Question Number 101363 by bemath last updated on 02/Jul/20

The solution set of inequality  (((√((3x−7)^2 ))−2)/(x−3)) ≤ ((3−(√x^2 ))/(x−3)) is __  (A) (−∞, (1/2)]       (D) [(1/2), ∞)  (B) [(1/2),1 ]            (E) (−∞,(2/3)]  (C) (−∞,1 ]

Answered by 1549442205 last updated on 02/Jul/20

⇔((∣3x−7∣−2)/(x−3))−((3−∣x∣)/(x−3))≤0(1)  i)if  x≥(7/3) then (1)⇔((3x−7−2−(3−x))/(x−3))≤0  ⇔((4x−12)/(x−3))≤0⇔((4(x−3))/(x−3))≤0⇔4≤0   ⇒(1) has no solutions  ii)if  0≤x<(7/3) then (1)⇔((7−3x−2−(3−x))/(x−3))  ⇔((2−2x)/(x−3))≤0⇔((1−x)/(x−3))≤0⇔x∈{(−∞;1]∪(3;+∞)}∩[0;(7/3))  ⇔x∈[0;1]  iii)if  x<0 then (1)⇔((7−3x−2−(3+x))/(x−3))≤0  ⇔((2−4x)/(x−3))≤0⇔((1−2x)/(x−3))≤0⇔(−∞;(1/2)]∪(3;+∞)∩(−∞;0)  ⇔(−∞;0)  Combinating three  above cases we get  The solutions of given inequality is            x∈(−∞;1],so choose answer C

Commented by1549442205 last updated on 02/Jul/20

Thank you.You are welcome sir.

Commented bybemath last updated on 02/Jul/20

thank you. agree

Commented byRasheed.Sindhi last updated on 02/Jul/20

⇔((∣3x−7∣−2)/(x−3))−((3−∣x∣)/(x−3))≤0(1)  i)if  x≥(7/3) then (1)⇔((3x−7−2−(3−x))/(x−3))≤0  ⇔((4x−12)/(x−3))≤0⇔((4(x−3))/(x−3))≤0⇔4≤0   At x=(7/3),((4(x−3))/(x−3))≤0⇏4≤0  But x=(7/3),((4(x−3))/(x−3))≤0⇏4≥0                 ∵ x−3<0  ⇒(1) has no solutions  ii)if  0≤x<(7/3) then (1)⇔((7−3x−2−(3−x))/(x−3))        .....................                 ..........                     .....