MCQ 11 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 21 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 31 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 41 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 51 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 61 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 71 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 81 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 91 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 101 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 111 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 121 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 131 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 141 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 151 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 161 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 171 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→MCQ 181 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 191 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 201 Mark
If $x=\sqrt[3]{2+\sqrt{3}}$, then $x^3+\frac{1}{x^3}=$
View full question & answer→MCQ 211 Mark
If $x+\sqrt{15}=4$, then $x+\frac{1}{x}=$
View full question & answer→MCQ 221 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 231 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 241 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 251 Mark
If $x=\frac{2}{3+\sqrt{7}}$, then $(x-3)^2=$
View full question & answer→MCQ 261 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 271 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 281 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 291 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 301 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 311 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 321 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 331 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 341 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 351 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 361 Mark
If $\sqrt{2}=1.4142$, then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ is equal to
View full question & answer→MCQ 371 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 381 Mark
If $\sqrt{13-a \sqrt{10}}=\sqrt{8}+\sqrt{5}$, then $a=$
View full question & answer→MCQ 391 Mark
If $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b \sqrt{3}$, then
View full question & answer→MCQ 401 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 411 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 421 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 431 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 441 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→