`3(x-1) le 2 (x-3)`

1/3(8x-5)le1/2(5x-2), x in R

If the relation f(x)={(2x-3",",x le 2),(x^(3)-a",",x ge2):} is a function, then find the value of a.

Solve (x)/(x+2) le (1)/(|x|)

Solve the following : (a) 1 le |x-2| le 3 " (b) "0 lt |x-3|le 5 (c ) |x-2|+|2x-3|=|x-1| " (d) " |(x-3)/(x+1)| le1

If the function f(x) ={x+1 if x le 1 , 2x+1 if 1 lt x le 2 and g(x) = {x^2 if -1 le x le 2 , x+2 if 2 le x le 3 then the number of roots of the equation f(g(x))=2

Find the range of the function f(x) = abs(x - 1) + abs(x - 2) , -1 le x le 3

(x+1)/3le(2x-1)/4,x in R

Solve for x: ((x-1)^2(x-2)^3(x+1)^6)/(x^8(2x+3)^3)le0

Complete solution set of inequlaity ((x+2)(x+3))/((x-2)(x-3))le1

For x le 2," solve "x^(3)3^(|x-2|)+3^(x+1) = x^(3)*3^(x-2)+3^(|x-2|+3)