If nC8 = nC2. find nC2 Solution: We know that, nCa = nCb then a = b or n = a + b Therefore, nC8 = nC2, since 8≠2 , we have 8 + 2 = 10 nC2 = 10C2 = 10!/2!(10-2)! = 10!/2!8! = 10x9x8/2x1x8 = 45 Determine n if (i) (2nC3 :… Continue reading Permutation and Combination – Class XI – Exercise 7.4

# Tag: NCERT class 11 Mathematics

## Complex Numbers and Quadratic Equations – Class XI – Exercise 5.2

1.Find the modulus and the argument of the complex number z = -1 – i √3 Solution: z = -1 - i√3 Let r cosθ = -1 and r sinθ = -√3 On squaring and adding, we get (r cosθ)2 + ( r sinθ)2 = (-1)2 + (-√3)2 r2(cos2θ + sin2θ) = 1 +… Continue reading Complex Numbers and Quadratic Equations – Class XI – Exercise 5.2

## Complex Numbers and Quadratic Equations – Class XI – Exercise 5.1

1: Express the given complex number in the form a + ib: (5i)(-3/5 i) Solution: (5i)(-3/5 i) = -5 x 3/5 x i x i = -3i2 = -3(-1) = 3 2: Express the given complex number in the form a + ib: i9 + i 19 Solution: i9 + i 19 = i4x2+1 +… Continue reading Complex Numbers and Quadratic Equations – Class XI – Exercise 5.1

## Linear Inequalities – Class XI – Exercise 6.1 [1 0 – 26]

Solve the given inequality for real x: 3(x-2)/5 ≤ 5(2-x)/3 Solution: The given inequality is 3(x-2)/5 ≤ 5(2-x)/3 ⇒ 3(x-2)/5 ≤ 5(2-x)/3 ⇒ 9(x-2) ≤25(2-x) ⇒ 9x – 18 ≤ 50 – 25x ⇒ 34x – 18 ≤ 50 ⇒ 34x ≤ 68 ⇒ 34x/34 ≤ 68/34 ⇒ x ≤ 2 Thus, all real numbers… Continue reading Linear Inequalities – Class XI – Exercise 6.1 [1 0 – 26]

## Permutation and Combination – Class XI – Exercise 7.3

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Solution: 3-digit numbers have to be formed using the digits 1 to 9. Here, the order of the digits matters. Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits… Continue reading Permutation and Combination – Class XI – Exercise 7.3

## Permutation and Combination – Class XI – Exercise 7.1

1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) Repetition of the digits is allowed? (ii) Repetition of the digits is not allowed? Solution: (i) There will be number of ways of filling 3 vacant places in succession by the given five digits. Here,… Continue reading Permutation and Combination – Class XI – Exercise 7.1

## THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.5 – Class IX

Construct parallelograms ABCD with the following measurements: (a) AD = 4.2 cm, DC = 5.8 cm, ∟D = 43˚ (b) AB = 43. Cm, BC = 3.2 CM, ∟C= 120˚ (c) AD =4.2 cm, DC = 5.8 cm, AC = 3.4cm (d) AD = 4.3 cm, DC = 5.7 cm, DB = 7cm (e) AD… Continue reading THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.5 – Class IX

## Linear Inequalities – Class XI – Exercise 6.1 [1 – 10]

Solve 24x < 100, when (i) x is a natural number (ii) x is an integer Solution: We have the given inequality 24x < 100. then, 24x/24 < 100/24 Thus, x < 25/6 (i) We know that 1, 2, 3, and 4 are the only natural numbers less than 25/6. Thus, when x is a… Continue reading Linear Inequalities – Class XI – Exercise 6.1 [1 – 10]

## Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 2]

11: Prove the following by using the principle of mathematical induction for all n€ N: 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + … +1/n(n+1)(n+2) = n(n+3)/4(n+1)(n+2) Solution: Let the given statement be P(n), i.e., P(n): 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + … +1/n(n+1)(n+2) = n(n+3)/4(n+1)(n+2) For n = 1, we have P(1): 1/1.2.3 = 1.(1+3)/4(1+1)(1+2) =… Continue reading Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 2]

## Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 1]

Prove the following by using the principle of mathematical induction for all n € N: 1 + 3 + 32+ ………..+3n-1 = (3n – 1)/2 Solution: Let the given statement be P(n) , i.e., P(n): 1 + 3 + 32+ ………..+3n-1 = (3n – 1)/2 For n = 1, we have P(1): (31 – 1)/2… Continue reading Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 1]