MATHS M.C.Q

$\text{x}=\frac{2}{3+\sqrt7}=\frac{2}{3+\sqrt7}\times\frac{3-\sqrt7}{3-\sqrt7}=\frac{2(3-\sqrt7)}{3-\sqrt7}\\ \ =\frac{6-2\sqrt7}{9-7}=\frac{6-2\sqrt7}{2}=3-2\sqrt7$
Now $(\text{x}-3)^2=(\not3-\sqrt7-\not3)^2=\big(-\sqrt7\big)^2=7$
Hence, correct option is $(d).$

$\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Multiplying numerator and denominator by $2\div\sqrt{3}$
So, $\frac{(2+\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}$
$=\frac{4+3+4\sqrt{3}}{4-3},$
$=7\div4\sqrt{3}$
Now equating $7\div4\sqrt{3}$ and $\text{a}\div\text{b}\sqrt{3}$
we get,
$a = 7$ and $b = 4$

$\sqrt[3]{500}=(500)^{\frac{1}{3}}=\Big(\frac{500\times2}{2}\Big)^{\frac{1}{3}}\\ \ =\Big(\frac{1000}{2}\Big)^{\frac{1}{3}}=(10^{\not3})^{\frac{1}{\not3}}.\frac{1}{2^{\frac{1}{3}}}\Rightarrow10.2^{\frac{-1}{3}}$
The simplest Rationalisation factor of $\sqrt[3]{500}$
After simplify it to $\Big(10.2^{\frac{-1}{3}}\Big)$ is $2^{\frac{1}{3}}$ or $\sqrt[3]{2}.$
Hence, correct option is $(a).$

When we multiply two different root number then only number is multiplied not the root because we have both number power equal,
$\sqrt{\text{a}}\times\sqrt{\text{b}}=\sqrt{\text{ab}}$
or, $\text{a}\frac{1}{2}\times\text{b}\frac{1}{2}=(\text{ab})\frac{1}{2}$

Let $\text{x}=1.\overline{27}=1.272727.....(\text{i})$
Multiplying equation $(i)$ with $100 [$since there is bar over $2$ & $7, ($two digits$)]$
$100x = 100 × (1.272727...)$
$100x = 127.2727 .......(ii)$
Subtracting $(i)$ from $(ii)$ i.e. $(ii) - (i)$
$100x - x = 127.2727 ... - 1.272727...$
$99x = 126.000$
$\text{x}=\frac{126}{99}$
$\text{x}=\frac{14}{11}$
Thus, $1.\overline{27}=\frac{14}{11}$

If two irrational numbers i.e. $\sqrt{2},\sqrt{5},2+\sqrt{3},2-\sqrt{3}$ etc. are added it is not necessary that sum comes out to be an irrational number always, or a rational nnumber always, or a rational number always...
Since $\sqrt{2}+\sqrt{5}=$ an irrational number
$2+\sqrt{\not\text{3}}+2-\sqrt{\not\text{3}}=4=$ a rational number
So we see that $\sqrt{2}$ and $\sqrt{5}$ are irrational numbers, and their sum is also irrational.
But $2+\sqrt{3}$ and $2-\sqrt{3}$ are also irrational numbers, and their sum is rational number $'4'.$
So sum of two irrational numbers can be either an irrational number or a rational number depending which numbers are being added.
So options $(a)$ and $(b)$ are totally wrong, because they are not 'always' true.
Option $(c)$ is correcrt because sum can be either irrational or rational and option $(c)$ is verifying this statement.
Option $(d) -$ again it is not always true, if we add two irrational numbers like $2+\sqrt{3}$ and $2-\sqrt{3}.$
Sum is an integer $= 4,$ but if we add $\sqrt{3}$ and $\sqrt{3},$ sum is $2\sqrt{3}$ which is not an integer but again an irrational number.
So option $(d)$ is also incorrect.
Hence, correct option is $(c).$

$\big(\sqrt[4]{2}\big)^2=\Big(2^{\frac{1}{4}}\Big)^2=2^{\frac{1}{4}\times2}=2^{\frac{1}{2}}=\sqrt{2},$ which is irrational.
$\big(\sqrt[4]{2}\big)^2=\Big(2^{\frac{1}{4}}\Big)^2=2^{\frac{1}{4}\times4}=2^1=2,$ which is rational.
Hence, the correct option is $(d).$

We have,
$\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{5}{3}=\frac{5\times2}{3\times2}=\frac{10}{6}$
And, $\frac{1}{2}=\frac{1\times3}{2\times3}=\frac{3}{6}$ and $\frac{2}{1}=\frac{2\times6}{1\times6}=\frac{12}{6}$
Also, $\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{4}{3}=\frac{4\times2}{3\times2}=\frac{8}{6}$
Since, $\frac{1}{6}<\frac{2}{6}<\frac{3}{6}\Big(\frac{1}{2}\Big)<\frac{4}{6}\Big(=\frac{2}{3}\Big)<\frac{5}{6}<\frac{7}{6}\\<\frac{8}{6}\Big(=\frac{4}{3}\Big)<\frac{10}{6}\Big(=\frac{5}{3}\Big)<\frac{12}{6}\Big(=\frac{2}{1}\Big)$
So, the two rational numbers between $\frac{2}{3}$ and $\frac{5}{3}$ are $\frac{5}{6}$ and $\frac{7}{6}.$
Hence, the correct opion is $(c).$

$3^\text{x+64}=2^6+(\sqrt{3})^8$
But we know that,
$2^6=64$
So, $2^6+3^\text{x}=2^6+(\sqrt{3})^{2\times4}$
$\Rightarrow2^6+3\text{x}=2^6+(3)^4$
Now by equating both
We get,
$\text{x}=4$

$\sqrt{20}\times\sqrt{5}$
$=2\sqrt{5}\times\sqrt{5}$
$=2\times5$
$=10$

After rationalising
$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{\sqrt{9}-\sqrt{8}}\times\frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}}$
$=\frac{\sqrt{9}+\sqrt{8}}{(\sqrt{9})^2-(\sqrt{8})^2}$
$=\frac{\sqrt{3\times3}+\sqrt{2\times2\times2}}{9-8}$
$=\frac{3+2\sqrt{2}}{1}$
$=3+2\sqrt{2}$

We have to find the value of x provided $\frac{3^{5\text{x}}\times81^2\times6561}{3^{2\text{x}}}=3^7$
So,
$\frac{3^{5\text{x}}\times81^2\times6561}{3^{2\text{x}}}=3^7$
By using law of rational exponents we get
$3^{5\text{x}+8+8-2\text{x}}=3^7$
By equating exponents we get
$5\text{x}+88-2\text{x}=7$
$3\text{x}+16=7$
$3\text{x}=7-16$
$3\text{x}=-9$
$\text{x}=\frac{-9}{3}$
$\text{x}=-3$
Hence the correct choice is $b.$

$\text{x}=\sqrt{5}+2$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt5+2}=\frac{1}{\sqrt5+2}\frac{\sqrt5-2}{\sqrt5-2}\\ \ =\frac{\sqrt5-2}{1}=\sqrt5-2$
Now, $\text{x}-\frac{1}{\text{x}}=\sqrt5+2-\big(\sqrt5-2\big)\\ \ =\sqrt5+2-\sqrt5+2=4$
Hence, correct option is $(b).$

$\big(\sqrt{11}-\sqrt{7}\big)\big(\sqrt{11}+\sqrt{7}\big)$
$\big(\sqrt{11}\big)^2-\big(\sqrt{7}\big)^2$
$11-7=4$
Hence, the correct answer is option $(b).$

Every rational number $(1,4.5,10,\frac{1}{2},-27,\frac{75}{5},0)$ is a real number.
However, not every real number is a rational number.
Although some numbers that appear to be irrational are actually rational because they can be reduced i.e. $\sqrt{25}=5$

We have to find $3^{2\Big(\frac{\text{n}}{4}-4\Big)}$
Given $\sqrt{2^\text{n}}=1024$
$\sqrt{2^\text{n}}=2^\text{10}$
$2^{\text{n}\times\frac{1}{2}}$
Equating powers of rational exponents we get
$\text{n}\times\frac{1}{2}=10$
$\text{n}=10\times2$
$\text{n}=20$
Substituting in $3^{2\Big(\frac{\text{n}}{4}-4\Big)}$ we get
$3^{2\Big(\frac{\text{n}}{4}-4\Big)}=3^{2\Big(\frac{20}{4}-4\Big)}$
$=3^{2(5-4)}$
$=3^{2\times1}$
$=9$
Hence the correct choice is $b.$

$\frac{2}{\sqrt{5}-\sqrt{3}}$
Multiplying nu nominator and denominator by $\sqrt{5}+\sqrt{3},$
We get,
$\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$
$=\frac{2(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{5}+\sqrt{3}$

Given $2\sqrt{3}+\sqrt{3}=(2+1)\sqrt{3}=3\sqrt{3}$
Hence, $(c)$ is the correct answer.

$\sqrt{5}\times\sqrt{2}=\sqrt{10},$ is irrational number
$\sqrt{5}\times\sqrt{5}=5,$ is retional number

The decimal expansion of $\sqrt{2}=1.41421356...,$ which is non-terminating, nonrecurring.
Hence, the correct opion is $(d).$

Since, $-4.3 < -3.4 < -3 < 0 < 1.101100110001... < 3$
But $1.101100110001...$ is an irrational number
So, the rational number between $-3$ and $3$ is $0.$
Hence, the correct option is $(a).$

Given: $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16,\ \frac{3^\text{p}}{2^\text{n}}=81$ and $\text{a}=2^{\frac{1}{10}}$
To find: $\frac{\text{a}^{2{\text{m}+\text{n}-\text{p}}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}$
Find: $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16$
By using rational components $\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^{\text{m}-\text{n}}$ We get
$2^{\text{m}+\text{n}-\text{n}+\text{m}}=16$
$2^{2\text{m}}=2^4$
By equating rational exponents we get
$2\text{m}=4$
$\text{m}=\frac{4}{2}$
$\text{m}=2$
Now, $\frac{\text{a}^{2\text{m}+\text{n}-\text{p}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})}=(\text{a}^{2\text{m}+\text{n}-\text{p}}).(\text{a}^{\text{m}-2\text{n}+2\text{p})}$ we get
$=\text{a}^{2\text{mn}+\text{n}+\text{p}+\text{m}-2\text{n}+2\text{p}}$
$=\text{a}^{3\text{m}-\text{n}+\text{p}}$
Now putting value of $\text{a}=2^\frac{1}{10}$ we get,
$=2^{\frac{3\text{m}-\text{n}+\text{p}}{10}}$
$=2^{\frac{6-\text{n}+\text{p}}{10}}$
Also, $\frac{3^\text{p}}{3^\text{n}}=81$
$3^{\text{p}-\text{n}}=3^4$
On comparing $LHS$ and $RHS$ we get, $p - n = 4.$
Now,
$\frac{\text{a}^{2{\text{m}+\text{n}-\text{p}}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}=\text{a}^{3\text{m}-\text{n}+\text{p}}$
$=2^{\frac{6+(\text{p}-\text{n})}{10}}$
$=2^{\frac{6+4}{10}}$
$=2^{\frac{10}{10}}=2^1$
$=2$
So, option $(a)$ is the correct answer.

The numbers of the form $\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are integers and $\text{q}\neq0,$ are known as rational numbers.
$\sqrt{2},$ $\sqrt{23}$ and $0.1010010001...$ are irrational numbers, since they cannot be expressed in the form $\frac{\text{p}}{\text{q}}.$
$\sqrt{225}=15=\frac{15}{1}$ is a rational number.
Hence, the correct opion is $(c).$

Find the value of $\Big\{8^{-\frac{4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}$
$\Big\{8^{-\frac{4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}=\Big\{2^{3\times\frac{-4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}$
$\big\{2^{-4}\div2^{-2}\big\}^{\frac{1}{2}}$
$\Big\{8^{-\frac{4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}=\Big\{2^{-4\times\frac{1}{2}}\div2^{-2\times\frac{1}{2}}\Big\}$
$=\Big\{2^{-2}\div2^{-1}\Big\}$
$=\Bigg\{\frac{\frac{1}{2^2}}{\frac{1}{2}}\Bigg\}$
$\Big\{8^{-\frac{4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}=\Big\{\frac{1}{2\times2}\times\frac{2}{1}\Big\}$
$=\frac{1}{2}$
Hence the correct choice is a.

The numbers of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ are known as rational numbers.
Decimal representation of a rational number is either terminating or a repeating decimal, since every decimal of this form can be expressed in the form $\frac{\text{p}}{\text{q}}.$
Hence, the correct opion is $(b).$

When we divide $1$ by $999$ result is $0.001001001001001.....$
So, $0.\overline{001}=\frac{1}{999}$

$\pi=3.14159265359...,$ which is non-terminating non-recurring.
Hence, it is an irrational number.
Hence, the correct opion is $(c).$

$\frac{15\sqrt{15}}{3\sqrt{5}}$
$=\frac{(3\times5)\sqrt{3}\times\sqrt{5}}{3\sqrt{3}}$
$=5\sqrt{3}$

$9^3+(-3)^3-6^3=729-27-216=486$
Hence, the correct answer is option (c).

Let $\text{x}=1.\overline{27}=1.272727...(1)$
Now, $100\text{x}=127.272727=127.\overline{27}...(2)$
Subtracting equation $(1)$ from $(2),$ we get
$99\text{x}=126$
$\Rightarrow\text{x}=\frac{126}{99}=\frac{14}{11}$
Hence, option $(b)$ is correct.

$0$ is an integer and all integers are rational numbers.

We know that, If $a$ and $b$ are two distinct positive rational numbers such that ab is not a perfect square of a rational number, Then $\sqrt{\text{ab}}$ is an irrational number lying between $a$ and $b,$
Here also we have $5$ and $6$ two distinct rational numbers and $5 × 6 = 30$ is not a perfect square,
So irrational number betweens and $6 =\sqrt{5\times6}$

Decimal representation of a rational number cannot be non $-$ terminating non$-$repeating because the decimal expansion of rational number is either terminating or non $-$ terminating recurring $($repeating$).$

$1\ cm = 75$
So, $\frac{3}{5}\times\frac{15}{15}$ and $\frac{2\times25}{3\times25}$
i.e. $\frac{45}{75}$ and $\frac{50}{75}$
So, $\frac{46}{75},\frac{47}{75},\frac{48}{75},\frac{49}{75}$

$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\Rightarrow\frac{(\sqrt{3}+\sqrt{2})^2+(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
$\Rightarrow\frac{(3+2+2\sqrt{6})+3+2-2\sqrt{6}}{3-2}$
$\Rightarrow10$

Consider, the irrational number, $\sqrt{3}.$
Product $=\sqrt{3}\times\sqrt{3}=3,$ which is a rational number.
Consider two irrational numbers, $\sqrt{2}$ and $\sqrt{3}.$
Product $=\sqrt{2}\times\sqrt{3}=\sqrt{6},$ which is an irrational number.
Hence, the product of two irrational numbers are sometimes rational and sometimes irrational.
Hence, the correct opion is $(d).$

$(5+\sqrt{8})+(3-\sqrt{2})(\sqrt{2}-6)$
$=(5+2\sqrt{2})+(3\sqrt{2}-18-2+6\sqrt{2})$
$=(5+2\sqrt{2})+(9\sqrt{2}-20)$
$=11\sqrt{2}-15$
And we know that the value of $11\sqrt{2}$ is greater than $15$ so it's value will be positive,
And also sum or differences of rational and irrational is irrational.