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The set of values of a for which each on of the roots of `x^(2) - 4ax + 2a^(2) - 3a + 5 = 0` is greater than 2, isA. `a in (1, oo)`B. `a = 1`C. `a in (-oo, 1)`D. `a in (9//2, oo)` |
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Answer» Correct Answer - D Let `f(x) = x^(2) - 4 ax + 2a^(2) - 3a + 5`. Clearly, y = f(x) represents a parabola opening upward. So, its both roots will be greater than 2, if (i) Discriminat `ge 0` (ii) x-coordinate of vertex `gt 2` (iii) 2 lies outside the roots i.e. `f(2) gt 0` Now, (i) Discriminat `ge 0` `rArr" "16a^(2) - 4(2a^(2)-3a + 5)ge 0` `rArr" "2a^(2) + 3a - 5 ge 0` `rArr" "(2a+5)(a-1) ge 0 rArr a le -(5)/(2) or, a ge 1" "...(i)` (ii) x-coordinate of vertex `gt 2` `rArr" "2a gt 2 rArr a gt 1" "...(ii)` (iii) `f(2) gt 0` `rArr" "2a^(2) - 11 a + 9 gt 0` `rArr" "(a-1)(2a-9) gt 0 rArr a lt 1 or, a gt (9)/(2)" "...(iii)` From (i), (ii) and (iii), we get `a gt (9)/(2)`. |
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