1.Algebra Booster

1.70 Algebra Booster
6. We have
1 1 2
- =
2
x - 1 x + 1 x - 1
2 2 4
- =
2 2 4
x - 1 x + 1 x -1
4 4 8
- =
4 4 8
x - 1 x + 1 x -1
o
n n n+
1
2 2 2
2n
2n
n 1
x - 1 - x + 1 =
2
x + - 1
1 Ê
n
1 2 2 ˆ
Thus, - + +º+
2 2n
x - 1
Á
x 1
˜
Ë + x + 1 x + 1¯
1
7. We have S1
= = 2
1 - (1/2)
2
S2
= = 3
1 - (1/3)
3
S3
= = 4
1 - (1/4)
Now,
S
o
2 n–1
(2n
- 1)
= = 2n
1 - (1/2 n)
2 2 2 2
1 + 2 + 3 +º+ 2n
1
S S S S -
Ê
n+
1
1 2 ˆ
= Á -
n 1
x 1 2
x + ˜
Ë - - 1¯
= 2 2 + 3 2 + 4 2 + … + (2n) 2
= (1 2 + 2 2 + 3 2 + … + (n) 2 – 1)
+ (n + 1) 2 + (n + 2) 2 + … + (2n) 2
Ênn
( + 1)(2n+
1) ˆ
= Á
-1
Ë
˜
6 ¯
+ S + S + S + … + S
8. We have,
sin 2 x + sin 4 x + sin 6 x + …
2
sin x
=
2
1 - sin x
= tan 2 x
2
tan xlog 2
2
tan x
Now, e
e
= 2
Now, x 2 – 9x + 8 = 0
fi (x – 1)(x – 8) = 0
fi x = 1, 8
2
tan 0
when 2 x = 1=
2
fi tan 2 x = 0
2 2 2 2
n n+ 1 n+ 2 2n-1
fi tan x = 0
2
when 2 x = 8=
2
tan 3
fi tan 2 x = 3
fi tan x = 3
Thus,
cos x
cos x + sin
x
1
=
1 + tan x
1
=
1 + 3
Ê 3 - 1ˆ
= Á
Ë
˜
2 ¯
9. The given series is in the form
1 + a + a 2 + a 3 + … to
Ê 2 - 1ˆ
where a = Á
Ë ˜
2 2 ¯
1
S n =
1 - a
1
=
Ê 2 - 1ˆ
1 - Á
Ë ˜
2 2 ¯
2 2
=
2 2 - 2 + 1
2 2
=
2 + 1
= 2 2( 2 -1)
10. Given f(1) = 2
f(2) = f(1)f(1) = 2 2
f(3) = f(2)f(1) = 2 3
o
f(n) = f(n – 1)f(1) = 2 n
Now,
n
Â
k = 1
f( a + k) = 16(2 -1)
= f(a + 1) + f(a + 2) + … + f(a + n)
= f(a)(f(1) + f(2) + … + f(n))
fi 2 a (2 + 2 2 + … + 2 n ) = 16(2 n – 1)
fi 2 a ◊ 2(2 n – 1) = 16(2 n – 1)
fi 2 a + 1 = 16 = 2 4
fi a + 1 = 4
fi a = 3
Hence, the value of a is 3.
11. From the given polynomial, x 1
+ x 2
+ x 3
= 1 …(i)
x 1
x 2
+ x 2
x 3
+ x 3
x 1
= b
…(ii)
x 1
x 2
x 3
= –g
…(iii)
From Eq. (i), we get
1
2x2+ x2= 1fi x2=
3
n