Wednesday, January 29, 2014
I
IMPORTANT FACTS AND FORMULAE I
1.
LAWS OF INDICES:
(i)
am x an = am
+ n
(ii) am / an = am-n
(iii)
(am)n = amn
(iv)
(ab)n = anbn
(v) ( a/ b )n = ( an / bn
)
(vi)
a0 = 1
2. SURDS: Let a be a rational
number and n be a positive integer such that a1/n = nsqrt(a)
is irrational. Then nsqrt(a) is called a surd of order n.
3. LAWS
OF SURDS:
(i) n√a
= a1/2
(ii) n √ab = n √a * n √b
(iii) n √a/b = n √a / n
√b
(iv) (n √a)n = a
(v) m√(n√(a)) = mn√(a)
(vi) (n√a)m = n√am
I SOLVED EXAMPLES
Ex. 1. Simplify : (i) (27)2/3 (ii) (1024)-4/5 (iii)( 8 / 125 )-4/3
Sol . (i)
(27)2/3 = (33)2/3 = 3( 3 * ( 2/ 3))
= 32 = 9
(ii) (1024)-4/5 = (45)-4/5 =
4 { 5 * ( (-4) / 5 )} = 4-4 = 1 / 44 = 1 / 256
(iii) ( 8 / 125 )-4/3 = {(2/5)3}-4/3
= (2/5){ 3 * ( -4/3)} = ( 2 / 5 )-4 = ( 5 / 2 )4 = 54 / 24 = 625 / 16
Ex. 2. Evaluate: (i) (.00032)3/5 (ii)l (256)0.16 x (16)0.18.
Sol. (i) (0.00032)3/5 = ( 32 / 100000
)3/5. = (25 / 105)3/5 = {( 2
/ 10 )5}3/5 = ( 1 / 5 )(5 * 3 / 5) = (1/5)3 = 1 /
125
(ii) (256)0. 16 * (16)0. 18 = {(16)2}0.
16 * (16)0. 18 = (16)(2 * 0. 16) * (16)0. 18
=(16)0.32
* (16)0.18 = (16)(0.32+0.18) = (16)0.5 = (16)1/2 =
4.
196
Ex. 3. What is the
quotient when (x-1 - 1) is divided by (x - 1) ?
Sol. x-1 -1 =
(1/x)-1 = _1 -x *
1 = -1
x
- 1 x - 1
x (x - 1) x
Hence, the required quotient is
-1/x
Ex. 4. If 2x - 1 + 2x + 1 = 1280,
then find the value of x.
Sol. 2x
- 1 + 2X+ 1 = 1280 ó 2x-1 (1 +22)
= 1280
ó 2x-1
= 1280 / 5 = 256 = 28 ó
x -1 = 8 ó
x =
9.
Hence, x =
9.
Ex. 5. Find the
value of [ 5 ( 81/3 + 271/3)3]1/ 4
Sol. [ 5 ( 81/3 + 271/3)3]1/
4 = [ 5 { (23)1/3 + (33)1/3}3]1/
4 = [ 5 { (23 * 1/3)1/3
+ (33 *1/3 )1/3}3]1/ 4
= {5(2+3)3}1/4
= (5 * 53)1/ 4 =5(4 * 1/ 4) = 51 = 5.
Ex. 6. Find the Value of
{(16)3/2 + (16)-3/2}
Sol. [(16)3/2 +(16)-3/2
= (42)3/2 +(42)-3/2 = 4(2 *
3/2) + 4{ 2* (-3/2)}
= 43
+ 4-3 = 43 + (1/43) = ( 64 + ( 1/64)) =
4097/64.
Ex. 7. If (1/5)3y = 0.008, then
find the value of(0.25)y.
Sol. (1/5)3y = 0.008 = 8/1000 =
1/125 = (1/5)3 ó 3y = 3 ó Y = 1.
\ (0.25)y =
(0.25)1 = 0.25.
Ex. 8. Find the value of
(243)n/5 ´ 32n +
1
9n
´
3n -1 .
Sol. (243)n/5
x32n+l = 3 (5 * n/5) ´ 32n+l
_ = 3n ´32n+1
(32)n ´ 3n
- 1 32n
´
3n - 1 32n ´ 3n-l
= 3n +
(2n + 1) = 3(3n+1) = 3(3n+l)-(3n-l)
= 32 = 9.
32n+n-1 3(3n-1)
Ex. 9. Find the value Of (21/4-1)(23/4+21/2+21/4+1)
Sol.
Putting 21/4 = x, we get
:
(21/4-1) (23/4+21/2+21/4+1)=(x-1)(x3+x2+x+1)
, where x = 21/4
=(x-1)[x2(x+1)+(x+1)]
=(x-1)(x+1)(x2+1) = (x2-1)(x2+1)
=(x4-1) = [(21/4)4-1]
= [2(1/4*4) –1] = (2-1) = 1.
Ex. 10.
Find the value of 62/3 ´ 3√67
3√66
Sol. 62/3 ´ 3√67 = 62/3
´ (67)1/3 = 62/3 ´ 6(7 * 1/3) = 62/3
´ 6(7/3)
3√66 (66)1/3 6(6 * 1/3) 62
=62/3
´
6((7/3)-2) = 62/3 ´ 61/3 = 61 = 6.
Ex. 11.
If x= ya, y=zb and z=xc,then find the value of
abc.
Sol. z1= xc =(ya)c [since x= ya]
=y(ac) = (zb)ac [since y=zb]
=zb(ac)= zabc
\
abc = 1.
= 24
Ex. 12. Simplify [(xa /
xb)^(a2+b2+ab)] * [(xb / xc
)^ b2+c2+bc)] * [(xc/xa)^(c2+a2+ca)]
Sol.
Given Expression
= [{x(o -
b)}^(a2 + b2 + ob)].['(x(b
- c)}^ (b2 + c2 + bc)].['(x(c
- a)}^(c2 + a2 + ca])
= [x(a - b)(a2 + b2 +
ab) . x(b - c) (b2 +c2+ bc).x(c-
a) (c2 + a2 + ca)]
= [x^(a3-b3)].[x^(b3-e3)].[x^(c3-a3)]
= x^(a3-b3+b3-c3+c3-a3)
= x0 = 1.
Ex. 13. Which is larger √2 or 3√3 ?
Sol. Given surds are of order 2 and 3. Their L.C.M.
is 6. Changing each to a surd of order 6, we get:
√2
= 21/2 = 2((1/2)*(3/2)) =23/6 = 81/6 = 6√8
3√3=
31/3 = 3((1/3)*(2/2)) = 32/6 = (32)1/6
= (9)1/6 = 6√9.
Clearly, 6√9 > 6√8
and hence 3√3 > √2.
Ex. 14. Find
the largest from among 4√6, √2 and 3√4.
Sol. Given surds are of order 4, 2 and 3
respectively. Their L.C,M, is 12, Changing each to a surd of order 12, we get:
4√6 = 61/4 = 6((1/4)*(3/3))
= 63/12 = (63)1/12 = (216)1/12.
√2 = 21/2 = 2((1/2)*(6/6)) = 26/12
= (26)1/12 = (64)1/12.
3√4 = 41/3 = 4((1/3)*(4/4)) = 44/12
= (44)1/12 =
(256)1/12.
Clearly,
(256)1/12 > (216)1/12 >
(64)1/12
