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If the roots of equation `(a + 1)x^2-3ax + 4a = 0` (a is not equals to -1) are greater than unity, thenA. `[-(10)/(7),1]`B. `[-(12)/(7),0]`C. `[-(16)/(7),-1)`D. `(-(16)/(7),0)` |
Answer» Correct Answer - C `(c )` `(a+1)x^(2)-3ax+4a=0` `D=9a^(2)-16a(a+1) ge 0` ………`(i)` `x_(1) gt 1`, `x_(2) gt 1` `:. (a+1)f(1) gt 0` and `(3a)/(2(a+1)) gt 1` `implies(a+1)(2a+1) gt 0` and `(a-2)/(a+1) gt 0` ……..`(ii)` `(2a+1)/(a+1) gt 0`………`(iii)` Solving `(i)`, `(ii)` and `(iii)`, we get `-(16)/(7) le a lt -1`. |
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