M.C.Q

Question 11 Mark

If $\sqrt2=1.414,$ then the value of $\sqrt6-\sqrt3$ upto three place of decimal is:

0.235
0.707
1.414
0.471

Answer

0.707

Solution:
$\sqrt6-\sqrt3$
$=\sqrt3\big(\sqrt2-1\big)$
Now, $\sqrt3=1.732$
$\sqrt2=1.414$
$\therefore\sqrt6-\sqrt3=1.732(1.414-1)\\ \ =1.732(0.414)=0.717$ (upto 3 decimal places)
Hence, correct option is (b).

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Question 21 Mark

$\sqrt{10}\times\sqrt{15}$ is equal to:

$5\sqrt6$
$6\sqrt5$
$\sqrt{30}$
$\sqrt{25}$

Answer

$5\sqrt6$

Solution:
10 = 5 × 2
15 = 5 × 3
$\therefore\sqrt{10}\times\sqrt{15}=\sqrt{5\times2}\times\sqrt{5\times3}$
$=\sqrt{5}\times\sqrt{2}\times\sqrt{5}\times\sqrt3$
$=\big(\sqrt5\times\sqrt5\big)\times\sqrt2\times\sqrt3$
$=5\sqrt6$
Hence, correct option is (a).

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Question 31 Mark

If $\text{x}=7+4\sqrt3$ and xy = 1, then $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=$

64
134
194
$\frac{1}{49}$

Answer

Solution:
$\text{x}=7+4\sqrt3,\ \text{xy}=1\Rightarrow\text{y}=\frac{1}{\text{x}}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}\\ \ =\frac{7-4\sqrt3}{(7)^2-\big(4\sqrt3\big)^2}=\frac{7-4\sqrt3}{49-48}=7-4\sqrt3$
Now, $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=\frac{\text{y}^2+\text{x}^2}{\text{x}^2\text{y}^2}=\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}$
$\text{x}^2=\big(7+4\sqrt3\big)^2=49+48+56\sqrt3=97+56\sqrt3$
$\text{y}^2=\big(7-4\sqrt3\big)^2=49+48-56\sqrt3=97-56\sqrt3$
$\therefore\text{x}^2+\text{y}^2=97+56\sqrt3+97-56\sqrt3=194$
$\text{xy} = 1$
$\therefore\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}=\frac{194}{(1)^2}=194$
Hence, correct option is (c).

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Question 41 Mark

If $\frac{\sqrt3-1}{\sqrt3+1}=\text{a}-\text{b}\sqrt3,$ then:

a = 2, b = 1
a = 2, b = -1
a = -2, b = 1
a = b = 1

Answer

a = 2, b = 1

Solution:
$\frac{\sqrt3-1}{\sqrt3+1}$
Multiplying and dividing by the rationalisation factor of denominator, we get
$\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}$
$=\frac{\big(\sqrt3-1\big)^2}{\big(\sqrt3\big)^2-1^2}$
$=\frac{3-2\sqrt3+1}{3-1}$
$=\frac{4-2\sqrt3}{2}$
$=\frac{2(2-\sqrt3)}{2}$
$=2-\sqrt3$
Comparing with $\text{a}-\text{b}\sqrt3,$ we get a = 2 and b = 1.
Hence, correct option is (a).

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Question 51 Mark

If $\text{x}=\sqrt6+\sqrt5,$ then $\text{x}^2+\frac{1}{\text{x}^2}-2=$

$2\sqrt6$
$2\sqrt5$
$24$
$20$

Answer

Solution:
$\text{x}^2+\frac{1}{\text{x}^2}-2=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2$
$\text{x}=\sqrt6+\sqrt5$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt6+\sqrt5}=\frac{1}{\sqrt6+\sqrt5}\times\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}\\ \ =\frac{\sqrt6-\sqrt5}{1}=\sqrt6-\sqrt5$
Now,
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\big[\sqrt6+\sqrt5-\big(\sqrt6-\sqrt5\big]^2\\ \ =\big(2\sqrt5\big)^2=4\times5=20$
Hence, correct option is (d).

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Question 61 Mark

The simplest rationalising factor of $2\sqrt5-\sqrt3,$ is:

$2\sqrt5+3$
$2\sqrt5+\sqrt3$
$\sqrt{5}+\sqrt3$
$\sqrt{5}-\sqrt3$

Answer

$2\sqrt5+\sqrt3$

Solution:
Rationalising factor of any number of kind $\text{a}\sqrt{\text{a}}\pm\sqrt{\text{b}}$ is $\sqrt{\text{a}}\mp\sqrt{\text{b}}$
So. for given number $2\sqrt5-\sqrt3.$ Rationalising factor would be $2\sqrt5+\sqrt3.$
Hence, correct option is (b).

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MCQ 71 Mark

If $\text{x}=\frac{2}{3+\sqrt7},$ then $(x - 3)^2 =$

A
$1$
B
$3$
C
$6$
✓
$7$

Answer

Correct option: D.

$7$

$\text{x}=\frac{2}{3+\sqrt7}$
$=\frac{2}{3+\sqrt7}\times\frac{3-\sqrt7}{3-\sqrt7}$
$=\frac{2(3-\sqrt7)}{3-\sqrt7}$
$\ =\frac{6-2\sqrt7}{9-7}$
$=\frac{6-2\sqrt7}{2}$
$=3-2\sqrt7$
Now $(\text{x}-3)^2$
$=(\not3-\sqrt7-\not3)^2$
$=\big(-\sqrt7\big)^2$
$=7$

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Question 81 Mark

The simplest rationalising factor of $\sqrt[3]{500},$ is:

$\sqrt[3]{2}$
$\sqrt[3]{5}$
$\sqrt{3}$
None of these

Answer

$\sqrt[3]{2}$

Solution:
$\sqrt[3]{500}=(500)^{\frac{1}{3}}=\Big(\frac{500\times2}{2}\Big)^{\frac{1}{3}}\\ \ =\Big(\frac{1000}{2}\Big)^{\frac{1}{3}}=(10^{\not3})^{\frac{1}{\not3}}.\frac{1}{2^{\frac{1}{3}}}\Rightarrow10.2^{\frac{-1}{3}}$
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).

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Question 91 Mark

If $\text{x}=\sqrt5+2,$ then $\text{x}-\frac{1}{\text{x}}$ equals:

$2\sqrt5$
$4$
$2$
$\sqrt5$

Answer

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

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Question 101 Mark

If $\text{x}+\sqrt{15}=4,$ then $\text{x}+\frac{1}{\text{x}}=$

Answer

Solution:
$\text{x}+\sqrt{15}=4$
$=\text{x}=4-\sqrt{15}\Rightarrow\frac{1}{\text{x}}=\frac{1}{4-\sqrt{15}}$
$\frac{1}{\text{x}}=\frac{1}{4-\sqrt{15}}\times\frac{4+\sqrt{15}}{4+\sqrt{15}}=\frac{4+\sqrt{15}}{(4)^2-\big(\sqrt{15}\big)^2}\\ \ =\frac{4+\sqrt{15}}{16-15}=4+\sqrt{15}$
Now, $\text{x}+\frac{1}{\text{x}}=4-\sqrt{15}+4+\sqrt{15}=8$
Hence, correct option is (c).

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Question 111 Mark

The positive square root of $7+\sqrt{48},$ is:

$7+2\sqrt3$
$7+\sqrt3$
$2+\sqrt3$
$3+\sqrt2$

Answer

$2+\sqrt3$

Solution:
$\sqrt{7+\sqrt{48}}$
$=\sqrt{7+2\sqrt{12}}$
$=\sqrt{4+3+2\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4\big)^2+\big(\sqrt3\big)^2+2\times\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4+\sqrt3\big)^2}$
$=\pm\big(\sqrt4+\sqrt3\big)$
Positive value is $\sqrt4+\sqrt3=2+\sqrt3$
Hence, correct option is (c).

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Question 121 Mark

The simplest rationalising factor of $\sqrt3+\sqrt5,$ is:

$\sqrt3-5$
$3-\sqrt5$
$\sqrt{3}-\sqrt5$
$\sqrt{3}+\sqrt5$

Answer

$\sqrt{3}-\sqrt5$

Solution:
Rationalising factor of any number of kind $\sqrt{\text{a}}\pm\sqrt{\text{b}}$ is $\sqrt{\text{a}}\mp\sqrt{\text{b}}$
So. for given number $\sqrt3+\sqrt5.$ Rationalising factor would be $\sqrt3-\sqrt5.$
Hence, correct option is (c).

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Question 131 Mark

The rationalisation factor of $2+\sqrt3,$ is:

$2-\sqrt3$
$\sqrt2+3$
$\sqrt2-3$
$\sqrt3-2$

Answer

$2-\sqrt3$

Solution:
Rationalisation factor of any number like ${\text{a}}\pm\sqrt{\text{b}}$ is ${\text{a}}\mp\sqrt{\text{b}}.$
So. Rationalisation factor of $2+\sqrt3$ will be $2-\sqrt3$
Hence, correct option is (a).

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Question 141 Mark

The rationalisation factor of $\sqrt3,$ is:

$-\sqrt3$
$\frac{1}{\sqrt3}$
$2\sqrt3$
$-2\sqrt3$

Answer

$\frac{1}{\sqrt3}$

Solution:
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).

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Question 151 Mark

The value of $\sqrt{5+2\sqrt6},$ is:

$\sqrt3-\sqrt2$
$\sqrt3+\sqrt2$
$\sqrt5+\sqrt6$
None of these

Answer

$\sqrt3+\sqrt2$

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

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Question 161 Mark

If $\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$ and $\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3},$ then $\text{x}+\text{y}+\text{xy}=$

Answer

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

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Question 171 Mark

$\sqrt[5]{6}\times\sqrt[5]{6}$ is equal to:

$\sqrt[5]{36}$
$\sqrt[5]{6\times0}$
$\sqrt[5]{6}$
$\sqrt[5]{12}$

Answer

$\sqrt[5]{36}$

Solution:
$\sqrt[5]{6}=(6)^{\frac{1}{5}}$
So $\sqrt[5]{6}\times\sqrt[5]{6}=(6)^{\frac{1}{5}}\times(6)^{\frac{1}{5}}$
$=(6\times6)^{\frac{1}{5}}$
$(36)^{\frac{1}{5}}$
$=\sqrt[5]{36}$
Hence, correct option is (a).

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Question 181 Mark

If $\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$ and $\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},$ then $\text{x}^2+\text{xy}+\text{y}^2=$

Answer

Solution:
$\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
$\therefore\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}\\ \ =\frac{\big(\sqrt3-\sqrt2\big)^2}{3-2}=5-2\sqrt6$
$\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},$
$\therefore\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ \ =\frac{\big(\sqrt3+\sqrt{2}\big)^2}{3-2}=5+2\sqrt6$
Now, $\text{x}^2+\text{xy}+\text{y}^2$
$=\big(5 -2\sqrt6\big)^2+\big(5-2\sqrt6\big)\big(5+2\sqrt6\big)+\big(5+2\sqrt6\big)^2$
$=\big(25+24-20\sqrt6\big)+(25-24)+\big(25+24+20\sqrt6\big)$
$=49-20\sqrt6+1+49+20\sqrt6$
$=99$
Hence, correct option is (b).

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Question 191 Mark

The value of $\sqrt{3-2\sqrt2},$ is:

$\sqrt2-1$
$\sqrt2+1$
$\sqrt3-\sqrt2$
$\sqrt3+\sqrt2$

Answer

$\sqrt2-1$

Solution:
$\sqrt{3-2\sqrt2}$
$=\sqrt{2+1-2\sqrt2}$
$=\sqrt{\big(\sqrt2\big)^2+(1)^2-2\big(\sqrt2\big)(1)}$
$=\sqrt{\big(\sqrt2-1\big)^2}$
$=\sqrt2-1$
Hence, correct option is (a).

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Question 201 Mark

If $\frac{5-\sqrt3}{2+\sqrt3}=\text{x}+\text{y}\sqrt3,$ then:

x = 13, y = -7
x = -13, y = 7
x = -13, y = -7
x = 13, y = 7

Answer

x = 13, y = -7

Solution:
$\frac{5-\sqrt3}{2+\sqrt3}$
$=\frac{5-\sqrt3}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{\big(5-\sqrt3\big)\big(2-\sqrt3\big)}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{10-5\sqrt3-2\sqrt3+3}{4-3}$
$=\frac{13-7\sqrt3}{1}$
$=13-7\sqrt3$
⇒ x = 13 and y = -7
Hence, correct option is (a).

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Question 211 Mark

The value of $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}},$ is:

$\frac{4}{3}$
$4$
$3$
$\frac{3}{4}$

Answer

$\frac{4}{3}$

Solution:
$\sqrt{48}=\sqrt{16\times3}=4\sqrt3$
$\sqrt{32}=\sqrt{16\times2}=4\sqrt2$
$\sqrt{27}=\sqrt{9\times3}=3\sqrt3$
$\sqrt{18}=\sqrt{9\times2}=3\sqrt2$
Now, $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}}=\frac{4\sqrt3+4\sqrt2}{3\sqrt3+3\sqrt2}$
$=\frac{4\big(\sqrt{\not3}+\sqrt{\not2}\big)}{3\big(\sqrt{\not3}+\sqrt{\not2}\big)}$
$=\frac{4}{3}$
Hence, correct option is (a).

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Question 221 Mark

If $\sqrt{13-\text{a}\sqrt{10}}=\sqrt8+\sqrt5,$ then a =

-5
-6
-4
-2

Answer

-4

Solution:
$\sqrt{13-\text{a}\sqrt10}=\sqrt8+\sqrt5$
Squaring both sides, we get
$13-\text{a}\sqrt{10}=8+5+2\sqrt{40}$
$={13}-\text{a}\sqrt{10}-{13}=2\times2\sqrt{10}$
$=-\text{a}\sqrt{10}=4\sqrt{10}$
$\Rightarrow\text{a}=-4$
Hence, correct option is (c).

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Question 231 Mark

If $\sqrt2=1.4142,$ then $\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}$ is equal to:

0.1718
5.8282
0.4142
2.4142

Answer

0.4142

Solution:
By Rationalising $\frac{\sqrt2-1}{\sqrt2+1},$ we get
$\frac{\sqrt2-1}{\sqrt2+1}\times\frac{\sqrt2-1}{\sqrt2-1}=\frac{\big(\sqrt2-1\big)^2}{\big(\sqrt2\big)^2-1^2}=\frac{\big(\sqrt2-1\big)^2}{1}$
So $\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}=\sqrt{\frac{\big(\sqrt2-1\big)^2}{1}}\\ \ =\big(\sqrt2-1\big)=1.4142-1=0.4142$
Hence, correct option is (c).

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Question 241 Mark

$\frac{1}{\sqrt9-\sqrt8}$ is equal to:

$3+2\sqrt2$
$\frac{1}{3+2\sqrt2}$
$3-2\sqrt2$
$\frac{3}{2}-\sqrt2$

Answer

$3+2\sqrt2$

Solution:
$\frac{1}{\sqrt9-\sqrt8}$
$=\frac{1}{\sqrt9-\sqrt8}\times\frac{{\sqrt9+\sqrt8}}{{\sqrt9+\sqrt8}}$
$\frac{{\sqrt9+\sqrt8}}{\big(\sqrt9\big)^2-\big(\sqrt8\big)^2}$
$=\frac{{\sqrt9+\sqrt8}}{9-8}$
$={\sqrt9+\sqrt8}$
$=3+2\sqrt2$
Hence, correct option is (a).

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Question 251 Mark

If $\text{x}=\sqrt[3]{2+\sqrt3},$ then $\text{x}^3+\frac{1}{\text{x}^3}=$:

Answer

Solution:
$\text{x}=\sqrt[3]{2+\sqrt3}+\big(2+\sqrt3\big)^{\frac{1}{3}}$
$\text{x}^3=\big\{\big({2+\sqrt3}\big)^{\frac{1}{\not3}}\big\}^{\not3}=\big(2+\sqrt3\big)$
$\Rightarrow\frac{1}{\text{x}^3}=\frac{1}{2+\sqrt3}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}\\ \ =\frac{2-\sqrt3}{4-3}=2-\sqrt3$
Now, $\text{x}^3+\frac{1}{\text{x}^3}=2+\sqrt3+2-\sqrt3=4$
Hence, correct option is (b).

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