inequalities and integration

• unclebutcher

  3 Sep 07, 23:17

  help me solve this qn:
  
  1/(a-x) less than or equal to 1/(a+x)
  
  a is more than zero
  
  hence solve 1/(1+2x) more than or equal to 1/(1-2x)
  
  also
  
  using the substitution x= 2secx
  
  find (the integrate sign) 1/(x^2(x^2 - 4)^0.5) dx
  
  thanks
• tanjun

  3 Sep 07, 23:57

  Originally posted by unclebutcher:

  help me solve this qn:
  
  1/(a-x) less than or equal to 1/(a+x)
  
  a is more than zero
  
  hence solve 1/(1+2x) more than or equal to 1/(1-2x)
  
  also
  
  using the substitution x= 2secx
  
  find (the integrate sign) 1/(x^2(x^2 - 4)^0.5) dx
  
  thanks

  In the qn below, <= denotes less than or equal to and >= denotes greater than or equal to.
  
  1. 1/(a-x) <= 1/(a+x)
  Multiply by (a-x)^2(a+x)^2
  (a-x)(a+x)(a+x) <= (a+x)(a-x)(a-x)
  (a-x)(a+x)(a+x) - (a+x)(a-x)(a-x) <=0
  (a^2 - x^2)(a+x - a + x) <=0
  (a^2 - x^2)(2x) <=0
  (a+x)(a-x)(2x) <=0
  
  Graph is a negative x^3 curve with 3 roots.
  
  After you have drawn the graph,
  Range of values of x which satsify the inequality above = -a<=x<=0 or x >=a
  
  Second part
  since 1/(1+2x) >= 1/(1-2x)
  1/(1-2x) <= 1/(1+2x)
  By comparison, a = 1, X = 2x
  Sub into the final inequality.
  (1+2x)(1-2x)(4x) <=0
  Range of values of x = -1/2<=x<=0 or x>=1/2
• eagle

  4 Sep 07, 00:00

  -- my careless mistake --
• ^tamago^

  4 Sep 07, 00:17

  Using superscripts ² and ³ for clarity. Method same as tanjun's.
  
  1/(a-x) <= 1/(a+x)
  Multiply by (a-x)²(a+x)²,
  (a-x)(a+x)² <= (a-x)²(a+x)
  (a-x)(a+x)² - (a-x)²(a+x) <= 0
  (a-x)(a+x)[(a+x)-(a-x)] <= 0
  (a²-x²)(2x) <= 0
  (2a²)x - 2x³ <= 0
  
  three roots are 0, a, -a. (a is +ve)
  
  soln set: x is R, -a <= x <= 0 or x >= a.
  
  1/(1-2x) <= 1/(1+2x)
  1/(½-x) <= 1/(½+x)
  sub a = ½,
  1/(a-x) <= 1/(a+x)
  
  Hence, the three roots are 0, ½ and -½.
  
  soln set: x is R, -½ <= x <= 0 or x >= ½.
  
  2. integral of 1/[x²•sqrt(x²-4)] dx...
  
  Let x = 2 sec u ==> dx = 2 tan u•sec u du
  integrate 1/[4sec²u•sqrt(4sec²u-4)] dx
  = int [2tan u•sec u/(8sec²u•tan u)] du (tan²u+1=sec²u)
  = 1/4 int (1/sec u) du
  = 1/4 int cos u du
  = 1/4 sin u + C
  = sqrt(¼x²-1)/4(½x) + C (since sec u = ½x and sin 0 = )
  = ½sqrt(x²-4)/2x + C
  = sqrt(x²-4)/4x + C
• unclebutcher

  4 Sep 07, 00:44

  Originally posted by ^tamago^:
  U
  Multiply by (a-x)²(a+x)²,
  
  soln set: x is [b]R, -½ <= x <= 0 or x >= ½.[/b]

  why must multiply by that number?
  
• eagle

  4 Sep 07, 00:44

  Originally posted by unclebutcher:

  why must multiply by that number?
  

  so that u will multiply by something that is always positive => inequality sign not changed
• ^tamago^

  4 Sep 07, 00:49

  Originally posted by eagle:

  so that u will multiply by something that is always positive => inequality sign not changed

  as above....
  
  another method is by adding an AND condition that retains the denominator if it is +ve or inverts if it is -ve, but that is usually much more tedious if both sides contain denominators.
• unclebutcher

  4 Sep 07, 00:59

  Originally posted by eagle:

  so that u will multiply by something that is always positive => inequality sign not changed

  (a-x)²(a+x)²,
  
  as in why not just make all common denominator
• ^tamago^

  4 Sep 07, 01:11

  Originally posted by unclebutcher:

  (a-x)²(a+x)²,
  
  as in why not just make all common denominator

  You will still have to cancel out the common denominator.
• unclebutcher

  4 Sep 07, 01:14

  Originally posted by ^tamago^:

  You will still have to cancel out the common denominator.

  hmm this is weird....but promo can use GC solve...so shouldn't have this prob i hope. this is the first ineq. qn that i dunno how to do sia
• eagle

  4 Sep 07, 01:15

  Originally posted by unclebutcher:

  hmm this is weird....but promo can use GC solve...so shouldn't have this prob i hope. this is the first ineq. qn that i dunno how to do sia

  ....
  that is a super standard question
• ^tamago^

  4 Sep 07, 01:20

  Originally posted by unclebutcher:

  hmm this is weird....but promo can use GC solve...so shouldn't have this prob i hope. this is the first ineq. qn that i dunno how to do sia

  never mind, learn lor. like me, i almost always stumble at P&C questions.
• eagle

  4 Sep 07, 01:21

  Originally posted by ^tamago^:

  never mind, learn lor. like me, i almost always stumble at P&C questions.

  Hwk Forum lor... Let all of us try...
• unclebutcher

  4 Sep 07, 01:26

  Originally posted by eagle:

  ....
  that is a super standard question

  i know ineq la....i just dunno how to do this particular qn
• ^tamago^

  4 Sep 07, 01:32

  Originally posted by eagle:

  Hwk Forum lor... Let all of us try...

  cannot lah, u sexprain liao i also cannot grasp. dunno my foundation which part got error. lucky it's not a major question, and i know enough to understand TOTO.
• lpx88

  5 Sep 07, 15:52

  Sorry ah butcher borrow yr thread for my qn...
  
  Integrate with respect to x
  
  I see trigo & fraction i tio stun Nt sure if u jus integrate normally or wad...
• eagle

  5 Sep 07, 17:35

  Originally posted by lpx88:

  Sorry ah butcher borrow yr thread for my qn...
  
  Integrate with respect to x
  
  I see trigo & fraction i tio stun Nt sure if u jus integrate normally or wad...

  edit: Suddenly thought of my mistake on the train this morning..... ignore previous post
  
  note that sec^2 is a derivative of tan2x. You can integrate immediately
• ^tamago^

  6 Sep 07, 10:44

  yeah, turn sec²2x dx = d(tan2x)...
• unclebutcher

  6 Sep 07, 15:17

  Originally posted by ^tamago^:

  yeah, turn sec²2x dx = d(tan2x)...

  how u make superscript ah?