let `f(x)` be a polynomial function of degree 2 and `f(x)gt0` for all `x inR.` if `g(x)=f(x)+f'(x)+f''(x),` then for any x show that `g(x)gt0`

Options

A. `g(x)lt0" for all "x`
B. `g(x)gt0" for all "x`
C. `g(x)=0" for all "x`
D. `g(x)ge0" for all "x.`

Answer

B

Explanation

Let `f(x)=ax^(2)+bx+c.` Then,
`f(x)gt0" for all "x in R`
`implies" "agt0" and "b^(2)-4aclt0.`
Now,
`g(x)=f(x)+f'(x)+f''(x)" for all "x`
`implies" "g(x)=ax^(2)+bx+c+2ax+b+2a`
`implies" "g(x)=ax^(2)+x(2a+b)+2a+b+c`
Let D be the discriminant of g(x). Then,
`D=(2a+b)^(2)-4a(2a+b+c)`
`implies" "D=-4a^(2)+b^(2)-4ac`
`implies" "D=-4a^(2)+(b^(2)-4ac)lt0" "[becauseb^(2)-4aclt0]`
Thus, g(x) is a quadratic polynomial such that cocfficient of `x^(2)gt0` and, `Dlt0`. Therefore, `g(x)gt0` for all `x in R.`