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 $|\frac{c}{b}|$ lies in the interval:
(a) $\bigg \lgroup 0,\ \frac{1}{2}\bigg \rgroup$
(b) $\bigg [\frac{1}{2},\ \frac{1}{\sqrt{2}}\bigg \rgroup$
(c) $\bigg [\frac{1}{\sqrt{2}},\ \sqrt{2}\bigg ]$
(d) (√2,∞)

Answer 1

f(x) = x² + 2bx + 2c²

differentiating with respect to x,

f'(x) = 2x + 2b = 0 =>x = -b

again differentiating with respect to x,

f"(x) = 2 >0 hence, at x = -b f(x) will be minimum.

so, min f(x) = f(-b) = b² - 2b² + 2c²

= -b² + 2c² ......(1)

now, g(x) = -x² - 2cx + b²

differentiating with respect to x,

g'(x) = -2x - 2c = 0 => x = -c

again differentiating with respect to x,

g"(x) = -2 < 0 hence g(x) will be maximum at x = -c

so, max g(x) = g(-c) = -c² + 2c² + b²

= c² + b²

now, min f(x) > max g(x)

or, -b² + 2c² > c² + b²

or, c² > 2b²

or, |c| > √2|b|

or, |c|/|b| > √2

hence, option (d) is correct.