Question paper, Set 9

Sol: $N = \{1, 2, 3, 4, \ldots\}$, so the smallest natural number is $1$.

$W = \{0, 1, 2, 3, 4, \ldots\}$, so the smallest whole number is $0$.

Answer: $1$ and $0$

Sol: (i) $\sqrt{92} = 9.5916\ldots$ is a non terminating non recurring decimal, so it is irrational.

(ii) $\displaystyle \frac{24}{5}$ is in the form $\displaystyle \frac{p}{q}$ with $q \ne 0$, so it is rational.

Answer: (i) Irrational (ii) Rational

Sol: $e = 2.71828\ldots$ is a non terminating non recurring decimal, so it is irrational.

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

$e$ cannot be written in the form $\displaystyle \frac{p}{q}$, so the statement is false.

Answer: False

Sol: $-1 + \sqrt{15} + 2 + 3\sqrt{15}$

$= (-1 + 2) + (1 + 3)\sqrt{15}$

$= 1 + 4\sqrt{15}$

Answer: $1 + 4\sqrt{15}$

Sol: $(6 - 8\sqrt{19}) - (-8 - 7\sqrt{19})$

$= 6 - 8\sqrt{19} + 8 + 7\sqrt{19}$

$= (6 + 8) + (-8 + 7)\sqrt{19}$

$= 14 - \sqrt{19}$

Answer: $14 - \sqrt{19}$

Sol: $\sqrt{14} \times \sqrt{3} = \sqrt{14 \times 3}$

$= \sqrt{42}$

Answer: $\sqrt{42}$

Sol: $8\sqrt{7} \times (-3\sqrt{19}) = 8 \times (-3) \times \sqrt{7} \times \sqrt{19}$

$= -24 \times \sqrt{7 \times 19}$

$= -24\sqrt{133}$

Answer: $-24\sqrt{133}$

Sol: (i) Positive terms: $40402\sqrt{15}$

Negative terms: $86843\sqrt{15} + 629\sqrt{15} + 96516\sqrt{15} = 183988\sqrt{15}$

$40402\sqrt{15} - 86843\sqrt{15} - 629\sqrt{15} - 96516\sqrt{15} = 40402\sqrt{15} - 183988\sqrt{15}$

$= -(183988 - 40402)\sqrt{15} = -143586\sqrt{15}$

(ii) Positive terms: $25004\sqrt{6}$

Negative terms: $103471\sqrt{6} + 3610\sqrt{6} = 107081\sqrt{6}$

$25004\sqrt{6} - 103471\sqrt{6} - 3610\sqrt{6} = 25004\sqrt{6} - 107081\sqrt{6}$

$= -(107081 - 25004)\sqrt{6} = -82077\sqrt{6}$

Answer: (i) $-143586\sqrt{15}$ (ii) $-82077\sqrt{6}$

Sol: (i) $8\sqrt{5} \times (-8\sqrt{5}) = 8 \times (-8) \times \sqrt{5} \times \sqrt{5}$

$= -64 \times 5 = -320$ (since $\sqrt{a} \times \sqrt{a} = a$)

(ii) $8\sqrt{28} \times (-2\sqrt{7}) = 8 \times (-2) \times \sqrt{28 \times 7}$

$= -16 \times \sqrt{196}$

$\sqrt{196} = 14$

$= -16 \times 14 = -224$

(iii) $-4\sqrt{45} \times 6\sqrt{125} = -4 \times 6 \times \sqrt{45 \times 125}$

$= -24 \times \sqrt{5625}$

Long division: $7 \times 7 = 49$, remainder $7$; bring down $25$ to get $725$; $145 \times 5 = 725$. So $\sqrt{5625} = 75$

$= -24 \times 75 = -1800$

Answer: (i) $-320$ (ii) $-224$ (iii) $-1800$

Sol: $18 = \sqrt{18} \times \sqrt{18}$

$\displaystyle \frac{18}{\sqrt{18}} = \frac{\sqrt{18} \times \sqrt{18}}{\sqrt{18}}$

$= \sqrt{18}$

Answer: $\sqrt{18}$

Sol: $\sqrt{117} = \sqrt{9 \times 13} = \sqrt{9} \times \sqrt{13} = 3\sqrt{13}$

Answer: $3\sqrt{13}$

Sol: $\sqrt{176} = \sqrt{16 \times 11} = \sqrt{16} \times \sqrt{11} = 4\sqrt{11}$

$\sqrt{99} = \sqrt{9 \times 11} = \sqrt{9} \times \sqrt{11} = 3\sqrt{11}$

$\sqrt{360} = \sqrt{36 \times 10} = \sqrt{36} \times \sqrt{10} = 6\sqrt{10}$, so $2\sqrt{360} = 2 \times 6\sqrt{10} = 12\sqrt{10}$

$\sqrt{160} = \sqrt{16 \times 10} = \sqrt{16} \times \sqrt{10} = 4\sqrt{10}$

$\therefore \sqrt{176} + \sqrt{99} + 2\sqrt{360} - \sqrt{160} = 4\sqrt{11} + 3\sqrt{11} + 12\sqrt{10} - 4\sqrt{10}$

$= (4 + 3)\sqrt{11} + (12 - 4)\sqrt{10}$

$= 7\sqrt{11} + 8\sqrt{10}$

Answer: $7\sqrt{11} + 8\sqrt{10}$

Sol: (i) $\displaystyle \frac{0}{7} = 0 \in R$, so it is true.

(ii) $\displaystyle \frac{22}{0}$ is not defined, so $\displaystyle \frac{22}{0} \notin R$; it is false.

Answer: (i) True (ii) False

Sol: $a + b = -7\sqrt{10} + (-6\sqrt{10}) = -13\sqrt{10}$

$a - b = -7\sqrt{10} - (-6\sqrt{10}) = -\sqrt{10}$

$a \times b = -7\sqrt{10} \times (-6\sqrt{10}) = 42 \times 10 = 420$

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

Answer: $a + b = -13\sqrt{10}$, $a - b = -\sqrt{10}$, $a \times b = 420$; each $\in R$

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

$\therefore 6\sqrt{22} \times (-4) = -4 \times 6\sqrt{22}$

Answer: $6\sqrt{22}$

Sol: $a - b = 42 - 6 = 36$

$b - a = 6 - 42 = -36$

$\therefore a - b \ne b - a$

$a \div b = 42 \div 6 = 7$

$\displaystyle b \div a = 6 \div 42 = \frac{1}{7}$

$\therefore a \div b \ne b \div a$

Answer: $a - b = 36 \ne b - a = -36$; $\displaystyle a \div b = 7 \ne b \div a = \frac{1}{7}$

Sol: (i) LHS $= 80 + (20 + 4) = 80 + 24 = 104$, RHS $= (80 + 20) + 4 = 100 + 4 = 104$

LHS $=$ RHS

(ii) LHS $= 80 - (20 - 4) = 80 - 16 = 64$, RHS $= (80 - 20) - 4 = 60 - 4 = 56$

LHS $\ne$ RHS

(iii) LHS $= 80 \times (20 \times 4) = 80 \times 80 = 6400$, RHS $= (80 \times 20) \times 4 = 1600 \times 4 = 6400$

LHS $=$ RHS

(iv) LHS $= (80 \div 20) \div 4 = 4 \div 4 = 1$, RHS $= 80 \div (20 \div 4) = 80 \div 5 = 16$

LHS $\ne$ RHS

Answer: (i) $104 = 104$ (ii) $64 \ne 56$ (iii) $6400 = 6400$ (iv) $1 \ne 16$

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

$\therefore -2 + 0 = -2$

Answer: $0$

Sol: (i) Additive inverse of $a$ is $-a$, since $a + (-a) = 0$. So it is $6\sqrt{15}$.

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

Answer: (i) $6\sqrt{15}$ (ii) $\displaystyle -\frac{1}{9}$

Sol: Additive inverse of $\displaystyle \frac{4}{3}$ is $\displaystyle -\frac{4}{3}$.

Multiplicative inverse of $\displaystyle \frac{4}{3}$ is $\displaystyle \frac{3}{4}$.

Check: $\displaystyle \frac{4}{3} + (-\frac{4}{3}) = 0$

Check: $\displaystyle \frac{4}{3} \times \frac{3}{4} = 1$

Answer: Additive inverse $\displaystyle -\frac{4}{3}$, multiplicative inverse $\displaystyle \frac{3}{4}$