Ncert Solutions Maths class 11th

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P
Payal Gupta

Contributor-Level 10

6. (i) (c)

(ii) (a)

(iii) (d)

(iv) (b)

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Payal Gupta

Contributor-Level 10

5. (i) A = {1, 3, 5, 7, …}

(ii) B = {0,1,2,3,4} (as 92 =4.5 and 12 = –0.5)

(iii) x2 ≤ 4

x2 ≤ 22

x ≤ ± 2

So, C = {–2, –1,0,1,2}

(iv) D = {L, O, Y, A}

(v) E = {February, April, June, September, November}

(vi) F = {b, c, d, f, g, h, j}

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Payal Gupta

Contributor-Level 10

4. (i) {3,6,9,12}= {3 * 1, 3 * 2, 3 * 3, 3 * 4}

= {x : x = 3n, n is natural number and 1≤ n ≤ 4}

(ii) {2,4,8,16,32}= {21, 22, 23, 24, 25}

= {x : x = 2n, n is natural number and 1 ≤ n ≤ 5}

(iii) {5,25,125,625}= {51, 52, 53, 54}

= {x : x = 5n, n is natural number and 1 ≤ n ≤ 4}

(iv) {2,4,6, }= {2 * 1, 2 * 2, 2 * 3, …}

= {x : x = 2n, n is a natural number}

(v) {1,4,9, …, 100}= {12, 22, 32, …, 102}

= {x : x = n2, x is a natural number and 1 ≤ n ≤ 10}

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Payal Gupta

Contributor-Level 10

(i) A = {– 3, –2, –1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = {1, 2, 3, 4, 5}

(iii) C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv) x = Prime number which are divisor of 60

Factors of 60 are 1,2,3,4,5,6,10,12,15,20,30,60

Hence, x = 2, 3, 5

?D = {2, 3, 5}

(v) E = {T, R, I, G, O, N, M, E, Y}

(vi) F = {B, E, T, R}

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Payal Gupta

Contributor-Level 10

2. (i) 5  A

(ii) 8  A

(iii) 0  A

(iv) 4  A

(v) 2 A

(vi) 10  A

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Payal Gupta

Contributor-Level 10

1. (i) The collection of all months of a year with J as initial are January, June and July. Hence, it is a well-defined and is therefore a set January, June, July .

(ii) The collection of ten most talented writers of India is not well-defined as it may vary from one person to another. Hence, it is not a set.

(iii) The team of 11 best-cricket batsmen of the world is not well-defined as it may vary from one parson to another as they may vary from one person to another. Hence, it is not a set.

(iv) The collection of all boys in your class is well-defined as your-class is fixed. Hence, it is a set.

(v) The collection of all natural numbersless than

...more

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A
alok kumar singh

Contributor-Level 10

32.

Since P (a, b) is the mid-point of the line segment say AB with points A (0, y) and B (x, 0) we can write,

(a, b)= (x+02, y+02)

a=x2

b=y2

x=2a  
y=2b

So, the equation of line with x and y intercept 2a and 2b using intercept form is

x2a+y2b=1

xa+yb=2

Hence, proved

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A
alok kumar singh

Contributor-Level 10

31. 

Assuming the price per litre say P in x-axis and the corresponding demand say D in y-axis, we have two point (14, 980) and (16, 1220) in xy plane. Then the points (P, D) will satisfy the equation.

⇒D980=12209801614 (P14)

⇒D980=2402 (P14)

⇒D120 (P14)+980

⇒D120P1680+980

⇒D=120P700

Which is the required relation

Where P = 17, we have

D = 120 * 17 – 700

D = 1340

Hence, the owner can sell 1340 litres of milk weekly at 17/litre

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