JEE MAIN - Mathematics (2009 - No. 19)
If the roots of the equation $$b{x^2} + cx + a = 0$$ imaginary, then for all real values of $$x$$, the expression $$3{b^2}{x^2} + 6bcx + 2{c^2}$$ is :
less than $$4ab$$
greater than $$-4ab$$
less than $$-4ab$$
greater than $$4ab$$
Explanation
Given that roots of the equation
$$b{x^2} + cx + a = 0$$ are imaginary
$$\therefore$$ $${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
Let $$y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$$
$$ \Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$$
As $$x$$ is real, $$D \ge 0$$
$$ \Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0$$
$$ \Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0$$
$$ \Rightarrow {c^2} + y \ge 0$$
$$ \Rightarrow y \ge - {c^2}$$
But from eqn. $$(i),$$ $${c^2} < 4ab$$
or $$ - {c^2} > - 4ab$$
$$\therefore$$ we get $$y \ge - {c^2} > - 4ab$$
$$y > - 4ab$$
$$b{x^2} + cx + a = 0$$ are imaginary
$$\therefore$$ $${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
Let $$y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$$
$$ \Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$$
As $$x$$ is real, $$D \ge 0$$
$$ \Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0$$
$$ \Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0$$
$$ \Rightarrow {c^2} + y \ge 0$$
$$ \Rightarrow y \ge - {c^2}$$
But from eqn. $$(i),$$ $${c^2} < 4ab$$
or $$ - {c^2} > - 4ab$$
$$\therefore$$ we get $$y \ge - {c^2} > - 4ab$$
$$y > - 4ab$$
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