1.Algebra Booster

Binomial Theorem 6.63
Ê 2 ˆ Ê 2 - 1ˆ
= 2+ Á -1˜
logÁ1-
˜
Ë 2 - 1 ¯ Ë 2 ¯
1 Ê 1 ˆ Ê 1 ˆ
= 2+ logÁ ˜
= 2 + 1
Á
( 2 1)
2 2 1 ˜
- Ë ¯ Ë - ¯
1
= 2 + ( 2 + 1) ¥ - log(2)
2
( 2 + 1)
= 2 - ¥ log(2)
2
Integer Type Questions
1. We have,
(1 – x) 5 (1 + x + x 2 + x 3 ) 4
= (1 – x) 5 (1 + x) 4 (1 + x 2 ) 4
= (1 – x) 4 (1 + x) 4 (1 + x 2 ) 4 (1 – x)
= (1 – x 2 ) 4 (1 + x 2 ) 4 (1 – x)
= (1 – x 4 ) 4 ¥ (1 – x)
= (1 – 4 C 1
◊ x 4 + 4 C 2
x 8 – 4 C 3
x 12 + …) × (1 – x)
Hence, the co-efficients of x 13 is 4 C 3
= 4.
2. We have,
-5
2 40Ê
2 1 ˆ
Áx
Ë 2 ˜
x ¯
2
- 5
Ê
2 40 Ê 1ˆ
ˆ
(1 + x ) + 2 +
= (1 + x ) ÁÁx
+ ˜ ˜
ËË
x¯
¯
-10
2 40Ê
1ˆ
= (1 + x ) Áx
+
Ë
˜
x¯
2
-10
2 40Ê
x + 1ˆ
= (1 + x ) Á
Ë
˜
x ¯
= (1 + x 2 ) 30 ¥ x 10
Co-efficient of x 20 in (1 + x 2 ) 30 ¥ x 10
= Co-efficient of x 10 in (1 + x 2 ) 30
= 30 C 5
= 30 C 25
Thus, m = 30 and n = 25.
Hence, the value of
m – n + 2 = 30 – 25 + 2
= 7
3. We have,
(1 + 5x – 3x 2 + 4x 3 – 7x 4 + x 5 ) 2017
Putting x = 1 in the given expression, we get
Sum of the co-efficients = (1 + 5 – 3 + 4 0 7 + 1) 2017
= (11 – 10) 2017
= 1
4. We have,
3 5 3 5
( x + x - 1) + ( x - x -1)
5 5 3
= ( x + a) + ( x - a) , a = x -1
5 5 5 3 3 5 3 2
= 2( Cx 0 + Cx 2 ( x - 1) + Cxx 4 ( -1) )
Thus, the degree of the polynomial is 7.
5. We have t 9
= t 8+1
x -1
1 x -1
10 log log(5 1)
3 25 + 7 2 - +
8
8
= C ¥ ((3 ) (3 ) )
10
10
8
8
x-1 x-1
log 3(25 + 7) - log 3(5 + 1)
= C ¥ ((3 )(3 ))
x-1 x-1
log 3(25 + 7) - log 3(5 + 1)
= C8
¥ ((3 ))
25
1
10 Ê
x -
Ê Ê + 7ˆˆˆ
= C log
8 ¥ 3Á
(5
x - 1 ˜
Ë 1)
3
+ ¯
Ë Á Ë Á
¯ ˜
¯ ˜
x 1
10
Ê
-
25 + 7ˆ
= C8 ¥ Á x-1
˜
Ë(5 + 1) ¯
Ê
2( x-1)
5 + 7ˆ
= 45 ¥ Á x-1
˜
Ë (5 + 1) ¯
Thus,
Ê
2( x-1)
5 + 7ˆ
45 ¥ Á
180
x-1
˜ =
Ë (5 + 1) ¯
Ê
2( x-1)
5 + 7ˆ
fi Á
4
x-1
˜ =
Ë (5 + 1) ¯
fi 5 2(x–1) + 7 = 4(5 x–1 + 1)
fi a 2 – 4a + 3 = 0, a = 5 x–1
fi (a – 1)(a – 3) = 0
fi a = 1, 3
fi 5 x–1 = 1, 3
fi 5 x–1 = 1 = 5 0
fi x – 1 = 0
fi x = 1
Hence, the value of x is 1.
6. We have,
n-1Ê
n
C ˆ
r
Fn ( ) = Â Á n n ˜
r = 0Ë
Cr + Cr+
1 ¯
n-1Ê
n
C ˆ
r
= Â Á n+
1 ˜
r = 0Ë
Cr
+ 1 ¯
n-1Ê
n
C ˆ
r
= Â Á
r
1
= 0
n +
˜
Ê ˆ n
C
Á
r
Ë
Á
Ër
+ 1˜
¯ ˜
¯
n-1
Êr
+ 1ˆ
= Â Á
Ën
+ 1˜
¯
\
r = 0
n -1
1
= Â ( r + 1)
n + 1
r = 0
(1 + 2 + 3 + + n)
=
n + 1
nn ( + 1)/2
=
n + 1
n
=
2
16
F (16) = = 8
2