Binomial theorem class 11 miscellaneous exercise

NCERT Solutions for Class 11 Maths Chapter 8 - Binomial Theorem Exercise 8.1 * According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7. The NCERT Solutions of the first exercise of Class 11 Chapter 8 are available here. These solutions are present in PDF format to help students with their studies. Exercise 8.1 of • Introduction to Binomial Theorem • Binomial Theorem for Positive Integral Indices • Pascal’s Triangle • Binomial theorem for any positive integer n • Some special cases NCERT textbook contains numerous questions which are intended for the students to solve and practise. To score high marks in the Class 11 examination, solving and practising the Download PDF carouselExampleControls112 Previous Next Solutions for Class 11 Maths Chapter 8 – Exercise 8.1 Expand each of the expressions in Exercises 1 to 5. 1. (1 – 2x) 5 Solution: From the binomial theorem expansion, we can write as (1 – 2x) 5 = 5C o (1) 5 – 5C 1 (1) 4 (2x) + 5C 2 (1) 3 (2x) 2 – 5C 3 (1) 2 (2x) 3 + 5C 4 (1) 1 (2x) 4 – 5C 5 (2x) 5 = 1 – 5 (2x) + 10 (4x) 2 – 10 (8x 3) + 5 ( 16 x 4) – (32 x 5) = 1 – 10x + 40x 2 – 80x 3 + 80x 4– 32x 5 Solution: From the binomial theorem, the given equation can be expanded as 3. (2x – 3) 6 Solution: From the binomial theorem, the given equation can be expanded as Solution: From the binomial theorem, the given equation can be expanded as Solution: From the binomial theorem, the given equation can be expanded as 6. (...

Transcript Misc 3 Find the value of (a2 + √(a2 −1))4 + (a2 – √(a2 −1))4 . We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn Hence, (a + b)4 = 4C0 a4 + 4C1 a3 b1 + 4C2 a2 b2 + 4C3 a1b3 + 4C4 b4 = 4!/0!(4 − 0)! a4 + 4!/(1 × (4 − 1)!) a3b1 + 4!/2!(4 − 2)! a2b2 + 4!/(3!(4 − 3)!) a3 b1 + 4!/4!(4 −4)! b4 = 4!/(1 × 4!) a4 + 4!/(1 × 3!) a3 b + 4!/(2! × 2!) a2 b2 + 4!/(3! × 1!) ab3 + 4!/(4! 0!) b4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Hence, (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Replacing b with –b (a + (–b))4 = a4 + 4a3 (–b) + 6a2 (–b)2 + 4a (–b)3 + (–b)4 (a – b)4 = a4 – 4a3 b + 6a2b2 – 4ab3 + b4 We need to find (a + b)4 + (a – b)4 (a + b)4 + (a – b)4 = (a4 + 4a3b + 6a2b2 + 4ab3 + b4 ) + (a4 – 4a3b + 6a2b2 – 4ab3 + b4 ) = a4 + 4a3b + 6a2b2 + 4ab3 + b4 + a4 − 4a3b + 6a2b2 − 4ab3 + b4 = 2a4 + 12a2b2 + 2b4 = 2(a4 + 6 a2b2 + b4) Thus, (a + b)4 + (a – b)4 = 2(a4 + 6 a2b2 + b4) We need to find ("a2 + " √(𝑎2−1))^2 + ("a2 – " √(𝑎2−1))^2 ("a2 + " √(𝒂𝟐−𝟏))^𝟐 + ("a2 – " √(𝒂𝟐−𝟏))^𝟐 = 2("(a2)4 + 6(a2)2 " (√(𝑎2−1) )^2 " + " (√(𝑎2−1) )^4 ) = 2("a8 + 6a4 " (𝑎2−1)^(1/2 × 2) " +" (𝑎2−1)^(1/2 × 4) ) = 2("a8 + 6a4 " (𝑎2−1)^1 " +" (𝑎2−1)^2 ) = 2a8 + 12a4 (𝑎2−1) + 2 (a4 + 1 – 2a2)a = 2a8 + 12a6 – 12a4 + 2a4 + 2 – 4a2 = 2a8 + 12a6 – 10a4 – 4a2 + 2 So,("a2 + " √(𝑎2−1))^2 + ("a2 – " √(𝑎2−1))^2 = 2a8 + 12a6 – 10a4 – 4a2 + 2 Show More

Rajasthan Board 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 T 1 = a n = 729 .......... (1) 2nd term = T 2 = nC 1 a n - 1. b = 7290 ................ (2) 3rd term = T 3 = nC 2 a n - 2. b 2 = 30375 ......... (3) Dividing equation (i) by (ii) ⇒ 12n - 12 = 10n ⇒ 2n = 12 ⇒ n = 6 Putting value of n in equation (1) a 6 = 729 ⇒ a 6 = 3 6 On comparing a = 3 Putting values of n and a in equation (4) \(\frac\right)^4\), x ≠ 0. Answer: Question 10. Find the expansion of (3x 2 - 2ax + 3a 2) 3 using binomial theorem. Answer: Given expression = (3x 2 - 2ax + 3a 2) 3 Let, y = 3x 2 - 2ax Then by binomial theorem

Binomial expression is an algebraic expression with two terms only, e.g. 4x 2+9. When such terms are needed to expand to any large power or index say n, then it requires a method to solve it. Therefore, a theorem called Binomial Theorem is introduced which is an efficient way to expand or to multiply a binomial expression. Binomial Theorem is defined as the formula using which any power of a binomial expression can be expanded in the form of a series. Binomial theorem Binomial Theorem is used to solve binomial expressions in a simple way. It gives an expression to calculate the expansion of (a+b) n for any positive integer n. The Binomial theorem is stated as: (a + b) n = nC 0 a n + nC 1 a n-1 b 1 + nC 2 a n-2 b 2 + …. + nC r a n-r b r + …. + nC n b n Now, as it is understood that what is a binomial expression and the purpose of the binomial theorem, so try to expand (a+b) n for large values of n (e.g. n = 10. 11, 12,…) using the above statement as: • (a + b) 10 = a 10 + 10a 9b + 45a 8b 2 + 120a 7b 3 + 210a 6b 4 + 252a 5b 5 + 210a 4b 6+120a 3b 7+ 45a 2b 8+ 10ab 9+ b 10 • (a + b) 11 = a 11 + 11a 10b + 55a 9b 2 + 165a 8b 3 + 330a 7b 4 + 462a 6b 5 + 462a 5b 6 + 330a 4b 7 + 165a 3b 8 + 55a 2b 9 + 11ab 10 + b 11 • (a + b) 12 = a 12 + 12a 11b + 66a 10b 2 + 220a 9b 3 + 495a 8b 4 + 792a 7b 5 + 924a 6b 6 + 792a 5b 7 + 495a 4b 8 + 220a 3b 9 + 66a 2b 10 + 12ab 11 + b 12 and so on. In case of the expansion of smaller powers, it is difficult to calculate the coefficients of the binomia...

NCERT Solutions for Class 11 Maths provided here are given by experts adhering to the Latest CBSE Syllabus Guidelines. NCERT Solutions for Class 11 Maths The NCERT solutions for class 11 maths cover all the solutions of exercises given in chapters like a binomial theorem, trigonometric function, statistics, and many more chapters in the class 11 maths syllabus. The students just need to have access to the internet to go through the Class 11 Maths NCERT solutions. They can access these solutions anytime and from anywhere. Below is the overview of the chapters that we are providing for you to select the topics. You can choose from whichever medium you are comfortable with and learn the respective chapters easily. • • NCERT Solutions for Class 11 Maths Chapter 1 • • • • • • • NCERT Solutions for Class 11 Maths Chapter 2 • • • • NCERT Solutions for Class 11 Maths Chapter 3 • • • • • NCERT Solutions for Class 11 Maths Chapter 4 • NCERT Solutions for Class 11 Maths Chapter 5 • • • • NCERT Solutions for Class 11 Maths Chapter 6 • • • • NCERT Solutions for Class 11 Maths Chapter 7 • • • • • NCERT Solutions for Class 11 Maths Chapter 8 • • • NCERT Solutions for Class 11 Maths Chapter 9 • • • • • NCERT Solutions for Class 11 Maths Chapter 10 • • • • NCERT Solutions for Class 11 Maths Chapter 11 • • • • • NCERT Solutions for Class 11 Maths Chapter 12 • • • • NCERT Solutions for Class 11 Maths Chapter 13 • • • NCERT Solutions for Class 11 Maths Chapter 14 • • • • • • NCERT Solutions f...

Transcript Misc 6 Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem. We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn Hence (a + b)3 = 3C0 a3 + 3C1 a2b1 + 3C2 a1 b2 + 3C3 b3 = a3 + 3!/(1! (3 −1) !) a2 b + 3!/2!(3 −1)! ab2 + b3 = a3 + 3a2b + 3b2a + b3 Now, (3x2 – 2ax + 3a2)3 = (3x2 – (2ax – 3a2))3 = (3x2 – a(2x – 3a))3 Putting a = 3x2 and b = –a( 2x – 3a) in (1) (a + b)3 = a3 + 3a2b + 3b2a + b3 (3x2 – a (2x – 3a2))3 = (3x2)3 + 3(3x2)2 (–a(2x – 3a)) + 3 (–a(2x – 3a))2 (3x2) + (–a(2x – 3a))3 = 27x6 – 27x4a (2x – 3a) + 9x2a2 (2x – 3a)2 – a3 (2x – 3a)3 = 27x6 – 27x4 (2xa – 3a2) + 9x2a2 (4x2 + 9a2 − 12ax) – a3 ((2x)3 + (−3a)3 + 3(2x)2 (−3a) + 3 (2x) (−3a)2] = 27x6 – 54x5a + 81 x4a2 + 36x4a2 + 81x2a4 – 108x3a3 – a3 (8x3 – 27a3 – 36x2a + 54xa2) = 27x6 – 54x5a + 81 x4a2 + 36x4a2 + 81x2a4 – 108x3a3 – 8x3 a3 + 27a6 + 36x2a4 – 54a5x = 27x6 – 54ax5 + 81a2x4 + 36a2x4 – 108x3a3 – 8x3 a3 + 36x2a4 + 81x2a4 – 54a5 x + 27a6 = 27x6 – 54ax5 + 117a2x4 – 116a3x3 + 117x2a4 – 54a5 x + 27a6 Thus, (3x2 – 2ax + 3a2)3 = 27x6 – 54ax5 + 117a4x2 – 116a3x3 – 54a5 x + 27a6 Show More

Get here the revised and modified class 11 Maths chapter 7 solutions based on rationalised textbooks issued by NCERT for academic year 2023-24. NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem in Hindi and English Medium updated for CBSE session 2023-2024. Class 11 Maths Chapter 7 Solutions in English Medium Class 11 Maths Chapter 7 Solutions in Hindi Medium Related Links The Middle Term: In the expansion of (a + b)^n, the total number of terms are (n + 1). The middle term in the expansion of (a + b)^n depend on n. 1. When n is even: Let n = 2m, where m is positive integer. The total number of terms will be 2m + 1. Hence, the middle term of the expansion (a + b)^n will be 1/2[(2m + 1) + 1], i. e. when n is even then (m + 1)th term or (n/2 + 1)th will be the middle term. 1. Binomial Expression: Any expression containing two terms combined by + or – is called 2. In the expansion of (a + b)^n, the coefficient of first term = coefficient of last term, coefficient of second term = coefficient of second term from last. Thus we get that in the expansion of (a + b)^n, the terms from first term and from the last term at equal distance have the same coefficients. By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)10000 can be obtained as (1.1)^10000 = (1 + 0.1)^10000 = C(10000, 0) + C(10000, 1) (10000)^1 (0.1)^1 + Other positive terms. = 1 + 10000×0.1 + Other positive terms. = 1001 + Other positive terms. > 1000 Hence,(1.1)^10000 > 1000....

Class 11 Maths Chapter 1 Sets • • • • • • • Class 11 Maths Chapter 2 Relations and Functions • • • • Class 11 Maths Chapter 3 Trigonometric Functions • • • • • Class 11 Maths Chapter 4 Principle of Mathematical Induction • Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations • • • • Class 11 Maths Chapter 6 Linear Inequalities • • • • Class 11 Math Chapter 7 Permutations and Combinations • • • • • Class 11 Maths Chapter 8 Binomial Theorem • • • Class 11 Maths Chapter 9 Sequences and Series • • • • • Class 11 Maths Chapter 10 Straight Lines • • • • Class 11 Maths Chapter 11 Conic Sections • • • • • Class 11 Maths Chapter 12 Introduction to three Dimensional Geometry • • • • Class 11 Maths Chapter 13 Limits and Derivatives • • • Class 11 Maths Chapter 14 Mathematical Reasoning • • • • • • Class 11 Maths Chapter 15 Statistics • • • • Class 11 Maths Chapter 16 Probability • • • • Filed Under: Primary Sidebar