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a month ago

Sets Class 11th

The number of elements in the set {x ∈ R : (|x| - 3)|x + 4| = 6} is equal to :

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Raj Pandey

Contributor-Level 9

(|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.

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New answer posted

a month ago

Sets Class 11th

A natural number has prime factorization given by n = $2^{x} 3^{y} 5^{z},$ where y and z are such that y + z = 5 and $y^{- 1} + z^{- 1} = \frac{5}{6}, y > z .$ Then the number of odd divisors of n, including 1, is:

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Vishal Baghel

Contributor-Level 10

Given n = 2^x. 3^y. 5^z . (i)

$y + z = 5 & \frac{1}{y} + \frac{1}{z} = \frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

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

Hence no. of divisors including 1 = 12

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New answer posted

2 months ago

Sets Class 11th

Let x denote the total number of one-one function from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions form the set A to the set A * B. Then

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Payal Gupta

Contributor-Level 10

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.

$=^{5} C_{3} . 3! = 60$

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

$=^{15} C_{3} . 3! = 15 * 14 * 13 = 2730$

$\frac{Y}{x} = \frac{2730}{60} \Rightarrow 2 y = 91 x$

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New answer posted

2 months ago

Sets Class 11th

The number of elements in the set S = ${x \in R : 2 c o s (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} i s :$

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Payal Gupta

Contributor-Level 10

2cos $(\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$

$- 2 \leq L H S \leq 2 L H S = 2 & R H S = 2 \Rightarrow x = 0 o n l y \to t h e n L H S = 2 a l s o$

RHS $\geq$ 2

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New answer posted

3 months ago

Sets Maths NCERT Exemplar Solutions Class 12th Chapter One

Let $⋃_{i = 1}^{50} X_{i} = ⋃_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y i$ contains 5 elements. If each element of the set $T$ is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's then $n$ is equal to:

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alok kumar singh

Contributor-Level 10

$f (x) = \int_{1}^{3} ? \frac{\sqrt[]{x} d x}{(1 + x)^{2}} = \int_{1}^{\sqrt[]{3}} ? \frac{t . 2 t d t}{{(1 + t^{2})}^{2}}$ (put $\sqrt[]{x} = t$ )

$= {(- \frac{t}{1 + t^{2}})}_{1}^{\sqrt[]{3}} + {({t a n}^{- 1} ? t)}_{1}^{\sqrt[]{3}}$ [Applying by parts]

$\begin{array}{r} = - (\frac{\sqrt[]{3}}{4} - \frac{1}{2}) + \frac{π}{3} - \frac{π}{4} \\ = \frac{1}{2} - \frac{\sqrt[]{3}}{4} + \frac{π}{12} \end{array}$

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New answer posted

4 months ago

Sets Sets

66. 1 * 2 * 3 + 2 * 3 * 4 + 3 * 4 * 5 + .

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Payal Gupta

Contributor-Level 10

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

a_n = (n^th term of A. P. 1, 2, 3, …) ´* (n^th 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.

S_n = ∑n³ + 3 ∑n² + 2 ∑n

$= {[\frac{m (n + 1)}{2}]}^{2} + \frac{3 n (n + 1) (2 n + 1)}{6} + \frac{2 n (n + 1)}{2}$

= $\frac{n (n + 1)}{2} [\frac{n (n + 1)}{2} + \frac{3}{3} (2 n + 1) + 2]$

= $\frac{n (n + 1)}{2} [\frac{n^{2} + n}{2} + 2 n + 1 + 2]$

$= \frac{n (n + 1)}{2} [\frac{n^{2} + n + 2 * 2 n + 2 * 3}{2}]$

$= \frac{n (n + 1)}{2} [\frac{x^{2} + n + 4 x + 6}{2}]$

$= \frac{n (n + 1)}{2} * \frac{[n^{2} + 5 n + 6]}{2}$

$= \frac{n (n + 1)}{4} [n^{2} + 2 n + 3 n + 6]$

$= \frac{n (n + 1)}{4} [n (n + 2) + 3 (n + 2)]$

$= \frac{n (n + 1) (n + 2) (n + 3)}{4}$

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New answer posted

4 months ago

Sets Sets

43. Evaluate $\sum_{k = 1}^{11} (2 + 3^{x}) .$

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Payal Gupta

Contributor-Level 10

43. $\sum_{}^{11} (2 + 3^{x}) = (2 + 3^{1}) + (2 + 3^{2}) + . . . . . . . . . + (2 + 3^{11})$

$x = 1$ [2+2+ ….+ (11lines)] + (31+ 32+……+311)

= $11 * 2 + \frac{3 (3^{11} - 1)}{3 - 1} [? s_{n} = \frac{a (r^{n} - 1)}{r - 1}, r \geq 1]$

= $22 + \frac{3 (3^{11} - 1)}{2}$

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New answer posted

4 months ago

Sets Class 11th

64. In a survey it was found that 21 people liked product A, 26 liked product B and29 liked product C. If 14 people liked products A and B, 12 people liked productsC and A, 14 people liked products B and C and 8 liked all the three products.Find how many liked product C only.

0 Follower 10 Views

Payal Gupta

Contributor-Level 10

64. Let A, B and C be the set of people who like product A, B and C respectively.

Then,

Number of people who like product A, n (A) = 21

Number of people who like product B, n (B) = 26

Number of people who like product C, n (C) = 29.

Number of people likes both product A and B, n (A $\cap$ B) = 14

Number of people likes both product A and C, n (A $\cap$ C) = 12

Number of people likes both product B and C, n (B $\cap$ C) = 14.

No. of people who likes all product, n (A $\cap$ B $\cap$ C) = 8

a→n (A $\cap$ B)

b→n (A $\cap$ C)

d→n (B $\cap$ C)

c→n (A $\cap$ B $\cap$ C)

From the above venn diagram we can see that,

Number of people who likes product C only

= n (C) - b - d + c

= n (C) - n (A $\cap$ C) - n (B $\cap$ C) + n (A $\cap$ B $\cap$ C)

= 29 - 12 - 14 +

...more

64. Let A, B and C be the set of people who like product A, B and C respectively.

Then,

Number of people who like product A, n (A) = 21

Number of people who like product B, n (B) = 26

Number of people who like product C, n (C) = 29.

Number of people likes both product A and B, n (A $\cap$ B) = 14

Number of people likes both product A and C, n (A $\cap$ C) = 12

Number of people likes both product B and C, n (B $\cap$ C) = 14.

No. of people who likes all product, n (A $\cap$ B $\cap$ C) = 8

a→n (A $\cap$ B)

b→n (A $\cap$ C)

d→n (B $\cap$ C)

c→n (A $\cap$ B $\cap$ C)

From the above venn diagram we can see that,

Number of people who likes product C only

= n (C) - b - d + c

= n (C) - n (A $\cap$ C) - n (B $\cap$ C) + n (A $\cap$ B $\cap$ C)

= 29 - 12 - 14 + 8

= 11

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New answer posted

4 months ago

Sets Sets

63. In a survey of 60 people, it was found that 25 people read newspaper H, 26 readnewspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T,8 read both T and I, 3 read all three newspapers. Find:

(i) The number of people who read at least one of the newspapers.

(ii) The number of people who read exactly one newspaper

0 Follower 17 Views

Payal Gupta

Contributor-Level 10

63. Let H, T and I be of people who reads newspaper H, T and I respectively.

Then,

number of people who reads newspaper H, n (H) = 25.

number of people who T, n (T) = 26.

number of people who I, n (I) = 26

number of people who both H and T, n (H $\cap$ I) = 9

number of people who both H and T, n (H $\cap$ T) = 11

number of people who both T and I, n (T $\cap$ I) = 8

number of people who reads all newspaper, n (H $\cap$ T $\cap$ I) = 3.

Total no. of people surveyed = 60

The given sets can be represented by venn diagram

(i) The number of people who reads at least one of the newspaper.

in (H∪T $\cup$ I) = n (H) + n (T) + n (I) n (H $\cap$ T) n (H $\cap$ I) n (T $\cap$ I) + n (H $\cap$ T $\cap$ I)

= 25 + 26 + 26 11 9 8 + 3

= 80 2

...more

63. Let H, T and I be of people who reads newspaper H, T and I respectively.

Then,

number of people who reads newspaper H, n (H) = 25.

number of people who T, n (T) = 26.

number of people who I, n (I) = 26

number of people who both H and T, n (H $\cap$ I) = 9

number of people who both H and T, n (H $\cap$ T) = 11

number of people who both T and I, n (T $\cap$ I) = 8

number of people who reads all newspaper, n (H $\cap$ T $\cap$ I) = 3.

Total no. of people surveyed = 60

The given sets can be represented by venn diagram

(i) The number of people who reads at least one of the newspaper.

in (H∪T $\cup$ I) = n (H) + n (T) + n (I) n (H $\cap$ T) n (H $\cap$ I) n (T $\cap$ I) + n (H $\cap$ T $\cap$ I)

= 25 + 26 + 26 11 9 8 + 3

= 80 28

= 52

(ii) From venn diagram,

a = number of people who reads newspapers H and T only.

b = number of people who reads newspapers H and I only.

c = number of people who reads newspapers T and I only.

d = number of people who reads all newspaper.

So, n (H $\cap$ T) = a + d.

n (H $\cap$ I) = b + d

n (T $\cap$ I) = c + d

So, a + d + c + d + b + d = n (H $\cap$ T) + n (H $\cap$ I) + n (T $\cap$ I)

a + d + c + b + 2d = 11 + 9 + 8

a + b + c +d = 28 - 2d

= 28 2 3 [d = n (H $\cap$ T $\cap$ I) = 3]

= 28 - 6

= 22

Number of people who reads exactly one newspaper

= Total no. of people - No. of people who reads more than one newspaper

= 52 - (a + b +c + d)

= 52 - 22

= 30

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