* * * * *

.

sq rt (1 - cos2x) = 2^1/2 sinx

                  * * * * *

.

we can also write the equn as

let tanx = t

sec²x dx = dt

so equn becomes

I = ln t  + (t²/2) + c

* * * * *

.

let,

e^x + x^e = t

(e^x  + ex^e-1 )dx  = dt

e ( e^x-1 +  x^e-1) dx = dt

so the equn becomes

* * * * * *

.

i think there is a misprint.

in 1st equn limit must be from 0 to T not from 6 to T.

ans then comes to be *(a)*1.5 I

let  4^sinx = a &  3^secx = b

so equation becomes,

5a²-2b = 2

and

a+b = 11

eleminating 'b' from both equn.

5a²+2a-24 = 0

a = 4 , (-24/5)

as a is always positive

4^sinx = 4

sinx = 1

  n belongs to integers.

pls rate if correct.............

as tan(x/2) = cosec x  -  cot x

given, tan(x/2) = cosec x  -  sin x

so =>   cot x = sin x

sin²x = cosx

=>  cos²x + cosx -1 = 0

=>

using half angle formula

=>

pls verify the answer.............................................................

.......

20(1-cos²@)) + 21cos@ - 24 = 0

solving quad. in cos@ we get,

cos@  =  1/4,   4/5

given     .

=>    

so cos@ = 4/5

sin@ = (-3/5) ......(sin will be -ve in this range of @)

using half angle formula

cot(@/2) = -3 , (-1/3)

so cot @/2 = -1/3

is it right ans.......................

culb the first & last term

arey, is that bowl fixed or not

it's a standard result.....

as f(x).f(1/x) = f(x) + f(1/x)

so f(x) must be of the form f(x) = 1 + xⁿ

f(3) = 28

1+ 3ⁿ  = 28

=> n = 3

therefore, f(x) = 1+ x³

f(4) = 1+ 4³  = 65

f(4) = 65

RATE IF USEFUL................................................

by its graph........

r

multiply & divide the equn by sin(pi/7)

we get,

pls. rate if useful

2*(2.5 * 1000) cos@ = 4*1000

=>  cos@ = 4/5

=> @ = 37 ⁰

AD = 0.4 m

=>  AB = 0.5 m

so length of rope = AB + BC = 1m

is the ans,

where

PLEASE VERIFY THE ANS.  if wrong tell the correct ans....

@ vignesh

arey, how can the initial velocity of rocket is zero.

we have to give some initial velocity to project the rocket upwards

length of angle bisector =

and

      where s = (a+b+c)/2

eleminate cos(A/2) u will get the ans.

nudge me if ny problem ...............

let velocity of ball (solid sphere) just after collision be 'u' m/s

u = ev₀.................(i)

just after collision the ball will have 'u' in opposite to the direction of  initial motion & angular velocity in same direction to that of initial motion

as wall is smooth, there is no loss in energy due to friction,

let 'v' be the velocity of the ball when it starts pure rolling in opposite dir.

friction = ma

=> 

equation of torque..

mar = (0.4)mr²@          (@ = ang. acc)

a = 0.4r@........(ii)

equation of motion

v = u - at ...........(iii)

v/r = -v₀/r + @t..............(iv)    ( t = time after which ball starts pure rolling after collision)

from (ii),(iii),(iv) eleminating 't'

7v = 5u - 2v₀

from (i)

substituting the values,

v = 1.5 m/s

PLEASE RATE.......