MCQ 11 Mark
If $\sqrt2=1.414,$ then the value of $\sqrt6-\sqrt3$ upto three place of decimal is:
Answer$\sqrt6-\sqrt3$
$=\sqrt3\big(\sqrt2-1\big)$
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 $(b)$.
View full question & answer→MCQ 21 Mark
$\sqrt{10}\times\sqrt{15}$ is equal to:
- ✓
$5\sqrt6$
- B
$6\sqrt5$
- C
$\sqrt{30}$
- D
$\sqrt{25}$
AnswerCorrect option: A. $5\sqrt6$
$10 = 5 \times 2$
$15 = 5 \times 3$
$\therefore\sqrt{10}\times\sqrt{15}=\sqrt{5\times2}\times\sqrt{5\times3}$
$=\sqrt{5}\times\sqrt{2}\times\sqrt{5}\times\sqrt3$
$=\big(\sqrt5\times\sqrt5\big)\times\sqrt2\times\sqrt3$
$=5\sqrt6$
Hence, correct option is $(a)$.
View full question & answer→MCQ 31 Mark
If $\text{x}=7+4\sqrt3$ and $xy = 1$, then $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=$
- A
$64$
- B
$134$
- ✓
$194$
- D
$\frac{1}{49}$
Answer$\text{x}=7+4\sqrt3,\ \text{xy}=1\Rightarrow\text{y}=\frac{1}{\text{x}}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}\\ \ =\frac{7-4\sqrt3}{(7)^2-\big(4\sqrt3\big)^2}=\frac{7-4\sqrt3}{49-48}=7-4\sqrt3$
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 $(c)$.
View full question & answer→MCQ 41 Mark
If $\frac{\sqrt3-1}{\sqrt3+1}=\text{a}-\text{b}\sqrt3,$ then:
- ✓
$a = 2, b = 1$
- B
$a = 2, b = -1$
- C
$a = -2, b = 1$
- D
$a = b = 1$
AnswerCorrect option: A. $a = 2, b = 1$
$\frac{\sqrt3-1}{\sqrt3+1}$
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 $(a)$.
View full question & answer→MCQ 51 Mark
If $\text{x}=\sqrt6+\sqrt5,$ then $\text{x}^2+\frac{1}{\text{x}^2}-2=$
- A
$2\sqrt6$
- B
$2\sqrt5$
- C
$24$
- ✓
$20$
Answer$\text{x}^2+\frac{1}{\text{x}^2}-2=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2$
$\text{x}=\sqrt6+\sqrt5$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt6+\sqrt5}=\frac{1}{\sqrt6+\sqrt5}\times\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}\\ \ =\frac{\sqrt6-\sqrt5}{1}=\sqrt6-\sqrt5$
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 $(d)$.
View full question & answer→MCQ 61 Mark
The simplest rationalising factor of $2\sqrt5-\sqrt3,$ is:
- A
$2\sqrt5+3$
- ✓
$2\sqrt5+\sqrt3$
- C
$\sqrt{5}+\sqrt3$
- D
$\sqrt{5}-\sqrt3$
AnswerCorrect option: B. $2\sqrt5+\sqrt3$
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 $(b)$.
View full question & answer→MCQ 71 Mark
If $\text{x}=\frac{2}{3+\sqrt7},$ then $(x - 3)^2$ =
Answer$\text{x}=\frac{2}{3+\sqrt7}=\frac{2}{3+\sqrt7}\times\frac{3-\sqrt7}{3-\sqrt7}=\frac{2(3-\sqrt7)}{3-\sqrt7}\\ \ =\frac{6-2\sqrt7}{9-7}=\frac{6-2\sqrt7}{2}=3-2\sqrt7$
Now $(\text{x}-3)^2=(\not3-\sqrt7-\not3)^2=\big(-\sqrt7\big)^2=7$
Hence, correct option is $(d)$.
View full question & answer→MCQ 81 Mark
The simplest rationalising factor of $\sqrt[3]{500},$ is:
- ✓
$\sqrt[3]{2}$
- B
$\sqrt[3]{5}$
- C
$\sqrt{3}$
- D
AnswerCorrect option: A. $\sqrt[3]{2}$
$\sqrt[3]{500}=(500)^{\frac{1}{3}}=\Big(\frac{500\times2}{2}\Big)^{\frac{1}{3}}\\ \ =\Big(\frac{1000}{2}\Big)^{\frac{1}{3}}=(10^{\not3})^{\frac{1}{\not3}}.\frac{1}{2^{\frac{1}{3}}}\Rightarrow10.2^{\frac{-1}{3}}$
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 $(a)$.
View full question & answer→MCQ 91 Mark
If $\text{x}=\sqrt5+2,$ then $\text{x}-\frac{1}{\text{x}}$ equals:
- A
$2\sqrt5$
- ✓
$4$
- C
$2$
- D
$\sqrt5$
Answer$\text{x}=\sqrt{5}+2$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt5+2}=\frac{1}{\sqrt5+2}\frac{\sqrt5-2}{\sqrt5-2}\\ \ =\frac{\sqrt5-2}{1}=\sqrt5-2$
Now, $\text{x}-\frac{1}{\text{x}}=\sqrt5+2-\big(\sqrt5-2\big)\\ \ =\sqrt5+2-\sqrt5+2=4$
Hence, correct option is $(b)$.
View full question & answer→MCQ 101 Mark
If $\text{x}+\sqrt{15}=4,$ then $\text{x}+\frac{1}{\text{x}}=$
Answer$\text{x}+\sqrt{15}=4$
$=\text{x}=4-\sqrt{15}\Rightarrow\frac{1}{\text{x}}=\frac{1}{4-\sqrt{15}}$
$\frac{1}{\text{x}}=\frac{1}{4-\sqrt{15}}\times\frac{4+\sqrt{15}}{4+\sqrt{15}}=\frac{4+\sqrt{15}}{(4)^2-\big(\sqrt{15}\big)^2}\\ \ =\frac{4+\sqrt{15}}{16-15}=4+\sqrt{15}$
Now, $\text{x}+\frac{1}{\text{x}}=4-\sqrt{15}+4+\sqrt{15}=8$
Hence, correct option is $(c)$.
View full question & answer→MCQ 111 Mark
The positive square root of $7+\sqrt{48},$ is:
- A
$7+2\sqrt3$
- B
$7+\sqrt3$
- ✓
$2+\sqrt3$
- D
$3+\sqrt2$
AnswerCorrect option: C. $2+\sqrt3$
$\sqrt{7+\sqrt{48}}$
$=\sqrt{7+2\sqrt{12}}$
$=\sqrt{4+3+2\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4\big)^2+\big(\sqrt3\big)^2+2\times\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4+\sqrt3\big)^2}$
$=\pm\big(\sqrt4+\sqrt3\big)$
Positive value is $\sqrt4+\sqrt3=2+\sqrt3$
Hence, correct option is $(c)$.
View full question & answer→MCQ 121 Mark
The simplest rationalising factor of $\sqrt3+\sqrt5,$ is:
- A
$\sqrt3-5$
- B
$3-\sqrt5$
- ✓
$\sqrt{3}-\sqrt5$
- D
$\sqrt{3}+\sqrt5$
AnswerCorrect option: C. $\sqrt{3}-\sqrt5$
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 $(c)$.
View full question & answer→MCQ 131 Mark
The rationalisation factor of $2+\sqrt3,$ is:
- ✓
$2-\sqrt3$
- B
$\sqrt2+3$
- C
$\sqrt2-3$
- D
$\sqrt3-2$
AnswerCorrect option: A. $2-\sqrt3$
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 $(a)$.
View full question & answer→MCQ 141 Mark
The rationalisation factor of $\sqrt3,$ is:
- A
$-\sqrt3$
- ✓
$\frac{1}{\sqrt3}$
- C
$2\sqrt3$
- D
$-2\sqrt3$
AnswerCorrect option: B. $\frac{1}{\sqrt3}$
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 $(b)$.
View full question & answer→MCQ 151 Mark
The value of $\sqrt{5+2\sqrt6},$ is:
- A
$\sqrt3-\sqrt2$
- ✓
$\sqrt3+\sqrt2$
- C
$\sqrt5+\sqrt6$
- D
AnswerCorrect option: B. $\sqrt3+\sqrt2$
$\sqrt{5+2\sqrt6}$
$=\sqrt{3+2+2\big(\sqrt3\big)\big(\sqrt2\big)}$
$=\sqrt{\big(\sqrt3\big)^2+\big(\sqrt2\big)^2-2\big(\sqrt3\big)\big(\sqrt2\big)}$
$=\sqrt{\big(\sqrt2+\sqrt2\big)^2}$
$=\sqrt3+\sqrt2$
Hence, correct option is $(b)$.
View full question & answer→MCQ 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$\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$
$\therefore\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}\\ \ =\frac{\big(\sqrt5+\sqrt3\big)^2}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}$
$\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}$
$\therefore\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}\\ \ =\frac{\big(\sqrt5-\sqrt3\big)^2}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}=\frac{8-2\sqrt{15}}{2}=4-\sqrt{15}$
$\text{xy}=\big(4+\sqrt{15}\big)\big(4-\sqrt{15}\big)=16-15=1$
Now, $\text{x}+\text{y}+\text{xy}=4+\sqrt{15}+4-\sqrt{15}+1\\ \ =4+4+1=9$
Hence, correct option is $(a)$.
View full question & answer→MCQ 171 Mark
$\sqrt[5]{6}\times\sqrt[5]{6}$ is equal to:
- ✓
$\sqrt[5]{36}$
- B
$\sqrt[5]{6\times0}$
- C
$\sqrt[5]{6}$
- D
$\sqrt[5]{12}$
AnswerCorrect option: A. $\sqrt[5]{36}$
$\sqrt[5]{6}=(6)^{\frac{1}{5}}$
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 $(a)$.
View full question & answer→MCQ 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$\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
$\therefore\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}\\ \ =\frac{\big(\sqrt3-\sqrt2\big)^2}{3-2}=5-2\sqrt6$
$\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},$
$\therefore\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ \ =\frac{\big(\sqrt3+\sqrt{2}\big)^2}{3-2}=5+2\sqrt6$
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 $(b)$.
View full question & answer→MCQ 191 Mark
The value of $\sqrt{3-2\sqrt2},$ is:
- ✓
$\sqrt2-1$
- B
$\sqrt2+1$
- C
$\sqrt3-\sqrt2$
- D
$\sqrt3+\sqrt2$
AnswerCorrect option: A. $\sqrt2-1$
$\sqrt{3-2\sqrt2}$
$=\sqrt{2+1-2\sqrt2}$
$=\sqrt{\big(\sqrt2\big)^2+(1)^2-2\big(\sqrt2\big)(1)}$
$=\sqrt{\big(\sqrt2-1\big)^2}$
$=\sqrt2-1$
Hence, correct option is $(a)$.
View full question & answer→MCQ 201 Mark
$\frac{1}{\sqrt9-\sqrt8}$ is equal to:
- ✓
$3+2\sqrt2$
- B
$\frac{1}{3+2\sqrt2}$
- C
$3-2\sqrt2$
- D
$\frac{3}{2}-\sqrt2$
AnswerCorrect option: A. $3+2\sqrt2$
$\frac{1}{\sqrt9-\sqrt8}$
$=\frac{1}{\sqrt9-\sqrt8}\times\frac{{\sqrt9+\sqrt8}}{{\sqrt9+\sqrt8}}$
$\frac{{\sqrt9+\sqrt8}}{\big(\sqrt9\big)^2-\big(\sqrt8\big)^2}$
$=\frac{{\sqrt9+\sqrt8}}{9-8}$
$={\sqrt9+\sqrt8}$
$=3+2\sqrt2$
Hence, correct option is $(a)$.
View full question & answer→MCQ 211 Mark
If $\frac{5-\sqrt3}{2+\sqrt3}=\text{x}+\text{y}\sqrt3,$ then:
Answer$\frac{5-\sqrt3}{2+\sqrt3}$
$=\frac{5-\sqrt3}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{\big(5-\sqrt3\big)\big(2-\sqrt3\big)}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{10-5\sqrt3-2\sqrt3+3}{4-3}$
$=\frac{13-7\sqrt3}{1}$
$=13-7\sqrt3$
$\Rightarrow x = 13$ and $y = -7$
Hence, correct option is $(a)$.
View full question & answer→MCQ 221 Mark
The value of $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}},$ is:
- ✓
$\frac{4}{3}$
- B
$4$
- C
$3$
- D
$\frac{3}{4}$
AnswerCorrect option: A. $\frac{4}{3}$
$\sqrt{48}=\sqrt{16\times3}=4\sqrt3$
$\sqrt{32}=\sqrt{16\times2}=4\sqrt2$
$\sqrt{27}=\sqrt{9\times3}=3\sqrt3$
$\sqrt{18}=\sqrt{9\times2}=3\sqrt2$
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 $(a)$.
View full question & answer→MCQ 231 Mark
If $\sqrt{13-\text{a}\sqrt{10}}=\sqrt8+\sqrt5,$ then $a =$
Answer$\sqrt{13-\text{a}\sqrt10}=\sqrt8+\sqrt5$
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 $(c)$.
View full question & answer→MCQ 241 Mark
If $\sqrt2=1.4142,$ then $\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}$ is equal to:
AnswerBy 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 $(c)$.
View full question & answer→MCQ 251 Mark
If $\text{x}=\sqrt[3]{2+\sqrt3},$ then $\text{x}^3+\frac{1}{\text{x}^3}=$:
Answer$\text{x}=\sqrt[3]{2+\sqrt3}+\big(2+\sqrt3\big)^{\frac{1}{3}}$
$\text{x}^3=\big\{\big({2+\sqrt3}\big)^{\frac{1}{\not3}}\big\}^{\not3}=\big(2+\sqrt3\big)$
$\Rightarrow\frac{1}{\text{x}^3}=\frac{1}{2+\sqrt3}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}\\ \ =\frac{2-\sqrt3}{4-3}=2-\sqrt3$
Now, $\text{x}^3+\frac{1}{\text{x}^3}=2+\sqrt3+2-\sqrt3=4$
Hence, correct option is $(b)$.
View full question & answer→MCQ 261 Mark
Which of the following is the value of $(\sqrt{11}+\sqrt{7})(\sqrt{11}-\sqrt{7})$ ?
- ✓
- B
- C
$\sqrt{7}$
- D
$\sqrt{11}$
Answer(a)
Using $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$, we obtain
$(\sqrt{11}+\sqrt{7})(\sqrt{11}-\sqrt{7})=11-7=4$
View full question & answer→MCQ 271 Mark
The value of $\frac{15 \sqrt{15}}{3 \sqrt{3}}$ is
- A
$3 \sqrt{5}$
- B
$5 \sqrt{3}$
- ✓
$5 \sqrt{5}$
- D
$3 \sqrt{3}$
AnswerCorrect option: C. $5 \sqrt{5}$
(c)
Using $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$, we obtain
$\frac{15 \sqrt{15}}{3 \sqrt{3}}=\frac{15}{3} \sqrt{\frac{15}{3}}=5 \sqrt{5}$
View full question & answer→MCQ 281 Mark
The value of $\sqrt{5-2 \sqrt{6}}$ is
- ✓
$\sqrt{3}-\sqrt{2}$
- B
$\sqrt{2}-\sqrt{3}$
- C
$\sqrt{5}-\sqrt{6}$
- D
$\sqrt{5}+\sqrt{6}$
AnswerCorrect option: A. $\sqrt{3}-\sqrt{2}$
(a)
$\sqrt{5-2 \sqrt{6}}=\sqrt{3+2-2 \sqrt{3 \times 2}}=\sqrt{(\sqrt{3})^2+(\sqrt{2})^2-2 \sqrt{3} \sqrt{2}}=\sqrt{(\sqrt{3}-\sqrt{2})^2}=\sqrt{3}-\sqrt{2}$
View full question & answer→MCQ 291 Mark
If $x=\frac{2}{\sqrt{10}-\sqrt{8}}$ and $y=\frac{2}{\sqrt{10}+2 \sqrt{2}}$, then $(x-y)^2=$
- A
$4 \sqrt{2}$
- ✓
- C
$8 \sqrt{2}$
- D
Answer(b)
We have,$x=\frac{2}{\sqrt{10}-\sqrt{8}}$ and $y=\frac{2}{\sqrt{10+2 \sqrt{2}}}$
$\begin{array}{ll}\Rightarrow & x=\frac{2}{\sqrt{10}-2 \sqrt{2}} \text { and } y=\frac{2}{\sqrt{10}+2
\sqrt{2}} \\ \Rightarrow & x=\frac{2(\sqrt{10}+2 \sqrt{2})}{(\sqrt{10}+2 \sqrt{2})(\sqrt{10}-2 \sqrt{2})} \text {
and } y=\frac{2(\sqrt{10}-2 \sqrt{2})}{(\sqrt{10}+2 \sqrt{2})(\sqrt{10}-2 \sqrt{2})} \\
\Rightarrow & x=\frac{2(\sqrt{10}+2 \sqrt{2})}{10-8} \text { and } y=\frac{2(\sqrt{10}-2 \sqrt{2})}{10-8} \\
\Rightarrow & x=\sqrt{10}+2 \sqrt{2} \text { and } y=\sqrt{10}-2 \sqrt{2}
\Rightarrow x-y=4 \sqrt{2} \Rightarrow(x-y)^2=32\end{array}$
View full question & answer→MCQ 301 Mark
If $\sqrt{2}=1.4142$, then $\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}$ is equal to
Answer(a)
We find that $\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{(\sqrt{2}+1)^2}{2-1}=(\sqrt{2}+1)^2$
$\therefore \quad \sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}=\sqrt{2}+1=1.4142+1=2.4142$
View full question & answer→MCQ 311 Mark
If $x=7-4 \sqrt{3}$, then $x+\frac{1}{x}=$
Answer(d)
We have, $x=7-4 \sqrt{3}$.
$\therefore \quad \frac{1}{x}=\frac{1}{7-4 \sqrt{3}}=\frac{7+4 \sqrt{3}}{(7+4 \sqrt{3})(7-4 \sqrt{3})}=\frac{7+4 \sqrt{3}}{49-48}=7+4 \sqrt{3}$
$x+\frac{1}{x}=7-4 \sqrt{3}+7+4 \sqrt{3}=14$
View full question & answer→MCQ 321 Mark
If $x=2+\sqrt{3}$, then $x+\frac{1}{x}$ is equals
- A
- ✓
- C
$-2 \sqrt{3}$
- D
$4-2 \sqrt{3}$
Answer(b)
We have,$x=2+\sqrt{3}$
$\therefore \quad \frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$
Hence, $x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$.
View full question & answer→MCQ 331 Mark
If $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$, then the value of $\sqrt{5-2 \sqrt{6}}$ is
Answer(a)
$\begin{aligned} \sqrt{5-2 \sqrt{6}} & =\sqrt{3+2-2 \sqrt{3 \times 2}}=\sqrt{(\sqrt{3})^2+(\sqrt{2})^2-2 \sqrt{3} \times \sqrt{2}}=\sqrt{(\sqrt{3}-\sqrt{2})^2}=\sqrt{3}-\sqrt{2} \\ & =(1.732)-1.414=0.318\end{aligned}$
View full question & answer→MCQ 341 Mark
If $\sqrt{2}=1.414$, then $\sqrt{3+2 \sqrt{2}}$ is equal to
Answer(b)
$\sqrt{3+2 \sqrt{2}}=\sqrt{2+1+2 \sqrt{2}}=\sqrt{(\sqrt{2})^2+1^2+2 \sqrt{2}}=\sqrt{(\sqrt{2}+1)^2}=\sqrt{2}+1=1.414+1=2.414$
View full question & answer→MCQ 351 Mark
The simplest rationalising factor $\sqrt[3]{500}$ is
- A
$\sqrt{5}$
- B
- C
$\sqrt[3]{5}$
- ✓
$\sqrt[3]{2}$
AnswerCorrect option: D. $\sqrt[3]{2}$
(d)
We find that
$\sqrt[3]{500} \times \sqrt[3]{2}=\sqrt[3]{500 \times 2}=\sqrt[3]{10^3}=10$, which is a rational number.
Hence, $\sqrt[3]{2}$ is a rationalising factor of $\sqrt[3]{500}$.
ALITER $\sqrt[3]{500}=\sqrt[3]{125 \times 4}=\sqrt[3]{5^3 \times 4}=5\left(\sqrt[3]{2^2}\right)$
A rationalising factor of $\sqrt[3]{2^2}$ is $\sqrt[3]{2}$. Hence, the simplest rationalising factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}$.
View full question & answer→MCQ 361 Mark
If $\sqrt{3}=1.732$ and $\sqrt{2}=1.414$, then the value of $\frac{1}{\sqrt{3}-\sqrt{2}}$ is
Answer(b)
A rationalising factor of $\sqrt{3}-\sqrt{2}$ is $\sqrt{3}+\sqrt{2}$.
$\therefore \frac{1}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\frac{\sqrt{3}+\sqrt{2}}{3-2}=\sqrt{3}+\sqrt{2}=1.732+1.414=3.146$
View full question & answer→MCQ 371 Mark
If $\sqrt{2}=1.414$, then $\frac{3}{\sqrt{2}}$ is equal to
Answer(c)
A rationalising factor of $\sqrt{2}$ is $\sqrt{2}$.
$\therefore$ $ \frac{3}{\sqrt{2}}=\frac{3 \sqrt{2}}{\sqrt{2} \times \sqrt{2}}=\frac{3}{2} \sqrt{2}=(1.5) \sqrt{2}=1.5 \times 1.414=2.121 $
View full question & answer→MCQ 381 Mark
$2 \frac{\sqrt[3]{3}}{\sqrt[3]{25}}$ when written with a rational denominator is
AnswerCorrect option: B. $\frac{2}{5} \sqrt[3]{15}$
(b)
We have,
$2 \frac{\sqrt[3]{3}}{\sqrt[3]{25}}=2 \frac{\sqrt[3]{3}}{\sqrt[3]{5^2}}=2 \frac{\sqrt[3]{3}}{\sqrt[3]{5^2}} \times \frac{\sqrt[3]{5}}{\sqrt[3]{5}} \quad\left[\because \sqrt[3]{5}\right.$ is a rationalising factor of $\left.\sqrt[3]{5^2}\right]$
$=2 \frac{\sqrt[3]{3 \times 5}}{\sqrt[3]{5^3}}=2 \frac{\sqrt[3]{15}}{5} \quad\left[\because \sqrt[n]{a^n}=a\right]$
View full question & answer→MCQ 391 Mark
A rationalising factor $3 \sqrt{3}$ is
- A
$\sqrt{3}$
- ✓
$3^{2 / 3}$
- C
$\frac{1}{3}$
- D
$3^{1 / 3}$
AnswerCorrect option: B. $3^{2 / 3}$
(b)
A rationalising factor of $\sqrt[n]{a}$ is $\sqrt[n]{a^{n-1}}$. Therefore, a rationalising factor of $\sqrt[3]{3}$ is $\sqrt[3]{3^{3-1}}=\sqrt[3]{3^2}=3^{2 / 3}$.
View full question & answer→MCQ 401 Mark
If $x=1+\sqrt{7}$, then $-\frac{6}{x}$ is equal to
- A
$1+\sqrt{7}$
- ✓
$1-\sqrt{7}$
- C
$\sqrt{7}$
- D
$-\frac{6}{\sqrt{7}}$
AnswerCorrect option: B. $1-\sqrt{7}$
(b)
We have, $x=1+\sqrt{7}$
$\therefore$ $ -\frac{6}{x}=\frac{6}{1+\sqrt{7}}=\frac{-6(1-\sqrt{7})}{(1+\sqrt{7})(1-\sqrt{7})}=\frac{-6(1-\sqrt{7})}{1-7}=1-\sqrt{7} $
View full question & answer→MCQ 411 Mark
If $\sqrt{2}=1.414$, then the value of $\frac{1}{1+\sqrt{2}}$ is
Answer(d)
A rationalising factor of $1+\sqrt{2}$ is $1-\sqrt{2}$.
$\therefore$ $ \frac{1}{1+\sqrt{2}}=\frac{1-\sqrt{2}}{(1-\sqrt{2})(1+\sqrt{2})}=\frac{1-\sqrt{2}}{1-2}=\sqrt{2}-1=1.414-1=0.414 $
View full question & answer→MCQ 421 Mark
The number obtained by rationalising the denominator of $\frac{1}{\sqrt{5}-2}$ is
- ✓
$\sqrt{5}+2$
- B
$\frac{\sqrt{5}+2}{3}$
- C
$\frac{\sqrt{5}-2}{3}$
- D
$\frac{\sqrt{5}-2}{21}$
AnswerCorrect option: A. $\sqrt{5}+2$
(a)
Rationalising the denominator, we obtain
$ \frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{(\sqrt{5}+2)(\sqrt{5}-2)}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2
$
View full question & answer→MCQ 431 Mark
A rationalising factor of $\frac{1}{\sqrt{9}-\sqrt{7}}$ is
- A
$\sqrt{9}+7$
- B
$9+\sqrt{7}$
- ✓
$3+\sqrt{7}$
- D
$3-\sqrt{7}$
AnswerCorrect option: C. $3+\sqrt{7}$
(c)
We know that a rationalising factor of $\sqrt{a} \pm \sqrt{b}$ is $\sqrt{a} \mp \sqrt{b}$. So, a rationalising factor of $\sqrt{9}-\sqrt{7}$ is $\sqrt{9}+\sqrt{7}=3+\sqrt{7}$.
View full question & answer→MCQ 441 Mark
The value of $\frac{\sqrt{32}+\sqrt{68}}{\sqrt{8}+\sqrt{12}}$ is
Answer(b)
$ \frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16 \times 2}+\sqrt{16 \times 3}}{\sqrt{4 \times 2}+\sqrt{4 \times 3}}=\frac{\sqrt{16} \sqrt{2}+\sqrt{16} \sqrt{3}}{\sqrt{4} \sqrt{2} \times \sqrt{4} \sqrt{3}}=\frac{4 \sqrt{2}+4 \sqrt{3}}{2 \sqrt{2}+2 \sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}=2
$
View full question & answer→MCQ 451 Mark
$\sqrt{14} \times\sqrt{21}$ is equal to
- ✓
$7 \sqrt{6}$
- B
$6 \sqrt{7}$
- C
$5 \sqrt{7}$
- D
$\sqrt{147}$
AnswerCorrect option: A. $7 \sqrt{6}$
(a)
Using: $\sqrt{a} \times \sqrt{b}=\sqrt{a b}$, we obtain
$\sqrt{14} \times \sqrt{21}=\sqrt{14 \times 21}=\sqrt{7 \times 2 \times 7 \times 3}=\sqrt{7^2 \times 6}=\sqrt{7^2} \times \sqrt{6}=7 \sqrt{6}$
View full question & answer→MCQ 461 Mark
The value of $\sqrt{20} \times\sqrt{5}$ is
- ✓
- B
$2 \sqrt{5}$
- C
$20 \sqrt{5}$
- D
$4 \sqrt{5}$
Answer(a)
Using: $\sqrt{a} \times \sqrt{b}=\sqrt{a b}$, we obtain$\sqrt{20} \times \sqrt{5}=\sqrt{20 \times 5}=\sqrt{100}=\sqrt{10^2}=10$
View full question & answer→MCQ 471 Mark
If $\sqrt{13-a \sqrt{10}}=\sqrt{8}+\sqrt{5}$, then $a=$
View full question & answer→MCQ 481 Mark
If $x=\sqrt{6}+\sqrt{5}$, then $x^2+\frac{1}{x^2}-2=$
- A
$2 \sqrt{6}$
- B
$2 \sqrt{5}$
- C
- ✓
View full question & answer→MCQ 491 Mark
The positive square root of $7+\sqrt{48}$, is
- A
$7+2 \sqrt{3}$
- B
$7+\sqrt{3}$
- ✓
$2+\sqrt{3}$
- D
$3+\sqrt{2}$
AnswerCorrect option: C. $2+\sqrt{3}$
View full question & answer→MCQ 501 Mark
If $\sqrt{2}=1.414$, then the value of $\sqrt{6}-\sqrt{3}$ upto three places of decimal is
View full question & answer→MCQ 511 Mark
If $\sqrt{2}=1.4142$, then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ is equal to
View full question & answer→MCQ 521 Mark
The value of $\sqrt{5+2 \sqrt{6}}$, is
- A
$\sqrt{3}-\sqrt{2}$
- ✓
$\sqrt{3}+\sqrt{2}$
- C
$\sqrt{5}+\sqrt{6}$
- D
AnswerCorrect option: B. $\sqrt{3}+\sqrt{2}$
View full question & answer→MCQ 531 Mark
The value of $\sqrt{3-2 \sqrt{2}}$, is
- ✓
$\sqrt{2}-1$
- B
$\sqrt{2}+1$
- C
$\sqrt{3}-\sqrt{2}$
- D
$\sqrt{3}+\sqrt{2}$
AnswerCorrect option: A. $\sqrt{2}-1$
View full question & answer→MCQ 541 Mark
If $x=\sqrt[3]{2+\sqrt{3}}$, then $x^3+\frac{1}{x^3}=$
View full question & answer→MCQ 551 Mark
If $\frac{5-\sqrt{3}}{2+\sqrt{3}}=x+y \sqrt{3}$, then
- ✓
$x=13, y=-7$
- B
$x=-13, y=7$
- C
$x=-13, y=-7$
- D
$x=13, y=7$
AnswerCorrect option: A. $x=13, y=-7$
View full question & answer→MCQ 561 Mark
$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to
- ✓
$3+2 \sqrt{2}$
- B
$\frac{1}{3+2 \sqrt{2}}$
- C
$3-2 \sqrt{2}$
- D
$\frac{3}{2}-\sqrt{2}$
AnswerCorrect option: A. $3+2 \sqrt{2}$
View full question & answer→MCQ 571 Mark
If $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, then $x^2+x y+y^2=$
View full question & answer→MCQ 581 Mark
If $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ and $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, then $x+y+x y=$
View full question & answer→MCQ 591 Mark
If $x+\sqrt{15}=4$, then $x+\frac{1}{x}=$
View full question & answer→MCQ 601 Mark
If $x=7+4 \sqrt{3}$ and $x y=1$, then $\frac{1}{x^2}+\frac{1}{y^2}=$
View full question & answer→MCQ 611 Mark
If $x=\frac{2}{3+\sqrt{7}}$, then $(x-3)^2=$
View full question & answer→MCQ 621 Mark
The simplest rationalising factor of $2 \sqrt{5}-\sqrt{3}$, is
- A
$2 \sqrt{5}+3$
- ✓
$2 \sqrt{5}+\sqrt{3}$
- C
$\sqrt{5}+\sqrt{3}$
- D
$\sqrt{5}-\sqrt{3}$
AnswerCorrect option: B. $2 \sqrt{5}+\sqrt{3}$
View full question & answer→MCQ 631 Mark
The simplest rationalising factor of $\sqrt{3}+\sqrt{5}$, is
- A
$\sqrt{3}-5$
- B
$3-\sqrt{5}$
- ✓
$\sqrt{3}-\sqrt{5}$
- D
$\sqrt{3}+\sqrt{5}$
AnswerCorrect option: C. $\sqrt{3}-\sqrt{5}$
View full question & answer→MCQ 641 Mark
If $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b \sqrt{3}$, then
View full question & answer→MCQ 651 Mark
If $x=\sqrt{5}+2$, then $x-\frac{1}{x}$ equals
- A
$2 \sqrt{5}$
- ✓
- C
- D
$\sqrt{5}$
View full question & answer→MCQ 661 Mark
The rationalisation factor of $2+\sqrt{3}$, is
- ✓
$2-\sqrt{3}$
- B
$\sqrt{2}+3$
- C
$\sqrt{2}-3$
- D
$\sqrt{3}-2$
AnswerCorrect option: A. $2-\sqrt{3}$
View full question & answer→MCQ 671 Mark
The rationalisation factor of $\sqrt{3}$, is
- A
$-\sqrt{3}$
- B
$\frac{1}{\sqrt{3}}$
- C
$2 \sqrt{3}$
- ✓
View full question & answer→MCQ 681 Mark
$\sqrt[5]{6} \times \sqrt[5]{6}$ is equal to
- ✓
$\sqrt[5]{36}$
- B
$\sqrt[5]{6 \times 0}$
- C
$\sqrt[5]{6}$
- D
$\sqrt[5]{12}$
AnswerCorrect option: A. $\sqrt[5]{36}$
View full question & answer→MCQ 691 Mark
$\sqrt{10} \times \sqrt{15}$ is equal to
- ✓
$5 \sqrt{6}$
- B
$6 \sqrt{5}$
- C
$\sqrt{30}$
- D
$\sqrt{25}$
AnswerCorrect option: A. $5 \sqrt{6}$
View full question & answer→