Question

If the minimum value of f(x) = x² + 2bx + 2c² is greater than the maximum value of g(x) = – x² – 2cx + b² then

(x being real) -

• A. | c | >$\frac{|b|}{\sqrt{3}}$
• B. $\frac{|c|}{\sqrt{2}}$> | b |
• C. – 1 < c <$\sqrt{2}$b
• D. No real value of b and c exist.

Answer 1

Answer: (B)

$\frac{|c|}{\sqrt{2}}$> | b |

Answer 2

\ f(x)(x + b)²+(2c² – b²)

and g(x) = (b² + c²) –(x – c)²

\ min. value of f(x) = 2c²– b²

max. value of g(x) = b²+ c²

\ 2c² – b² > b² + c² Ž c² > b²

Ž c > $\sqrt{2}$b Ž $\frac{|c|}{\sqrt{2}}$ > | b |