﻿ Class 11 NCERT Math Solution
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TOPICS
Miscellaneous

Question-1 :-  Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Solution :-
```  The (r + 1)th term of the expansion (a + b)n is given by Tr+1 =  nCr an-r br.
The first three terms of the expansion are given as 729, 7290, and 30375 respectively.
Therefore, we obtain
T1 = nC0 an-0 b0 = nC0 an = an = 729 ....(i)
T2 = nC1 an-1 b1 = nan-1 b = 7290 .....(ii)
T3 = nC2 an-2 b2 = n(n-1)/2 . an-2 . b2 = 30375 .....(iii)
Dividing (ii) by (i), we obtain
(nan-1 b)/an = 7290/729
nb/a = 10 .....(iv)
Dividing (iii) by (ii), we obtain
[n(n-1)/2 . an-2 . b2]/(nan-1 b) = 30375/7290
[(n-1)b]/2a = 30375/7290
[(n-1)b]/a = 25/3
nb/a - b/a = 25/3
10 - b/a = 25/3    [using (iv)]
b/a = 10 - 25/3
b/a = 5/3 .....(v)
From (iv) and (v), we obtain
n x 5/3 = 10
n = 30/5
n = 6
Substituting n = 6 in equation (i), we obtain a6 = 729; a6 = 36; a = 3
From (v), we obtain
b/3 = 5/3
b = 5
Thus, a = 3, b = 5, and n = 6.
```

Question-2 :-  Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 =  nCr xn-r yr.
Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9,
we obtain Tr+1 = 9Cr 39-r (ax)r = 9Cr 39-r ar xr
Comparing the indices of x in x2 and in Tr+1, we obtain r = 2
Thus, the coefficient of x2 is T2+1 = 9C2 39-2 a2 = 36 37 a2

Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9,
we obtain Tk+1 = 9Ck 39-k (ax)k = 9Ck 39-k ak xk
Comparing the indices of x in x3 and in Tk+1, we obtain k = 3
Thus, the coefficient of x3 is T3+1 = 9C3 39-3 a3 = 84 36 a3
It is given that the coefficients of x2 and x3 are the same.
84 36 a3 = 36 37 a2
84a = 36 x 3
a = 108/84
a = 9/7
Thus, the required value of a is 9/7.
```

Question-3 :-  Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.

Solution :-
```  (1 + 2x)6
= 6 C0(1)6 + 6 C1(1)5 (2x) + 6 C2(1)4 (2x)2 + 6 C3(1)3 (2x)3 + 6 C4(1)2 (2x)4 + 6 C5(1)(2x)5  + 6 C6(2x)6
= 1 + 6(2x) + 15(2x)2  + 20(2x3 + 15(2x)4 + 6(2x)5 + (2x)6
= 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6

(1 - x)7
= 7C0(1)7 - 7C1(1)6(x) + 7C2(1)5(x)2 - 7C3(1)4(x)3 + 7C4(1)3(x)4 - 7C5(1)2(x)5 + 7C6(1)(x)6 - 7C7(x)7
= 1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7

Therefore, (1 + 2x)6 . (1 - x)7
= (1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6) . (1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7)
The complete multiplication of the two brackets is not required to be carried out.
Only those terms, which involve x5, are required.
The terms containing x5 are
1(-21x5) + (12x)(35x4) + (60x2)(-35x3) + (160x3)(21x2) + (240x4)(-7x) + (192x5)(1)
= 171x5
Thus, the coefficient of x5 in the given product is 171.
```

Question-4 :-  If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer.

Solution :-
```  In order to prove that (a – b) is a factor of (an – bn), it has to be proved that an – bn = k (a – b),
where k is some natural number
It can be written that, a = a – b + b
This shows that (a – b) is a factor of (an – bn), where n is a positive integer. ```

Question-5 :-  Evaluate (√3 + √2)6 - (√3 - √2)6 .

Solution :-
```  (a + b)6 = 6C0(a)6 + 6C1(a)5(b) + 6C2(a)4(b)2 + 6C3(a)3(b)3 + 6C4(a)2(b)4 + 6C5(a)(b)5 + 6C6(b)6
(a - b)6 = 6C₀(a)6 - 6C1(a)5(b) + 6C2(a)4(b)2 - 6C₃(a)3(b)3 + 6C4(a)2(b)4 - 6C5(a)(b)5 + 6C6(b)6
Now, (a + b)6 - (a - b)6
= 6C0(a)6 + 6C1(a)5(b) + 6C2(a)4(b)2 + 6C3(a)3(b)3 + 6C4(a)2(b)4 + 6C5(a)(b)5 + 6C6(b)6 - [6C0(a)6 - 6C1(a)5(b) + 6C2(a)4(b)2 - 6C3(a)3(b)3 + 6C4(a)2(b)4 - 6C5(a)(b)5 + 6C6(b)6]
= 6C0(a)6 + 6C1(a)5(b) + 6C2(a)4(b)2 + 6C3(a)3(b)3 + 6C4(a)2(b)4 + 6C5(a)(b)5 + 6C6(b)6 - 6C0(a)6 + 6C1(a)5(b) - 6C2(a)4(b)2 + 6C3(a)3(b)3 - 6C4(a)2(b)4 + 6C5(a)(b)5 - 6C6(b)6
= 2(6C1(a)5(b) + 6C3(a)3(b)3 + 6C5(a)(b)5)
= 2(6a5b + 20a3b3 + 6ab5)
By putting a = √3 and b = √2, we obtain
= 2(6(√3)5(√2) + 20(√3)3(√2)3 + 6(√3)(√2)5)
= 2(54√6 + 120√6 + 24√6)
= 2 x 198 x √6
= 396√6
```

Question-6 :- .

Solution :-
```  (x + y)4 = 4C0(x)4 + 4C1(x)3.y + 4C2(x)2.y2 + 4C3(x).y3 + 4C4.y4
(x - y)4 = 4C0(x)4 - 4C1(x)3.y + 4C2(x)2.y2 - 4C3(x).y3 + 4C4.y4
Now, (x + y)4 + (x - y)4
= 4C0(x)4 + 4C1(x)3.y + 4C2(x)2.y2 + 4C3(x).y3 + 4C4.y4 + 4C0(x)4 - 4C1(x)3.y + 4C2(x)2.y2 - 4C3(x).y3 + 4C4.y4
= 2(4C0(x)4 + 4C2(x)2.(y)2 + 4C4.(y)4)
= 2(x4 + 6x2y2 + y4)
By putting x = a2 and y = √a2 - 1, we obtain ```

Question-7 :-  Find an approximation of (0.99)5 using the first three terms of its expansion.

Solution :-
```  0.99 = 1 – 0.01
(0.99)5 = (1 - 0.01)5
= 5C0 (1)5 - 5C1 (1)4 (0.01) + 5C2 (1)3 (0.01)2  [approx.]
= 1 - 5(0.01) + 10(0.01)2
= 1 - 0.05 + 0.001
= 0.001 - 0.05
= 0.951
Thus, the value of (0.99)5 is approximately 0.951.
```

Question-8 :-  Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of Solution :-
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Question-9 :-  Expand using Binomial Theorem (1 + x/2 - 2/x)4, x ≠ 0.

Solution :-
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Question-10 :-  Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.

Solution :-
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