If f ( x ) = x 2 + 2 b x + 2 c 2 and g ( x ) = - x 2 - 2 c x + b 2 such that min f ( x ) max g ( x ) , then the relation between b and c is [2003]

Question

If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ such that $\min f (x) > \max g (x),$ then the relation between $b$ and $c$ is [2003]

Smart Achievers · Accepted Answer

(4)

$f (x) = x^{2} + 2 b x + 2 c^{2}$ $\Rightarrow f (x) = {(x + b)}^{2} + 2 c^{2} - b^{2}$

$\Rightarrow f_{\min} = 2 c^{2} - b^{2}$

and $g (x) = - x^{2} - 2 c x + b^{2}$

$\Rightarrow g (x) = - {(x + c)}^{2} + b^{2} + c^{2}$ $\Rightarrow g_{\max} = b^{2} + c^{2}$

For $f_{\min} > g_{\max}$ $\Rightarrow 2 c^{2} - b^{2} > b^{2} + c^{2}$

$\Rightarrow c^{2} > 2 b^{2}$

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

Solution

Q.
If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ such that $\min f (x) > \max g (x),$ then the relation between $b$ and $c$ is [2003]

1 no real value of $b$ & $c$

2 $0 < c < b \sqrt{2}$

3 $| c | < | b | \sqrt{2}$

4 $| c | > | b | \sqrt{2}$

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Solution

Q. If f(x)=x2+2bx+2c2 and g(x)=-x2-2cx+b2 such that min f(x)>max g(x), then the relation between b and c is [2003]

1 no real value of b & c

2 0<c<b2

3 |c|<|b|2

4 |c|>|b|2