Question paper, Set 4

Sol: $Z = \{\ldots, -2, -1, 0, 1, 2, 3, \ldots\}$

The integers between $-10$ and $-3$ are $-9, -8, -7, -6, -5, -4$.

Answer: $-9, -8, -7, -6, -5, -4$

Sol: (i) $\sqrt{83} = 9.1104\ldots$ is a non terminating non recurring decimal, so it is irrational.

(ii) $9.\overline{8}$ is a non terminating recurring decimal, so it is rational.

(iii) $\displaystyle \sqrt{49} = 7 = \frac{7}{1}$, so it is rational.

(iv) $0.913$ is a terminating decimal, so it is rational.

Answer: (i) irrational (ii) rational (iii) rational (iv) rational

Sol: $\displaystyle Q = \{\frac{p}{q}$ where $p, q$ are integers and $q \ne 0\}$

$19$ can be written in the form $\displaystyle \frac{p}{q}$, so the statement is true.

Answer: True

Sol: $-8\sqrt{6} - 6\sqrt{3} + 9\sqrt{6} - 11\sqrt{3}$

$= (-8 + 9)\sqrt{6} + (-6 - 11)\sqrt{3}$

$= \sqrt{6} - 17\sqrt{3}$

Answer: $\sqrt{6} - 17\sqrt{3}$

Sol: $(-9\sqrt{7} - 5\sqrt{2}) - (-11\sqrt{7} + 7\sqrt{2})$

$= -9\sqrt{7} - 5\sqrt{2} + 11\sqrt{7} - 7\sqrt{2}$

$= (-9 + 11)\sqrt{7} + (-5 - 7)\sqrt{2}$

$= 2\sqrt{7} - 12\sqrt{2}$

Answer: $2\sqrt{7} - 12\sqrt{2}$

Sol: $\sqrt{17} \times \sqrt{5} = \sqrt{17 \times 5}$

$= \sqrt{85}$

Answer: $\sqrt{85}$

Sol: $11\sqrt{23} \times (-10\sqrt{23}) = 11 \times (-10) \times \sqrt{23} \times \sqrt{23}$

$= -110 \times 23$

$= -2530$

Answer: $-2530$

Sol: (i) Positive terms: $680455\sqrt{11} + 9111\sqrt{11} = 689566\sqrt{11}$

Negative terms: $80487\sqrt{11} + 80582\sqrt{11} = 161069\sqrt{11}$

$-80487\sqrt{11} + 680455\sqrt{11} - 80582\sqrt{11} + 9111\sqrt{11} = 689566\sqrt{11} - 161069\sqrt{11}$

$= (689566 - 161069)\sqrt{11} = 528497\sqrt{11}$

(ii) Positive terms: $7502\sqrt{7}$

Negative terms: $105838\sqrt{7} + 4869\sqrt{7} = 110707\sqrt{7}$

$7502\sqrt{7} - 105838\sqrt{7} - 4869\sqrt{7} = 7502\sqrt{7} - 110707\sqrt{7}$

$= -(110707 - 7502)\sqrt{7} = -103205\sqrt{7}$

Answer: (i) $528497\sqrt{11}$ (ii) $-103205\sqrt{7}$

Sol: (i) $2\sqrt{21} \times (-6\sqrt{21}) = 2 \times (-6) \times \sqrt{21} \times \sqrt{21}$

$= -12 \times 21 = -252$ (since $\sqrt{a} \times \sqrt{a} = a$)

(ii) $7\sqrt{2} \times (-4\sqrt{50}) = 7 \times (-4) \times \sqrt{2 \times 50}$

$= -28 \times \sqrt{100}$

$\sqrt{100} = 10$

$= -28 \times 10 = -280$

(iii) $-3\sqrt{24} \times 2\sqrt{96} = -3 \times 2 \times \sqrt{24 \times 96}$

$= -6 \times \sqrt{2304}$

Long division: $4 \times 4 = 16$, remainder $7$; bring down $04$ to get $704$; $88 \times 8 = 704$. So $\sqrt{2304} = 48$

$= -6 \times 48 = -288$

Answer: (i) $-252$ (ii) $-280$ (iii) $-288$

Sol: $19 = \sqrt{19} \times \sqrt{19}$

$\displaystyle \frac{19}{\sqrt{19}} = \frac{\sqrt{19} \times \sqrt{19}}{\sqrt{19}}$

$= \sqrt{19}$

Answer: $\sqrt{19}$

Sol: $\sqrt{240} = \sqrt{16 \times 15} = \sqrt{16} \times \sqrt{15} = 4\sqrt{15}$

Answer: $4\sqrt{15}$

Sol: $\sqrt{320} = \sqrt{64 \times 5} = \sqrt{64} \times \sqrt{5} = 8\sqrt{5}$

$\sqrt{243} = \sqrt{81 \times 3} = \sqrt{81} \times \sqrt{3} = 9\sqrt{3}$, so $2\sqrt{243} = 2 \times 9\sqrt{3} = 18\sqrt{3}$

$\sqrt{80} = \sqrt{16 \times 5} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5}$, so $3\sqrt{80} = 3 \times 4\sqrt{5} = 12\sqrt{5}$

$\sqrt{108} = \sqrt{36 \times 3} = \sqrt{36} \times \sqrt{3} = 6\sqrt{3}$

$\therefore \sqrt{320} + 2\sqrt{243} - 3\sqrt{80} + \sqrt{108} = 8\sqrt{5} + 18\sqrt{3} - 12\sqrt{5} + 6\sqrt{3}$

$= (8 - 12)\sqrt{5} + (18 + 6)\sqrt{3}$

$= -4\sqrt{5} + 24\sqrt{3}$

Answer: $-4\sqrt{5} + 24\sqrt{3}$

Sol: $a + b = -10 + 7 = -3$

$a - b = -10 - 7 = -17$

$a \times b = -10 \times 7 = -70$

All three are real numbers: real numbers are closed under addition, subtraction and multiplication.

Answer: $a + b = -3$, $a - b = -17$, $a \times b = -70$; all belong to $R$

Sol: $a + b = 3\sqrt{22} + (-7\sqrt{22}) = -4\sqrt{22}$

$a - b = 3\sqrt{22} - (-7\sqrt{22}) = 10\sqrt{22}$

$a \times b = 3\sqrt{22} \times (-7\sqrt{22}) = -21 \times 22 = -462$

All three are real numbers, so $a + b, a - b, a \times b \in R$.

Answer: $a + b = -4\sqrt{22}$, $a - b = 10\sqrt{22}$, $a \times b = -462$; each $\in R$

Sol: Real numbers are commutative under multiplication: $a \times b = b \times a$.

$\therefore 9\sqrt{5} \times \sqrt{6} = \sqrt{6} \times 9\sqrt{5}$

Answer: $9\sqrt{5}$

Sol: $a + b = 2\sqrt{21} + 7\sqrt{21} = 9\sqrt{21}$

$b + a = 7\sqrt{21} + 2\sqrt{21} = 9\sqrt{21}$

$a \times b = 2\sqrt{21} \times 7\sqrt{21} = 14 \times 21 = 294$

$b \times a = 7\sqrt{21} \times 2\sqrt{21} = 14 \times 21 = 294$

So $a + b = b + a$ and $a \times b = b \times a$.

Answer: $a + b = b + a = 9\sqrt{21}$, $a \times b = b \times a = 294$

Sol: (i) LHS $= 72 + (12 + 2) = 72 + 14 = 86$, RHS $= (72 + 12) + 2 = 84 + 2 = 86$

LHS $=$ RHS

(ii) LHS $= 72 - (12 - 2) = 72 - 10 = 62$, RHS $= (72 - 12) - 2 = 60 - 2 = 58$

LHS $\ne$ RHS

(iii) LHS $= 72 \times (12 \times 2) = 72 \times 24 = 1728$, RHS $= (72 \times 12) \times 2 = 864 \times 2 = 1728$

LHS $=$ RHS

(iv) LHS $= (72 \div 12) \div 2 = 6 \div 2 = 3$, RHS $= 72 \div (12 \div 2) = 72 \div 6 = 12$

LHS $\ne$ RHS

Answer: (i) $86 = 86$ (ii) $62 \ne 58$ (iii) $1728 = 1728$ (iv) $3 \ne 12$

Sol: $a + 0 = a$, where $0$ is the additive identity.

$\therefore -\sqrt{23} + 0 = -\sqrt{23}$

Answer: $0$

Sol: (i) Additive inverse of $a$ is $-a$, since $a + (-a) = 0$. So it is $\displaystyle \frac{4}{3}$.

(ii) Multiplicative inverse of $a$ is $\displaystyle \frac{1}{a}$, since $\displaystyle a \times \frac{1}{a} = 1$. So it is $\displaystyle \frac{11}{9}$.

Answer: (i) $\displaystyle \frac{4}{3}$ (ii) $\displaystyle \frac{11}{9}$

Sol: Additive inverse of $9\sqrt{21}$ is $-9\sqrt{21}$.

Multiplicative inverse of $9\sqrt{21}$ is $\displaystyle \frac{1}{9\sqrt{21}}$.

Check: $9\sqrt{21} + (-9\sqrt{21}) = 0$

Check: $\displaystyle 9\sqrt{21} \times \frac{1}{9\sqrt{21}} = 1$

Answer: Additive inverse $-9\sqrt{21}$, multiplicative inverse $\displaystyle \frac{1}{9\sqrt{21}}$