Show that :

|z-w|² <= (|z| - |w|)² +(arg z - arg w)²

z and w are complex nos...

please please send it to :(a bit tricky)

 

1)No. of ways of selecting 1st couple = ¹⁵C_1.¹⁵C₁

No. of ways of selecting 2nd couple =¹⁴C₁¹⁴C₁

and so on... therefore total ways = ¹⁵C_1.¹⁵C₁+¹⁴C1_.¹⁴C₁ +and so on till ¹C₁.¹C₁

= 15^2+14^2+13^2..............1^2 = n.(n+1)(2*n+1)/6 = 15.16.31/6 = 1240 (a)

 

2) No. of students answering atleast n questions wrong = 2ⁿ

No. of students answering atleast n-1 questions wrong = 2^n-1

No. of students answering atleast n-2 questions wrong = 2^n-2

No. of students answering atleast 1 questions wrong = 2¹

 

No. of students answering 1 questions wrong = 2ⁿ-2^n-1*1...................similarly for other till n question wrong.

 

 

therefore total no. of wrong ans. = On simplifying......2ⁿ+2^n-1+2^n-2+........2+1

solving as G.P we get n = 13

A signum function is one which is defined as follows:

 

f(x) = |x|/x  for  x not equal to 0

     =  0 for x = 0 

 

it range lies within {0,1,-1) .It therefore returns 0 for 0  input,1 for +ne nos. and -1 for -ve nos.

 

 

1)

_{[ x][0}]  x sin 1/x

 

x                  1/x                   sin1/x

inf.                0                         0 

2/pi             pi/2                       1

1/pi             pi                          0

2/3 pi          3pi/2                      -1

1/2pi           2pi                          0

2/5pi           5pi/2                       1

 

 

from above info . we conclude dat there's no particular value for sin1/x it oscillates between -1 and 1.......

 

Left hand limit:   

_{[ x][0^-]} xsin1/x = (x tends to 0)* (sin1/x oscillating between -1 and 1) = 0

 

similarly for R.H.L which comes out to be equal to L.H.L = 0

Hence the ans. = 0

ⁿ_n = n!/(n-n)! 

 

 ⁿ_n = n!/(0)!                               (ⁿ_n = n!)

 

 n! = n!/0!

1 = 1/0!

0! = 1...............

C₂H₅OH+4I₂+6NaOH      CHI₃            + 5NaI+ 5H₂O + HCOONa

                                   pale yellow

                                     crystals

 

 

 

3 is a super prime........coz it is the 2nd prime no.(2 itself is a prime no.) 

5 is a super prime ........coz it is the 3rd prime no.(3 itself is a prime no.)

7 is not a super prime ..coz it is the 4th prime no.( where 4 is not  a prime no.)

 

i hope im clear enugh.... 

 

put the expansion for log (x+1) = x - x^2/2+x^3/3............

therefore the given question can be written as:

 

{(1+x)log(1+x)}/x-1/x = 

 

 

_{[ x][0 ]}{(1+x)(x-x^2/2+x^3/3-----)}/x -   _{[x ][0 ]}   1/x   solve the first part:

_{[x ][0 ]}(1+x)(1-x+x^2/3........) = 1

 

solve the 2nd part: .......????   

 

try to study these cones...parabola,ellipse nd hyperbola using a table. Make a table nd then try to compare the coordinates,eqns etc of each cone eg:

                  Focus    Vertex etc...

y2=4ax        (a,0)         (0,0)

y2=-4ax       (-a,0)        (0,0) and so on........