If `a` and `b` are randomly chosen from the set `{1,2,3,4,5,6,7,8,9}`, then the probability that the expression `ax^(4)+bx^(3)+(a+1)x^(2)+bx+1` has positive values for all real values of `x` is

Options

A. `(34)/(81)`
B. `(31)/(81)`
C. `(32)/(81)`
D. `(10)/(27)`

Answer

C

Explanation

`(c )` The expression `ax^(4)+ax^(2)+bx^(3)+bx+x^(2)+1`
`=(x^(2)+1)(ax^(2)+bx+1)`
`:.x^(2)+1` is positive for all real `x`
For `ax^(2)+bx+1` to be positive for all real `x`
`a gt 0`, `b^(2)-4a lt 0`
If `b=1`, a can take `9` value from `1` to `9`
`b=2`, a can take `8` value from `2` to `9`
`b=3`, a can take `7` value from `3` to `9`
`b=4`, a can take `5` value from `5` to `9`
`b=5`, a can take `3` value from `7` to `9`
`b` cannot take the values `6,7,8,9`.
`:.` Number of exhaustive cases `=9+8+7+5+3=32`
For each of the `9` values of `a`, there are `9` corresponding values for `h`.
`:.` Number of exhaustive cases `=9xx9=81`
`:.` The required probability `=(32)/(81)`