Sets

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New answer posted•a year ago
A
a year ago

2x=tan (π/9)+tan (7π/18)
=sin (π/9+7π/18) / cos (π/9)cos (7π/18)
=sin (π/2) / cos (π/9)cos (7π/18)
=1 / cos (π/9)cos (7π/18)
=1 / cos (π/9)sin (π/2−7π/18)
=1 / cos (π/9)sin (π/9)
⇒x=1 / 2cos (π/9)sin (π/9)
=1 / sin (2π/9)=cosec (2π/9)

Again 2y=tan (π/9)+tan (5π/18)
⇒2y=sin (π/9+5π/18) / cos (π/9)cos (5π/18)
=

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New answer posted•a year ago
New answer posted•a year ago
A
a year ago

n (C) = 73, n (T) = 65, n (C ∩ T) = x
n (C ∪ T) ≤ 100
⇒ n (C) + n (T) - n (C ∩ T) ≤ 100
⇒ x ≥ 38
n (C ∩ T) ≤ min (n (C), n (T) ⇒ x ≤ 65
⇒ 38 ≤ x ≤ 65

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
A
a year ago

Yes, the Bihar BEd CET Answer Key is released separately for all paper sets – A, B, C, and D. Candidates must download the key according to their question paper set.

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New answer posted•a year ago
A
a year ago

A = {2,4,6,8, . . .50} ⇒ 25 element
A = {7,14,21, . . . .49} ⇒ 7 elements
A ∩ B = {14,28,42} = 3 elements
Required number of elements = 25 + 7 - 3 = 29

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New answer posted•a year ago
A
a year ago

Sum of elements A ∩ ( B ∪ C ) = 2 7 4 × 4 0 0  

In set B numbers of the form 9k + 2 are {101, 109, .992}

∴ s u m = 1 0 0 2 ( 1 0 1 + 9 9 2 ) = 1 0 0 × 1 0 9 3 2 . . . . . . ( i )          

Another possible number is 9k + 5 forms are {104, .995}

∴ s u m = 1 0 0 2 ( 1 0 4 + 9 9 5 ) = 1 0 0 2 × 1 0 9 9 . . . . . . . . . ( i i )  

∴ T o t a l = 1 0 0 2 × [ 1 0 9 3 + 1 0 9 9 ] = 1 0 0 × 1 0 9 6 = 2 7 4 × 4 × 1 0 0 = 2 7 4 × 4 0 0           

∴  possible value of l = 5

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New answer posted•a year ago
A
a year ago

2040 = 23 × 3 × 5 × 17

If HCF between {n & 2040} = 1

∴   n can not be multiple of 2, 3, 5, 17.

Let S(n) denote sum of numbers divisible by n.

S ( 2 ) = 2 + 4 + − − − − + 1 0 0 = 2 × 5 0 × 5 1 2 = 5 0 × 5 1              

S ( 3 ) = 3 + 6 + − − − + 9 9 = 3 × 3 3 × 3 4 2 = 3 3 × 5 1             

S ( 5 ) = 5 + 1 0 − − − + 1 0 0 = 5 × 2 0 × 2 1 2 = 5 0 × 2 1            

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New answer posted•a year ago
A
a year ago

  1 1 n − 9 n > 1 0 n             

⇒ ( 1 0 + 1 ) n − ( 1 0 − 1 ) n > 1 0 n           

->For  n ≥ 5     the inequality Satisfies.

For n= 4.             2 [4 × 103 + 4 × 10] > 104

->8 × 1010 > 104 which is a contradiction

∴ n ≥ 5 all values

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