1.Algebra Booster

6.22 Algebra Booster
LEVEL IV
COMPREHENSIVE LINK PASSAGES
Passage-I: 1. (b) 2. (a) 3. (c) 4. (c) 5. (d)
Passage-II: 1. (c) 2. (b) 3. (c) 4. (a) 5. (d)
Passage-III: 1. (b) 2. (d) 3. (a) 4. (a) 5. (a)
Passage-IV: 1. (c) 2. (c) 3. (d) 4. (b) 5. (d)
Passage-V: 1. (c) 2. (c) 3. (b) 4. (c) 5. (b)
Passage-VI: 1. (c) 2. (d) 3. (b) 4. (a) 5. (d)
MATCH MATRIX
1. A Æ (P, Q, T), B Æ (P), C Æ (R, S)
2. A Æ (Q, R), B Æ (P, R, T), C Æ (S)
3. A Æ (R), B Æ (S), C Æ (Q), D Æ (P)
4. A Æ (R), B Æ (R), C Æ (P), D Æ (Q)
ASSERTION AND REASON
1. (D) 2. (D) 3. (C) 4. (B)
5. (B) 6. (D) 7. (B) 8. (D)
INTEGER TYPE QUESTIONS
1. 4 2. 7 3. 1 4. 7 5. 1
6. 8 7. 4 8. 7 9. 8 10. 4
11. 6 12. 7 13. 5 14. 6 15. 3
HINTS AND SOLUTIONS
LEVEL I
1. We have,
(x + 1) 6
6 6 6 5 6 4 6 3
Cx 0 Cx 1 Cx 2 Cx 3
6 2 6 6 0
+ Cx 2 + Cx 1 + Cx 0
6 5 4 3 2
= + + +
= x + 6x + 15x + 20x + 15x + 6x
+ 1
2. We have,
5
Ê x 1ˆ
Á +
Ë
˜
5 x¯
5 0 4 1 3 2
5 Ê xˆ Ê1ˆ 5 Ê xˆ Ê1ˆ 5 Ê xˆ Ê1ˆ
C0Á ˜ Á ˜ C1Á ˜ Á ˜ C2Á ˜ Á ˜
= + +
Ë5¯ Ë x¯ Ë5¯ Ë x¯ Ë5¯ Ë x¯
2 3 1 4 0 5
5 Ê xˆ Ê1ˆ 5 Ê xˆ Ê1ˆ 5 Ê xˆ Ê1ˆ
C3Á ˜ Á ˜ C4Á ˜ Á ˜ C5Á ˜ Á ˜
+ + +
Ë5¯ Ë x¯ Ë5¯ Ë x¯ Ë5¯ Ë x¯
3. We have,
( x+ x - 1) + ( x- x -1)
2 6 2 6
6 6 2
= ( x + a) + ( x - a) , a = x -1
= 2( + + + )
= 2( + + + )
= 2[ + ( -1)
6 6 0 6 4 2 6 2 4 6 0 6
Cxa 0 Cxa 2 Cxa 4 Cxa 6
6 6 6 4 2 6 2 4 6 6
Cx 0 Cxa 2 Cxa 4 Ca 6
6 6 6 4 2
Cx 0 Cx 2 x
6 2 2 2 6 2 3
+ Cx 4 ( x - 1) + C6( x -1) ]
Hence, the degree of the polynomial is 6.
4. We have, the number of terms in
(i) (1 + x) 2013 is 2014.
(ii) (1 + 2x + x 2 ) 1007
= ((1 + x) 2 ) 1007 = (1 + x) 2014 is 2015.
(iii) (1 + 3x + 3x 2 + x 3 ) 668
= ((1 + x) 3 ) 668 = (1 + x) 2014 is 2005.
(iv) (1 + 4x + 6x 2 + 4x 3 + x 4 ) 503
= ((1 + x) 4 ) 503 = (1 + x) 2012 is 2013.
(v) (1 + x)(1 + x 2 ) 10 is 12.
(vi) (1 – x) 11 (1 + x + x 2 ) 10
= (1 – x){(1 – x)(1 + x + x 2 )} 10
= (1 – x)(1 – x 3 ) 10 is 12.
5. The total number of terms in the expansion of
Ê50 ˆ
(a + b) 50 + (a – b) 50 = Á + 1
Ë
˜
2 ¯ = 26.
6. The total number of terms in the expansion of
99 99
Ê 1ˆ Ê 1ˆ
Ê99 + 1ˆ Áx+ ˜ + Áx-
Ë
˜ = 50
x¯ Ë x¯
Á ˜ = .
Ë 2 ¯
7. The total number of terms in the expansion of
20 20
Ê 2 1 ˆ Ê 2
x
x
1 ˆ
Á +
2˜ -Á -
2˜
=
Ê 20
Á
ˆ = 10
Ë x ¯ Ë x ¯ Ë
˜
2 ¯
8. The total number of terms in the expansion of
2015 2015
Ê2 3ˆ Ê2
3ˆ
Ê2015 + 1ˆ Á + x ˜ -Á - x
Ë
˜ =
1008
x ¯ Ë x ¯
Á =
Ë
˜
2 ¯
9. We have
t 13
= t 12+1
12
18 18-12
Ê 1 ˆ
= C12
¥ (9 x)
¥ Á -
Ë ˜
3 x ¯
18 6 Ê 1 ˆ
= C12 ¥ (9 x)
¥ Á 12 6
Ë3
x ˜
¯
= 18 C 12
10. We have, 4th term from the end = (7 + 1) – (4 – 1)
= 5th term from the beginning.
7-
4 3
4
7 Ê 3 ˆ Ê x ˆ
Thus, t5= t4+
1= C4¥ Á ¥ -
Ë 2 ˜
x ¯
Á
Ë
˜
6 ¯
3 3
4
7 Ê 3 ˆ Ê x ˆ
= C4 ¥ Á
Ë 2 ˜
x ¯
¥Á
Ë
˜
6 ¯
6
7
Ê x ˆ 35 6
= C4
¥ Á = ◊x
Ë
˜
48¯
48