Answer 1

Clearly, f(x) = ax² + bx + c represent a parabola opening downwards. Therefore, `a < 0`

` y= ax^2 + bx + c `  cuts y-axis at P which lies on `OY`. Putting x = 0 in ` y = ax^2 + bx + c `, we get y =c. So the coordinates P are `(0,c)`. Clearly, P lies on `(OY)`. Therefore  `c > 0`

The vertex  `(-b)/(2a), (-D)/(4a)` of the parabola is in the second quadrant. Therefore `-b /(2a)`, `b < 0`

Therefore `a < 0,b>0` and `c > 0`

Hence, the correct choice is `(b)`