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Question Number 161854 by mnjuly1970 last updated on 23/Dec/21

Commented by cortano last updated on 23/Dec/21

yx^2 −yx−6y=x^2 −ax+a (y−1)x^2 +(a−y)x−(6y+a)=0 Δ≥0 (a−y)^2 +4(y−1)(6y+a)≥0 y^2 −2ay+a^2 +4(6y^2 +(a−6)y−a)≥0 25y^2 +(2a−24)y+a^2 −4a≥0 since R_f =R then ⇒(2a−24)^2 −4.25(a^2 −4a)≥0 ⇒a^2 −24a+144−25a^2 +100a≥0 ⇒−24a^2 +76a+144≥0 ⇒6a^2 −19a−36≤0 ⇒(2x−9)(3x+4)≤0 ⇒−(4/3)≤x≤(9/2)

yx2−yx−6y=x2−ax+a(y−1)x2+(a−y)x−(6y+a)=0Δ⩾0(a−y)2+4(y−1)(6y+a)⩾0y2−2ay+a2+4(6y2+(a−6)y−a)⩾025y2+(2a−24)y+a2−4a⩾0sinceRf=Rthen⇒(2a−24)2−4.25(a2−4a)⩾0⇒a2−24a+144−25a2+100a⩾0⇒−24a2+76a+144⩾0⇒6a2−19a−36⩽0⇒(2x−9)(3x+4)⩽0⇒−43⩽x⩽92

Answered by mr W last updated on 23/Dec/21

y=((x^2 −ax+a)/(x^2 −x−6)) (y−1)x^2 −(y−a)x−(6y+a)=0 (y−a)^2 +4(y−1)(6y+a)≥0 ⇒25y^2 +2(a−12)y+a^2 −4a≥0 (a−12)^2 −25(a^2 −4a)≤0 ⇒(2a−9)(3a+4)≥0 a≤−(4/3) a≥(9/2) i.e. a∈(−∞,−(4/3)] ∧ [(9/2),+∞)

y=x2−ax+ax2−x−6(y−1)x2−(y−a)x−(6y+a)=0(y−a)2+4(y−1)(6y+a)⩾0⇒25y2+2(a−12)y+a2−4a⩾0(a−12)2−25(a2−4a)⩽0⇒(2a−9)(3a+4)⩾0a⩽−43a⩾92i.e.a∈(−∞,−43]∧[92,+∞)

Commented by mnjuly1970 last updated on 23/Dec/21

grateful mr W

gratefulmrW

Commented by Tawa11 last updated on 23/Dec/21

Am sorry sir.

Amsorrysir.

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