RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 11 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 11. Students can also read RBSE Class 11 Maths Important Questions for exam preparation. Students can also go through RBSE Class 11 Maths Notes to understand and remember the concepts easily.

RBSE Class 11 Maths Solutions Chapter 8 Binomial Theorem Miscellaneous Exercise

Question 1.
Find a, b and n in the expansion of (a + b)n first three terms of the expansion are 729, 7290, and 30375 respectively.
Answer:
In the expansion of (a + b)n
1st term T1 = an = 729 .......... (1)
2nd term = T2 = nC1 an - 1. b = 7290 ................ (2)
3rd term = T3 = nC2 an - 2. b2 = 30375 ......... (3)
Dividing equation (i) by (ii)
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 1
⇒ 12n - 12 = 10n
⇒ 2n = 12 ⇒ n = 6
Putting value of n in equation (1)
a6 = 729
⇒ a6 = 36
On comparing
a = 3
Putting values of n and a in equation (4)
\(\frac{3}{6 \times b}=\frac{1}{10}\)
⇒ b = 5
Thus, a = 3, b = 5, n = 6

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Question 2.
Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.
Answer:
Tr + 1 = 9Cr 39 - r . ar xr
Putting r = 2
Coefficient of x2 = 9C2 37 . a2 = 36.37 a2
Putting r = 3
Coefficient of x3 = 9C3 36 a3 = 84.36.a3
∵ Coefficient of x2 = Coefficient of x3 (given)
Thus, 36.37 . a2 = 84(3)6 a3
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 2

Question 3.
Find the coefficient of x5 in product (1 + 2x)6 (1 - x)7 using binomial theorem.
Answer:
Given expansion = (1 + 2x)6 (1 - x)7
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 3
Now, (1 + 2x)6 (1 - x)7
= (1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + ...... ]
[1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + ........]
Coefficient of x5 in above expansion
= [- 21 + 12 × 35 - 60 × 35 + 160 × 21 + 240(- 7) + 192 × 1]
= [- 21 + 420 - 2100 + 3360 - 1680 + 192]
= [- 3801 + 3972] = 171

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

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:
∵ an - bn = [(a - b) + b]n - bn
Expanding by binomial theorem,
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 4
We see that (a - b) is a factor in R.H.S.
So, we can say that
an - bn = (a - b)k
[Where k = nC0 (a - b)n - 1 + ....... nCn - 1bn - 2]
It is clear that (a - b) is a factor of a - bn
Hence proved

Question 5.
Evaluate (√3 + √2)6 - (√3 - √2)6
Answer:
By expansion of binomial theorem,
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 5

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Question 6.
Find the value of
(a2 + \(\sqrt{a^2-1}\))4 + (a2 - \(\sqrt{a^2-1}\))4.
Answer:
By Binomial theorem,
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 6
Adding (1) and (2).
(a2 + \(\sqrt{a^2-1}\))4 + (a2 - \(\sqrt{a^2-1}\))4.
= 2a8 + 12a4 (a2 - 1) + 2(a2 - 1)2
= 2a8 + 12a4 (a2 - 1) + 2(a4 - 2a2 + 1)
= 2a8 + 12a6 - 12a4 + 2a4 - 4a2 + 2
= 2a8 + 12a6 - 10a4 - 4a2 + 2
Thus, (a2 + \(\sqrt{a^2-1}\))4 + (a2 - \(\sqrt{a^2-1}\))4
= 2a8 + 12a6 - 10a4 - 4a2 + 2
= 2(a8 + 6a6 - 5a4 - 2a2 + 1)

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Question 7.
Find an approximation of (0.99)5 using the first three terms of its expansion.
Answer:
∵ (0.99)5 = (1 - 0.01)5
Thus, by Binomial theorem
(1 - 0.01)5 = 5C0 - 5C1 (0.01) + 5C2(0.01)2
(upto first 3 terms)
= 1 - 5(0.01) + 10(0.01)2
= 1 - 0.05 + 10 × 0.000 1
= 0.95 + 0.0010 = 0.9510
Thus, value of (0.99)5 = 0.9510

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 \(\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^n\) is √6 : 1
Answer:
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 7

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Question 9.
Expand using binomial theorem
\(\left(1+\frac{x}{2}-\frac{2}{x}\right)^4\), x ≠ 0.
Answer:
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 8

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Question 10.
Find the expansion of (3x2 - 2ax + 3a2)3 using binomial theorem.
Answer:
Given expression = (3x2 - 2ax + 3a2)3
Let, y = 3x2 - 2ax
Then by binomial theorem
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 9

Bhagya
Last Updated on Nov. 7, 2023, 9:48 a.m.
Published Nov. 6, 2023