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NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem Miscellaneous Exercise

Page No 175:

Question 1:

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

Answer:

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .
The first three terms of the expansion are given as 729, 7290, and 30375 respectively.
Therefore, we obtain
Dividing (2) by (1), we obtain
Dividing (3) by (2), we obtain
From (4) and (5), we obtain
Substituting n = 6 in equation (1), we obtain
a6 = 729
From (5), we obtain
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.

Answer:

It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .
Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9, we obtain
Comparing the indices of x in x2 and in Tr + 1, we obtain
r = 2
Thus, the coefficient of x2 is
Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9, we obtain
Comparing the indices of x in x3 and in Tk+ 1, we obtain
= 3
Thus, the coefficient of x3 is
It is given that the coefficients of x2 and x3 are the same.
Thus, the required value of a is.

Question 3:

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

Answer:

Using Binomial Theorem, the expressions, (1 + 2x)6 and (1 – x)7, can be expanded as
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
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.
[Hint: write an = (a – b b)n and expand]

Answer:

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.

Answer:

Firstly, the expression (a + b)6 – (a – b)6 is simplified by using Binomial Theorem.
This can be done as

Question 6:

Find the value of.

Answer:

Firstly, the expression (x + y)4 + (x – y)4 is simplified by using Binomial Theorem.
This can be done as

Question 7:

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

Answer:

0.99 = 1 – 0.01
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 

Answer:

In the expansion, ,
Fifth term from the beginning 
Fifth term from the end 
Therefore, it is evident that in the expansion of, the fifth term from the beginning is and the fifth term from the end is.
It is given that the ratio of the fifth term from the beginning to the fifth term from the end is. Therefore, from (1) and (2), we obtain
Thus, the value of n is 10.

Page No 176:

Question 9:

Expand using Binomial Theorem.

Answer:

Using Binomial Theorem, the given expression  can be expanded as
Again by using Binomial Theorem, we obtain
From (1), (2), and (3), we obtain

Question 10:

Find the expansion of using binomial theorem.

Answer:

Using Binomial Theorem, the given expression  can be expanded as
Again by using Binomial Theorem, we obtain
From (1) and (2), we obtain

Courtesy : CBSE