InterviewSolution
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InterviewSolution
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Column I contains rational algebraic expressions and Column II contains possible integers of a. |
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Answer» Correct Answer - `(A)to(q,r,s,t),(B)to(q,r),(C)to(p,q)` `Ato(q,r,s,t), Bto(q,r,):Cto(p,q)` (A) We have `y=(ax^(2)+3x-4)/(3x-4x^(2)+a)` `impliesx^(2)(a+4y)+3(1-y)x-(ay+4)=0` As `x epsilon R ` we get `Dge0` `implies9(1-y)^(2)+4(a+4y)(ay+4)ge0` `implies(9+16a)y^(2)+(4a^(2)+46)y+(9+16a)ge0,AA y espsilon R` `implies` If `9+16agt0` then `Dle0` Now `D le0` `implies(4a^(2)+46)^(2)-4(9+16a)^(2)le0` `implies4[(2a^(2)+23)^(2)-(9+16a)^(2)]le0` `implies[(2a^(2)+23)+(9+16)][(2a^(2)+23)-(9+16a)]le0` ltbRgt `implies(2a^(2)+16a+32)(2a^(2)-16a+14)le0` `implies4(a+4)^(2)(a^(2)-8a+8)le0` `impliesa^(2)-8a+8le0` `implies(a-1)(a-8)le0` ltbr `implies1leale7` `:.9+16ag0` and 1leale7` `implies1leale7` (B) we have `y=(ax^(2)+x-2)/(a+x-2x^(2))` `impliesx^(2)(a+2y)+x(1-y)-(2+ay)=0` As `x epsilonR` we get `Dge0` `implies(1-y)^(2)+4(2+ay)(a+2y)ge0` `implies(1+8a)y^(2)+(4a^(2)+14)y+(1+8a)ge0` `implies` If `1+8agt0` then `Dle0` `implies(4a^(2)+14)^(2)-4(1+8a)^(2)le0` `implies4[(2a^(2)+7)^(2)-(1+8a)^(2)]le0` `implies[(2a^(2)+7)+(1+8a)][(2a^(2)+7)-(1+8a)]le0` `implies(2a^(2)+8a+8)(2a^(2)-8a+6)le0` `implies4(a+2)^(2)(a^(2)-4a+3)le0` `impliesa^(2)-4a+3le0` `implies(a-1)(a-3)le0` `implies1leale3` Thus `1+8agt0` and `1leale3` `implies1leale3` (C)We have `y=(x^(2)+2x+a)/(x^(2)+4x+3a)` `impliesx^(2)(y-1)+2(2y-1)x+a(3y-1)=0` As `x epsilon R` We get `Dge0` ltbRgt `implies4(2y-1)^(2)-4(y-1)a(3y-1)ge0` `implies(4-3a)y^(2)-(4-4a)y+(1-a)ge0` `implies` If `4-3agt0` then `Dle0` `implies(4-4a)^(2)-4(4-3a)(1-a)le0` `implies4(2-2a)^(2)-4(4-3a)(1-a) le 0` `implies4+4a^(2)-8a-(4-8a+3a^(2))le0` `implies a^(2)-ale0` `impliesa(a-1)le0` ltbRgt `implies0leale1` |
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