SELINA Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1D

Question 1

Simplify : ${\displaystyle \frac{\sqrt{18}}{5\sqrt{18}+3\sqrt{72}-2\sqrt{162}}}$

Sol:

${\displaystyle \frac{\sqrt{18}}{5\sqrt{18}+3\sqrt{72}-2\sqrt{162}}}$

= ${\displaystyle \frac{\sqrt{9\times 2}}{5\sqrt{9\times 2}+3\sqrt{36\times 2}-2\sqrt{81\times 2}}}$

= ${\displaystyle \frac{3\sqrt{2}}{15\sqrt{2}+18\sqrt{2}-18\sqrt{2}}}$

= ${\displaystyle \frac{3\sqrt{2}}{15\sqrt{2}}}$

= ${\displaystyle \frac{1}{5}}$

Question 2

Simplify:

${\displaystyle \frac{\sqrt{x^2+y^2}-y}{x-\sqrt{x^2-y^2}}÷\frac{\sqrt{x^2-y^2}+x}{\sqrt{x^2+y^2}+y}}$

Sol :

Simplify:

${\displaystyle \frac{\sqrt{x^2+y^2}-y}{x-\sqrt{x^2-y^2}}÷\frac{\sqrt{x^2-y^2}+x}{\sqrt{x^2+y^2}+y}}$

Question 3

Evaluate, correct to one place of decimal, the expression ${\displaystyle \frac{5}{\sqrt{20}-\sqrt{10}}}$, if ${\displaystyle \sqrt{5}}$ = 2.2 and ${\displaystyle \sqrt{10}}$ = 3.2.

Sol :

Given - ${\displaystyle \sqrt{5}}$ = 2.2 and ${\displaystyle \sqrt{10}}$ = 3.2.

${\displaystyle \frac{5}{\sqrt{20}-\sqrt{10}}}$

${\displaystyle =\frac{5}{\sqrt{20}-\sqrt{10}}\times \frac{\sqrt{20}+\sqrt{10}}{\sqrt{20}+\sqrt{10}}}$

${\displaystyle =\frac{5(\sqrt{20}+\sqrt{10})}{20-10}}$

${\displaystyle =\frac{{\cancel{5}}^1(\sqrt{20}+\sqrt{10})}{{\cancel{10}}_2}}$

${\displaystyle =\frac{\sqrt{20}+\sqrt{10}}{2}}$

${\displaystyle =\frac{\sqrt{4\times 5}+\sqrt{10}}{2}}$

${\displaystyle =\frac{2\sqrt{5}+\sqrt{10}}{2}}$

${\displaystyle =\frac{2\left(2.2\right)+3.2}{2}}$

${\displaystyle =\frac{4.4+3.2}{2}}$

${\displaystyle =\frac{7.6}{2}}$

= 3.8

Question 4

If x = ${\displaystyle \sqrt{3}-\sqrt{2}}$, find the value of:

(i) ${\displaystyle x+\frac{1}{x}}$

(ii) ${\displaystyle x^2+\frac{1}{x^2}}$

(iii) ${\displaystyle x^3+\frac{1}{x^3}}$

(iv) ${\displaystyle x^3+\frac{1}{x^3}-3(x^2+\frac{1}{x^2})+x+\frac{1}{x}}$

Sol :

x = ${\displaystyle \sqrt{3}-\sqrt{2}}$

${\displaystyle \frac{1}{x}=\frac{1}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

${\displaystyle \frac{1}{x}=\frac{\sqrt{3}+\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}}$

${\displaystyle \frac{1}{x}=(\sqrt{3}+\sqrt{2})}$

(i) ${\displaystyle x+\frac{1}{x}}$

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

${\displaystyle =\sqrt{3}-\cancel{\sqrt{2}}+\sqrt{3}+\cancel{\sqrt{2}}}$

${\displaystyle =\sqrt{3}+\sqrt{3}}$

= ${\displaystyle 2\sqrt{3}}$   

(ii) ${\displaystyle x^2+\frac{1}{x^2}}$

${\displaystyle ={(x+\frac{1}{x})}^2-2\cdot x\cdot \frac{1}{x}}$    ...[a² + b² = (a + b)² - 2ab]

${\displaystyle ={\left(2\sqrt{3}\right)}^2-2}$

${\displaystyle =4\times 3}$ - 2

= 10

(iii) ${\displaystyle x^3+\frac{1}{x^3}}$

${\displaystyle ={(x+\frac{1}{x})}^3-3\cdot x\cdot \frac{1}{x}(x+\frac{1}{x})}$  ...[a³ + b³ = (a + b)³ - 3 · a · b (a + b)]

${\displaystyle ={\left(2\sqrt{3}\right)}^3-3\times \left(2\sqrt{3}\right)}$

${\displaystyle =8\times 3\sqrt{3}-6\sqrt{3}}$

${\displaystyle =24\sqrt{3}-6\sqrt{3}}$

${\displaystyle =18\sqrt{3}}$

(iv) ${\displaystyle x^3+\frac{1}{x^3}-3(x^2+\frac{1}{x^2})+x+\frac{1}{x}}$

${\displaystyle =18\sqrt{3}-3\left(10\right)+2\sqrt{3}}$

${\displaystyle =20\sqrt{3}-30}$

${\displaystyle =10(2\sqrt{3}-3)}$

Question 5 (i)

Show that Negative of an irrational number is irrational.

Sol :

Let us assume that x is an irrational number such that - x is rational.

So, - x = ${\displaystyle \frac{\text{a}}{\text{b}}}$ where a, b are integer and b ≠ 0

x = ${\displaystyle \frac{\text{- a}}{\text{b}}}$

Since, - a, b is also integer and b ≠ 0.

So x is a rational number it contradict our assumption.

∴ - x is irrational.

Question 5 (ii)

Show that the product of a non-zero rational number and an irrational number is an irrational number.

Sol :

Let x is an irrational number and y is non zero rational number.

Let us assume that xy is rational.

Since y is rational then y = ${\displaystyle \frac{\text{a}}{\text{b}}}$ where a and b are integers and b ≠ 0

Since x is irrational so x can be written as fraction form.

∵ xy is rational. Let xy = ${\displaystyle \frac{\text{c}}{\text{d}}}$ where c and dare integers and

${\displaystyle \Rightarrow x\times \frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}}$

${\displaystyle \Rightarrow x=\frac{\text{c}}{\text{d}}\times \frac{\text{a}}{\text{b}}=\frac{\text{ac}}{\text{bd}}}$

Since a, b,c and d are integers so ac and bd are also integers and bd ≠ 0

⇒ x is rational number.

It contradicts our assumptions.

The product of x and y is irratioanl.

Question 6

Draw a line segment of length ${\displaystyle \sqrt{5}}$ cm.

Sol :

Construct a right-angled triangle OAB with

OA = 2 cm,

∠OAB = 90° and

AB = 1 cm

Using OB² = OA² + AB²

OB² = 2² + 1²

OB² = 4 + 1

OB² = 5

OB = ${\displaystyle \sqrt{5}}$

Question 7

Draw a line segment of length ${\displaystyle \sqrt{3}}$ cm.

Sol :

Construct a right angled triangle OAB, in which

∠A = 90°, OB = 2 cm and AB = 1 cm

Using OA² + AB² = OB²

we get: OA = ${\displaystyle \sqrt{3}}$ cm

Question 8

Draw a line segment of length ${\displaystyle \sqrt{8}}$ cm.

Sol :

${\displaystyle \sqrt{8}=\sqrt{3^2-1}}$

Question 9

Show that: ${\displaystyle \frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{2}{5-\sqrt{3}}=\frac{52}{11}}$

Sol :

${\displaystyle \frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{2}{5-\sqrt{3}}}$

${\displaystyle =\frac{4-\sqrt{5}}{4+\sqrt{5}}\times \frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}}\times \frac{5-\sqrt{3}}{5-\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}}\times \frac{4+\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5-\sqrt{3}}\times \frac{5+\sqrt{3}}{5+\sqrt{3}}}$

${\displaystyle =\frac{{(4-\sqrt{5})}^2}{{\left(4\right)}^2-{\left(\sqrt{5}\right)}^2}+\frac{2(5-\sqrt{3})}{{\left(5\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{{(4+\sqrt{5})}^2}{{\left(4\right)}^2-\left(\sqrt{5}\right)}+\frac{2(5+\sqrt{3})}{{\left(5\right)}^2-{\left(\sqrt{3}\right)}^2}}$

${\displaystyle =\frac{16+5-8\sqrt{5}}{16-5}+\frac{10-2\sqrt{3}}{25-3}+\frac{16+5+8\sqrt{5}}{16-5}+\frac{2(5+\sqrt{3})}{25-3}}$

${\displaystyle =\frac{21-8\sqrt{5}}{11}+\frac{10-2\sqrt{3}}{22}+\frac{21+8\sqrt{5}}{11}+\frac{{\cancel{2}}^1(5+\sqrt{3})}{{\cancel{22}}_{11}}}$

${\displaystyle =\frac{21-8\sqrt{5}}{11}+\frac{{\cancel{2}}^1(5-\sqrt{3})}{{\cancel{22}}_{11}}+\frac{21+8\sqrt{5}}{11}+\frac{5+\sqrt{3}}{11}}$

${\displaystyle =\frac{21-\cancel{8\sqrt{5}}+5-\cancel{\sqrt{3}}+21+\cancel{8\sqrt{5}}+5+\cancel{\sqrt{3}}}{11}}$

${\displaystyle =\frac{21+5+21+5}{11}}$

${\displaystyle =\frac{52}{11}}$

Question 10 (i)

Show that: ${\displaystyle x^3+\frac{1}{x^3}=52}$, if x = 2 + ${\displaystyle \sqrt{3}}$

Sol :

Given: x = 2 + ${\displaystyle \sqrt{3}}$

${\displaystyle \frac{1}{x}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}}$

${\displaystyle =\frac{2-\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}}$

${\displaystyle =\frac{2-\sqrt{3}}{4-3}}$

${\displaystyle =2-\sqrt{3}}$

Now,

${\displaystyle x+\frac{1}{x}=2+\cancel{\sqrt{3}}+2-\cancel{\sqrt{3}}}$

${\displaystyle x+\frac{1}{x}=2+2}$

${\displaystyle x+\frac{1}{x}}$= 4

${\displaystyle \therefore x^3+\frac{1}{x^3}}$

${\displaystyle ={(x+\frac{1}{x})}^3-3\cdot \cancel{x}\cdot \frac{1}{\cancel{x}}(x+\frac{1}{x})}$

${\displaystyle ={\left(4\right)}^3-3\times 4}$

= 64 - 12

= 52

Question 10 (ii)

Show that: ${\displaystyle x^2+\frac{1}{x^2}=34,}$ if x = 3 + ${\displaystyle 2\sqrt{2}}$

Sol :

Given: x = 3 + ${\displaystyle 2\sqrt{2}}$

${\displaystyle \frac{1}{x}=\frac{1}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}}$

${\displaystyle \frac{1}{x}=\frac{3-2\sqrt{2}}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}}$

${\displaystyle \frac{1}{x}=\frac{3-2\sqrt{2}}{9-8}}$

${\displaystyle \frac{1}{x}=3-2\sqrt{2}}$

Now, ${\displaystyle x+\frac{1}{x}=3+\cancel{2\sqrt{2}}+3-\cancel{2\sqrt{2}}}$

${\displaystyle x+\frac{1}{x}=6}$

Squaring on both sides

${\displaystyle {(x+\frac{1}{x})}^2={\left(6\right)}^2}$

${\displaystyle \Rightarrow x^2+\frac{1}{x^2}+2\times \cancel{x}\times \frac{1}{\cancel{x}}=36}$

${\displaystyle =x^2+\frac{1}{x^2}=36-2}$

${\displaystyle =x^2+\frac{1}{x^2}=34}$

Question 10 (iii)

Show that: ${\displaystyle \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11}$

Sol :

${\displaystyle \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}}$

${\displaystyle \Rightarrow \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}\times \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

${\displaystyle \Rightarrow \frac{{(3\sqrt{2}-2\sqrt{3})}^2}{{\left(3\sqrt{2}\right)}^2-{\left(2\sqrt{3}\right)}^2}+\frac{2\sqrt{3}(\sqrt{3}+\sqrt{2})}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}}$

${\displaystyle \Rightarrow \frac{{\left(3\sqrt{2}\right)}^2+{\left(2\sqrt{3}\right)}^2-2\times 3\sqrt{2}\times 2\sqrt{3}}{(9\times 2)-(4\times 3)}+\frac{6+2\sqrt{6}}{3-2}}$

${\displaystyle \Rightarrow \frac{18+12-12\sqrt{6}}{18-12}+6+2\sqrt{6}}$

${\displaystyle \Rightarrow \frac{30-12\sqrt{6}}{6}+6+2\sqrt{6}}$

${\displaystyle \Rightarrow \frac{\cancel{6}(5-2\sqrt{6})}{\cancel{6}}+6+2\sqrt{6}}$

${\displaystyle \Rightarrow 5-\cancel{2\sqrt{6}}+6+\cancel{2\sqrt{6}}}$

= 5 + 6

= 11

Question 11 (i)

Show that x is irrational, if x² = 6.

Sol :

x² = 6

x = ${\displaystyle \sqrt{6}}$

It is irrational.

Question 11 (ii)

Show that x is irrational, if x² = 0.009.

Sol :

x² = 0.009

x = ${\displaystyle \sqrt{\frac{9}{1000}}}$

x = ${\displaystyle \frac{3}{10\sqrt{10}}}$

It is a irrational.

Question 11 (iii)

Show that x is irrational, if x² = 27.

Sol :

x² = 27

x = ${\displaystyle \sqrt{27}}$

x = ${\displaystyle \sqrt{3\times 3\times 3}}$

x = ${\displaystyle 3\sqrt{3}}$

It is a Irrational.

Question 12 (i)

Show that x is rational, if x² = 16.

Sol :

x² = 16

x = ${\displaystyle \sqrt{16}}$

x = ${\displaystyle \sqrt{4\times 4}}$

x = 4

It is rational.

Question 12 (ii)

Show that x is rational, if x² = 0.0004.

Sol :

x² = 0.0004

x = ${\displaystyle \frac{0.0004}{10000}}$

x = ${\displaystyle \sqrt{\frac{4}{10000}}}$

x = ${\displaystyle \frac{2}{100}}$

x = 0.02

It is a Rational.

Question 12 (iii)

Show that x is rational, if x² = ${\displaystyle 1\frac{7}{9}}$

Sol :

x² = ${\displaystyle 1\frac{7}{9}}$

x² = ${\displaystyle \frac{16}{9}}$

x = ${\displaystyle \sqrt{\frac{16}{9}}}$

x = ${\displaystyle \frac{4}{3}}$

It is a Rational.

Question 13

Using the following figure, show that BD = ${\displaystyle \sqrt{x}}$.

Sol :

AB = x and BC = 1

AC = AB + BC

= x + 1

diameter = x + 1

radius OA = OD = OC = OB = OC - BC

${\displaystyle =\frac{x+1}{2}-1=\frac{x+1-2}{2}=\frac{x-1}{2}}$

Using pythagoras in ΔBOD

P² + B² = H²

${\displaystyle {\text{P}}^2+{\left(\frac{x-1}{2}\right)}^2={\left(\frac{x+1}{2}\right)}^2}$

${\displaystyle {\text{P}}^2={\left(\frac{x+1}{2}\right)}^2-{\left(\frac{x-1}{2}\right)}^2}$

= ${\displaystyle \frac{{(x+1)}^2-{(x-1)}^2}{4}}$

${\displaystyle =\frac{(x^2+1+2x)-(x^2+1-2x)}{4}}$

${\displaystyle =\frac{x^2+1+2x-x^2-1+2x}{4}}$

${\displaystyle {\text{P}}^2=\frac{4x}{4}}$

P² = x

P = ${\displaystyle \sqrt{x}}$