inequalities and integration

unclebutcher SG

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 SG

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 SG

~~- my careless mistake -~~

^tamago^ SG

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/[x²sqrt(x²-4)] dx...

Let x = 2 sec u ==> dx = 2 tan usec u du
integrate 1/[4sec²usqrt(4sec²u-4)] dx
= int [2tan usec u/(8sec²utan 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 SG