RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 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 8.1

Expand each of the expressions in Exercise 1 to 5.

Question 1.
(1 - 2x)5
Answer:
(1 - 2x)5 = [1 + (- 2x)]5
Using binomial theorem, 
(1 - 2x)5 = 5C0 (1)5 (- 2x)0 + 5C1 (1)4 (- 2x)1 + 5C2(1)3 (- 2x) + 5C3(1)2 (- 2x)3 + 5C4(1)1 (- 2x)4 + 5C5 (1)0 (- 2x)5
= 1 + 5(- 2x) + 10(- 2x)2 + 10 (- 2x)3 + 5(- 2x)4 + (- 2x)5
= 1 - 10x + 40x2 - 80x3 + 80x4 - 32x5
Thus, (1 - 2x)5 = 1 - 10x + 40x2 - 80x3 + 80x4 - 32x5

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1

Question 2.
\(\left(\frac{2}{x}-\frac{x}{2}\right)^5\)
Answer:
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 1

Question 3.
(2x - 3)6
Answer:
(2x - 3)6 = [2x + (3)]6
Using binomial theorem,
(2x - 3)6 = 6C0 (2x)6 (- 3)0 + 6C1 (2x)5 (- 3) + 6C2 (2x)4 (- 3)2 + 6C3(2x)3 (- 3)3 + 6C4(2x)2 (- 3)4 + 6C5(2x) (- 3)5 + 6C6(2x)0 (- 3)6
= 1 × 64x6 × 1 + 6 × 32x5 × (- 3) + 15 × 16x4 × 9 + 20 × 8x3 × (- 27) + 15 × 4x2 × 81 + 6 × 2x (- 243) + 1 × 1 × 729
= 64x6 - 576x5 + 2160x4 - 4320x3 + 4860x2 - 2916x + 729
Thus, (2x - 3)6 = 64x6 - 576x5 + 2160x4 - 4320x3 + 4860x2 - 2916x + 729

Question 4.
\(\left(\frac{x}{3}+\frac{1}{x}\right)^5\)
Answer:
By Binomial theorem,
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 2

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1

Question 5.
\(\left(x+\frac{1}{x}\right)^6\)
Answer:
Using Binomial theorem,
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 3

Question 6.
(96)3
Answer:
96 = 100 - 4
Thus, (96)3 = (100 - 4)3 = [100 + (- 4)]3
Using binomial theorem,
(96)3 =(100 + (- 4))3
= 33C0(100)3(- 4)0 + 3C1 (100)2 (- 4)1 + 3C2 (100) (- 4)2 + 3C3(100)0 (- 4)3
= 1 × 100000 × 1 + 3 × 10000 × (- 4) + 3 × 100 × 16 + 1 × 1 × (- 64)
= 10000Q - 120000 + 4800 - 64
= 1004800 - 120064 = 884736
Thus, (96)3 = 884736

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1

Question 7.
(102)5
Answer:
102 = 100 + 2
Thus, (102)5 = (100 + 2)5
Using binomial theorem.
(102) = (100 + 2)5
= 5C0 (100)5 (2)0 + 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 + 5C3 (100)2 (2)3 + 5C4 (100)(2)4 + 5C5(100)025
= 1 × 10000000000 × 1 + 5 × 100000000 × 2 + 10 × 1000000 × 4 + 10 × 10000 × 8 + 5 × 100 × 10 + 1 × 1 × 32
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032
Thus, (102)5 = 11040808032

Question 9.
(99)5
Answer:
99 = (100 - 1)
Thus, (99)5 = (100 - 1)5 = [100 + (- 1)]5
Using binomial theorem,
995 = 5C0 (100)5 (- 1)0 + 5C1 (100)4 (- 1)1 + 5C2 (100)3 (- 1)2 + 5C3 (100)2 (- 1)3 + 5C4 (100) (- 1)4 + 5C5 (100)0 (- 1)5
= 1 × 10000000000 × 1 + 5 × 100000000 × (- 1) + 10 × 1000000 × 1 + 10 × 10000 × (- 1) + 5 × 100 × 1 + 1 × 1 × (- 1)
= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1
= 10010000500 - 500100001
= 9509900499
Thus, (99)5 = 9509900499

Question 10.
Using binomial theorem, find which number is larger (1.1)10000 or 1000.
Answer:
(1.1)10000 = (1 + 0.1)10000
= 10000C0 (1)10000 (0.1)0 + 10000C1 (1)9999 (0.1)1 + ................
= 1 × 1 × 1 + 10000 × 1 × 01 + .......
= 1 + 1000 + ........ other positive numbers
= 1001 + .... other positive numbers
Thus, (1.1)10000 = 1001 + .... other positive numbers
But 1001 > 1000
Then. (1 .1)10000 >1000.
(1.1)10000 is greater

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1

Question 11.
Find (a + b)4 - (a - b)4
Hence, evaluate (√3 + √2)4 - (√3 - √2)4.
Answer:
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 4

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1

Question 12.
Find (x + 1)6 + (x - 1)6. Hence, or otherwise, evaluate (√2 + 1)6 + (√2 - 1)6.
Answer:
By Binomial theorem.
(x + 1)6 = 6C0 x6 10 + 6C1 x5 (1)1 + 6C2 x4 (1)2 + 6C3 x3 (1)1 + 6C4 x2 (1)4 + 6C5 x1 (1)5 + 6C6 x0 (1)6
= 6C0 x6 + 6C1x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6
(x + 1)6 = x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1 ....... (1)
Similarly,
(x - 1)6 = x6 - 6x5 + 15x4 - 20x3 + 15x2 - 6x + 1 ...... (2)
Adding equation (1) and (2)
(x + 1)6 + (x - 1)6 = 2(x6 + 15x4 + 15x2 + 1)
Putting x = √2 in above equation,
(√ 2 + 1)6 + (√2 - 1)6
= 2[(√ 2)6 + 15(√2)4 + 15(√2)2 + 1]
= 2[23 + 15(2)2 + 15(2) + 1]
= 2[8 + 60 + 30 + 1]
= 198
Thus, (√2 + 1)6 + (√2 - 10)6 = 198

Question 13.
Show that 9n + 1 - 8n - 9, is divisible by 64. whenever n is a positive integer.
Answer:
By binomial theorem,
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 5
We see that In R.H.S. one factor of 64. Thus R.H.S. is divisible by 64 then 9n + 1 - 8n - 9 will also be divisible by 64.

RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1

Question 14.
Prove that \(\sum_{i=0}^n\) 3r . nCr = 4n
Answer:
By Binomial theorem,
(1 + x)n = nC0 x0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + ........ + nCn xn
Putting x = 3 in both sides,
(1 + 3)n = nC0 30 + nC1 31 + nC2 32 + nC3 33 + nC4 34 + ......... + nCn 3n
RBSE Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 6

Bhagya
Last Updated on Nov. 7, 2023, 9:47 a.m.
Published Nov. 6, 2023