11th maths part 2 exercise 1.1 answers

Balbharti Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2 Question 1. Expand: (i) (√3 + √2) 4 Solution: Here, a = √3, b = √2 and n = 4. Using binomial theorem, ∴ (√3 + √2) 4 = 1(9) (1) + 4(3√3) (√2) + 6(3)(2) + 4(√3) (2√2) + 1(1)(4) = 9 + 12√6 + 36 + 8√6 + 4 = 49 + 20√6 (ii) (√5 – √2) 5 Solution: Here, a = √5, b = √2 and n = 5. Using binomial theorem, Question 2. Expand: (i) (2x 2 + 3) 4 Solution: Here, a = 2x 2, b = 3 and n = 4. Using binomial theorem, (ii) \(\left(2 x-\frac\) and n = 6. Using binomial theorem, Question 3. Find the value of (i) (√3 + 1) 4– (√3 – 1) 4 Solution: (ii) (2 + √5) 5 + (2 – √5) 5 Solution: Adding (i) and (ii), we get ∴ (2 + √5 ) 5 + (2 – √5) 5 = (32 + 80√5 + 400 + 200√5 + 250 + 25√5) + (32 – 80√5 + 400 – 200√5+ 250 – 25√5 ) = 64 + 800 + 500 = 1364 Question 4. Prove that: (i) (√3 + √2) 6 + (√3 – √2) 6 = 970 Solution: (ii) (√5 + 1) 5– (√5 – 1) 5 = 352 Solution: Question 5. Using binomial theorem, find the value of (i) (102) 4 Solution: (ii) (1.1) 5 Solution: Question 6. Using binomial theorem, find the value of (i) (9.9) 3 Solution: (ii) (0.9) 4 Solution: Question 7. Without expanding, find the value of (i) (x + 1) 4– 4(x + 1) 3 (x – 1) + 6(x + 1) 2 (x – 1) 2– 4(x + 1) (x – 1) 3 + (x – 1) 4 Solution: (ii) (2x – 1) 4 + 4(2x – 1) 3 (3 – 2x) + 6(2x – 1) 2 (3 – 2x) 2 + 4(2x – 1) 1 (3 – 2x) 3 + (3 – 2x) 4 Solution: Question 8. Find the value of (1.02) 6, correct upto four places of decimals. Soluti...

It is obvious that you are searching for 1st year math exercise 1.1 solution notes in PDF format. For that reason, we have uploaded high-quality 11th-class math exercise notes. This class 11 math notes include detailed notes for exercise 1.1. Using these notes, you will learn the simplest method to solve a question. This is a complete solution to all the questions in your really challenging book. These math notes are easy to download. Like Our Facebook Page For Educational Updates These 11th Class math exercise 1.1 notes were prepared according to the syllabus of all Punjab Boards. Other boards other than Punjab do not follow 1st year math notes. These Punjab boards are Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Bahawalpur Board Sargodha Board, DG Khan Board, and Sahiwal Board. Finally words, we made our finest efforts to make these notes beneficial for you. But if you find any mistake, any suggestion for further precision is however invited. And if you find that our efforts help you, share it with your mates because “Sharing is Caring”.

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2 Get Free NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2 PDF in Hindi and English Medium. Sets Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. Sets Exercise 1.2 Class 11 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 1 Class 11 Sets Ex 1.2 provided in NCERT Textbook. • • • • • • • • • • • • • • • • • • Chapter 1: • अध्याय 1: Free download NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20. Q.1: Which of the following given below is null set? (i). Set of odd natural numbers which is divisible by 2. (ii). Set of even numbers which are prime (iii). हल: यहाँ समुच्चय B और D के अवयव 1, 2, 3, 4, हैं। B = D तथा समुच्चय E और G में -1, 1 अवयव समान हैं। E = G Filed Under:

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Balbharti Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Question 1. Simplify: (i) √-16 + 3√-25 + √-36 – √-625 Solution: = 4i + 3(5i) + 6i – 25i = 25i – 25i = 0 (ii) 4√-4 + 5√-9 – 3√-16 Solution: Question 2. Write the conjugates of the following complex numbers (i) 3 + i Solution: Conjugate of (3 + i) is (3 – i). (ii) 3 – i Solution: Conjugate of (3 – i) is (3 + i). (iii) √-5 – √7 i Solution: Conjugate of (√-5 – √7 i) is (√-5 + √7 i). (iv) -√-5 Solution: -√-5 = -√5 × √-1 = -√5 i Conjugate of (-√-5) is √5 i (v) 5i Solution: Conjugate of (5i) is (-5i). (vi) √5 – i Solution: Conjugate of (√5 – i) is (√5 + i). (vii) √2 + √3 i Solution: Conjugate of (√2 + √3 i) is (√2 – √3 i) (viii) cos θ + i sin θ Solution: Conjugate of (cos θ + i sin θ) is (cos θ – i sin θ) Question 3. Find a and b if (i) a + 2b + 2ai = 4 + 6i Solution: a + 2b + 2ai = 4 + 6i Equating real and imaginary parts, we get a + 2b = 4 …..(i) 2a = 6 ……(ii) ∴ a = 3 Substituting, a = 3 in (i), we get 3 + 2b = 4 ∴ b = \(\frac(1+i)\) Solution: (2x + 3y + 1) + (8 – 3x + 2y)i = 9 + 9i Equating real and imaginary parts, we get 2x + 3y + 1 = 9 and 8 – 3x + 2y = 9 2x + 3y = 8 ……(i) and 3x – 2y = – 1 ……(ii) Equation (i) × 2 + equation (ii) × 3 gives 13x = 13 ∴ x = 1 Substituting x = 1 in (i), we get 2(1) + 3y = 8 3y = 6 ∴ y = 2 ∴ x = 1 and y = 2 (iv) If x(1 + 3i) + y(2 – i) – 5 + i 3 = 0, find x + y Solution: x(1 + 3i) + y(2 – i) – 5 + i 3 = 0 x + 3xi + 2y – yi – 5 – i = 0 ……[∵ i 3 = -i] (x + 2y –...

Here you will see the complete notes of Maths 11th Class Exercise 1.1 Notes (Solution), Chapter # 01 (Numbers Systems). These notes are valid for all BISE Punjab boards (Lahore Board, Faisalabad Board, DG Khan Board, Gujranwala Board, Sahiwal Board, Sargodha Board, Multan Board, Bahawalpur Board). According to the latest syllabus of PTB (Punjab Text Book Board), we provide the HSSC/ class-11 / FSc 1st year complete maths notes. These notes are in PDF format and easily readable and understandable. You can also download these maths 1st-year notes with a single click. See Also: The notes below consist of the exercise’s solution, and the explanations of examples before Exercise # 1.1 are also available. The other solved exercises of Chapter # 01, Number System, are also available

We know you are looking for FSC Part 1 or 1st Year Math solution notes in pdf to download. That’s why we have uploaded the best quality notes of 11th class Math notes. These Math notes include the solution of the complete book of Math 1st Year. In addition, you can easily download these Math notes or view them online. Click on the required exercise notes to download them. We uploaded the best quality ICS / FSC Part 1 Math notes that cover the requirements of the new syllabus of all Punjab boards. These notes are in proper format and written by qualified teachers, and these notes will significantly help you in your exam. Furthermore, These notes are in the best format so that you can print them and prepare yourself for exams. Chapter MCQs 1st Year All Subjects Notes in PDF Key Features of our 1st Year Math Solution Notes The notes we provide for the first year are specifically designed to meet the demands of the updated syllabus. Even the most talented students sometimes struggle with marking out the amount and what they must write in response to a question set in an exam. As a result of these Math Notes 1st Year, students will be able to give accurate answers to the questions in the examinations. Moreover, we provide all sorts of study material that can help you in the exam to secure high marks. That’s why we uploaded complete notes of all books of FSC Part 1. Additionally, You can get these notes from the given below link. Moreover, We divided these FSC Part 1 Math notes ...

CBSE, Department of Pre-University Education, Karnataka Class 11 Maths Solutions Guide Shaalaa provides solutions for CBSE, Department of Pre-University Education, Karnataka Class 11 Maths Solutions Digest. Shaalaa is undoubtedly a site that most of your classmates are using to perform well in exams. You can solve Class 11 Maths Book Solutions CBSE, Department of Pre-University Education, Karnataka textbook questions by using Shaalaa.com to verify your answers. This will help you practise better and become more confident. NCERT Solutions for Class 11 Maths Chapterwise List | Class 11 Maths Digest The answers to the NCERT books are the best study material for students. The following links are • • • • • • • • • • • • • • • • Appears in • • • • We also provide solutions for other subjects to help you top the exams. These NCERT solutions are specially curated with respect to the exam pattern and old papers. Find the best questions and solutions here. Click now to access it. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Concepts covered in Sets are Complement of a Set, Difference of Sets, Disjoint Sets, Element Count Set, Empty Set (Null or Void Set), Equal Sets, Finite and Infinite Sets, Intersection of Sets, Intrdouction of Operations on Sets, Open and Close Intervals, Power Set, Practical Problems on Union and Intersection of Two Sets, Proper and Improper Subset, Sets and Their Representations, Subsets, Union of Sets, Universal S...

Search for: • Balbharati Solutions for Class 12 • Balbharati Solutions for Class 11 • Balbharati Solutions for Class 10 Expand / Collapse • Balbharati Solutions for Class 9 • Balbharati Solutions for Class 8 • Balbharati Solutions for Class 7 • Balbharati Solutions for Class 6 • Balbharati Solutions for Class 5 • Maharashtra State Board Books • Balbharati Solutions for Class 12 • Balbharati Solutions for Class 11 • Balbharati Solutions for Class 10 • Balbharati Solutions for Class 9 • Balbharati Solutions for Class 8 • Balbharati Solutions for Class 7 • Balbharati Solutions for Class 6 • Balbharati Solutions for Class 5 • Maharashtra State Board Books Balbharti Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.1 Question 1. If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB. Solution: Let P(x, y) be any point on the required locus. Given, A(1, 3), B(2, 1) and PA = PB ∴ PA 2 = PB 2 ∴ (x – 1) 2 + ( y – 3) 2 = (x – 2) 2 + (y – 1) 2 ∴ x 2– 2x + 1 + y 2– 6y + 9 = x 2– 4x + 4 + y 2– 2y + 1 -2x – 6y + 10 = -4x – 2y + 5 ∴ 2x – 4y + 5 = 0 ∴ The required equation of locus is 2x – 4y + 5 = 0. Question 2. A(- 5,2) and B(4,1). Find the equation of the locus of point P, which is equidistant from A and B. Solution: Let P(x, y) be any point on the required locus. P is equidistant from A(- 5, 2) and B(4, 1). ∴ PA = PB ∴ PA 2 = PB 2 ∴ (x + 5) 2 + (y – 2) 2 = (x – 4) 2 + (y – 1) 2 ∴ x 2 + 10x + 25 + y 2 — 4y + 4 = x 2– 8x + ...