Question

Let f(x) be a quadratic expression which is +ve for all real values of x. If g(x) = f(x) + f'(x) + f"(x), then for any real x,

(a) g(x) < 0

(b) g(x) > 0

(c) g(x) = 0

(d) g(x) ≥ 0

Answer 1

Let f(x) = ax2 + bx + c, a > 0 (since f(x) is + ve for all real values of (x)

f'(x) = 2ax + b

f"(x) 2a

Now, g(x) = f(x) + f'(x) + f"(x)

= (ax2 + bx + c) + (2ax + b) + 2a

= a x2 + (2 + b)x + (2a + b + c)

Now, D = (2a + b)2 – 4a(2a + b + c)

= 4a2 + 4ab + b2 – 8a2 – 4ab – 4ac

= – 4a2 – 4ac + b2

= – 4a2 + (b2 – 4ac) < 0 for all x in R.

Thus, g(x) > 0, " x ∈ R