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| **** 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