MATH A2

[preity]

could someone help me solve this ques...it's from may/june 09 paper...

solve the equation ln(2+e^-x)=2

                           2+e^-x=e^2
                          
    
                           e^-x=e^2-2 pls help me to cotinue..im stuck..n i also need help for Q3,5 and 8...

[astarmathsandphysics]

Log both sides -x=ln(e^2-2) then times both sides by -1. Will got up soon to post more

[astarmathsandphysics]

Which paper specicifically is the q from

[loggerhead]

your really close pretty much just have to log both sides
e^-x = e^2 - 2
-x = ln (e^-2 - 2)
-x = 1.68
x = -1.68

3i)
L.H.S. = cosec 2x + cot 2x
L.H.S. =   1       + cos 2x
            sin 2x      sin 2x
L.H.S. = 1 + cos 2x
              sin 2x
L.H.S. = 1 + 2cos^2 x - 1
               2sin x cos x
L.H.S. = cos x
            sin x
L.H.S. = cot x
L.H.S. = R.H.S.

ii) cosec 2x + cot 2x = 2
therefore cot x = 2
tan x = 1/2
x = 26.6, 26.6+180
x = 26.6, 206.6

5i) (1 + 2x)(1+ax)^2/3 = (1 + 2x)(1 + (2/3)ax + ...)
                               = 1 + (2/3)ax + ...
                                     +    2x     + (4/3)ax^2 + ...
no coefficient of x
therefore (2/3)a + 2 = 0
(2/3)a = -2
a = -3

ii)(1 + 2x)(1 - 3x)^2/3
= (1 + 2x)(1 - 2/3 * 3x + 2/3 * -1/3 * 1/2! * (-3x)^2 + 2/3 * -1/3 * -4/3 * 1/3! * (-3x)^3 + ...)
= (1 + 2x)(1 - 2x - x^2 - 4/3x^3 + ...)
= 1 - 2x - x^2 - 4/3x^3
    + 2x - 4x^2 - 2x^3 + ...
therefore coefficient of x^3 = -4/3 - 2
                                      = -10/3

[loggerhead]

and finally 8, its a biggy

i)     100        =  A +   B   +     C
  x^2(10 - x)     x    x^2    (10 - x)
therefore 100 = Ax(10 - x) + B(10-x) = Cx^2

let x = 10 =>  100c = 100
                        c = 1
x = 0 =>   10B = 100
                  B = 10
coefficient of x^2 => -A + C = 0
                                 A = 1

therefore f(x) = 1 +  10 +       1   
                      x    x^2    (10 - x)

ii)
dx = x2(10-x)
dt       100

im going to use '|' as the intergral sign
|     100       dx = |1dt
  x^2(10-x)
| 1 +  10  +      1      dx = |1dt
  x    x^2    (10 - x)
lnx - ln(10 - x) - 10x^-1 = t + c
     x = 1 where t = 0
ln 1 - ln 9 - 10 = 0 + c
c = -ln 9 - 10

therefore t = ln x - ln (10 - x) - 10x^-1 + ln 9 + 10
t = ln (    9x     ) - 10  + 10
         (10 - x)       x

and there we go all answered
good luck for the exam

[astarmathsandphysics]

thanks loggerhead

[preity]

thanks a lot to both of u...will post sm more up soon..(: