Question

If non-zero real numbers b and c are such that min f(x) > max g(x), where f(x) = x² + 2bx + 2c² and g(x) = − x² − 2cx + b² (x∈R), then |c/b| lies in the interval

(a)  (o,1/2)

(b)  [1/2, 1/√2)

(c)  [1/√2,√2]

(d)  (√2,∞)

Answer 1

Correct option (d)  (√2,∞)

f(x) = x² + 2b x + 2c² = x² + 2bx + b² + 2c² − b² = (x + b)² + 2c² − b²

Therefore, min f(x) = 2 c² − b² at x = − b.

g(x) = − x² − 2cx + b² = − {x² + 2cx − b²}

= − {x² + 2cx + c² − b² − c²}

= − (x + c)² + b² + c²

Therefore, max g(x) = b² + c² when x = − c.

Now according to question,

2c² − b² > b² + c² ⇒ c² > 2b²