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?

Options

A. AB = 3
B. ` 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

Answer

1,3,4

Explanation

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.