Maths M.C.Q

 We have to find which statement must be true?
Given $0<\text{y}<\text{x},$
Option (a):
Left hand side:
$\sqrt{\text{x}}-\sqrt{\text{y}}=\sqrt{\text{x}\text{}}=\sqrt{\text{y}}$
Right Hand side:
$\sqrt{\text{x}-\text{y}}=\sqrt{\text{x}-\text{y}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (b):
$\sqrt{\text{x}}+\sqrt{\text{x}}=\sqrt{2\text{x}}$
Left hand side:
$\sqrt{\text{x}}+\sqrt{\text{x}}=2\sqrt{\text{x}}$
Right Hand side:
$\sqrt{2\text{x}}=\sqrt{2\text{x}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (c):
$\text{x}\sqrt{\text{y}}=\text{y}\sqrt{\text{x}}$
Left hand side:
$\text{x}\sqrt{\text{y}}=\text{x}\sqrt{\text{y}}$
Right Hand side:
$​​\text{y}\sqrt{​​\text{x}}=​​\text{y}\sqrt{​​\text{x}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (d):
$\sqrt{\text{xy}}=\sqrt{\text{x}}\sqrt{\text{y}}$
Left hand side:
$\sqrt{\text{xy}}=\sqrt{\text{xy}}$
Right Hand side:
$\sqrt{\text{x}}\sqrt{\text{y}}=\sqrt{\text{x}}\times\sqrt{\text{y}}$
$=\sqrt{\text{xy}}$
Left hand side is equal to right hand side
The statement is true.
Hence the correct choice is $d.$

$\sqrt[4]{(81)^{-2}}=\sqrt[4]{\Big(\frac{1}{81}\Big)^2}=\sqrt[4]{\Big\{\Big(\frac{1}{9}\Big)^2\Big\}^2}=\sqrt[4]{\Big(\frac{1}{9}\Big)^4}=\Big(\frac{1}{9}\Big)^{4\times\frac{1}{4}}=\frac{1}{9}$
Hence, $(a)$ is the correct answer.

 $\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$
$\therefore\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}\\ \ =\frac{\big(\sqrt5+\sqrt3\big)^2}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}$
$\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}$
$\therefore\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}\\ \ =\frac{\big(\sqrt5-\sqrt3\big)^2}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}=\frac{8-2\sqrt{15}}{2}=4-\sqrt{15}$
$\text{xy}=\big(4+\sqrt{15}\big)\big(4-\sqrt{15}\big)=16-15=1$
Now, $\text{x}+\text{y}+\text{xy}=4+\sqrt{15}+4-\sqrt{15}+1\\ \ =4+4+1=9$
Hence, correct option is $(a).$

 We have to find the product $($say $L)$ of the square root of $x$ with the cube root of $x$ is.
So,
$\text{L}=\sqrt[2]{\text{x}}\times\sqrt[3]{\text{x}}$
$=\text{x}^\frac{1}{2}\times\text{x}^\frac{1}{3}$
$=\text{x}^{\frac{1}{2}+\frac{1}{3}}$
$=\text{x}^{\frac{1\times3}{2\times3}+\frac{1\times2}{3\times2}}$
$=\text{x}^{\frac{3+2}{6}}=\text{x}^\frac{5}{6}$
The product of the square root of $x$ with the cube root of $x$ is $\text{x}^{\frac{5}{6}}$
Hence the correct alternative is $b.$

Every point on a number line represents a unique number.
Hence, the correct opion is $(d).$

 We have to find $x^3$ provided $x^2 = 2$ So,
By raising both sides to the power $\frac{1}{2}$
$\text{x}^{2\times\frac{1}{2}}=2^{\frac{1}{2}}$
$\text{x}^{2\times\frac{1}{2}}=\sqrt{2}$
$\text{x}=\sqrt{2}$
By substituting $\text{x}=\sqrt{2}$ in $x^3$ we get
$\text{x}^3\big(\sqrt{2}\big)^3$
$=\sqrt{2}\times\sqrt{2}\times\sqrt{2}$
$=2\sqrt{2}$
The value of $x^3$ is $2\sqrt{2}$
Hence the correct choice is $b.$

 $\text{x}^{\frac{1}{12}}=49^{\frac{1}{24}},$
$\Rightarrow\text{x}^{\frac{1}{12}}=7\frac{2}{24}=7^\frac{1}{12}$
Equating both, $x = 7$

 $\frac{7}{17}=\frac{7\times3}{17\times3}=\frac{21}{57}$
Hence, the correct opion is $(d).$

 A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number.
So, decimal representation of an irrational number is neither terminating nor a repeating decimal.
Hence, the correct opion is $(d).$

 We have to find the value of $4^{2 x-2}$ provided $(16)^{2 x+3}=(64)^{x+3}$
So,
$(16)^{2 x+3}=(64)^{x+3}$
$\left(2^4\right)^{2 x+3}=\left(2^6\right)^{x+3}$
$2^{8 x+12}=2^{6 x+18}$
Equating the power of exponents we get
$8x +12 = 6x +18$
$8x - 6x = 18 - 12$
$2x = 6$
$\text{x}=\frac{6}{2}$
$x = 3$
The value of $4^{2 x-2}$ is
$=4^{2 x-2}$
$=4^{2 \times 3-2}$
$=4^{6-2}$
$=4^4$
$=256$
Hence the correct alternative is $b.$

 $(243)^{\frac{1}{5}}$
$=(3^5)^{\frac{1}{5}}$
$=3$

 $\sqrt{\text{p}^{-1}\text{q}}\times\sqrt{\text{q}^{-1}\text{r}}\times\sqrt{\text{r}^{-1}\text{p}}$
$=\sqrt{\frac{\text{p}}{\text{q}}}\times\sqrt{\frac{\text{r}}{\text{q}}}\times\sqrt{\frac{\text{p}}{\text{r}}}$
$=\sqrt{\frac{\text{q}}{\text{p}}\times\frac{\text{r}}{\text{q}}\times\frac{\text{p}}{\text{r}}}$
$=1$

 Let $\text{x}=0.\overline{23}=0.232323...(1)$
Now, $\text{y}=0.\overline{22}=0.22222...(2)$
Adding equation $(1)$ and $(2),$ we get
$\text{x}+\text{y}=0.454545=0.\overline{45}$
$\Rightarrow0.\overline{23}+0.\overline{22}=0.\overline{45}$
Hence, option $(a)$ is correct.

 $3+2^\text{x}=(64)^{\frac{1}{2}}+(27)^{\frac{1}{3}}$
$\Rightarrow3+2^\text{x}=\sqrt{64}+\sqrt[3]{27}$
$\Rightarrow2^\text{x}=8=2^3$
Equating both,
$\text{x}=3$

 $\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}}=\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}\times\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}-1\big)}}$
$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{\big(\sqrt{2}\big)^2-(1)^2}}$
$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{1}}$
$=\sqrt{\big(\sqrt{2}-1\big)^2}$
$=\sqrt{2}-1$
$=1.414-1$
$=0.414$
Hence, the correct option is $(c).$

Let $\text{x}=0.\overline{32}$
Then, $\text{x}=0.3222 \ ...(\text{i})$
$\therefore10\text{x}=3.222 \ ...(\text{ii})$
and $100\text{x}=32.222 \ ...(\text{iii})$
On subtracting $(ii)$ from $(iii),$ we get
$90\text{x}=29$
$\Rightarrow\text{x}=\frac{29}{90}$
Hence, the correct option is $(c).$

$2^x=4^y=8^z$
$⇒ 2^x=2^{2 y}=2^{3 z}$
$⇒ x = 2y = 3z$
$\Rightarrow\text{y}=\frac{\text{x}}{2}$ and $\Rightarrow\text{z}=\frac{\text{x}}{3}$
So, $\frac{1}{2\text{x}}+\tfrac{1}{4\text{y}}+\frac{1}{4\text{z}}=4$
$\Rightarrow\frac{1}{2\text{x}}+\frac{2}{4\text{x}}+\frac{3}{4\text{x}}=4$
$\Rightarrow\frac{7}{4\text{x}}=4$
$\Rightarrow\text{x}=\frac{7}{16}$

$\text{x}+\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}$
Now put, $\text{x}=2+\sqrt{3}$
We have,
$\frac{(2+\sqrt{3})^2+1}{2+\sqrt{3}}$
$\Rightarrow\frac{4+3+2(2\sqrt{3})+1}{2+\sqrt{3}}$
$\Rightarrow\frac{8+4\sqrt{3}}{2+\sqrt{3}}$
$\Rightarrow\frac{4(2+\sqrt{3})}{2+\sqrt{3}}$
$=4$

$\frac{1}{\sqrt{3}-\sqrt{2}}$
$\Rightarrow\frac{1}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\Rightarrow\frac{\sqrt{3}+\sqrt{2}}{3-2}=\sqrt{3}+\sqrt{2}$
$\Rightarrow1.732+1.414$
$\Rightarrow3.146$

Decimal Expansion of a Rational number is not only terminating,
It can be either terminating like $\frac{1}{2}=0.5$ or non-terminating Repeating like $\frac{1}{3}=0.3333333......$ So option $(a)$ is not true alone.
Now we know that Non-Terminating numbers are of two types:
One is Non-Terminating Repeating and other is Non-Terminating Non-Repeating.
The Decimal expansion of a Rational number matches one of it's kind i.e Non-Terminating Repeating of Non-Terminating numbers.
So Rational number does not consist both the kinds of Non-Terminating numbers.
Hence, they are not Non-Terminating numbers.
An irrational number is always Non-Terminating in nature, but again not of both of it's kinds.
The decimal Expansion of an irrational number is Non-Terminating Non-Repeating in Nature.
So from all above points and theory we can conclude an Irrational number is Non-Terminating but Non-Repeating in nature
i.e. $\sqrt{2}=1.4142135623730...$
So, option $(d)$ is correct.

$(625)^{0.16} \times(625)^{0.09}$
$(625)^{0.16+09}$
$(625)^{0.25}$ or $(625)^{\frac{1}{4}}$
But $625=5^4$
So, $(5^4)^{\frac{1}{4}}=5$

$0.318564318564318564 \ ...=0.\overline{318564}$ is a Non-terminating repeating Number.
Hence, it is a rational number.
So, correct option is $(c).$

$x^{p-q} \cdot x^{q-r} \cdot x^{r-p}$
$=x^{p-q+q-r+r-p}$
$=x^0$
$=1$
Hence, the correct option is $(b).$

$\Big(\frac{2}{3}\Big)^{\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{81}{16}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{3^4}{2^4}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}+2\text{x}}=\Big(\frac{3}{2}\Big)^4$
$\Rightarrow\Big(\frac{3}{2}\Big)^{\text{x}}=\Big(\frac{3}{2}\Big)^4$
$\Rightarrow\text{x}=4$
Hence, the correct option is $(d).$