maths ipe imp q & ans

8/20/2019 maths ipe imp q & ans

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1. f: A→ B, g: B→ C are two b!e"to#, t$e# %ro&e t$at gof: A →

C ' a('o b!e"to#.

So(: Given: f: A→ B, g: B →C are bijection.

Part 1: To %ro&e t$at gof: A → C ' o#e ) o#e.

Now f: A→ B, g: B →C are one – one functions. gof: A→ C is a function.

Let a1, a2 ∈A

u!!ose t"at #gof$ #a1$ % #gof$ #a2$

g &f#a1$'% g&f#a2$'

f #a1$ % f#a2$ ( g is one – one

a1% a2 ( f is one – one

(gof: A→ C is one – one.

Part *: To %ro&e t$at gof: A→C ' o#to.

Now f: A→ B, g: B →C are onto functions.

gof: A→ C is a function.

Let c ∈ C,

ince g: B →C is onto, t"ere e)ists at *east one e*e+ent b ∈ B suc" t"at

g #b$ %c.

ince f: A→ B is a*so onto, t"ere e)ists at *east one e*e+ents a ∈ A

suc" t"at f #a$ %b.

Now #gof$ #a$ %g &f #a$'

%g #b$%c.

for c ∈ C, t"ere is an e*e+ent a ∈ A suc" t"at #gof$ #a$ % co gof: A→ C is onto.ince gof: A→ C is bot" one –one an- onto, "ence gof: A→ C is a

bijection.

1


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*. If f: A→B, g: B→C are b!e"to#, t$e# %ro&e t$at+gof-1f -1og-1.A: given t"at f: A→B, g: B→C are bijection.f 1: B →A, g1: C →B.Now gof: A →C is a*so a bijection.#g o f$1:C→AA*so g1: C →B, f 1: B →A, f 1 o g1: C →A

/"us #gof$1 an- f 1og1 bot" t"e functions e)ist an- "ave sa+e -o+ain

an- t"e sa+e co -o+ain.

Let c be an0 e*e+ent in C.ince g: B→ C is onto, t"ere e)ists at *east one e*e+ent b ∈ B suc" t"atg #b$ %cb%g1#c$ ( g is a bijection

ince g: A→ B is onto, t"ere e)ists at *east one e*e+ent a ∈ A suc" t"atf #a$ %ba%f 1#b$ f ' a b!e"to#/. +1

Co#'0er

#gof$ #a$ %g &f #a$' %g #b$

#gof$ #a$ %ca % #gof$1#c$ .. #1$ gof ' a b!e"to#/+*

A*so #f 1og1$ #c$ %f 1&g1#c$'% f 1#b$

#f 1 o g1$ #c$ % a.#2$

ro+ #1$ 3 #2$

#gof$1#c$ %f 1og1$ #c$ 4 c ∈C.He#"e +gof-1f -1og-1.

2


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. If f: A→ B ' a b!e"to#, t$e# S.T fof -1IB a#0 f -1ofIA.

A: given t"at f: A→ B, bijection f 1: B →A.

Part 1: To '$ow t$at fof-1 IB

Now f 1: B →A, f: A→ B fof 1:B→B.A*so 5B: B→B /"us fof 1 an- 5B bot" t"e functions e)ist an- "ave sa+e -o+ain

an- t"e sa+e co -o+ain .Let a be an0 e*e+ent in A.ince f: A→B, t"ere is uni6ue e*e+ent b ∈B.uc" t"at f #a$ %b

a%f 1#b$ ( f is a bijectionConsi-er #fof 1$ #b$ %f &f 1#b$' %f #a$ %b

%5B #b$ ( 5B:B→B

5B #b$ %b

#fof1$ #b$ %5B #b$ 4 b ∈B.

/"us fof 1%5B.

Part *: To '$ow t$at f-1 ofIA

Now, f: A→ B, f 1: B →A f 1of:A→A.A*so 5A: A→A /"us f 1of an- 5A bot" t"e functions e)ist an- "ave sa+e -o+ain

A an- t"e sa+e co -o+ain A.

Now #f 1

of$ #a$ %f1

&f #a$'

%f 1#b$ %a %5A #a$ (5A:A→A 5A #a$ %a

#f 1of$ #a$ %5A #a$ 4 a ∈A.

f 1of%5A. 7ence t"at fof 1%5B an- f 1of%5A.

8. Let f : A→ B, IA a#0 IB are 0e#tt2 f3#"to#' o# A a#0 B re'%e"t&e(2. T$e#

foIAfIB of.A: given t"at f: A→ B,5A: A→A is -e9ne- b0 5A #a$ %a 4 a ∈A.5B: B→B is -e9ne- b0 5B #b$ %b 4 b ∈B


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Part1: to %ro&e t$at foIAf Now 5A: A→A, f: A→ B fo5A: A→ B,

A*so f: A→ B, /"us fo5A an- f bot" t"e functions e)ist an- "ave sa+e -o+ain A an- t"e sa+e co

-o+ain B.Let a ∈A.ince f: A→B, t"ere is uni6ue e*e+ent b ∈ B.uc" t"at f #a$ %bConsi-er #fo5A$ #a$ %f &5A #a$'

%f #a$

#fo5A$ #a$ %f #a$ 4 a ∈A.

7ence fo5A %f.. #1$.

Part*: to %ro&e t$at foIAf

Now f: A→ B, 5B: B→B, 5B of: A→ B,

A*so f: A→ B,

/"us 5Aof an- f bot" t"e functions e)ist an- "ave sa+e -o+ain A an- t"e

sa+e co -o+ain B.

Let b ∈B.Consi-er #5B of$ #a$ % &5B #f #a$$'

5B #b$%f #a$

#5B of$ #a$ %f #a$ 4 a ∈A.

7ence 5B of %f.. #2$.

4ro5 +1 6 +* foIAfIB of.

7. If f: A→ B, g: B→ A are two b!e"to# '3"$ t$at gofIA a#0 fog IB t$e# %ro&e

t$at gf -1.

8


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So(: Part 1: /o !rove t"at gof: A→ C is one – one.

Now f: A→ B, g: B →A are one – one functions. gof: A→ A is a function.

Let a1, a2 ∈A f #a1$

u!!ose t"at #gof$ #a1$ % #gof$ #a2$

g &f#a1$'% g&f#a2$'

f #a1$% f#a2$ ( g is one – one

a1% a2 ( f is one – one

gof: A→ C is one – one.

Part *: /o !rove t"at gof is ontoLet b ∈B ,

g: B→ A, t"ere e)ists a uni6ue e*e+ents a ∈ A

uc" t"at g#b$%aNow f #a$ %f &g #b$'; #fog$ #b$IB +b b

o f: A→ B is ontoince f is bot" oneone an- onto, so f is a bijection.;f 1:B→A

A*so g: B→A /"us bot" t"e functions f 1 an- g "ave t"e sa+e -o+ain B an- sa+e co -o+ain A.

Part : to s"ow t"at gf -1.ro+ !revious !art, f#a$%b;a%f 1#b$A*so g #b$ %a

g #b$ %f 1#b$ ∀ b∈B.henceg=¿ f 1

1. U'#g 5at$e5at"a( #03"to#, S$ow t$at

12+ 22+32+……… ..+n2=

n (n+1 )(2 n+1)6 .


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o*: let p (n )=12+ 22+32+..+n2=

n (n+1)(2 n+1)6 .

Ste%-1: !ut n%1

L.7. =.7.n

2 n (n+1 )(2n+ 1)6 .

% #1$2 %1.2.3

6

%1 %1L.7.%=.7. ! #1$ is true for n%1.

Ste%- 2: let us assumethat p (k ) istrue for n=k .

⇒ p (k )=12+22+32+..+k 2=k (k +1 )(2 k +1)

6

Ste%-: adding bothsides (k +1)term

p (k +1)=12+22+32+..+k 2+(k +1 )2 ¿ k ( k +1)(2 k +1)

6 >(k +1)2

1

¿ k ( k +1 ) (2 k +1 )+6 (k +1)2

6

%( k +1 ) [k (2 k +1 )+ 6 (k +1 )]

6

%

( k +1 )[2k 2+k +6 k +6]6

%

( k +1 )[2k 2+7 k +6]6

?

t n=n2

t k =k 2

t k +1=(k +1)2


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%

( k +1 )[2k 2+4 k +3 k +6]6

%

( k +1 )[2k (k +2)+3(k + 2)]6

%

( k +1 )(k +2)(2 k +1)6

∴Thus p (k +1 ) is true for n=k +1 .

by the principle of mathematical induction, p (n) is true for a** n ∈ N.

*. U'#g 5at$e5at"a( #03"to#, S$ow t$at

13+ 23+ 33+……… ..+n3=

n2(n+ 1)2

4 .

o*: *et

! #n$ % 13+23+33+……… ..+n3=

n2(n+1)2

4 .

Ste%-1: !ut n%1L.7. =.7.

¿n3=n

2 (n+1)2

4

% #1$ %1. 2

2

4

%1 % 1L.H.SR.H.S 8% +1 ' tr3e for #1.


@

t n=n3

t k =k 3

3


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; p (k )=¿ 13+23+33+……… ..+n3=

n2(n+1)2

4


p (k +1 )=¿ 13+23+33+……… ..+k 3+(k +1 )

3=

k 2(k +1)2

4 >(k +1)3

1

%

k 2(k +1)2+4 (k +1)3

4

%(k +1)2{k 2+4 (k +1 ) }

4

%(k +1)2{k 2+4 k + 4 }

4

%(k +1)2(k +2)2

4

∴Thus p (k +1 )is true for n=k +1 .


. 1.*.9*..9..79///..3% to # ter5'n ( n+ 1 ) (n+ 2 )(n+3)

4 .

o*: *et


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p (n )=1.2.3+2.34+…+n (n+1 ) (n+2)=n (n+1 ) (n+2)(n+3)

4 .

Ste%-1: !ut n%1L.7. =.7.

n (n+1)(n+2)n (n+1) (n+2 )(n+3)

4 .

%1.2. %1.2.3.4

4

%? %?L.H.SR.H.S 8% +1 ' tr3e for #1.


; p (k )=1.2.3+2.3.4+…+k (k +1 ) (k +2 )=n (n+1 ) (n+2)(n+3)

4


#D>1$%1.2.>2..8>.> k (k +1)(k +2)+(k +1)(k +2)( K +3)

% k (k +1 ) (k +2)(k +3)4 >

(k +1)(k +2)(k +3)1

%

k (k +1) (k +2) (k +3)+ 4(k +1)( k +2)( k +3)4

%( k +1 ) ( k +2 ) ( k +3 ) {k +4 }

4


E


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. *.9.9.79/3% to # ter5'n(n2 +6 n+11)

3 .

o*: let

p (n )=2.3+3.4 +…+(n+1 ) (n+2 )=n(n2+6 n+11)

3 .

Ste%-1: !ut n%1L.7. =.7.

n

n(¿¿ 2+6 n+11)3

(n+1 )(n+2)¿

%2. %1.(1+6+11)

3

%? %18

3=6

L.H.SR.H.S 8% +1 ' tr3e for #1.

Ste%- 2: let us assumethat p (k ) istrue forn=k .

; p (k )=¿

k

k (¿¿ 2+6 k +11)3

2.3+3.4+…+(k +1 )(k +2)=¿


#D>1$ % 2.3+3.4+…+(k +1 ) (k +2)+( k +2 ) ( k +3)

1F


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k

k (¿¿2+6 k +11)3

+(k +2 )(k +3)

¿ ¿

%

k

k (¿¿ 2+6 k +11)3

+ (k +2 )(k +3)

1¿

%

k

k (¿¿ 2+6 k +11)+3 ( k +2 )(k +3)3¿

%k 3+6 k 2+11k +3 (k 2+5 k +6)

3

%k

3+6 k 2+11k +3 k 2+ 15 k +183

%k

3+9 k 2+26 k + 183

%( k +1 )(k 2+8 k +18)

3

%( k +1 )(k 2 +2 k +1+6 k +6+11)

3

%

k +1¿¿¿

(k +1 ) ¿¿



7. a+( a+d )+….+up ¿nterms=n[2 a+(n−1)d ]

2 .

11


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sol : let p (n )=a+( a+d )+….+a+( n−1) d=n[2 a+(n−1)d ]

2 .

Ste%-1: !ut n%1

L.7. =.7.a+(n−1 ) d

n[2 a+(n−1)d ]2

%a>F %1. [2 a+0 ]

2 %2 a

2

%a % aL.H.SR.H.S 8%+1 ' tr3e for #1.


; p (k )=a+(a+d )+…+a+(k −1 ) d=k

2{2 a+(k −1)d }


; p (k +1 )=a+ (a+d )+…+ {a+(k −1 ) d }+(a+kd )

¿ k 2

{2 a+(k −1)d } > (a+kd)

¿ k {2 a+(k −1 ) d }

2+{a+kd }

1

%2 ak +k 2 d−kd+2 a+2 kd

2

%

2 a (k +1 )+k 2 d+kd

2

%2 a (k +1 )+kd(k +1)

2

%( k +1 ){2 a+kd }

2

12


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%( k +1 ){2 a+(k +1−1)d }

2


by the principle of mathematical induction, p (n) 5s true for a** n ∈ N.

?. a+ar+a r* +…up¿nterms=a (rn−1 )(r−1 )

sol : let

p (n )=a+ar+a r2 +…a.rn−1

=a(r n−1)

(r−1)

Ste%-1: !ut n%1L.7. =.7.

a . rn−1

a(rn−1)(r−1)

%a.rF % a r−1r−1

%a % a

L.H.SR.H.S 8% +1 ' tr3e for #1.


; p (k )=a+ar+a r2 +…a.rk −1=a

(rk −1 )(r−1 )

.


p (k +1)=a+ar+a r2 +…a.rk −1+a . rk

¿a( rk −1 )(r−1)

+a . rk

1


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%a ( rk −1 )(r−1 )

+r

k

1 H

% a( rk −1+rk .r −rk

r−1 )

% a(r

k +1−1r−1

) .



;. 1*9

1¿¿¿

9** 9 1

2+ 22+¿¿

*9 /3% to # ter5'

n ( n+ 1)2(n+2)12 .

o*: *et ! #n$ % 12>

1¿¿¿

>22$ > 1

2+ 22+¿¿

2$ > ..

+n (n+1 )(2 n+1)6 %

n ( n+1 )2(n+2)12 .

Ste%-1: !ut n%1L.H.S R.H.S

n ( n+ 1)(2 n+1)

6

n ( n+ 1 )2(n+2)

12.

%1.2.3

6 %1 (1+1 )2 (1+2 )

12 %12

12

%1 %1L.H.SR.H.S 8% +1 ' tr3e for #1.

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%( k +1 )(2 k 2+13 k +24)

24

%

( k +1 )(2 k 2+4 k +2+9 k + 9+13)

24

%( k +1 ){2 ( k +1 )2+9(k +1)+13 }

24

#D>1$ is true7ence, b0 t"e !rinci!*e of +at"e+atica* in-uction, t"e given

state+ent is true for a** n ∈ N.

=. S$ow t$at1

1.4+

1

4.7+

1

7.10+…uptonterms=

n

3 n+1 .

o*: let

p (n )= 1

1.4+

1

4.7+

1

7.10+…+

1

(3 n−2)(3 n+1)=

n

3 n+1 .

Ste%-1: !ut n%1L.H.S R.H.S

1(3 n−2)(3 n+1)

n3 n+1

%1

(3−2 )(3+1)1

3+1

%1

1.4 %1

1.4

L.H.SR.H.S 8% +1 ' tr3e for #1.

Ste%-

2: let us assumethat p (k ) istrue forn=k .

; p (k )=

1

1.4+

1

4.7+

1

7.10+…+

1

(3 k −2)(3 k +1)=

k

3 k + 1 .


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p (k +1 )= 1

1.4+

1

4.7+

1

7.10+…+

1

(3 k −2 ) (3 k +1 )+

1

(3 k +1 ) (3 k +4 )

¿ k

3 k +1 >

1

(3 k + 1)(3 k +4 )

%1

(3 k +1){k

1+

1

3 k +4}

%1

(3 k +1){

3 k 2+4 k +13 k + 4

}

% 1(3 k +1) { 3 k 2

+3 k +1 k +13 k + 4 }

%1

(3 k +1) {

3 k (k +1)+1(k +1)3 k +4

}

%1

(3 k +1){(3 k +1)( k +1)

3 k + 4}

%(k +1)(3 k +4)

%(k +1)

{3 (k +1 )+1 }

#D>1$ is true7ence, b0 t"e !rinci!*e of +at"e+atica* in-uction, t"e given

state+ent is true for a** n ∈ N.

1>. S$ow t$at 49n+16 n−1 isdivisible by 64 for all+ve integersn .

o*: let p (n )=bethestatement

Ste%-1: !ut n%1

#1$ % 491+16 (1)−1 %8E>1?1

%?8

1


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%?8#1$Thus p (1)isdivisibleby 64.

Ste%-*: let us assumethat p (k )istrueforn=k .

; p (k )=49k +16 k −1=64 m, for some m∈ N

? 49k =(64 m−16 k +1) /. +1

Ste%-: f #@91

; p (k +1 )=49k +1+16(k +1)−1

% 49k .49+16 k + 16−1

%8E 64 m−16 k +1 H >1?D>1<

¿49.64 k −16 k .49+49+16 k +15

¿49.64 k −16 k .48+64

¿49.64 k −16 k .12 × 4+64

¿ 49.64 k −12 k .64 × 4 +64

%?88E+12D>1H

t"us t"e state+ent is -ivisib*e b0 ?8

7ence, by the principleof mathematicalinduction,

49n+16 n−1 isdivisible by64 for alln∈ N .

11. S$ow t$at 3.52 n+1+23 n+1 is divisibleby 17 for all +veintegers n .

1E


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o*: let p (n )=bethestatement

Ste%-1: !ut n%1

#1$ % 3.52 n+1+23 n+1 % 3.5

3+24

% 3(125)+16

¿375+16

¿ 391=17 ×23

Thus p (1)isdivisibleby 17.

Ste%-*: let us assumethat p (k )istrueforn=k .

; p (k )=3.52 n+1+ 23n +1=17 m,for somem∈ N

? 3.52 n+1=(17 m−23 n+ 1) //.. +1

Ste%-: if n%D>1

; p (k +1)=3.52 n+1+23 n+1

% 3.52 k +1

.52+23 k +1. 23

% 25(17 m−23 n+1)+23 k + 1 .8

%2


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3.52 n+1+23 n+1 isdivisibleby17 for alln∈ N .

1*. S.T *. 42 n+1+33 n+1 ' 0&'b(e b2 11, for a(( n∈ N .

o*: o*: let p (n )=bethestatement

Ste%-1: !ut n%1

#1$ % 2.42 n +1+33 n+1 % 2.4

3+34

¿2(64)+81

¿ 128+81

¿201=11×19

Thus p (1)isdivisibleby 11.

Ste%-*: letus assumethat p (k )istrue for n=k .

; p (k )=2.42 k +1 +33 k +1=11m, forsomem∈ N

? 2.42 k +1=(11 m−23 k +1 ) //.. +1

Ste%-: if n%D>1

; p (k +1 )=2.42 (k +1 )+1+33 (k +1)+1

% 2.42 k +1

. 42+33 k +1 .33

% 16(11m−33 k +1)+33 k +1 .27

%1?.11+ 1?. 33 n+1+27.33 n+1

%1?.11+ >11. 33 n+1

%11#2?+ 33 n+1

$

21


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t"us t"e state+ent is -ivisib*e b0 11


2. 42 n+1+33 n+1 isdivisibleby 11 for alln∈ N .

1. *9.*9 . 22+…upto n terms=n. 2n

o*: *et ! #n$ % 2>.2> . 22+…+(n+1 ) 2n−1=n . 2n.

Ste%-1: !ut n%1L.7. =.7.

2 n .2n.

% 2 % 1.2

%2 %2

L.H.SR.H.S 8 % +1 ' tr3e for #1.


;! #D$ % 2>.2> . 22+…+(k +1 ) 2k −1=k . 2k .


;! #D>1$ % 2>.2> . 22

+…+(k +1 ) 2k −1

+ (k +2 ) 2k

¿k . 2k + (k +2 ) 2k

% 2k (k +k +2)

22


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% 2k (2 k +2)

% 2k .2(k +1)

% 2k +1

.(k +1)



LAQ (2 ×714 )

1. .T |b+c c+a a+bc+a a+b b+ca+b b+c c+a|=2|

a b c

b c a

c a b|

o*: L.7.

|b +c c+a a+bc+a a+b b+ca+b b+c c+a|

!1→ !

1+ !

2+ !

3

¿|2(a+b+c ) 2 (a+b+c ) 2 ( a+b+c )

c+a a+b b +ca+b b+c c+a |

¿2|(a+b+c ) (a+b+c ) (a+b+c )

c+a a+b b+ca+b b+c c+a |

2


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!2→ !

2− !

1: !

3→!

3− !

1

¿2|(a+b+c ) (a+b+c ) (a+b+c )−b −c −a−c −a −b |

!1→ !

1+ !

2+ !

3

¿ 2| a b c

−b −c −a−c −a −b|

%

−¿¿

−¿|a b c

b c a

c a b|2 ¿

% 2

|a b c

b c ac a b|

=.7.

2. |a b c

b c a

c a b|

2

=|2 bc−a2 c2 b2

c2

2 ac−b2 a2

b2

a2

2 ab−c2|

¿(a2+b2+c2−3 abc)2

!

(¿¿ 2" !3)

#. $ .|a b c

b c a

c a b|2

=|a b c

b c a

c a b||a b c

b c a

c a b|¿

28


25/72

Jr.MATHEMATICS

AIMS TUTO I P E

−¿|a b c

c a b

b c a|

¿|a b c

b c a

c a b|¿

¿|a b c

b c a

c a b||

−a −b −cc a b

b c a |

¿|−a

2

+bc+bc −ab+ab+c2

−ac+b2

+ac−ab+c2+ab −b2+ac+ac −bc+bc+a2

−ac+ac+b2 −bc+a2+bc −c2+ab+ab|

¿|2 bc−a2

c2

b2

c2

2ac−b2 a2

b2

a2

2 ab−c2|….(1)

no%|a b cb c a

c a b|2

¿ [a (bc−a2 )−b (b2−ac)+c (ab−c2)]2

¿(abc−a3−b3+abc +abc−c3)2

¿(3 abc−a3

−b3

−c3

)2

¿(a3+b3+c3−3 abc )2…… (2)

¿ (1 )∧(2)

2


26/72

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AIMS TUTO I P E

#. $ .= ! .$ .

. &f |a a2

1+a3

b b2

1+b3

c c2

1+c3|=0∧|a a

21

b b2

1

c c2

1|'0,

sho% thatabc=−1.

ol :(iven

|

a a2

1+a3

b b2

1+b3

c c2

1+c3

|=0

⇒|a a2

1

b b2

1

c c2

1|+|a a

2a

3

b b2

b3

c c2

c3|=0

⇒

|

a a2

1

b b2

1

c c2

1

|+abc

|

1 a a2

1 b b2

c c c2

|=0 {)

1")

2}

⇒|a a2

1

b b2

1

c c2

1|−abc|a 1 a

2

b 1 b2

c 1 c2|=0 {) 2") 3}

⇒

|

a a2

1

b b2

1

c c2 1

|+abc

|

a a2

1

b b2

1

c c2 1

|=0

⇒|a a2

1

b b2

1

c c2

1|(1+abc)=0 [∴|a a

21

b b2

1

c c2

1|' 0]

2?


28/72

Jr.MATHEMATICS

AIMS TUTO I P E

¿ 2(a+b+c)|1 a b

0 (a+b+c) 00 0 (a+b +c)|

¿2(a+b+c)3|1 a b

0 1 0

0 0 1| *+pandingalong) 1

¿ 2 ( a+b+c )3 [1(1−0 )+0+0 ]

¿ 2(a+b+c)3


30/72

Jr.MATHEMATICS

AIMS TUTO I P E

¿|0 a2−b2 a3−b3

0 b2−c2 b3−c3

1 c2

c3 | ¿|0 (a−b )(a+b) (a−b )(a

2+ab+b2)

0 (b−c )(b+c) (b−c )(b2+bc+c2)1 c

2c

3 | ¿(a−b)(b−c )|0 (a+b) (a

2+ab+b2)

0 (b+c) (b2+bc +c2)1 c

2c

3 | !

2→ !

2− !

1:❑

¿(a−b)(b−c )

|0 a+b a2 +ab+b2

0

(c−a) b

2

+bc+c

2

−a

2

−ab−b

2

1 c2

c3

|

b2+bc+c2−a2−ab−b2

¿bc+c2−a2−ab

⇒b(c−a)+(c−a)(c+a)⇒(c−a)(a+b+c )

¿(a−b)(b−c )

|0 a+b a2+ab+b2

0 (c−a) (c−a)(a+b+c)1 c

2c

3 |

¿(a−b)(b−c )(c−a)|0 a+b a2+ab+b2

0 1 a+b+c1 c

2c

3 | *+pandingalong) 1 ¿(a−b)(b−c )( c−a ) 1[ (a+b ) (a+b+c )−a2−ab−b2]

¿(a−b)(b−c )( c−a ) {a2+ab+ac+ab+b2+bc

−a2−ab−b2}

¿(a−b)(b−c )(c−a)( ab+bc+ca) .

F


31/72

Jr.MATHEMATICS

AIMS TUTO I P E

@.

.T

|a b c

a2

b2

c2

a3 b3 c3

|=abc (a−b)(b−c)( c−a)

ol :|a b c

a2

b2

c2

a3

b3

c3|

¿abc

|

1 1 1

a b c

a2

b2

c2

|

) 1→)

1−)

2:)

2→)

2−)

3

¿abc| 0 0 1a−b b−c ca

2−b2 b2−c2 c2|

¿abc| 0 0 1

a−b b−c c(a−b )(a+b) (b−c )(b+c) c2|

¿abc (a−b)(b−c)| 0 0 1

1 1 c

(a+b) (b+c) c2| *+pandingalong !

1

¿abc (a−b)(b−c)1 {b+c−a−b }

¿abc (a−b)(b−c)( c−a)

1


32/72

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AIMS TUTO I P E

. find + if | +−2 2 +−3 3 +−4 + −4 2 +−9 3 +−16 +−8 2 +−27 3 +−64|=0.

sol :| +−2 2 +−3 3 +−4 +−4 2 + −9 3 +−16 +−8 2 +−27 3 +−64|=0

!2→ !

2− !

1: !

3→!

3− !

1

⇒| +−2 2 +−3 3 +−4−2 −6 −12−6 −24 −60 |=0.

⇒| +−2 2 +−3 3 +−4

1 3 6

1 4 10 |=0.

⇒ ( +−2) [ 30−24 ]−(2 +−3) [10−6 ] +(3 +−4 ) [ 4−3 ]=0

⇒ ( +−2) [ 6 ]−(2 +−3) [ 4 ] +(3 +−4 ) [1 ]=0

⇒6 +−12−8 ++12+3 +−4=0

⇒ +−4=0

2


33/72

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AIMS TUTO I P E

∴ +=4.

E. &f A=[

a1 b1 c1a

2 b

2 c

2

a3

b3

c3]isanon−singular marti+,

theni+ , t provethat A is invertible∧ A−1=adA

detA .

ol : A=

[

a1

b1

c1

a2

b2

c2

a3 b3 c3

]∧adA=

[

A1

A2

A3

B1

B2

B3

) 1 ) 2 ) 3

]

No% A.adA=[a

1 b

1 c

1

a2

b2

c2

a3

b3

c3

].[ A

1 A

2 A

3

B1

B2

B3

) 1

) 2

) 3

]

¿

[

a1 A

1+b

1B

1+c

1)

1

a2 A1+b2 B1+c2) 1

a3 A1+b3 B1+c3) 1

a1 A

2+b

1B

2+c

1)

2

a2 A2+b2 B2+c2) 2

a3 A2+b3 B2+c3 ) 2

a1 A

3+b

1B

3+c

1)

3

a2 A3+b2 B3+c2 ) 3

a3 A3+b3 B3+c3 ) 3

]

¿[detA 0 0

0 detA 0

0 0 detA ]


34/72

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AIMS TUTO I P E

¿detA [1 0 0

0 1 0

0 0 1]=detA.&

∴ A . adA

detA = &

imilarly %ecan provethat adA

detA = &

∴ A−1=

adA

detA

.

1F. 5f A% [ 1 −2 30 −1 4

−2 2 1 ] , t"en 9n- #AI$1.o*: A% [

1 −2 30 −1 4

−2 2 1 ]

;AI% [ 1 0 −2−2 −1 2

3 4 1 ]

JAIJ% 1|−1 2

4 1|+0

|−2 2

3 1|+3

|−2 −1

3 4 |% 1 (−1−8 )+0 (−2−6 )+3(−8+3)

% −9+¿ F >1F%1

8


35/72

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AIMS TUTO I P E

A-j #AI$ %

[

−1 24 1

−2 −13 4

0 −2−1 2

1 0

−2 −1

]% [

(−1−8) (6 +2) (−8+3)(−8+0) (1+6) (0−4)(0−2) (4−2) (−1−0)]

% [−9 8 −5−8 7 −4−2 2 −1]

% [−9 −8 −2

8 7 2

−5 −4 −1]A1%

1

| A- |ad ( A- )=1

1 [−9 −8 −2

8 7 2

−5 −4 −1]

¿[−9 −8 −2

8 7 2

−5 −4 −1]

11.

olve the follo%ing euationsbyusing )ramer / srule method .

ol : let A=[3 4 5

2 −1 85 −2 7] ,0 =[

+

y

1 ]∧B=[18

13

20]


36/72

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AIMS TUTO I P E

2=|3 4 5

2 −1 85 −2 7| ¿ 3|−1 8−2 7|−4|2 85 7|+5|2 −15 −2|

¿ 3(−7+16)−4(14−40)+5(−4+ 5)

¿3(9)−4(−26)+5(1)

¿ 27+10+5

¿136 '0 cramer- sruleaplicable

21=|

18 4 5

13 −1 820 −2 7| ¿18|−1 8−2 7|−4|13 820 7|+5|13 −120 −2|

¿18(−7+16)−4(91−160)+5(−26+20)

¿18(9)−4(−69)+5(−6)

¿ 162+ 276−30

¿408

22=|

3 18 5

2 13 8

5 20 7| ¿ 3|13 820 7|−18|2 85 7|+5|2 135 20|

¿3(91−160)−18(14−40)+5(40−65)

¿ 3(−69)−18(−26)+5(−25)

¿−207+468−125

¿ 136

23=

|

3 4 18

2 −1 135 −2 20

|=3

|

−1 13

−2 20|−4

|

2 13

5 20

|+18

|

2 −1

5 −2|

¿ 3(−20+26)−4( 40−65)+5 (−4+5)

¿3(6)−4(−25)+5(1)

¿ 18+100+18

?


37/72

Jr.MATHEMATICS

AIMS TUTO I P E

¿136

⇒ +=2

1

2

=408

136

, y=2

2

2

=136

136

, 1=2

3

2

=136

136

∴ +=3, y=1∧ 1=1.

12. ol : let A=[2 −1 31 1 1

1 −1 1] , 0 =[ +

y

1 ]∧B=[9

6

2]

2=

|2 −1 31 1 1

1 −1 1

|=2

| 1 1

−1 1|+1

|1 1

1 1

|+3

|1 1

1 −1|

¿2(1+1)+1(1−1)+3(−1−1)

¿ 2(2)+1( 0)+3 (−2)

¿4−6=−2'cramer - srule aplicable.

21=

|9 −1 3

6 1 12 −1 1|

=6| 1 1

−1 1|+1

|6 1

2 1|+3

|6 1

2 −1|

¿ 9(1+1)+1(6−2)+3(−6−2)

¿9(2)+1(4 )+3(−8)

¿ 18+4−24=22−24=−2

22=

|2 9 3

1 6 1

1 2 1|=2|

6 12 1|−

9

|1 11 1|+

3

|1 61 2|

¿ 2(6−2)−9(1−1)+3(2−6)

¿2(4 )+1(0)+3 (−24)

@


38/72

Jr.MATHEMATICS

AIMS TUTO I P E

¿8−12=−4

23=

|2 −1 9

1 1 61 −1 2|

=2| 1 6

−1 2|+1

|1 6

1 2

|+9

|1 1

1 −1|

¿2(2+6)+1(2−6)+9 (−1−1 )

¿2(8)+1(−4)+9 (−2)

¿16−4−18

¿ 16−22=−6

⇒ +=2

1

2 =−2

−2=1, y=

22

2 =−4

−2 =2, 1=

23

2 =−6

−2=3

∴ +=1, y=2∧ 1=3.

1. A=[1 1 1

2 5 7

2 1 −1] , 0 =[ +

y

1 ]∧B=[ 9

52

0 ]

2=|1 1 1

2 5 7

2 1 −1|=1|5 71 −1|−1|2 72 −1|+1|2 52 1| ¿ 1(−5−7)−1(−2−14)+1(2−10)

¿1(−12)−1(−16)+1(−8)

¿−12+16−8=−4 'cramer - sruleaplicable .

21=|

9 1 1

52 5 7

0 1 −1|=9|5 71 −1|−1|52 70 −1|+1|52 50 1|


39/72

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AIMS TUTO I P E

¿9(−5−7)−1(−52−0)+1(52−0)

¿ 9(−12)−1(−52)+1(52)

¿−108+52+52

¿−108+104=−4

¿22=|

1 9 1

2 52 7

2 0 −1|=1|52 70 −1|−9|2 72 −1|+1|2 522 0 | ¿ 1(−52−0)−9 (−2−14)+1(0−104)

¿1(−52)−9(−16)+1(−104) ¿−52+144−104

¿−156+144=−12

23=|1 1 9

2 5 52

2 1 0|=1|5 521 0 |−1|2 522 0 |+9|2 52 1|

¿1(0−52)−1(0−104)+9(2−10)

¿ 1(−52)−1(−104)+9 (−8)

¿−52+104−72

¿−124 +104=−20

⇒ +=2

1

2 =

−4−4

=1, y= 2

2

2 =

−12−4

=3, 1= 2

3

2 =

−20−4

=5

∴

+=1,

y=3

∧ 1=5.

18. A=[ 2 −1 3−1 2 1

3 1 −4] , 0 =[ +

y

1 ]∧B=[8

4

0]

E


40/72

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AIMS TUTO I P E

2=| 2 −1 3−1 2 1

3 1 −4|=2|2 11 −4|+ 1|−1 13 −4|+3|−1 23 1| ¿2(−8−1)+1(4−3)+3(−1−6)

¿2(−9)+1(1)+3(−7)

¿−−18+1−21

¿−38

21=

|

8 −1 34 2 1

0 1 −4

|=8

|

2 1

1 −4|+1

|

4 1

0 −4|+3

|

4 2

0 1

|

¿8(−8−1)+1(−16−0)+3(4−0)

¿ 8(−9)+1(−16)+3 (4)

¿−72−16+12 ¿−76

22=

| 2 8 3

−1 4 13 0 −4

|=2

|4 1

0 −4|−8

|−1 1

3 −4|+3

|−1 4

3 0

| ¿ 2(−16−0)−8( 4−3)+3 (0−12)

¿2(−16)−8(1)+3(−12)

¿−32−8−36 ¿−76

23=

|

2 −1 8−1 2 4

3 1 0

|=2|2 41 0|+1|−1 43 0|+8|−1 23 1|

0−12+8(−1−6)0−4+1¿

¿2¿

¿ 2(−4)+1(−12)+ 8(−7)

8F


41/72

Jr.MATHEMATICS

AIMS TUTO I P E

¿−8−12−56 ¿76

⇒ +=2

1

2

=−76

−38=2, y =

22

2

=−76

−38=2, 1=

23

2

=−76

−38=2

∴ +=2, y=2∧ 1=2.

¿method :

1


42/72

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AIMS TUTO I P E

¿136

A−1=adA

detA =

1

136

[

9 −38 3726 −4 −14

1 26 −11

]

⇒ 0 =¿ A1 .B

¿ 1

136 [ 9 −38 3726 −4 −141 26 −11] [

18

13

20]

¿ 1

136

[

162−494+740468−52−280

18+ 338+220

]

¿ 1

136 [408

136

136]∴ +=3, y=1∧ 1=1.

(b) .ol : let A=[2 −1 31 1 1

1 −1 1] ,0 =[ +

y

1 ]∧B=[9

6

2]

A0 =B⇒ 0 = A−1. B

¿ A1 ¿ adA

detA

detA=|2 −1 31 1 1

1 −1 1|=2| 1 1−1 1|+1|1 11 1|+3|1 11 −1| ¿2(1+1)+1(1−1)+3(−1−1)

¿ 2(2)+1( 0)+3 (−2)

¿4−6=−2' 0 A−1∃ .

82


43/72

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AIMS TUTO I P E

AdA=[ A

1 B

1 )

1

A2 B2 ) 2 A

3 B

3 )

3](

1 1

−1 11 1

1 −1−1 3

1 1

2 −11 1

)cofactor ¿[

(1+ 1) (1−1) (−1−1)(−3+1) (2−3) (−1+2)(−1−3) (3−2) (2+1) ]

¿[ 2 0 −2−2 −1 1−4 1 3 ]

AdA=[cofactor ]T =[ 2 −2 −40 −1 1−2 1 3 ]

A1 ¿ adA

detA =

1

−2 [ 2 −2 −40 −1 1

−2 1 3 ]⇒ 0 = A−1 .B

¿ 1

−2 [ 2 −2 −40 −1 1

−2 1 3 ][9

6

2]

¿ 1

−2 [18−12−8

0−6 +2−18+ 6+6 ] ¿ 1−2 [

−2−4−6]

∴ +=1, y=2∧ 1=3

8


44/72

Jr.MATHEMATICS

AIMS TUTO I P E

1?. A=[1 1 1

2 5 7

2 1 −1] , 0 =[ +

y

1 ]∧B=[ 9

52

0 ]

detA=|1 1 1

2 5 7

2 1 −1|=1|5 71 −1|−1|2 72 −1|+1|2 52 1| ¿1(−5−7)−1(−2−14)+1(2−10)

¿ 1(−12)−1(−16)+1(−8)

¿−12+16−8=−4 ' 0. A−1 ∃ .

AdA=

[ A

1 B

1 )

1

A2 B2 ) 2 A

3 B

3 )

3](5 7

1 −1

2 5

2 11 1

5 7

1 1

2 5 )cofactor ¿[

(−5−7) (14 +2) (2−10)(1+1) (−1−2) (2−1)(7−5) (2−7) (5−2) ]

¿

[−12 16 −8

2 −3 12 −5 3 ]

AdA=[cofactor ]T =[−12 2 2

16 −3 −5−8 1 3 ]

A1 ¿ adA

detA =

1

−4

[

−12 2 216 −3 −5

−8 1 3

]

⇒ 0 =¿ A1 .B

¿ 1

−4 [−12 2 2

16 −3 −5−8 1 3 ][

9

52

0 ]

88


45/72

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AIMS TUTO I P E

¿ 1

−2 [−108+104+0144−156+0−72+52+0 ]

¿ 1

−2 [ −4−12−20]∴ +=1, y=3∧ 1 =5.

1@. A=[ 2 −1 3−1 2 1

3 1 −4] , 0 =[ +

y

1 ]∧B=[8

4

0]

detA=| 2 −1 3−1 2 1

3 1 −4|=2|2 11 −4|+1|−1 13 −4|+3|−1 23 1| ¿2(−8−1)+1(4−3)+3(−1−6)

¿2(−9)+1(1)+3(−7)

¿−18+1−21

¿−38' 0. A−1∃ . AdA =

[ A

1 B

1 )

1

A2 B2 ) 2 A

3 B

3 )

3],cofactor ¿

( 2 1

1 −4

−1 2

3 1−1 3

2 1

2 −1−1 2 )

cofactor ¿[(−8−1) (3−4) (−1−6)(3−4) (−8−9) (−3−2)

(−1−6) (−3−2) ( 4−1) ]

¿

[−9 −1 −7−1 −17 −5−7 −5 3 ]

8


46/72

Jr.MATHEMATICS

AIMS TUTO I P E

AdA =[cofactor ]T =[−9 −1 −7−1 −17 −5−7 −5 3 ]

A 1 ¿ adA

detA =

1

−38 [−9 −1 −7−1 −17 −5−7 −5 3 ]

⇒ 0 =¿ A1 .B

¿ 1

−38

[

−9 −1 −7−1 −17 −5

−7 −5 3

][

8

4

0

]

¿ 1

−38 [ −72−4+ 0−8−68+0

−56−20+0 ]

¿ 1

−38 [−76−76−76 ]∴ +=2, y=2∧ 1=2.

1;. S$ow t$at t$e fo((ow#g '2'te5 of e3ato#' '

"o#''te#t a#0 'o(&e t "o5%(ete(2. ++ y+ 1=3,2 ++2 y− 1=3, ++ y− 1=1 K

sol : theaugmented ¿

[ A3 ]=[1 1

2 2

1 1

1 3

−1 3−1 1]

onapplying !2→!

2−2 !

1,!

3→ !

3− !

1

8?


47/72

Jr.MATHEMATICS

AIMS TUTO I P E

[1 1

0 0

0 0

1 3

−3 −3−2 −2]

anapplying !3→3 !

3−2 !

2

[1 1

0 0

0 0

1 3

−3 −30 0

] .. #1$Co+!aring wit" ec"e*on for+

Nu+ber of nonero rows in A are 2ranD #A$ %2

Nu+ber of nonero rows in AM are 2 ranD #AM$ %2

=anD #A$ %ranD #AM$ %2 /"e s0ste+ is consistent an- "as in9nite*0 +an0 so*utions.e write e6uiva*ent set of e6uations fro+ #1$

++ y+ 1=3 . #2$

−3 1=−3 .. #$

4 =1

$ence1=1, ++ y=2

#et +=k ⇒ y=2−k , 1=1, D ∈ ! is the solution set .

1


48/72

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AIMS TUTO I P E

[ A3 ]=[ 2 −1−1 2

3 1

3 8

1 4

−4 0 ] O [ 2 −1−1 2

3 1

3 8

1 4

−4 0 ]oninterchanging !

1∧ !

2

%etransformtheabove¿

anupper triangular ¿

onapplying !2→!

2+2 !

1 , !3→ !3+3 !1

[−1 2

0 3

0 7

1 4

5 16

−1 12] anapplying !

3→3 !

3−7 !

2

[−1 2

0 3

0 0

1 4

5 16

−38 −76] #1$Co+!aring wit" ec"e*on for+

Nu+ber of nonero rows in A are ranD #A$ %Nu+ber of nonero rows in AM are ranD #AM$ %

7ence ranD #A$ %ranD AM$' %. /"us t"e s0ste+ "as a uni6ue so*ution.e write t"e s0ste+ of e6uations fro+ #1$

− ++2 y+ 1=4 #2$

3 y+5 1=16 .. #$

−38 1=−76 . #8$

¿ (4 )

1=2¿ (3 )

8


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AIMS TUTO I P E

⇒3 y=16−10=6

y=2 y=2, 1=2∈(1 )⇒ +=2

+= y= 1=2is the solution.

1=. So(&e t$e fo((ow#g '2'te5 of e3ato#' ++ y+ 1=1,2 ++2 y+3 1=6, ++4 y+9 1=3 5

o*: the augmented ¿

[ A3 ]=[1 1

2 2

1 4

1 1

3 6

9 3] 6[

1 0

0 1

0 0

0 +

0 y

1 1 ]

onapplying !2 →!2−2 !1 ,!3→ !3− !1

¿[ 1 1

2−2 2−21−1 4−1

1 1

3−2 6−29−1 3−1]

% [1 1

0 0

0 3

1 1

1 4

8 2] !2" !3

% [1 1

0 3

0 0

1 1

8 2

1 4] !1→ 3 !1− !2

% [3−0 3−3

0 3

0 0

3−8 3−28 2

1 4 ]

% [3 0

0 3

0 0

−5 18 2

1 4]

!1→ !

1+5 !

3,!

2→!

2−8 !

3

8E


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AIMS TUTO I P E

¿[3+ 0 0+00−0 3−0

0 0

−5+5 1+ 208−8 2−32

1 4 ]

% [3 0

0 3

0 0

0 21

0 −301 4

] !1" [ !13 ] , !2" [ !23 ]

% [1 0

0 1

0 0

0 7

0 −101 4

] +=7, y=−10, 1=4

*>. +− y+3 1=5,4 ++2 y− 1=0,− ++3 y+ 1=5 5

o*: theaugmented¿

[ A3 ]=[ 1 −14 2

−1 3

3 5

−1 01 5

] 6[1 0

0 1

0 0

0 +

0 y

1 1 ]onapplying !2 →!2−4 !1 ,!3→ !3+ !1

¿[ 1 −14−4 2+ 4

−1+1 3−1

3 5

−1−12 0−201+3 5+5 ]

% [1 −10 6

0 2

3 5

−13 −204 10

] !2" [ !32 ]

% [1 −10 6

0 1

3 5

−13 −202 5

] !2→ !2−5 !3


51/72

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AIMS TUTO I P E

% [ 1 −1

0−0 6−50 1

3 5

−13−10 −20−252 5

]

% [1 −10 1

0 1

3 5

−23 −452 5

] !1→ !1 + !2 ,!3 → !3− !2

¿[1+ 0 −1+1

0 1

0−0 1−1

3−23 5−45−23 −45

2+23 5+45 ] % [1 0

0 1

0 0

−20 −40−23 −45

25 50 ]

!3" [

!3

25 ]

% [1 0

0 1

0 0

−20 −40−23 −45

1 2 ] !1→ !1 +20 !35 !2→!2+ 23 !3

¿

[1+0 0+0

0 10 0

−20+20 −40+40

−23+23 −45+461 2 ] %

[1 0

0 10 0

0 0

0 11 2]

+=0, y=1, 1=2

*1. 2 +− y+ 3 1=9, ++ y + 1=6, +− y+ 1=2 5

o*: theaugmented¿

[ A3 ]=

[2 −1

1 11 −1

3 9

1 61 2]

6

[1 0

0 10 0

0 +

0 y1 1 ]

1−¿ !1" !¿

!2


52/72

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AIMS TUTO I P E

¿[2−1 −1−1

1 1

1 −1

3−1 9−61 6

1 2 ] ¿[

1 −21 1

1 −1

2 3

1 6

1 2]

onapplying !2→!

2− !

1, !

3→!

3− !

1

¿[ 1 −2

1−1 1+21−1 −1+2

2 3

1−2 6−31−2 2−3 ] ¿[

1 −20 3

0 1

2 3

−1 3−1 −1]

!2→ !

2−2 !

3

¿[ 1 −2

0−0 3−20 1

2 3−1+2 3+2

−1 −1 ] ¿[1 −20 1

0 1

2 31 5

−1 −1]

!1→ !

1+2 !

2, !

3→!

3− !

2

¿

[1+ 0 −2+2

0 10−0 1−1

2+2 3+10

1 5−1−1 −1−5]

¿

[1 0

0 10 0

4 13

1 5−2 −6]

!3" [ !32 ]

¿[1 0

0 1

0 0

4 13

1 5

1 3 ]

!1→ !

1−4 !

35!

2→ !

2− !

3


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AIMS TUTO I P E

¿[1 0

0 1

0 0

4−4 13−121−1 5−3

1 3 ] % [

1 0

0 1

0 0

0 1

0 2

1 3] +=1, y=2, 1=3

1. &f A , B ,) are anglesof a triangle , prove that

cos2 A +cos2 B+cos2 ) =−1−4 cosAcosBcos).

o*: L7% cos2 A+cos2 B+cos2 )

{∴cos) +cos3=2cos () + 3

2) cos (

) − 32

) H

%2cos ( 2 A+ 2 B2 )cos(2 A−2 B

2 )+cos 2)

%2cos ( A +B ) cos ( A−B )+cos 2)

A+B+) =8

A +B=8 −)

Cos ( A+B )=cos (8 −) )

Cos ( A +B )=−cos)

% 2 cos)cos ( A−B )+cos 2)

cos2 ) =2cos2

) −1

¿−2cos)cos ( A−B )+2cos2 ) −1

% −2 cos) [cos ( A−B )−cos) ]−1


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AIMS TUTO I P E

% −2 cos) [cos ( A−B )+cos ( A +B) ]−1

∴cos ( A−B )+cos ( A+B )=2 cosAcosB

% −2 cos) [ 2cosAcosB ]−1

%18 cosAcosBcos) !$

*. &f A +B+) =2700, 9 . T

cos2 A +cos2 B+cos2 c=1−4 sinAsinBsin).

o*: L7% cos2 A+cos2 B+cos2 )

{∴cos) +cos3=2cos () + 3

2) cos (

) − 32

) H

%2cos ( 2 A+ 2 B2 )cos(2 A−2 B

2 )+cos 2)

%2cos ( A +B ) cos ( A−B )+cos 2)

A +B+) =3 8

2

A +B= 3 8 2

−)

Cos

( A +B )=cos (3 8

2−) )


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AIMS TUTO I P E

Cos ( A +B )=−sin)

% 2 sin)cos ( A−B )+cos2 )

cos2 ) =1−2sin2)

¿−2 sin)cos ( A−B )+ 1−2sin2)

% −2

sin) [cos

( A−B )+sin) ]+1

% −2 sin) [cos ( A−B )−cos ( A +B) ]+1

∴cos ( A−B )−cos ( A +B )=2 sinAsinB

% −2 sin) [2 sinAsinB ]+1

% 1−4 sinAsinBsin) !$

. If A, B, C are angles of a triangle, provethat

sin2 A

2+sin2 B

2−sin2 )

2=1−2cos A

2cos B

2sin )

2 .

o*: L7% sin2 A

2+sin2

B

2−sin2

)

2


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AIMS TUTO I P E

% sin2 A

2+1−cos2

B

2−sin2

)

2

%1# cos2 B2 −sin

2 A2 $ sin2 ) 2

∴cos2 B

2−sin2

A

2 %cos ( A+B2 )cos ( A−B2 )

¿1−cos( A+B2 )cos( A−B2 )−sin2 ) 2

A +B+) =8

A+B2

= 8

2−

)

2

Cos( A+B )

2=cos( 8 2 −) 2 )=sin ( c2 )

¿ 1−sin (c

2)cos( A−B2 )−sin2 ) 2

% 1−sin

)

2[cos( A−B2 )+cos( A+B2 )]

% 1−sin

)

2[2cos

A

2cos

B

2]

% 1−2cos A

2cos

B

2sin

)

2 .=7


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AIMS TUTO I P E

. If A, B, C are angles of a triangle , prove that

cos2 A+cos2 B−cos2 ) =1−2 sinAsinBcos) .

o*: cos2 A+cos2 B−cos2 )

¿ cos2 A +1−sin2 B−cos2 )

% 1+cos2 A −sin2 B−cos2 )

cos2 A−sin2 B=cos ( A +B) cos ( A−B)

%1> cos ( A+B ) cos ( A−B )−cos2)

A +B+) =8

A+B=8 −)

Cos ( A +B )=cos (8 −) )

Cos ( A +B )=−cos)

%1 cos) cos ( A−B )−cos2)

%1) [¿cos ( A−B )+cos) ]

cos¿

% 1−cos) [cos ( A−B )−cos ( A+B)]


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AIMS TUTO I P E

cos ( A−B )+cos ( A+B )=2 sinAsinB

%% 1−cos) [ 2 sinAsinB ]

%12 sinAsinBcos) !$

7. If A, B, C are angles of a triangle, provethat

sinA+sinB−sin) =4sin A

2sin

B

2cos

)

2 .

o*: L7% sinA+sinB−sin)

{∴sin) +sin3=2sin () + 3

2)cos (

) − 32

) H

%2 sin

( A+B

2

)cos

( A−B

2

)−sin)

A +B+) =8

A+B2

= 8

2−

)

2

∴sin ( A+B )

2=sin( 8 2 −) 2 )

¿ cos (c

2 )

% 2cos

)

2cos( A−B2 )−sin) &s ¿ 2 A=2 sinAcosA '


59/72

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AIMS TUTO I P E

% 2cos

)

2cos( A−B2 )−2sin ) 2 cos ) 2

% 2cos )

2[cos(

A−B2 )−

sin ) 2]

% 2cos

)

2[cos( A−B2 )−sin ) 2 ]

∴cos( A+B )

2=sin (

c

2) H

% 2cos

)

2[cos ( A−B2 )−cos (

A +B )2

]

% 2cos

)

2[2sin

A

2sin

B

2]

% 4 sin

A

2sin

B

2cos

)

2

=7

1. &f a

=13,b

=14, c

=15 , .T

!=65

8,r=4, r1=

21

2, r2=12,

r3=14

sol : Given a=13,

b=14,

c=15

a+b+c

2=

13+14+152 %

42

2=21


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AIMS TUTO I P E

2=√ (s ) ( s−a ) (s−b )(s−c)

% √ (21) (21−13 ) (21−14)(21−15)

% √ (21 ) (8 ) (7 )(6)

% √ 7.3.4.2.7.3.2

% √ (7.7 ) . (3.3 ) .(4.4)

% √ (7.3.4)2

%8

!=abc

4 2 =

13.1.4.15

4.84=

65

8

r=2

s =

84

21=4

r1= 2

(s−a )= 84

21−13=84

8 =21

2

r2=

2

(s−b )=

84

21−14=

84

7=12

r3= 2

(s−c )= 84

21−15= 84

6=14

*. If r1,r

1=2, r

2=3, r

3=6 finda,b,c.

o*: given r%1,r

1=2, r

2=3, r

3=6

e Dnow t"at

¿√ 1.2.3.6=√ 6.6=√ 62 %?

%6

1=6

¿6

1 %?

a=6−6

2

=6−3=3

b=6−6

3=6−2=4.

c=6−6

6=6−1=5.

?F

b=s−2

r2

a=s− 2

r

= 2r

2=√ r .r 1 . r2. r3


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AIMS TUTO I P E

. 5fr

1=8, r

2=12, r

3=24 find a, b, c.

o*: givenr

1=8, r

2=12, r

3=24

e Dnow t"at1

r=

1

r1

+ 1

r2

+ 1

r3

1

r=

1

8+

1

12+

1

24

1

r=

3+2+124

= 6

24=

1

4⇨r=4

¿√ 4.8.12.24 ¿√ 4.4.2.12.12.2

¿√ 42. 22 . 122 %8.2.12%.12%E?

=96

4

¿ 96

4 %28

a=24−96

8=24−12=12

b=24−96

12=24−8=16.

c ¿ 24−

96

24 =24−4=20

?1

b=s−2

r2

a=s− 2

r

=2

2=√ r .r1 . r2. r3


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AIMS TUTO I P E

8. . /r

1

bc+

r1

ca+

r1

ab =

1

r−

1

2 !

o*:r

1

bc+

r1

ca+

r1

ab

% :r

1

bc

¿ :ar

1

abc

¿ :2 ! sin A . . tan A

2

abc

%

A

2cos

A

2

2sin ¿¿2 !

abc :¿

%4 !

abc :(sin2

A

2)

%

2:(

1−cos A2

)

%1

2 r [ 1−cos A+1−cos B+1−cos ) ]

%

)

cos A+cos B+ cos ¿3−¿

1

2 r ¿

%

1

2 r [3−(1+4sin

A

2sin

B

2sin

)

2)]

% 12 r

[2−

4 ! sin A

2

sin B

2

sin )

2 !

]

?2

a%

r1=s . tan

tan A

2=

sin

cossin A=2sin

A

2cos

A

2

r=4 ! sin A

sin B

sin

)

2=

1

rsin

2 A

2=

1−c2

4 !=


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AIMS TUTO I P E

%1

2 r [2− r ! ] % 12 r .2 − 12 r . r ! ¿ 1r

− 1

2 !


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AIMS TUTO I P E

21. −4 ! cos A

2sin

B

2cos

)

2

22. %

4 ! cos )

2{sin

A

2cos

B

2−cos

A

2sin

B

2}

2.

28.

2


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AIMS TUTO I P E


66/72

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AIMS TUTO I P E

. %(a+b+c )2

a2+b2+c2

E.

EF. &f 9

1, 9

2,9

3are thealtitudes dra%n ¿

E1.

vertices A, B , c ¿ theopposite sidesof a

E2.

triangethen .T ( i ) 1

91

+ 1

92

+ 1

93

=1

r

E. ( ii )

1

91 +

1

92 −

1

93=

1

r3

(iii) 93.9

3.9

3=

8 23

abc =

(abc)2

8 !3

E8.E


67/72

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AIMS TUTO I P E

121. %

4 !2

sin A sin B−[4 !2 cos2 ) 2 ](2sin A2 cos

122. %

4 !2

sin A sin B−[4 !2 cos2 ) 2 ] (sin A ) (sin B)

12. % 4 !

2sin A sin B [1−cos2 ) 2 ]

128. % 4 !2

sin A sin B

[sin

2 )

2 ]122

!

1. o*:18. L.7. a

cos A

2+b cos

B

2+c cot

)

2

1


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AIMS TUTO I P E

18?.

+sin2 B+sin 2 Bsin2 A ¿

s+ !

2¿

18@.

18.

sinBsin)

4 sinA ¿

s + !

2¿

18E.

1


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AIMS TUTO I P E

1E.1E8.1E


70/72

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AIMS TUTO I P E

28F. ¿ !a2

S [sin (2 B )+sin (2 ) ) ]

281. ¿ !a2

S

[ 2 sinBcosB+ 2 sin)cos) ]

282. ¿ a2

S

[ (2 !sinB ) cosB+(2 !sin) )cos) ]

28.

288. ¿ a2

S [ (b ) cosB+() )cos) ]

28

b2 [ccos) +acosA ]+c2 [acosA +b cosB]

28@.

28. % ab [ acosB+bcosA ]

>

bc [bcos) +c cosB ]+ca [acos) +ccosA ]

28E.

2 bc [a ]+ca [b ]

2


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AIMS TUTO I P E

2F.

21.

22.

2.

28.

2


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