﻿ Class 11 NCERT Math Solution
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TOPICS
Exercise - 8.1

Question-1 :-  Expand : (1 - 2x)5

Solution :-
```  (1 - 2x)5
= 5C0(1)5 - 5C1(1)4(2x) + 5C2(1)3(2x)2 - 5C3(1)2(2x)3 + 5C4(1)(2x)4 - 5C5 (2x)5
= 1 - 5(2x) + 10(4x2) - 10(8x3) + 5(16x4) - 32x5
= 1 - 10x + 40x2 - 80x3 + 80x4 - 32x5
```

Question-2 :- Expand : (2/x - x/2)5

Solution :-
```  (2/x - x/2)5
= 5C0(2/x)5 - 5C1(2/x)4(x/2) + 5C2(2/x)3(x/2)2 - 5C3(2/x)2(x/2)3 + 5C4(2/x)(x/2)4 - 5C5 (x/2)5
= 32/x5 - 5(16/x4)(x/2) + 10(8/x3)(x2/4) - 10(4/x2)(x3/8) + 5(2/x)(x4/16) - x5/32
= 32/x5 - 40/x3 + 20/x - 5x + 5x3/8 - x5/32
```

Question-3 :-  Expand : (2x - 3)6

Solution :-
```  (2x - 3)6
= 6C0(2x)6 - 6C1(2x)5(3) + 6C2(2x)4(3)2 - 6C3(2x)3(3)3 + 6C4(2x)2(3)4 - 6C5(2x)(3)5 + 6C6(3)6
= 64x6 - 6(32x5)(3) + 15(16x4)(9) - 20(8x3)(27) + 15(4x2)(81) - 6(2x)(243) + 729
= 64x6 - 576x5 + 2160x4 - 4320x3 + 4860x2 - 2916x + 729
```

Question-4 :- Expand : (x/3 + 1/x)5

Solution :-
```  (x/3 + 1/x)5
= 5C0(x/3)5 + 5C1(x/3)4(1/x) + 5C2(x/3)3(1/x)2 + 5C3(x/3)2(1/x)3 + 5C4(x/3)(1/x)4 - 5C5(1/x)5
= x5/243 + 5(x4/81)(1/x) + 10(x3/27)(1/x2) + 10(x2/9)(1/x3) + 5(x/3)(1/x4) + 1/x5
= x5/243 + 5x3/81 + 10x/27 + 10/9x + 5/3x3 + 1/x5
```

Question-5 :- Expand : (x + 1/x)6

Solution :-
```  (x + 1/x)6
= 6C0(x)6 + 6C1(x)5(1/x) + 6C2(x)4(1/x)2 + 6C3(x)3(1/x)3 + 6C4(x)2(1/x)4 + 6C5(x)(1/x)5 + 6C6(1/x)6
= x6 + 6(x5)(1/x) + 15(x4)(1/x2) + 20(x3)(1/x3) + 15(x2)(1/x4) + 6(x)(1/x5) + 1/x6
= x6 + 6x4 + 15x2 + 20 + 15/x2 + 6/x4 + 1/x6
```

Question-6 :-  Using binomial theorem, evaluate (96)3

Solution :-
```  963 = (100 - 4)3
= 3C0(100)3 - 3C1(100)2.4 + 3C2(100).42 - 3C3 43
= 1000000 - 3 x 10000 x 4 + 3 x 100 x 16 - 64
= 1000000 - 120000 + 4800 - 64
= 1004800 - 120064
= 884736
```

Question-7 :-  Using binomial theorem, evaluate (102)5

Solution :-
```  1025 = (100 + 2)5
= 5C0(100)5 + 5C1(100)4.2 + 5C2(100)3.22 + 5C3(100)2.23 + 5C4(100).24 + 5C5(2)5
= 100000000000 + 5 x 100000000 x 2 + 10 x 1000000 x 4 + 10 x 10000 x 8 + 5 x 100 x 16 + 32
= 100000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032
```

Question-8 :-  Using binomial theorem, evaluate (101)4

Solution :-
```  1014 = (100 + 1)4
= 4C0(100)4 + 4C1(100)3.1 + 4C2(100)2.12 + 4C3(100).13 + 4C4.14
= 100000000 + 4 x 1000000 + 6 x 10000 + 4 x 100 + 1
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401
```

Question-9 :-  Using binomial theorem, evaluate (99)5

Solution :-
```  995 = (100 - 1)5
= 5C0(100)5 - 5C1(100)4.1 + 5C2(100)3.12 - 5C3(100)2.13 + 5C4(100).14 - 5C5(1)5
= 100000000000 - 5 x 100000000 x 1 + 10 x 1000000 x 1 - 10 x 10000 x 1 + 5 x 100 x 1 - 1
= 100000000000 - 500000000 + 10000000 - 100000 + 500 - 1
= 10010000500 - 500100001
= 9509900499
```

Question-10 :-  Using binomial theorem, , indicate which number is larger (1.1)10000 or 1000.

Solution :-
```  (1.1)10000 = (1 + 0.1)10000
= 10000C0 + 10000C1(0.1) + other positive terms
= 1 + 10000 × 0.1 + other positive terms
= 1 + 1000 + other positive terms
> 1000
Hence (1.1)10000 > 1000
```

Question-11 :-  Find (a + b)4 – (a – b)4. Hence, evaluate (√3 + √2)4 - (√3 - √2)4 .

Solution :-
```  (a + b)4 = 4C0(a)4 + 4C1(a)3.b + 4C2(a)2.b2 + 4C3(a).b3 + 4C4.b4
(a - b)4 = 4C0(a)4 - 4C1(a)3.b + 4C2(a)2.b2 - 4C3(a).b3 + 4C4.b4
Now, (a + b)4 - (a + b)4
= 4C0(a)4 + 4C1(a)3.b + 4C2(a)2.b2 + 4C3(a).b3 + 4C4.b4 - 4C0(a)4 + 4C1(a)3.b - 4C2(a)2.b2 + 4C3(a).b3 - 4C4.b4
= 2.4C1(a)3.b + 2.4C3(a).b3
= 2(4C1(a)3.b + 4C3(a).b3)
= 2(4(a)3.b + 4(a).b3)
= 8ab(a2 + b2)
By putting a = √3 and b = √2, we obtain
= 8 x √3 x √2[(√3)2 + (√2)2]
= 8√6(3 + 2)
= 40√6
```

Question-12 :-  Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate (√2 + 1)6 + (√2 – 1)6.

Solution :-
```  (x + 1)6 = 6C0(x)6 + 6C1(x)5(1) + 6C2(x)4(1)2 + 6C3(x)3(1)3 + 6C4(x)2(1)4 + 6C5(x)(1)5 + 6C6(1)6
(x - 1)6 = 6C0(x)6 - 6C1(x)5(1) + 6C2(x)4(1)2 - 6C3(x)3(1)3 + 6C4(x)2(1)4 - 6C5(x)(1)5 + 6C6(1)6
Now, (x + 1)6 + (x - 1)6
= 6C0(x)6 + 6C1(x)5(1) + 6C2(x)4(1)2 + 6C3(x)3(1)3 + 6C4(x)2(1)4 + 6C5(x)(1)5 + 6C6(1)6 + 6C0(x)6 - 6C₁(x)5(1) + 6C2(x)4(1)2 - 6C3(x)3(1)3 + 6C4(x)2(1)4 - 6C5(x)(1)5 + 6C6(1)6
= 2(6C0(x)6 + 6C2(x)4 + 6C4(x)2 + 6C6)
= 2(x6 + 15x4 + 15x2 + 1)
By putting x = √2, we obtain
= 2[(√2)6 + 15(√2)4 + 15(√2)2 + 1]
= 2(8 + 15 x 4 + 15 x 2 + 1)
= 2(8 + 60 + 30 + 1)
= 2 x 99
= 198
```

Question-13 :-  Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.

Solution :-
```  We prove that 9n+1 – 8n – 9 = 64k, where k is some natural number.
We have (1 + a)n = nC0 + nC1a + nC2a2 + ... + nCnan
For a = 5, we get
(1 + 8)n+1 = n+1C0 + n+1C18 + n+1C282 + ... + n+1Cn+18n+1 i.e.
(9)n+1 = 1 + 8(n+1) + 82.n+1C2 + 83.n+1C3 + ... + n+1Cn+18n+1 i.e.
(9)n+1 = 1 + 8(n+1) + 82(n+1C2 + n+1C38 + ... + n+1Cn+18n-1) or
(9)n+1 = 9 + 8n + 64(n+1C2 + 8 .n+1C3 + ... + n+1Cn+18n-1) or
9n+1 - 8n - 9 = 64k
where k =  n+1C2 + 8 .n+1C3 + ... + n+1Cn+18n-1
Thus, 9n+1 - 8n - 9 is divisible by 64.
```

Question-14 :-

Solution :-
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