Ok, dudes but i like maths i can't c anyone doing such a thing. these r so much imp. ques. and wat the he is doing and on dat he is commenting. He is lucky dat he is not ahead of me, otherwise he would have simply gone. even i am not interested in fight wid this stupid guys.

1. Put the given equation equal to t. Cubing both sides, u'll get

    40 + 6t = t³. So, t³ - 6t - 40 = 0. Now solve the equation to get t=4.

2. x + y + 90 / xy = 1

or, x - xy + y = -90

or, 1 + (x-1) (1-y) = -90

or, (1-x) (y-1) = -91 = (-13) . ( 7)

So, x -1 = -13 and y - 1 = 7 and therefore, x = 14 and y = 8.

So, x + y + 2 / 4 = 14 + 8 + 2 / 4 = 6

3. I am not sure about the third question. I tried it but sry. Yeah, but i solved it in 2 sec using hit and trial method, i.e. x and y = 1 and 2.

And so ur answer would cum out to be 9.

If you **** maths, i bet u can even think of clearing IIT or even AIEEE and it very much seems like as if u don't wanna do it and then u say u wanna be an engineer. i **** you on that. U can just be selected in For ii believe landra, doaba etc etc. if u could answer these questions then only reply, otherwise jus stop commenting!!!

1. If w = z / [ 1 - (1/3)i ] and |w| = 1, then z lies on:
(a) circle (b) ellipse (c) parabola (d) straight line

2. sin^-1{(z-1)/i}, where z is non-real, can be the angle of a triangle if
(a) Re(z) = 1, Im(z) = 2
(b) Re(z) = 1, -1 < Im(z) < 1
(c) Re(z) + Im(z) = 0
(d) None of these

3. If A + iB = C tan (x + iy), then tan2x =

(a) 2CA / C² - A² - B²   (b) 2CA / C² + A² + B²   (c) 2CA / C + A + B   (d) none of these

So, none could solve this. Answer to the question was no roots. Check it out now!

HINT - USE LOG

if u r preparing for JEE and other engineering entrance examinations then TATA MC'GRAW IS BEST.

z is a complex number....

how do u solve it?

           it is

       (e)^{-1/z^2} = 0

       yes, z is a complex number.

           it is

       (e)^{-1/z^2} = 0

how do you solve it?

V = kQ/R where Q = P V'' = P ( V' - V' / 8 ) = 7 PV' / 8           (V = Potential at O, V'' = Final Volume and V' = Initial Volume)

So, V = k (7PV') / 8R 

As, V' = 4πR³/ 3

So, V = 7PR² / 24ε_o 

bi + aj

When n = 1, tan x1 = 1.414 and similarly when n=2 tanx1 + tanx2 = 2

which implies tan x2 = 0.586.

from these values of tan cos x1 = 0.577 and cos x2 = 0.862

since x belogs to 0 to pi/2 , thus cosx can have a maximum of value 1.

add cos x1+ cos x2 > 1 which implies the least positive value of k would be 0.577. reply if still not satisfied!

Calculate the force between the two spheres using F = GMm/r^2

After calculating the force, now u can find out the acceleration using a = F/M.

Now use eq. of motion S = ut + (1/2)at^2 and calculate the time (remember u = 0).

1. Electric field at the center of shell is zero (sice charge enclosed is zero). If a charge is brought to close to the shell, field at center still remains zero. This was the case of conducting. But if the shell is non-conducting it is non-zero.

2. zero........When it is deshaped still the field remains zero only (use gauss's law).

3. No, basically bcoz the charge distribution takes place on the surface.

4. If the balloon is spherical field field inside would be zero even if it is not spherical but oval.