Class 11 Chapter 1 Set Exercise 1.1 NCERT Solutions

Q1. Which of the following are sets? Justify your answer.

(i) The collection of all the months of a year beginning with the letter J.

(ii) The collection of ten most talented writers of India.

(iii) A team of eleven best-cricket batsmen of the world.

(iv) The collection of all boys in your class.

(v) The collection of all natural numbers less than 100.

(vi) A collection of novels written by the writer Munshi Prem Chand.

(vii) The collection of all even integers.

(viii) The collection of questions in this Chapter.

(ix) A collection of most dangerous animals of the world.

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Similar Questions for you

R
Raj Pandey

(|x| - 3)|x + 4| = 6

Case (i) x < -4
(-x - 3) (- (x + 4) = 6
(x + 3) (x + 4) = 6 ⇒ x² + 7x + 12 = 6 ⇒ x² + 7x + 6 = 0
(x + 1) (x + 6) = 0 ⇒ x = -6 (since x < -4)

Case (ii) -4 ≤ x < 0
(-x - 3) (x + 4) = 6
⇒ -x² - 7x - 12 = 6
⇒ x² + 7x + 18 = 0
The discriminant is D = 7² - 4 (1) (18) = 49 - 72 < 0, so no real solution.

Case (iii) x ≥ 0
(x - 3) (x + 4) = 6
⇒ x² + x - 12 = 6
⇒ x² + x - 18 = 0
x = [-1 ± √ (1² - 4 (1) (-18)] / 2 = [-1 ± √73] / 2
Since x ≥ 0 ⇒ x = (√73 - 1) / 2
Only two solutions.

V
Vishal Baghel

Given    n = 2x. 3y. 5z . (i)

y + z = 5 & 1 y + 1 z = 5 6

On solving we get y = 3, z = 2

So, n = 2x. 33. 52

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

P
Payal Gupta

Let A = {a, b, c}, B = {1, 2, 3, 4, 5} n (A × B) = 15

x = number of one-one functions from A to B.

=5C3.3!=60

y = number of one-one functions for A to (A × B)

=15C3.3!=15×14×13=2730

Yx=2730602y=91x

P
Payal Gupta

2cos  (x2+x6)=4x+4x

2LHS2LHS=2&RHS=2x=0onlythenLHS=2also

RHS  2

P
Payal Gupta

66. Given series is 1× 2× 3 + 2× 3 ×4 + 3× 4 ×5 + … to n term

an = (nth term of A. P. 1, 2, 3, …) ´× (nth terms of A. P. 2, 3, 4) ×

i e, a = 1, d = 2- 1 = 1i e, a = 2, d = 3- 2 = 1

(nth term of A. P. 3, 4, 5)

i e, a = 3, d = 3 -4 = 1.

= [1 + (n -1) 1] ×[2 + (n -1):1]× [3 + (n- 1) 1]

= (1 + n -1)×(2 + n -1)×(3 + n -1)

= n (n + 1)(n + 2)

= n(n2 + 2n + n + 2)

=n3 + 2n2 + 2n.

Sn = ∑n3 + 3 ∑n2 + 2 ∑n

=[m(n+1)2]2+3n(n+1)(2n+1)6+2n(n+1)2

n(n+1)2[n(n+1)2+33(2n+1)+2]

n(n+1)2[n2+n2+2n+1+2]

=n(n+1)2[n2+n+2×2n+2×32]

=n(n+1)2[x2+n+4x+62]

=n(n+1)2×[n2+5n+6]2

=n(n+1)4[n2+2n+3n+6]

=n(n+1)4[n(n+2)+3(n+2)]

=n(n+1)(n+2)(n+3)4

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