Quadratic EquationHard
Question
Let f (x) be a quadratic expression which is positive for all real values of x.If g (x) = f(x) + f′(x) + f′′(x), then for any real x
Options
A.g (x) < 0
B.g (x) > 0
C.g (x) = 0
D.g(x) ≥ 0
Solution
Let f (x) = ax2 + bx + c > 0 for all x ∈ R
⇒ a > 0
and b2 - 4ac < 0 .....(i)
∴ g(x) = f(x) + f′(x) + f′′(x)
⇒ g(x) = ax2 + bx + c + 2ax + b + 2a
⇒ g(x) = ax2 + x(b + 2a) + (c + b + 2a)
whose discriminant
= (b + 2a)2 - 4a(c + b + 2a)
= b2 + 4a2 + 4ab - 4ac - 4ab - 8a2
= b2 - 4a2 - 4ac
= (b2 - 4ac) - 4a2 < 0 [frol Eq. (i)]
∴ g(x) > 0 for all x as and discriminant < 0.
Thus, g(x) > 0 for all x ∈ R.
⇒ a > 0
and b2 - 4ac < 0 .....(i)
∴ g(x) = f(x) + f′(x) + f′′(x)
⇒ g(x) = ax2 + bx + c + 2ax + b + 2a
⇒ g(x) = ax2 + x(b + 2a) + (c + b + 2a)
whose discriminant
= (b + 2a)2 - 4a(c + b + 2a)
= b2 + 4a2 + 4ab - 4ac - 4ab - 8a2
= b2 - 4a2 - 4ac
= (b2 - 4ac) - 4a2 < 0 [frol Eq. (i)]
∴ g(x) > 0 for all x as and discriminant < 0.
Thus, g(x) > 0 for all x ∈ R.
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