Question

The set of all real x satisfying the inequality 3-|x|/4-|x| \geq 0 is?????

Answer 1

Hey

$\frac{3 - |x| }{4 - |x| } \geqslant 0$

$= > (3 - |x| ) \geqslant 0 \: or \: (4 - |x| ) < 0$


$= > x \in [ - 3, + 3] \cup(4, + \infty) \cup( - 4, - \infty)$

Answer 2

ANSWER:

The given inequality is-||x-1|

- 1|<=1Therefore, -1<=( |x-1| -1)

<=1Adding 1, we get -0<=( |x-1|)

<=2Again we can write,-2<=(x-1)<=2

Therefore, -1<=x<=3Since x is real number and it lies between -1<=x<=3 therefore all values of x lying between [-1,3] will satisfy the given inequality.

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