RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 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 Ex 8.2

Question 1
x5 in (x + 3)8
Answer:
Let x5 occurs in (r + 1)th term in the expansion of (x + 3)8.
Then, (r + 1)th term in expansion of (x + 3)8
Tr + 1 = 8Crx8 - r (3)r.
∵ Here power of x is 5, so x8 - r = x5
Thus, x8 - x5 (given)
Comparing powers
8 - r = 5
∴ r = 8 - 5 = 3
Then, (r + 1)th term, Tr + 1 = 8C3 x5 33
Thus, coefficient of x5 = 8C3 33
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 1
Thus, coefficient of x5 in the expansion of
(x + 3)8 = 1512

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Question 2.
a5b7 in (a - 2b)12.
Answer:
Let in expansion of (a - 2b)12 coefficient of (r + 1)th term will be a5b7
Then in expansion of (a - 2b)12, (r + 1)th term.
Tr + 1 = 12Cr (a)12 - r (- 2b)r
We have, a12 - r (- 2b)r = a5b7
Comparing power of a on both sides
12 - r = 5
or r = 7
Then (r + 1)th term., Tr + 1 = 12C7 a12 - 7(- 2b)7
= \(\frac{12 !}{7 !(12-7) !}\) a5(- 2)7 b7
= \(\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 !}{7 ! \times 5 \times 4 \times 3 \times 2 \times 1}\) × a5 (- 2)7 b7
= 792 (- 128)a5 b7
= - 101376 a5 b7
Thus, coefficient of a5b7 in (a - 2b)12 = - 101376

Write the general term in the expansion of

Question 3.
(x2 - y)6
Answer:
General term in expansion of (x2 - y)6
Tr + 1 = 6Cr (x2)6 - r (- y)r [∵ Tr + 1 = nCr xn - r a4]
= 6Cr x12 - 2r (- 1)r yr
= 6Cr (- 1)r x12 - 2r yr
Thus, general term in expansion of (x2 - y)6
= 6Cr x12 - 2r yr

Question 4.
(x2 - yx)12, x ≠ 0
Answer:
General term in expansion of (x2 - yx)12
Tr + 1 = 12Cr (x2)12 - r (- yx)r [∵ Tr + 1 = nCr xn - r a4]
= 12Cr x24 - 2r (- 1)r yr xr
= 12Cr x24 - 2r + r (- 1)r yr
= 12Cr (- 1)r x24 - r yr
Thus, general term in expansion of (x2 - yx)12
= 12Cr (- 1)r x24 - r yr

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Question 5.
Find the fourth term ¡n the expansion of (x - 2y)12
Answer:
4th term in expansion of (x - 2y)12
T4 or T3 + 1 = 12C3 x12 - 3 (- 2y)3 [∵ Tr + 1 = nCr xn - r a4]
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 2
= 220 (- 8)x9 y3 = - 1760 x9 y3
Thus, 4th term in expansion (x - 2y)12
= - 1760 x9 y3

Question 6.
Find the 13th term in the expansion of
\(\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}\), x ≠ 0.
Answer:
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 3

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Find the middle terms in the expansions of

Question 7.
\(\left(3-\frac{x^3}{6}\right)^7\).
Answer:
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 4

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Question 8.
\(\left(\frac{x}{3}+9 y\right)^{10}\)
Answer:
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 5

Question 9.
In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.
Answer:
By general term Tr + 1 = nCrxn - rar
(r +1) th term in the expansion of
(1 + a)m + n = m + nCrar
Putting r = m, coeff. of xm = m + nCm
Putting r = n, coeff. of xm = m + nCn
= m + nCm + n - n = m + nCm
Thus, coeff. of xm = coff. of xn
[nCr = nCn - r]
Hence Proved

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Question 10.
The coefficient of the (r - 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r.
Answer:
(x + 1)n = nC0 + nC1xn - 1 + nC2xn - 2 + ..... + nCrxn - r + .......... nCn
(r - 1) th term = T(r - 2) + 1
= Tr - 1 = nCr - 2 xn - (r - 2)
= nCr - 2 xn - r + 2
r th term = T(r - 1) + 1
= nCr - 1 xn - 1(r - 1)
= nCr - 1 xn - r + 1
(r + 1) th term = nCr xn - r and coefficient of the rth, (r - 1)th and (r + 1)th terms respectively will be nCr - 2, nCr - 1 and nCr.
Thus, nCr - 2 : nCr - 1 : nCr = 1 : 3 : 5
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 6
or 3r - 3 = n - r + 2
or 3r + r = n + 2 + 3
or 4r = n + 5
or n = 4r - 5
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 7
or 5r = 3n - 3r + 3
or 3n = 5r + 3r - 3
3n = 8r - 3
From equation (1) and (2)
3 × (4r - 5) = 8r - 3
or 12r - 15 = 8r - 3
or 12r - 8r = 15 - 3
or. 4r = 12
∴ r = 3
Putting value of r in equation (1)
n = 4r - 5
or n = 4 × 3 - 5
or n = 12 - 5
∴ n = 7
Thus, n = 7, r = 3

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Question 11.
Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n - 1.
Answer:
(r + 1)th term in the expansion of (1 + x)2n
Putting r = n, coefficient of xn = 2nCn
Similarly, coefficient of xn in expansion of (1 - x)2n - 1
= 2n - 1Cn
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 8
Thus in expansion of (1 + x)2n coefficient of xn
= 2 × coeff. of xn in expansion of (1 + x)2n - 1
Hence Proved

Question 12.
Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.
Answer:
By Binomial theorem.
(1 + x)m = mC0 +mC1x + mC2x2 + .....
Then, coefficient of of x2 in expansion of (1 + x)n = mC2
But given that coeff. of x2 = 6
Thus, mc2 = 6
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 9
or m(m - 1) = 2 × 6
or m2 - m - 12 = 0
or m2 - (4 - 3)m - 12 = 0
or m2 - 4m + 3m - 12 = 0
or m(m - 4) + 3(m - 4) = 0
or (m- 4) (m + 3) = 0
∴ m = 4
m = - 3
But value of m is not negative. Thus, m = 4

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