BMM10233 Chapter 5 Inequalities
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Transcript of BMM10233 Chapter 5 Inequalities
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W g po d ug , , d , u quFo g qu w = qud quo
There are certain Rules of inequality
W u ddd o ud o o d o qu g o qu o g
d o
d
W v u dvdd o upd o o d o qu g o qu
Fo g d
dx y
k k
>
W -v u dvdd o upd o o d o qu g o qu g vd
Fo g d
dx y
5y
>
Inequalities
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2Similarly if it is x 3x + 2 < 4x 8
(x 1 ) ( x 2 ) < 4x 8
( x 1 ) (x 2 ) < 4 (x 2 )
2
2
2
Hence we get x 1 < 4 is wrong because we do not
know the sign of x 2. So correct way is
x 3x + 2 < 4x 8
x 3x +2 4x + 8 < 0
x 7x
+ 10 < 0 and then solve for x
2
2
2 2
2 2
ax bx c 0 then
px + qx +r
ax + bx + c 0 and px + qx + r > 0
or ax + bx + c 0 and px + qx + r < 0
+ +
2
2
2 2
2 2
ax bx c 0 then
px + qx +r
ax + bx + c 0 and px + qx + r > 0
or ax + bx + c 0 and px + qx + r < 0
+ +
Inequality with algebraic expression
o w
w
ax b
0px q
+>+ d p q o d p q
Bu w gv 9 AND o vu o w qu u w o gg o o
gv qu
Howv ouo o 9 OR vu w g 9 , , up o d vu
w , , up o gv o ud
o AND 9 w w v 9 u AND intersectiond o OR 9 w w u u
OR union
Mathematical Statements
M po vovg o =, , , , , d
Inequality
A w o g o qu o o d qu o o qu , , , Io wod, po vovg o o , , , quEp: 7 v
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Identity
A op w d u o p o v o v d d
A op w d u o pu vu o v d quo
Fo p: = = 3
5x= v =
Replacement set and the solution set
Cod qu 7 T qu u, w p v u , , , , , , , -, -,
T o w vu o o d p I ov p o g
p T o vu o o p w gv d ouo
ouo = {, , , , }
Absolute value of a real number
L u o o o T po o u w opod o u
T d o po o po d ou vu o d w o-gv,
ou vu o u pov o gv w o- gv T ou vu o u w
| | ||= , | | = w o | |= , ; | |= , Tu ||= , , | |= = T ou vu o , d ou vu o o- zo u w pov
Rules for working with inequalities
Basic properties of inequalities
I ,
I > & > , h >
I > , h qui or sur o oh h sis, os o h iquli
I > , h posiiv qui mulipli or ivi o oh sis os o h iquli I > 0, h >
a bx x>
I > h giv qui mulipli or ivi o oh h sis woul rvrs h iquli I < 0, h <
a bx x , h > or + > 0 or >
I > , >
&
>
, h +
+
> +
+
I > & > , h + > +
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I > & < , h >
9 I < & > , h <
0 I > , h >
Ia bc d< h < , i , > 0
1
a 2 if a > 0a+
a b
ab , if a and b are both + ve2
+
I < r o sm sig h
1 1
a b
Thus > & < 0 i < < 0h s o ll possil vlus sisig h giv iquli is : 0 < : - < < 0 = { : - < < }Thus ll h pois o h umr li lig w will sis h giv iquli
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Example: 3.
Fi h s {R / | | < }Sol:
Cs i L 0 Thror | | = I his s | | < <
< Thus < i < Cs ii wh < 0 i <
I his s | | = =
| | < < < > Thus > < i < <
Th s o ll possil vlus o sisig h giv iquli is
{R/ < < }{R/ < } = {R/ < < }
Example: 4.
Fi h s susiuig { R / < | | < }Sol:Cs i wh < | |
I 0, h | | = < | | < > I < 0, h | | = < | | < > < Thus ll vlus o < ll vlus o > will sis h iquli < | |
Thus, h s o ll rl vlus o sisig < | | is {R/ < } U {R/ >}Cs ii wh | | <
I 0, h | | = | | < < 0 < Furhr i < 0, h | | = | | < - < > < < 0
Thus {R/ < < 0}{R/0 < } h s o ll vlus o sisig h iquli < | | < is giv {R/ > }{R/ < }{R/ < < }= {R/ < < }{ R/ < < }
Example: 5.
Solv: + | | =
Sol:
Cs i wh 0, I his s || = h giv quio is + = = = Bu 0 = is lso soluio iss i his s
Cs ii l < 0I his s | | =
So, h giv quio is = or =
or =
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Graphical representation on the number line
A umr li hs rl umrs mrk o i
i Th igrs r mrk o h li s show low
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
I , h > or = Soluio s = { / > } U { / = } = { / > or = } = { / }Thror, soluio s is {, , }
-4 -3 -2 -1 0 1 2 3 4
ii I < , soluio s = {/ < } = {, 0,,,,, }
iii I > <
soluio s = { / >}{ / < } = { / > < }= {/ < < }= {, 0, }
Example: 6.
Ii o h umr li, h pois , whih sis h oiio | |, whr is igr
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Sol:
Suppos, 0, h h iquli| | will ru o 0 This ms h ll h pois o h umr li ligw 0 iluig 0 , will sis his pr o h iquli
Now l < 0, h ||= I < + , h iquli is rvrs hgig h sigSi < 0, w hv < < 0
ll h pois lig w iluig h pois will sis h iquli ||< hsoluio s is {, , , , 0, , , , , }
Example: 7.
For h iquli < , whr N, h rplm s is h s o url umrs, i, NN = { , , , , }
Sol:
Th iquli < is ru, whr is rpl , , , or Thror {, , , , } is h soluio s
No: For h iquli >
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i I rplm s = { , , , , 9}, h soluio s = { , , 9}
ii I rplm s = {0, , , , }, h soluio s = {, }
Example: 8.
Lis h soluio o < giv h is posiiv igrSol:
< + <
<
<
8x 4 1x
8 8 2
> >
Th soluio s = {, , , , }N
Example: 9.
Fi h soluio s o h iquli + > 0 whr is giv igr
Sol:
-4 -3 -2 -1 0 1 2 3 4
126x 12 x x 2
6
> > >
Thror h soluio s is {-}
Example: 10.
I P is h soluio s o > + Q is h soluio s o + whr N Fi h s P QSol:
> + , > + , i, > or3
x , x 13> >
+ , i, + or + 0
20
x or x 54
x 1 and x 5> Th soluio s o P = { , , , , , , } h soluio s Q = { , , }
P Q {5, 6, 7, 8, 9, ...} N. =
Example: 11.
Solv: + < rprs h soluio grphill Z
Sol:
Th iquli + < hs wo prsi +
or 8
x, 4 x.2
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ii + <
<
< or <
, h giv iquli rus o < Thror, h soluio s = {,,,, 0, }
Th ollowig is h grphil rprsio o h soluio s
-5 -4 -3 -2 -1 0 1 2 3 4 5
Example: 12.
Fi h soluio s o h iquli1 3
x 3x 2 , where x N4 4+ +
Sol:
3 1x 3x 2
4 4
1 52x 2 or 2 x
2 2
5x
4
1or, x 1
4
Si hr is o url umr lss h1
14
, h soluio s is mp
Example: 13.
Fi h soluio s o h iqulix 2 x 3
, where x N3 2
>
Sol:
> > 9
>9 + , > or < Thror h soluio s is {, , , }
Example: 14.
I h rplm s is { : Z} Fi h soluio s o
i >
Rplm s is {,,,,,, 0, , , , , , }
11 1
2x 11 or x i.e. x 5 . The solution set is {6}2 2
> > >
ii + + +
- or or2
x3
Thror, h soluio s is {,,,,,}
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Example: 15.
I P is h soluio s o > + Q is h soluio s o + 9 , whr N, i h s P QSol:
S P : > +
> +
> , > 62
, > Thror, h soluio s is P = { , , , , }
s Q : + 9 + 9
or 184
, i, Q = {, , , }
Thror P Q = {, , , , 9 }
Example: 16.
I is giv igr, i h soluio s o2 1
(x 1) 0.5 5+ + >
Sol:
x 1 2 x 2 1or
5 5 5 5 5 5
x 3 3x 5, i.e., x 3. Therefore, the solution set is { 2, 1}
5 5 5
+ > >
> > >
Example: 17.
I W s o whol umrs, solv h iquli < 0Sol:
< 0, W = {0, , , , , , , , , }
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Cs :x 4 x
2.2 3+
5x 122 or 5x 12 12, 5x 24
6
24 4 4or x x 4 or 4 x5 5 5
4Hence 4 x 6.
5
This ms h lis w4
45
Thror, h soluio s is {, }
Example: 19.
I h rplm s = {, , , , , }, lis h soluio ss o i < ii 0 < Sol:
i <
This ms, lis w is ilu, u i is o i h rplm s { , , }ii 0 < 0 < , < ,
52
< or >52
+ or 9 i, 9
2
52
< 9
2
Thror lis w52
9
2
9is included.
2
Thror, h soluio s is {, }
Example: 20.
I h rplm s = { / Z}, lis h soluio ss o i + < ii Sol:
i 2+ <
< 9, < < soluio s = {,, 0, , }
ii i, < + 9
33
9
3
Thror Th soluio s is {, , }
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No: Th iquli < 0 whr < is sisi i < < h iquli > 0 is sisi
i h irvls < s wll s >
Example: 21.
I m hv rl vlu i whih is grr 3 21x 1 or x x.+ +
Sol:
( ) ( )3 2 3 2x 1 x x x x x 1 .+ + =
( )( )2x 1 x 1=
( ) ( )2x 1 x 1= +
Now ( )2
x 1 is posiiv,
3x 1+ > or 2x x+
Aorig + is posiiv or giv, h is orig s > or < I = h i quli om, quli
Example: 22.
Fi miimum vlu o(a x) (b x)
c x
+ ++
Pu c x y+ =
This will l us o
2(a c) (b c)
y a c b c 2 (a c) (b c)y
+ + +
squr rm = 0Wh (a c) (b c)=
miimum r vlu is a c b c 2 (a c) (b c) + + orrspoig vlu o x (a c) (b c) c=
Example: 23.
Fi h grs vlu o 2 3x y wh 3x 4y 5+ =
L 2 3p x y ,= lrl p, is h Prou o ors suh h wo o hm r qul o h rmiig r qul o
Now, 3x 4y 5+ =
3x 4y
2( ) 3( ) 52 9+ =
3x 3x 4y 4y 4y5.
2 2 3 3 3= + + + + =
usig wigh A M G M iquli
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1
52 3
3x 4y2 3 3x 4y2 3 ( ) ( )
5 2 3
+
12 3 2 3516 3( x y ) 1 x y .
3 16
or mimum o3 3 3x y .
16=
Example: 24.
Fi soluio s o tan x 1 tan x| 3 3 | 2
Sol:
3| y | 2,
y wh tan xy 3 0= >
3 3y 2 or y 2y y
2 2y 2y 3 0 or y 2y 3 0 +
si > 0, hror , y 3 or y 1
tan x tan x3 3 or 3 1
tan x 1 or tan x 0
m z 1 1m , m
4 2
+ +
1m , m 1
2
+ +
Example: 25.
I h quio 4 3 2x 4x ax bx 1 0 + + + = hs our posiiv roos, i
Sol:
L , , our roos o h giv quio
h 4 + + + =
1 =
A M o , , ,
G.M of , , , .=
= = =
1 = = = =[ ] 4 + + + =
4 3 2 1 3 2x 4x ax bx 1 x 4x bx 4x 1 + + + = + +
a b and b 4 = =
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I + < 0, h Ass Co - 0000
< < 1
2< <
4 5x
5 4<
5 4x
4 5> > No o hs
Solv or i /+ / 0 Ass Co - 0000
> Ass Co - 0000
< < / =/ 1 1
x or x 1
x2
0 > 0 No soluio
9 A Rl umr is si o lgri i i sis Polomil quio wih igrl Whih o h ollowig is olgri? Ass Co - 00009
2 3 2 0
Practice Exercise - 1
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0 I x [x], x( 0), ER,= Whr [] is h grs igr lss h or qul o , h h umr o soluio o +
1f ( ) 1
x = r Ass Co - 0000
0 iii
I p, q, r r rl umr, h Ass Co - 000 m p, q = m p, q, r
mi p, q =1
2p + q |p q|
m p, q =1
2p + q |p q|
m p, q < mi p, q, r
m p, q =1
2p + q |p q|
I is rl h prssio ks ll rl vlus p hos whih li w , h r Ass Co -000
, , , , ,
I1
x 0, 0 and x 1x
> > + is lws o giv, h h ls vlu o is Ass Co - 000
1
2 0
1
4
3
4
For posiiv vlu o 2 4sin cos + lis i h irvl; Ass Co - 000
[, ] 3
, 14
1 5
,4 16
[ , ] No o hs
Th lows irvl or whih 12 9 4x x x x 1 0 + + > is Ass Co - 000
4 x 0 < 0 x 1< < 100x 100 < 0 x< < 3 x 0< <
SCORE SEET
Use HB pencil only. Abide by the time-limit
9
0
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For o-giv rl umrs suh h 1 2 na a ... a p+ + + = i jq a a= , h i < j Ass Co - 000
21
q p
2
21
q p
4
> p
q
2
> 2
pq
2
> 21q p
8>
Fi h grs vlu o 3 4(a x) (a x)+ or Rl vlu o umrill lss h Ass Co - 000
3 4 7
8
6 8 a
7
2 2 7
2
6 8 a
7
3 4 7
9
6 8 a
7
3 4 8
8
6 8 a
7
3 4 7
7
6 8 a
7
I r posiiv quiis , > h Ass Co - 000
1 1
1 1 + > +
1 1
1 1 + = +
1 1
1 1 + < +
1 1
1 1 + +
No o hs
I , , r i P > h Ass Co - 0009
n n na c b+ > n n na c 2b+ > n n na c 2b+ = p, o , , n n na c 2b+
I > , whih o h ollowigs rss s irss? Ass Co - 0000
i 2x x+ ii 24x x ii1
1x
+
I ol II ol
III ol Boh I II All o h ov
I h rplm s = { , , 0, , , } lis h soluio s o 2 3x 24 x< < Ass Co - 000
{, {, , } , } , ]
2 3
4
(x 1) (x 1)0
x (x 2)
+
Ass Co - 000
1 x 2 < < 1 x 2 < 1 x 2 1 x 2 < 1 x
For posiiv rl umr , , whih o h ollowig hol Ass Co - 000
2 2 2a b c bc ca ab+ + > + +
(b c) (c a) (a b) abc+ + +
a b c 3b c a
+ +
3 3 3a b c abc+ + >
2 2 2a b c bc ca ab+ + = + +
Practice Exercise - 2
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9 I 2f (x) ax bx c, 0= + + < h
I o isuss
b 1 0+ >
is giv rl is posiiv rl
b 1 0+