Most Important Selected Qs for JEE AdvancedMathematicsComplex Number
Match the following array |c|l|c|c| & Column-I & & Column - II (A) & array l f(z) is a complex valued function f(z)=(a+i b) z where a, b R and |a+i b|= 1 2 . It has the property that f ( z ) is always equidistant from array & (p) & 5 (B) & array l 0 and z, then a - b is equal to The number of all positive integers n=2^a 3^b(a, b 0) such that n^6 does not divide 6^n is array & (q) & 8 (C) & array l A is the region of
Options
- A( A ) ( q ) ;( B ) ( p ) ;( C ) ( s ) ;( D ) ( r )
- B( A ) ( t ) ;( B ) ( r ) ;( C ) ( p ) ;( D ) ( s )
- C( A ) ( r ) ;( B ) ( s ) ;( C ) ( p ) ;( D ) ( q )
- D( A ) ( s ) ;( B ) ( q ) ;( C ) ( r ) ;( D ) ( p )
Correct answer
B. ( A ) ( t ) ;( B ) ( r ) ;( C ) ( p ) ;( D ) ( s )
Step-by-step solution
(A) aligned & |a+i b||z|=|z||(a-1)+i b| 1 2 = (a-1)^2+b^2 and a^2+b^2= 1 2 & 1-2 a=0 a= 1 2 and b^2= 1 4 b= 1 2 & a-b=0 aligned (B) We must have either 6 a 2^a 3^b or 6 b 2^a, 3^b If b=0 , then 6 a 2^a a=1.2,3,4 If a =0 , then 6 ~b 3^ b b =1,2 suppose a 0 and b 0 if 6 a 2^ a 3^ b thus 6^a 2^a .3 so 2 a 2^a (not possible) similarly if 6 ~b 2^ a .3^ b then 3 ~b 3^ b not possible only solutions are 2,4,8,16,3,9 (C) Re ( z 4 ) (0 . 1). lm ( z 4 ) [0,1) means that if z=a+i b than a, b (0,4) Now 4 a-i b = 4 a a^2+b^2 + 4