InterviewSolution
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InterviewSolution
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Graph of ` y = ax^(2) + bx + c ` is as shown in the figure . If ` PQ= 9,` OR = 5 and OB = 2.5, the which of the following is /are ture? A. AB = 3B. ` y (-1) lt 0 `C. ` y (ge 7 for all x ge 3 `D. `ax^(2) + bx + c = ` mx has real roots for all real m |
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Answer» Correct Answer - 1,3,4 OR = 5 ` therefore f(0) = - 5= e` PQ = 9 . ` therefore - (D)/(4a) = - 9` `rArr (b^(2) + 20a)/(4a) = 9 ` `rArr b^(2) = 16 a ` …(1) OB - 2.5 So, one root of equation is 2.5 ` therefore 25a + 10 b - 20 = 0` ` or 5a + 2b - 4 = 0` ...(2) From (1) and (2), ` 5b^(2) + 32 b - 64 = 0 ` `rArr b = -8 ,(8)/(5) ` (Not possible . Since a ` gt ` 0, we must have ` b lt 0 `) ` therefore a = 4 ` `rArr y = 4x^(2) - 8x- 5 = 4x^(2) - 10x + 2x - 5 = (2x - 5) (2 + 1)` So, x = 2.5, - 0.5 are the roots of y(x) = 0 . Clealy, ` y(-1) g t0 ` ` y ge 7 ` `rArr 4x^(2) - 8x - 5 ge 7 ` `rArr x^(2) - 2x - 3 ge 0 ` `rArr (x - 3) (x +1) ge 0 ` `rArr x le - 1 or x ge 3` `ax^(2) + bx + c = mx ` `rArr 4x^(2) - (8 + m) x - 5 = 0 ` Clearly, above equation has two distinct real roots for any real values of m. |
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