1}{\sqrt3}.$
Hence, correct option is $(b).$

$\frac{1}{\sqrt{2}}=\frac{1}{1.41}=0.705$

The decimal expansion of a rational number is either terminating or non-terminating recurring.
The decimal expansion of $0.5030030003...$ is non-terminating, non-recurring, which is not a property of a rational number.
Hence, the correct opion is $(d).$

We have to find the value of $(2\text{x})^\text{x}$ if $4\text{x}-4^{\text{x}-1}=24$
So,
Taking $4x$ as common factor we get
$4\text{x}(1-4^{-1})=24$
$4\text{x}\Big(1-\frac{1}{4}\Big)=24$
$4\text{x}\Big(\frac{1\times4}{1\times4}-\frac{1}{4}\Big)=24$
$4^4\Big(\frac{4-1}{4}\Big)=24$
$4^\text{x}\times\frac{3}{4}=24$
$4^\text{x}=24\times\frac{4}{3}$
$4^\text{x}=32$
$2^{2\text{x}}=2^{5}$
By equating powers of exponents we get
$2\text{x}=5$
$\text{x}=\frac{5}{2}$
By substituting $\text{x}=\frac{5}{2}$ in $(2\text{x})^\text{x}$ we get
$(2\text{x})^\text{x}=\Big(2\times\frac{5}{2}\Big)^{\frac{5}{2}}$
$=\Big(2\times\frac{5}{2}\Big)^{\frac{5}{2}}$
$=5^{\frac{5}{2}}$
$=5^{5\times\frac{1}{2}}$
$(2\text{x})^\text{x}=\sqrt[2]{5^5}$
$=\sqrt[2]{5\times5\times5\times5}$
$=5\times5\times^\sqrt[2]{5}$
$=25\sqrt{5}$
Hence the correct choice is $c.$

$\sqrt{5+2\sqrt6}$
$=\sqrt{3+2+2\big(\sqrt3\big)\big(\sqrt2\big)}$
$=\sqrt{\big(\sqrt3\big)^2+\big(\sqrt2\big)^2-2\big(\sqrt3\big)\big(\sqrt2\big)}$
$=\sqrt{\big(\sqrt2+\sqrt2\big)^2}$
$=\sqrt3+\sqrt2$
Hence, correct option is $(b).$

$27$ can be written as $(3)^3$
So, $(27)^{-\frac{2}{3}}=\{(3)^3\}^{-\frac{2}{3}}$
$=\frac{1}{(3)^2}=\frac{1}{9}$

$\Big(\frac{125}{216}\Big)^{\frac{-1}{3}}$
$\Rightarrow\Big(\frac{5}{6}\Big)^{3\times\frac{-1}{3}}$
$\Rightarrow\Big(\frac{5}{6}\Big)^{-1}$
$\Rightarrow\frac{6}{5}$

We have to find the value of $\sqrt{\text{a}^{-1}\text{b}}\times\sqrt{\text{b}^{-1}\text{c}}\times\sqrt{\text{c}^{-1}\text{a}}$ when $a, b, c$ are positive real numbers.
So,
$\sqrt{\text{a}^{-1}\text{b}}\times\sqrt{\text{b}^{-1}\text{c}}\times\sqrt{\text{c}^{-1}\text{a}}$
$=\sqrt{\frac{1}{\text{a}}\times\text{b}}\times\sqrt{\frac{1}{\text{b}}\times\text{c}}\times\sqrt{\frac{1}{\text{c}}\times}\text{a}$
$=\sqrt{\frac{\text{b}}{\text{a}}}\times\sqrt{\frac{\text{c}}{\text{b}}}\times\sqrt{\frac{\text{a}}{\text{c}}}$
Taking square root as common we get
$\sqrt{\text{a}^{-1}\text{b}}\times\sqrt{\text{b}^{-1}\text{c}}\times\sqrt{\text{c}^{-1}\text{a}}=\sqrt{\frac{\text{b}}{\text{a}}\times\frac{\text{c}}{\text{b}}\times\frac{\text{a}}{\text{c}}}$
$\sqrt{\text{a}^{-1}\text{b}}\times\sqrt{\text{b}^{-1}\text{c}}\times\sqrt{\text{c}^{-1}\text{a}}=1$
Hence the correct alternative is $a.$

$\sqrt{12}=\sqrt{3\times2^2}=2\sqrt{3}$ and $\sqrt{15}=\sqrt{5}\times\sqrt{3}$
So, $\sqrt{12}\times\sqrt{15}=2\sqrt{3}\times\sqrt{3}\times\sqrt{5}$
$=2\times3\sqrt{5}=6\sqrt{5}$

Let $x = 0.222.... (i)$
Multiply eq. $(i)$ by $10,$ we get
$10x = 2.222...  (ii)$
$10x - x = 2.222... -0.222...$
$9x = 2$
$\text{x}=\frac{2}{9}$

$\frac{2^\circ+7^\circ}{5^\circ}=\frac{1+1}{1}=\frac{2}{1}=2$
Hence, the correct answer is option $(b).$

Given two rational numbers are negative and $\frac{3}{10}$ is a positive rational number.
So, it does not lie between $-\frac{2}{3}$ and $-\frac{1}{5}.$
Hence, the correct opion is $(b).$

$5 × 2 = 10 ⇒$ one $5$ and one $2$ make one zero, so $5 × 2 × 5 × 2 = 100$
Numbers of pairs of $5$ and $2$ will be equal to the number of consecutive zeros in the given number.
In the given number, there are three $2's$ and four $5's.$
So number of pairs of $5$ and $2$ are only three.
So there will be three consecutive zeros in the given number
Hence, option $(a)$ is correct.

 $\frac{8}{10}$ or $\frac{4}{5}$

$\therefore\sqrt{8}=\sqrt{2\times2\times2}=2\sqrt{2}$

 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.

 $(\frac{81}{16})^{\frac{-3}{4}}\times\{(\frac{25}{9})^{\frac{-3}{2}}\div(\frac{5}{2})^{-3}\}$
$\Rightarrow(\frac{3}{2})^{4\times\frac{-3}{4}}\times\{(\frac{5}{3})^{2\times\frac{-3}{2}}\div(\frac{5}{2})^{-3}\}$
$\Rightarrow(\frac{3}{2})^{-3}\times\{(\frac{5}{3})^{-3}\div(\frac{5}{2})^{-3}\}$
$\Rightarrow(\frac{3}{2})^{-3}\times(\frac{5}{3}\times\frac{2}{5})^{-3}$
$\Rightarrow(\frac{3}{2})^{-3}\times(\frac{2}{3})^{-3}$
$\Rightarrow(\frac{3}{2}\times\frac{2}{3})^{-3}$
$\Rightarrow(1)^{-3}=1$

 Consider, $\big(2+\sqrt{3}\big)$ and $\big(2-\sqrt{3}\big)$ which are two irrational numbers.
$\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4,$ which is a rational number.
Consider, $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ which are two irrational numbers.
$\sqrt{3}\times\frac{1}{\sqrt{3}}=1,$ which is a rational number.
Every real number can either be a rational number or an irrational number.
Hence, the correct opion is $(d).$

 Given $\text{t}=64,\ \text{g}=\text{t}^{\frac{2}{3}}+4\text{t}^{\frac{1}{2}}.$ We have to find the value of $g$
So,
$\text{g}=\text{t}^{\frac{2}{3}}+4\text{t}^{-\frac{1}{2}}$
$\text{g}=64^{\frac{2}{3}}+4\times64^{\frac{1}{2}}$
$\text{g}=(64)^{\frac{2}{3}}+4\times\frac{1}{64^{\frac{1}{2}}}$
$\text{g}=2^{6\times\frac{2}{3}}+4\times\frac{1}{2^{6\times\frac{1}{2}}}$
$\text{g}=2^{2\times2}+4\times\frac{1}{2^3}$
$\text{g}=2^4+4\times\frac{1}{8}$
$\text{g}=16+\frac{1}{2}$
$\text{g}=\frac{16\times2}{1\times2}+\frac{1}{2}$
$\text{g}=\frac{32}{2}+\frac{1}{2}$
$\text{g}=\frac{32+1}{2}$
$\text{g}=\frac{33}{2}$
The value of $g$ is $\frac{33}{2}$
Hence the correct choice is $b.$

 Given $\text{x}=3+\sqrt{8},$
$\frac{1}{\text{x}}=\frac{1}{(3+\sqrt{8})}=\frac{1}{(3+\sqrt{8})}\times\frac{(3-\sqrt{8})}{(3-\sqrt{8})}$
$=\frac{(3-\sqrt{8})}{(3^2-(\sqrt{8})^2}=\frac{(3+\sqrt{8})}{(9-8)}=(3-\sqrt{8})$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=(3+\sqrt{8})+(3-\sqrt{8})=6$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=6^2=36$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{1}{\text{x}}=36$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+2=36$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big) =36-2=34$

$\Big(\frac{256\text{x}^{16}}{81\text{y}^{4}}\Big)^{-\frac{1}{4}}$
$=\Big(\frac{81\text{y}^{4}}{256\text{x}^{16}}\Big)^{\frac{1}{4}}$
$=\Big(\frac{3^4\text{y}^4}{4^4\text{x}^{16}}\Big)^{\frac{1}{4}}$
$=\big[\Big(\frac{3\text{y}}{4\text{x}^4}\Big)^4\big]^{\frac{1}{4}}$
$=\Big(\frac{3\text{y}}{4\text{x}^4}\Big)^{\frac{1}{4}\times4}$
$=\frac{3\text{y}}{4\text{x}^4}$