ML Aggarwal Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1.5

 Exercise 1.5

Question 1

(i) $\frac{3}{4 \sqrt{5}}$

Sol :

$\frac{3}{4 \sqrt{5}} \times \frac{4 \sqrt{5}}{4 \sqrt{5}}$

$=\frac{12 \sqrt{5}}{16 \cdot(\sqrt{5})^{2}}$

$=\frac{12 \sqrt{5}}{16 \times 5}$

$=\frac{3 \sqrt{5}}{20}$


(ii) $\frac{5 \sqrt{7}}{\sqrt{3}}$

Sol :

$\frac{5 \sqrt{7}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{5 \sqrt{7 \times 3}}{(\sqrt{3})^{2}}$

$=\frac{5 \sqrt{21}}{3}$


(iii) $\frac{3}{4-\sqrt{7}}$

Sol :

$\frac{3}{4-\sqrt{7}} \times \frac{4+\sqrt{7}}{4+\sqrt{7}}$

$=\frac{3(4+\sqrt{7})}{(4-\sqrt{7})(4+\sqrt{7})}$

$=\frac{12+3 \sqrt{7}}{16-(\sqrt{7})^{2}}$

$= \frac{12+3 \sqrt{7}}{16-7}$

$=\frac{12+3 \sqrt{7}}{9}$

$=\frac{3(4+ \sqrt{7})}{9}$

$=\frac{4+\sqrt{7}}{3}$


(iv) $\frac{17}{3 \sqrt{2}+1}$

Sol :

$\frac{17}{3 \sqrt{2}+1} \times \frac{3 \sqrt{2}-1}{3 \sqrt{2}-1}$

$=\frac{17(3 \sqrt{2}-1)}{(3 \sqrt{2}+1)(3 \sqrt{2}-1)}$

$=\frac{51 \sqrt{2}-17}{(3 \sqrt{2})^{2}-(1)^{2}}$

$=\frac{17(3 \sqrt{2}-1)}{9 \times 2-1}$

$=\frac{17(3 \sqrt{2}-1)}{18-1}$

$=\frac{17(3 \sqrt{2}-1)}{17}$

$=3 \sqrt{2}-1$


(v) $\frac{16}{\sqrt{41}-5}$

Sol :

$\frac{16}{\sqrt{41}-5} \times \frac{\sqrt{41}+5}{\sqrt{41}+5}$

$=\frac{16(\sqrt{41}+5)}{(\sqrt{41})^{2}-(5)^{2}}$

$=\frac{16(\sqrt{41}+5)}{41-25}$

$=\frac{16(\sqrt{41}+5)}{16}$

$=\sqrt{41}+5$


(vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$

Sol :

$\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$

$=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}$

$=\frac{\sqrt{7}+\sqrt{6}}{7-6}$

$=\sqrt{7}+\sqrt{6}$


(vii) $\frac{1}{\sqrt{5}+\sqrt{2}}$

Sol :

$\frac{1}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}$

$=\frac{\sqrt{5}-\sqrt{2}}{5-2}$

$=\frac{\sqrt{5}-\sqrt{2}}{3}$


(viii) $\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$

Sol :

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

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

$=\frac{(\sqrt{2})^{2}+(\sqrt{3})^{2}+2 \cdot \sqrt{2} \cdot \sqrt{3}}{2-3}$

$=\frac{2+3+2 \sqrt{6}}{-1}$

$=-(5+2 \sqrt{6})$

$=-5-2 \sqrt{6}$


Question 2

(i) $\frac{7+3 \sqrt{5}}{7-3 \sqrt{5}}$

Sol :

$\frac{7+3 \sqrt{5}}{7-3 \sqrt{5}} \times \frac{7+3 \sqrt{5}}{7+3 \sqrt{5}}$

$=\frac{(7+3 \sqrt{5})^{2}}{(7)^{2}-(3 \sqrt{5})^{2}}$

$=\frac{(7)^{2}+(3 \sqrt{5})^{2}+2 \times 7 \times 3 \sqrt{5}}{49-9 \times 5}$

$=\frac{49+45+14 \times 3 \sqrt{5}}{49-45}$

$=\frac{94+42 \sqrt{5}}{4}$

$=\frac{2(47+21 \sqrt{5})}{4}$

$=\frac{47+21 \sqrt{5}}{2}$


(ii) $\frac{3-2 \sqrt{2}}{3+2 \sqrt{2}}$

Sol :

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

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

$=\frac{(3)^{2}+(2 \sqrt{2})^{2}-2 \times 3 \times 2 \sqrt{2}}{9-4 \times 2}$

$=\frac{9+8-12 \sqrt{2}}{9-8}$

$=\frac{17-12 \sqrt{2}}{1} .$

$=17-12 \sqrt{2}$


(iii) $\frac{5-3 \sqrt{14}}{7+2 \sqrt{14}}$

Sol :

$=\frac{5-3 \sqrt{14}}{7+2 \sqrt{14}} \times \frac{7-2 \sqrt{14}}{7-2 \sqrt{14}}$

$=\frac{5 \times 7-5 \times 2 \sqrt{14}-7 \times 3 \sqrt{14}+2 \times 3 \times \sqrt{14} \cdot \sqrt{14}}{(7)^{2}-(2 \sqrt{14})^{2}}$

$=\frac{35-10 \sqrt{14}-21 \sqrt{14}+6 \times 14}{49-4 \times 14}$

$=\frac{35-31 \sqrt{14}+84}{49-56}$

$=\frac{119-31 \sqrt{14}}{-7}$

$=\frac{31 \sqrt{14}-119}{7}$


Question 3

(i) $\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}$

Sol :

$\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}=\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}} \times \frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}}$

$=\frac{7 \sqrt{30}-7 \times 3}{(\sqrt{10})^{2}-(\sqrt{3})^{2}}$

$=\frac{7 \sqrt{30}-21}{10-3}$

$=\frac{7(\sqrt{30}-3)}{7}$

$=\sqrt{30}-3$

$\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}=\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}} \times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}$

$=\frac{2 \sqrt{30}-2 \times 5}{(\sqrt{6})^{2}-(\sqrt{5})^{2}}$

$=\frac{2 \sqrt{30}-10}{6-5}$

$=2 \sqrt{30}-10$


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

$=\frac{3 \sqrt{30}-9 \times 2}{(\sqrt{15})^{2}-(3 \sqrt{2})^{2}}$

$=\frac{3 \sqrt{30}-18}{15-18}$

$=\frac{3(\sqrt{30}-6)}{-3}$

$=-\sqrt{30}+6$

$\therefore \frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}$

$=\sqrt{30}-3-(2 \sqrt{30}-10)-(-\sqrt{30}+6)$

$=\sqrt{30}-3-2 \sqrt{30}+10+\sqrt{30}-6$

=1


Question 4

(i) $\frac{3-\sqrt{5}}{3+2 \sqrt{5}}=\frac{-19}{11}+9 \sqrt{5}$

Sol :

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

$=\frac{(3-\sqrt{5})(3-2 \sqrt{5})}{(3)^{2}-(2 \sqrt{5})^{2}}$

$=\frac{9-6 \sqrt{5}-3 \sqrt{5}+2(5)}{9-4(5)}$

$=\frac{19-9 \sqrt{5}}{9-20}$

$=\frac{+19-9 \sqrt{5}}{-11}$

$=\frac{-19}{11}+\frac{9}{11} \sqrt{5}$

$\therefore \frac{-19}{11}+a\sqrt{5}=\frac{-19}{11}+\frac{9}{11} \sqrt{5}$

$\Rightarrow a=\frac{9}{11}$


(ii) $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=a-b \sqrt{6}$

Sol :

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

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

$=\frac{3(2)+2 \sqrt{6}+3 \sqrt{6}+2(3)}{9(2)-4(3)}$

$=\frac{6+5 \sqrt{6}+6}{18-12}$

$=\frac{12+5 \sqrt{6}}{6}$

$=2+\frac{5}{6} \sqrt{6}$

$=2-\left(-\frac{5}{6}\right) \sqrt{6}$

$\therefore a-b \sqrt{6}=2-\left(-\frac{5}{6}\right) \sqrt{6}$

$\Rightarrow a=2 ; b=-\frac{5}{6}$


(ii) $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} b \sqrt{5}$

Sol :

$\frac{7+\sqrt{5}}{7-\sqrt{5}} \times \frac{7+\sqrt{5}}{7+\sqrt{5}}$

$=\frac{(7+\sqrt{5})^{2}}{(7)^{2}-(\sqrt{5})^{2}}$

$=\frac{49+(\sqrt{5})^{2}+2 \cdot 7 \cdot \sqrt{5}}{49-5}$

$=\frac{49+5+14 \sqrt{5}}{44}$

$\frac{7-\sqrt{5}}{7+\sqrt{5}} \times \frac{7-\sqrt{5}}{7-\sqrt{5}}=\frac{(7-\sqrt{5})^{2}}{(7)^{2}-(\sqrt{5})^{2}}$

$=\frac{49+(\sqrt{5})^{2}-2 \times 7 \times \sqrt{5}}{49-5}$

$=\frac{49+5-14 \sqrt{5}}{44}$

$\therefore \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=\frac{54+14 \sqrt{5}}{44}-\frac{54-14 \sqrt{5}}{44}$

$=\frac{54+14 \sqrt{5}-54+14 \sqrt{5}}{44}$

$=\frac{28 \sqrt{5}}{44}$

$=\frac{7}{11}\times \sqrt{5}$

$\therefore a+\frac{7}{11} b \sqrt{5}=\frac{7}{11} \sqrt{5}$

$\Rightarrow a=0 ; b=1$


Question 5

(i) $\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}=p+q \sqrt{5}$

Sol :

$\frac{7+3 \sqrt{5}}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}}=\frac{(7+3 \sqrt{5})(3-\sqrt{5})}{(3)^{2}-(\sqrt{5})^{2}}$

$=\frac{21-7 \sqrt{5}+9 \sqrt{5}-3(5)}{9-5}$

$=\frac{21+2 \sqrt{5}-15}{4}$

$=\frac{6+2 \sqrt{5}}{4}$

$\frac{7-3 \sqrt{5}}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{(7-3 \sqrt{5})(3+\sqrt{5})}{(3)^{2}-(\sqrt{5})^{2}}$

$=\frac{21+7 \sqrt{5}-9 \sqrt{5}-3(5)}{9-5}$

$=\frac{21-2 \sqrt{5}-15}{4}$

$=\frac{6-2 \sqrt{5}}{4}$

$\therefore \frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}=\frac{6+2 \sqrt{5}}{4}-\frac{6-2 \sqrt{5}}{4}$

$=\frac{6+2 \sqrt{5}-6+2 \sqrt{5}}{4}$

$=\frac{4 \sqrt{5}}{4}$

$=\sqrt{5}$

$\therefore p+q \sqrt{5}=\sqrt{5}$

$\Rightarrow p=0 ; q=1$


Question 6

(i) $\frac{\sqrt{2}}{2+\sqrt{2}}$

Sol :

$\frac{\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{\sqrt{2}(2-\sqrt{2})}{(2)^{2}-(\sqrt{2})^{2}}$

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

$=\frac{2(\sqrt{2}-1)}{2}$

$=\sqrt{2}-1$

=1.414-1

=0.414


Question 7




Question 8

(i) $x=1-\sqrt{2}$

Sol :

Given : $x=1-\sqrt{2}$

$\therefore \frac{1}{x}=\frac{1}{1-\sqrt{2}}$

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

$=\frac{1+\sqrt{2}}{(1)^{2}-(\sqrt{2})^{2}}$

$=\frac{1+\sqrt{2}}{1-2}$

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

$\therefore\left(x-\frac{1}{x}\right)^{4}=(1-\sqrt{2}-(-1-\sqrt{2}))^{4}$

$=(1-\sqrt{2}+1+\sqrt{2})^{4}$

$=2^4$

=6


Question 9

(i) $x=5-2 \sqrt{6}$

Sol :

Given : $x=5-2 \sqrt{6}$

$\therefore \frac{1}{x}=\frac{1}{5-2 \sqrt{6}}=\frac{1}{5-2 \sqrt{6}} \times \frac{5+2 \sqrt{6}}{5+2 \sqrt{6}}$

$=\frac{5+2 \sqrt{6}}{(5)^{2}-(2 \sqrt{6})^{2}}$

$=\frac{5+2 \sqrt{6}}{25-24}$

$=5+2 \sqrt{6}$

$\therefore x+\frac{1}{x}=(5-2 \sqrt{6})+(5+2 \sqrt{6})$

=10

We know that $\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2$

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

$=(10)^{2}-2$

=100-2

=98


Question 10

(i) $p=\frac{2-\sqrt{5}}{2+\sqrt{5}} ; q=\frac{2+\sqrt{5}}{2-\sqrt{5}}$

Sol :

p+q$=\frac{2-\sqrt{5}}{2+\sqrt{5}}+\frac{2+\sqrt{5}}{2-\sqrt{5}}$

$=\frac{(2-\sqrt{5})^{2}+(2+\sqrt{5})^{2}}{(2)^{2}-(\sqrt{5})^{2}}$

$=\frac{(4+5-4 \sqrt{5})+(4+5+4 \sqrt{5})}{4-5}$

$\frac{18}{-1}$

∴p+q=-18


(ii) $p-q=\frac{2-\sqrt{5}}{2+\sqrt{5}}-\frac{2+\sqrt{5}}{2-\sqrt{5}}$

$=\frac{(2-\sqrt{5})^{2}-(2+\sqrt{5})^{2}}{(2)^{2}-(\sqrt{5})^{2}}$

$=\frac{(4+5-4 \sqrt{5})-(4+5+4 \sqrt{5})}{4-5}$

$=\frac{9-4 \sqrt{5}-9-4 \sqrt{5}}{-1}$

$=\frac{-8 \sqrt{5}}{-1}$

$=8 \sqrt{5}$


(iii) $p^{2}+q^{2}$

Sol :

We know that 

$(p+q)^{2}=p^{2}+q^{2}+2 p q$

$\therefore pq=\frac{2-\sqrt{5}}{2+\sqrt{5}} \times \frac{2+\sqrt{5}}{2-\sqrt{5}}=1$

∴p+q=-18

$p^{2}+q^{2}=(p+q)^{2}-2 p q$

$=(-18)^{2}-2 \times 1$

=324-2=322


(iv) $p^{2}-q^{2}$

Sol :

$\therefore p^{2}-q^{2}=(p+q)(p-q)$

$=(-18)(8 \sqrt{5})$

=-144√5

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more