Question paper, Set 7

Sol: $N = \{1, 2, 3, 4, \ldots\}$

The natural numbers between $-4$ and $3$ are $1, 2$.

Answer: $1, 2$

Sol: (i) $4.592$ is a terminating decimal, so it is rational.

(ii) $e = 2.71828\ldots$ is a non terminating non recurring decimal, so it is irrational.

Answer: (i) Rational (ii) Irrational

Sol: $\sqrt{15} = 3.8729\ldots$ is a non terminating non recurring decimal, so it is irrational.

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

$\sqrt{15}$ is not in this set, so the statement is false.

Answer: False

Sol: $-13 + 4\sqrt{15} + 12 - 9\sqrt{15}$

$= (-13 + 12) + (4 - 9)\sqrt{15}$

$= -1 - 5\sqrt{15}$

Answer: $-1 - 5\sqrt{15}$

Sol: $(1 + 3\sqrt{15}) - (1 + 4\sqrt{15})$

$= 1 + 3\sqrt{15} - 1 - 4\sqrt{15}$

$= (1 - 1) + (3 - 4)\sqrt{15}$

$= -\sqrt{15}$

Answer: $-\sqrt{15}$

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

$\therefore \sqrt{58} \times \sqrt{58} = 58$

Answer: $58$

Sol: $3\sqrt{19} \times 5\sqrt{3} = 3 \times 5 \times \sqrt{19} \times \sqrt{3}$

$= 15 \times \sqrt{19 \times 3}$

$= 15\sqrt{57}$

Answer: $15\sqrt{57}$

Sol: (i) Positive terms: $92105\sqrt{6} + 600574\sqrt{6} = 692679\sqrt{6}$

Negative terms: $625\sqrt{6} + 77242\sqrt{6} = 77867\sqrt{6}$

$-625\sqrt{6} + 92105\sqrt{6} + 600574\sqrt{6} - 77242\sqrt{6} = 692679\sqrt{6} - 77867\sqrt{6}$

$= (692679 - 77867)\sqrt{6} = 614812\sqrt{6}$

(ii) Positive terms: $555\sqrt{5}$

Negative terms: $420918\sqrt{5} + 2081\sqrt{5} + 71860\sqrt{5} = 494859\sqrt{5}$

$-420918\sqrt{5} + 555\sqrt{5} - 2081\sqrt{5} - 71860\sqrt{5} = 555\sqrt{5} - 494859\sqrt{5}$

$= -(494859 - 555)\sqrt{5} = -494304\sqrt{5}$

Answer: (i) $614812\sqrt{6}$ (ii) $-494304\sqrt{5}$

Sol: (i) $2\sqrt{17} \times (-7\sqrt{17}) = 2 \times (-7) \times \sqrt{17} \times \sqrt{17}$

$= -14 \times 17 = -238$ (since $\sqrt{a} \times \sqrt{a} = a$)

(ii) $9\sqrt{2} \times (-7\sqrt{18}) = 9 \times (-7) \times \sqrt{2 \times 18}$

$= -63 \times \sqrt{36}$

$\sqrt{36} = 6$

$= -63 \times 6 = -378$

(iii) $2\sqrt{294} \times 2\sqrt{6} = 2 \times 2 \times \sqrt{294 \times 6}$

$= 4 \times \sqrt{1764}$

Long division: $4 \times 4 = 16$, remainder $1$; bring down $64$ to get $164$; $82 \times 2 = 164$. So $\sqrt{1764} = 42$

$= 4 \times 42 = 168$

Answer: (i) $-238$ (ii) $-378$ (iii) $168$

Sol: $\sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} = 3\sqrt{7}$

$\displaystyle \frac{\sqrt{63}}{3\sqrt{7}} = \frac{3\sqrt{7}}{3\sqrt{7}} = 1$

Answer: $1$

Sol: $\sqrt{90} = \sqrt{9 \times 10} = \sqrt{9} \times \sqrt{10} = 3\sqrt{10}$

Answer: $3\sqrt{10}$

Sol: $\sqrt{216} = \sqrt{36 \times 6} = \sqrt{36} \times \sqrt{6} = 6\sqrt{6}$, so $3\sqrt{216} = 3 \times 6\sqrt{6} = 18\sqrt{6}$

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

$\sqrt{396} = \sqrt{36 \times 11} = \sqrt{36} \times \sqrt{11} = 6\sqrt{11}$, so $2\sqrt{396} = 2 \times 6\sqrt{11} = 12\sqrt{11}$

$\sqrt{150} = \sqrt{25 \times 6} = \sqrt{25} \times \sqrt{6} = 5\sqrt{6}$

$\therefore 3\sqrt{216} + \sqrt{99} - 2\sqrt{396} + \sqrt{150} = 18\sqrt{6} + 3\sqrt{11} - 12\sqrt{11} + 5\sqrt{6}$

$= (18 + 5)\sqrt{6} + (3 - 12)\sqrt{11}$

$= 23\sqrt{6} - 9\sqrt{11}$

Answer: $23\sqrt{6} - 9\sqrt{11}$

Sol: $a + b = 25 + (-24) = 1$

$a - b = 25 - (-24) = 49$

$a \times b = 25 \times (-24) = -600$

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

Answer: $a + b = 1$, $a - b = 49$, $a \times b = -600$; all belong to $R$

Sol: $a + b = 11\sqrt{22} + (-5\sqrt{22}) = 6\sqrt{22}$

$a - b = 11\sqrt{22} - (-5\sqrt{22}) = 16\sqrt{22}$

$a \times b = 11\sqrt{22} \times (-5\sqrt{22}) = -55 \times 22 = -1210$

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

Answer: $a + b = 6\sqrt{22}$, $a - b = 16\sqrt{22}$, $a \times b = -1210$; each $\in R$

Sol: $46 - 8 = 38$

$8 - 46 = -38$

$38 \ne -38$, so real numbers are not commutative under subtraction.

Answer: No: $46 - 8 = 38$ but $8 - 46 = -38$

Sol: $a - b = 63 - 9 = 54$

$b - a = 9 - 63 = -54$

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

$a \div b = 63 \div 9 = 7$

$\displaystyle b \div a = 9 \div 63 = \frac{1}{7}$

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

Answer: $a - b = 54 \ne b - a = -54$; $\displaystyle a \div b = 7 \ne b \div a = \frac{1}{7}$

Sol: (i) LHS $= 64 + (16 + 4) = 64 + 20 = 84$, RHS $= (64 + 16) + 4 = 80 + 4 = 84$

LHS $=$ RHS

(ii) LHS $= 64 - (16 - 4) = 64 - 12 = 52$, RHS $= (64 - 16) - 4 = 48 - 4 = 44$

LHS $\ne$ RHS

(iii) LHS $= 64 \times (16 \times 4) = 64 \times 64 = 4096$, RHS $= (64 \times 16) \times 4 = 1024 \times 4 = 4096$

LHS $=$ RHS

(iv) LHS $= (64 \div 16) \div 4 = 4 \div 4 = 1$, RHS $= 64 \div (16 \div 4) = 64 \div 4 = 16$

LHS $\ne$ RHS

Answer: (i) $84 = 84$ (ii) $52 \ne 44$ (iii) $4096 = 4096$ (iv) $1 \ne 16$

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

Answer: Additive identity

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

(ii) Multiplicative inverse of $a$ is $\displaystyle \frac{1}{a}$, since $\displaystyle a \times \frac{1}{a} = 1$. So it is $\displaystyle \frac{1}{\sqrt{23}}$.

Answer: (i) $6$ (ii) $\displaystyle \frac{1}{\sqrt{23}}$

Sol: Additive inverse of $\displaystyle \frac{29}{10}$ is $\displaystyle -\frac{29}{10}$.

Multiplicative inverse of $\displaystyle \frac{29}{10}$ is $\displaystyle \frac{10}{29}$.

Check: $\displaystyle \frac{29}{10} + (-\frac{29}{10}) = 0$

Check: $\displaystyle \frac{29}{10} \times \frac{10}{29} = 1$

Answer: Additive inverse $\displaystyle -\frac{29}{10}$, multiplicative inverse $\displaystyle \frac{10}{29}$