If x < a < 0 means that x and a are numbers such that $x$ is less than a and a is less than zero, then;

x^{2} < a x < 0

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x^{2} > a x > a^{2}

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x^{2} < a^{2} < 0

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x^{2} > a x

but ax< 0

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Solution

The correct option is A $x^{2} > a x > a^{2}$
x<0 $∴$ x $^{2}$ >0
x<a, x and a are negative.
$∴$ x $^{2}$ >a $^{2}$ , x $^{2}$ >ax, ax>a $^{2}$
$∴$ x $^{2}$ >ax>a $^{2}$

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Similar questions

l i m x \to 0 \frac{\sqrt{a^{2} - a x + x^{2}} - \sqrt{a^{2} + a x + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}}

f (x) = \frac{\sqrt{a^{2} - a x + x^{2}} - \sqrt{a^{2} + a x + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}}

x = 0

f (0) =

The value of ${lim}_{x \to 0} \frac{\sqrt{a^{2} - a x + x^{2}} - \sqrt{a^{2} + a x + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}}$ is

\frac{\sqrt{a^{2} + a x + x^{2}} - \sqrt{a^{2} - a x + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}},

x = 0

x^{2} - a x + b = 0

\frac{a^{2} - c^{2}}{4}

\frac{a^{2}}{4} .

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