JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q72

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MathematicsSingle correct+4 / -1Question 72 of 90

If non-zero real numbers b and c are such that min $f(x)$ > max $g(x)$ , where
$f(x)=x^2+2bx+2c^2\ \text{and}\ g(x)=-x^2-2cx+b^2\ (x\in\mathbb R);$
then $\left|\dfrac cb\right|$ lies in the interval :

\left(0,\dfrac12\right)

\left[\dfrac12,\dfrac1{\sqrt2}\right)

\left[\dfrac1{\sqrt2},\sqrt2\right]

\left(\sqrt2,\infty\right)

Answer and worked solution

Official answer: D

$\min f=2c^2-b^2$ (at $x=-b$ ).
$\max g=b^2+c^2$ (at $x=-c$ ).

$2c^2-b^2>b^2+c^2$ gives $c^2>2b^2$ , so $\left|\frac cb\right|>\sqrt2$ .

Topics:Quadratic expressions

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 72