Real Numbers: Key Formulae

REAL NUMBERS
This chapter comes Under paper-I. From this

chapter, 2 Marks Questions-1 (12 = 2M), 1 Marks
Questions-2, (21= 2M), 4 Marks Questions-2
(24=8M) and 1/2 mark questions-6 (61/2=3M)
altogether 15 Marks we can score easily by spending
less time for preparation. The following material will
be helped alot in the public exams for S.S.C students.
Key formulae:
a
m
.a
n
= a
m+n
;
a = 1, where a 0
1. Solve 2
x+3
= 4
y2
and 3
x2
= 9
3y2x
sol:
2
x+3
= 4
y2
2
x+3
= (2
2
)
y2
2
x+3
= (2)
2(y2)
2
x+3
= (2)
2y4
If bases are equal powers also equal
x+3=2y 4
x2y = 4 3
x2y = 7 ......... (1)
given 3
x2
= 9
3y2x
3
x2
= (3
2
)
3y2x
3
x2
= 3
2 (3y2x)
3
x2
= 3
6y 4x
BAEPSE
x 2 = 6y 4x
x+4x 6y = 2
5x6y = 2.......... (2)
from (1) & (2)
x 2y = 7 (3)
5x 6y = 2 (1)
______________
3x 6y = 21
5x 6y = 2
+
_____________
2x = 23
x = 23/2
sub x value in (1)
x 2y = 7
2. If a
2
= 0.04 find a
3
.
sol:
a
2
= 0.04
a=2/10
Cubing On Both Sides
a
3
= (2/10)
3
a
3
=8/1000
a
3
=0.008
3. If (x
2/3
)
p
= x
2
find p?
sol:
(x
2/3
)
p
= x
2
x
2p/3
= x
2
BAEPSE
4
a
100
2
4
a
100
23 37
x , y
2 4

37 37
y y
2 2 4

2y
37
2
14 23
2y
2

23
2y 7
2

23
x 2y 7
2

n
n
n
a a
b b
( )
n
n n n
a a; a a
( )
n
m mn
a a
n m mn
a a
n 1/ 2 1/ n m m/ n n
a a ; a a ; a a
n n n
a b ab; a b ab
m; m
m m
1 1
a a
a a

( )
m
m m
ab a .b
m
m
m
a a
b b
j \
, (
( ,
n m
1
if m n
a

<
m
m n
n
a
a if m n
a
>
www.sakshieducation.com
2p/3 = 2
p = 3
3. If a
x
=b, b
y
= c, c
z
=a, show that xyz = 1
sol:
a
x
=b, b
y
= c, c
z
=a
consider a
x
= b
(c
z
)
x
= b
(c
zx
) = b
(b
y
)
zx
= b
b
yzx
= b
1
yzx = 1
xyz = 1
4.
If then show that
sol:
SOBS
5. If a
1/3
+ b
1/3
+c
1/3
= 0 then show that (a+b+c)
3
= 27 abc.
Sol:
a
1/3
+ b
1/3
+ c
1/3
= 0
a
1/3
+ b
1/3
= c
1/3
........(1)
(a
1/3
+ b
1/3
)
3
= ( c
1/3
)
3
a + b + 3.a
1/3
b
1/3
( c
1/3
) = c
a + b 3a
1/3
b
1/3
c
1/3
= c
a + b + c = 3a
1/3
b
1/3
c
1/3
(a+b+c)
3
= (3a
1/3
b
1/3
c
1/3
)
3
(a+b+c)
3
= 27 abc
6. If show that 3y
3
9y = 10
sol:
y
3
= 3+3
1
+ 3. 3. (y)
3y
3
= 10 + 9y
3y
3
9y = 10
7. If lmn=1, Prove that
sol:
lmn=1 l=1/mn l
-1
=1/mn
mn
.............(1)
mn 1 n
+ +
mn 1
1
m
n
j \
, (
, (
, (
( ,
l
l
m
1
m 1
n

+ +
m
m m 1
+ + l
1
1
1 1 1
a
1
a 1 m
1
m
j \

, (
+ + ( ,
+ +
l
l
1 1 1
1 1 1
1
1 m 1 m n 1 n

+ +
+ + + + + + l l
3
9 1 9y
y
3
+ +
3
1
y 3 3y
3
+ +
3
1
y 3 3(1).y
3
+ +
from(1)
( ) ( )
3
3 3
a b a b 3ab a b + + + +
3 1/ 3
y 3
( )
3
1/ 3
3
+
( )
3
( )
1/3 1/3 1/3 1/3
3.3 .3 3 3

+ +
3
3 1/3 1/3
y 3 3
, ]
+
]
1/3 1/3
y 3 3 ...........(1)
+
1/3
1/3
1
y 3
3
+
3
3
1
y 3
3
+
3
3
1
y 3
3
+
( )
3
3 1/ 3
a b c 3 . a + +
( )
3
1/ 3
. b
( )
3
1/ 3
. c
( )
3
from(1)
( )
3
3 3
a b a b 3ab(a b) + + + +
1/ 3
a
( )
3
1/ 3
b +
( )
3
( )
1/3 1/3 1/3 1/3 1/ 3
3a .b a b c + +
( )
3
1
1
x a a
2
, ]

]
2
1 a 1
x
2 a a
, ]

, ]
]
2
1
x [a 1]
2a

2
2ax a 1
2
2ax 1 a
2 2
a 2ax x +
2
x 1 +
( )
( )
2
2
a x x 1 +
2
a x x 1 +
2
a x x 1 + +
( )
1
1
x a a
2

2
a x x 1 + +
2p 2 3
L.H.S.= (1)+(2)+(3)
= 1 = R.H.S.
Modules of a Real number:
The modules or absolute value of a real number ''x'' is
denoted by ''|x|'' and defined as
|x| = x if x > 0
= x if x < 0
= 0 if x = 0
e.g. |3| = 3
|7| = 7
Note: Modules of a real number is never negative
e.g. |x| = 5 solution doesn't exists.
Absolute Value equations & Inequations:
|x| = a solution is x = a or x = a
|x| a solution is a x a
|x| < a solution is a < x < a
|x| a solution is x a or x a
|x| > a solution is x > a or x < a
8. Solve |2x 5| = 7
Sol:
|2x 5| = 7
|x| = a solution is x = a or x = a
|2x 5| = 7 solution is 2x 5 = 7 or 2x 5 = 7
2x = 7 + 5 or 2x = 7 + 5
x = 6 or x = 1
9. Solve |5x1| 0.
Sol:
|5x1| 0
Since Modules of a real number is never negative
|5x1| 0 exists
|x| = 0 solution is x = 0
|5x1| = 0 solution is 5x 1 = 0
5x = 1
x = 1/5
10.Solve
Sol:
|x|a solution is a x a
solution is
subtract 5
multiply with 3
18 x 12
18 x 12
12 x 18
11. Solve |9 4x| > 4
Sol:
|9 4x| > 4
|x|>a solution is x > a or x < a
|9 4x| > 4 solution is 9 4x > 4 or 9 4x < 4
4x > 4 9 or 4x < 4 9
4x > 5 or 4x < 13
4x < 5 or 4x > 13
x < 5/4 or x > 13/4
x > 13/4 or x < 5/4
Limit of a function:
f(x) is any function as x a f (x) l then 'l' is
called as limit of f as x tends to 'a', This is written as
and read as "Limit of f as x tends to a is l''
If the limit of function approaches to zero, then the
function is called "infinitesimal"
12.Evaluate
Sol:
43 = 1
3
4
1

( )
2
3
4
1

2
x 1
3
Lt 4
x
x a
Lt f (x)
l
x
6 3 3 4 3
3

x
6 4
3

x
1 5 5 5 1 5
3

x
1 5 1
3

x
5 1
3

x
5 1
3

x
5 1
3

1
2
6
x 12
1
or 2
1
x 2
mn n 1 + +
mn 1 n + +
mn n 1
mn 1 n n mn 1 1 n mn
+ +
+ + + + + +
1
mn 1
1
mn

j \
, (
, (

, (
( ,
l
l
l
1
1 1
............(3)
1 n mn 1 n

+ + + + l
n
..............(2)
n mn 1
+ +
1
1 1
1
1 m n
1 m
n

+ +
+ +
13.Evaluate
Sol:
which is indeterminate form
14.Find the limit of the series
Sol:
is an infinitesimal Geometric
Progression
.
Here a = 1,
15.Evaluate
Sol:
which is indeterminate form
16. Show that
Sol:
= Which is indeterminate form.
( )
2 1 3 3
2 2 1 2 1 2
+

( )
x 2
x 1
Lt
x x 1
+
( )
x 2
x 2
Lt
( ) ( )
(x 1)
x x 1 x 2
+

, ]
, ]
, ]
]
( )
( )( )
x 2
x x 2 1(x 2)
Lt
x x 1 x 2
, ] +
, ]

, ]
]
( )( )
2
x 2
x 2x x 2
Lt
x x 1 x 2
, ]
+
, ]

, ]
]
( )( )
2
x 2
x x 2
Lt
x x 1 x 2
, ]

, ]

, ]
]
( )
( )( )
x 2
x x 1 2
Lt
x x 1 x 2
, ]
, ]

, ]
]
( )( )
x 2
1 2
Lt
x 2 x x 1 x 2
, ]

, ]

, ]
]
1 2
0 0
1 2
0 0
( )( )
1 2
2 2 2 2 1 2 2

( )( )
x 2
1 2
Lt
x 2 x x 1 x 2
, ]
, ]

, ]
]
( )( )
x 2
1 2
3
Lt
2
x 2 x x 1 x 2
, ]

, ]

, ]
]
1 1 1
1 1 2 1 1

+ +
2
1 0
1 0 0 1
+
+ + +
( )
( )
x 0 2
1 x
Lt
1 x x 1
+
+ + +
x 0
x
Lt
( ) 1 x
x
+
( )
2
1 x x 1 + + +
x 0
1
Lt
2
x x 1 + +
( )
2
x 1 x x 1 + + +
( )
2
2 2
x 0 2
1 x x 1
Lt
x 1 x x 1
, ]
+ +
, ]
]
+ + +
2 2
2 x 0
1 x x 1 1 x x 1
Lt
x
1 x x 1
+ + + + +

+ + +
2
1 0 0 1 0 0 0
0 0 0
+ +

2
x 0
1 x x 1
Lt
x
+ +
2
x 0
1 x x 1
Lt
x
+ +
1 1 2
1
3/ 2 3
1
2

+
1
1
1
2
j \
, (
( ,
a
S
1 r
2
1
t 1/ 2 1
r 1
t 1 2
<
1 1 1
1 ......
2 4 8
+ +
1 1 1
1 ......
2 4 8
+ +
2
x 0 x 0
x 5x x
Lt Lt
x
+

(x 5)
x
+
x 0
Lt x 5 0 5 5
+ +
2
0 5(0) 0
0 0
+
2
x 0
x 5x
Lt
x
+
2
x 0
x 5x
Lt
x
+
17. Evaluate
Sol:
which is indeterminate form.
Note:
1.
2.
18. Evaluate
Sol:
0/0 which is indeterminate form.
Note:
as
19.Evaluate
which is indeterminate forms
Note:
1.
2.
3.
4. The limiting position of a secant of a circle is tangent
5.
6.
7.
8. 16
1.25
= 32
9. (a
1/3
b
1/3
)(a
2/3
+a
1/3
.b
1/3
+b
2/3
) = ab
10.|3x+1| = 0 solution is x = 1/ 3
( )
4
2/3 1/3 1/ 4 2
a a . a a
, ]
, ]
]
3
x 3
x 27
Lt 27
x 3
x 0
2x 3
Lt 2 / 3
3x 5
+
+
( )
x
x x
If x x x then x 9 / 4
( )
1/ n
x 0
1 x 1
Lt 1/ n
x
+
( )
n
x 0
1 x 1
Lt n
x
+
( ) 4 3. 0 4
2 3(0)

+ 2
1
2
1
as x 0 0
x

x
1
4 3.
x
Lt
1
2 3.
x
+
x
x
Lt
( ) 4 3/ x
x
( ) 2 3/ x +
x
4x 3
Lt
2x 3
+
( )
( )
4 3
3
2 3 3

+ +
x
4x 3
Lt
2x 3
+
n
1 1
x 0 0
x x

15/ 4
1
4a
15
4
1
. a
4
1 16
4
1
.a
4
n n
n 1
x a
x a
Lt n.a
x a
1
1/ 4 1/ 4
4
4
4 4
x a
x a 1
Lt .a
4 x a
1/ 4 1/ 4
4 4
a a
a a
1/ 4 1/ 4
4 4
x a
x a
Lt
x a
1/ 4 1/ 4
4 4
x a
x a
Lt
x a
m m
m n
n n
x a
x a m
Lt .a
n x a
n n
n 1
x a
x a
Lt n.a
x a
1
2 2a
1
2a 2a
+
1
a a 2a
+ +
x a
1
Lt
x a 2a
+ +
x a
x a
Lt
( ) x a
( )
x a 2a + +
( )
( )
x a
x a 2a
Lt
x a x a 2a
+ +
( ) ( )
( )
( )
2 2
x a
x a 2a
Lt
x a x a 2a
+ +
x a
x a 2a x a 2a
Lt
x a x a 2a
+ + +

+ +
2a 2a 0
0 0
a a 2a
a a
+
x a
x a 2a
Lt
x a
x a
x a 2a
Lt
x a
Assignment:
1. If x
1/2
= 0.2 then find x
3/2
2. If 64
x
= 1/256
y
then find the value of 3x+4y.
3. Solve |312x| = 0
4. If a
x
=b
y
=c
z
= d
w
and ab = cd then S.T
5. If a
x
= b
y
= c
z
;
then show that
6. Solve
7. Solve
8. Evaluate
9. Evaluate
10.Evaluate
( )
( )
4
3
x 0
1 x 1
Lt
1 x 1
+
+
a a
x p
x p
Lt
x p
x 0
1 x 1
Lt
x
+
m
1 9
3

2x 1
5
5
y 2z
x x z
+
b c
a b
1 1 1 1
x y z w
+ +

Real Numbers: Key Formulae

Uploaded by

Copyright:

Available Formats

Real Numbers: Key Formulae

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Real Numbers: Key Formulae

Uploaded by

Copyright:

Available Formats

REAL NUMBERS

This chapter comes Under paper-I. From this

You might also like