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Let S 1 = { z 1 ∈ C : | z 1 − 3 | = 1 2 }  and S 2 = { z 2 ∈ C : | z 2 − | z 2 + 1 | | = | z 2 + | z 2 − 1 | | } .  Then, for z 1 ∈ S 1 and s 2 ∈ S 2 ,  the least value of   | z 2 − z 1 | i s :

Option 1 -

0

Option 2 -

1 2

Option 3 -

3 2

Option 4 -

5 2

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

      S 1 = { z , ∈ c : ⌊ z 1 − 3 ⌋ = 1 2 } a n d     S 2 = { z 2 ∈ c : | z 2 − | z 2 + 1 | | = | z 2 + | z 2 − 1 | | }

    Then, for z 1 ∈ S 1  and z 2 ∈ S 2 ,  the least value of |z2 – z1|

    | z 2 + | z 2 − 1 | | 2 = | z 2 − | z 2 + 1 | | 2      

    ⇒ ( z 2 + z ¯ 2 ) ( | z 2 − 1 | + | z 2 + 1 | − 2 ) = 0      

    ∴ z 2 + z ¯ 2 = 0     o r     | z 2 − 1 | + | z 2 + 1 | − 2 = 0          

    ∴ z2 lies on imaginary axis or on real axis with in [-1, 1]

    also | z 1 − 3 | = 1 2  lie on circle having centre 3 and radius   1 2

                 

    Clearly | z 1 − z 2 | m i n = 5 2 − 1 = 3 2

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  l = ∫ − π 2 π 2 ( [ x ] + [ − s i n x ] ) d x . (i)

I = ∫ − π 2 π 2 ( [ − x ] + [ s i n x ] ) d x . (ii)

by using property ( ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x )

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Let J n , m = ∫ 0 1 2 x n x m − 1 d x , ∀ n > m and n, m ∈ N . Consider a matrix A = [ a i j ] 3 × 3 where a i j = { J 6 + i , 3 − i + 3 , 3 , i ≤ j 0 , i > j .       T h e n     | a d j     A − 1 | i s :  
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If α + β + = 2π, then the system of equations

x + ( c o s γ ) y + ( c o s β ) z = 0        

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If the system of linear equations

2x + y – z = 3

x – y – z = a

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Given 2x + y – z = 3         . (i)

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