Respuesta :
Answer:
A. X < 3.
Step-by-step explanation:
We have the equation, [tex]3 + \frac{x-2}{x-3} \leq 4[/tex]
Adding -4 on both sides, we get,
[tex]\frac{x-2}{x-3} -1 \leq 0[/tex]
i.e. [tex]\frac{x-2-x+3}{x-3} \leq 0[/tex]
i.e. [tex]\frac{1}{x-3} \leq 0[/tex]
So, we have the inequality [tex]\frac{1}{x-3} \leq 0[/tex].
Its critical points are given by x-3 = 0 which implies x = 3.
Now, we will find the region among the two [tex]( - \infty , 3 )[/tex] and [tex]( 3 , \infty )[/tex] that satisfies the inequality [tex]\frac{x-2}{x-3} -1 \leq 0[/tex]
Substitute x = 2 in the above inequality, we get -1 < 0, which is true. Again, substitute x=4 gives 1 < 0, which is false.
As x=4 lie in the region [tex]( 3 , \infty )[/tex] , it cannot be the solution of the given inequality.
So, the solution of the given inequality is the region [tex]( - \infty , 3 )[/tex] .i.e. x < 3.
Hence. option A is correct.
