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All possible values of a, so that 6 lies between the roots of the equation `x^2 + 2(a-3)x +9 =0`A. `a in [-3//4, oo)`B. `a in (oo, -3//4)`C. `a in (-oo, 0) uu(6, oo)`D. `a in (-3//4, 6)` |
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Answer» Correct Answer - B Let `f(x) = x^(2) + 2(a-3) x + 9`. If 6 lies between the roots fo f(x) = 0, then we must have (i) `"Disc" gt 0, and" "(ii) f(6) lt 0" "[therefore "Coeff of" x^(2) "is positive"]` Now, Disc `gt` 0 `rArr" "4(a-3)^(2) - 36 gt 0` `rArr" "(a-3)^(2) - 9 gt 0` `rArr" "a^(2) - 6a gt 0` `rArr" "a(a-6) gt 0 rArr a lt 0 or a gt 6" "...(i)` and, `f(6) lt 0` `rArr" "36 + 12(a-3) + 9 lt 0 rArr a lt -(3)/(4)" "....(ii)` From (i) and (ii), we get `a lt - 3//4 i.e. a in (-oo, -3//4)`. |
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