Lim x - 1 - -1-cosc x2-b x-a-1-x -2 where - is a root of a x2-b x-c-0- is equal to

The correct option is C 4ac b^2/2α^2x→ 1/αlim1 cos(cx^2+bx+a)/(1 xα)^2 If α is root of ax^2+bx+c=0 1/α is root of ax^2+bx+α=0 This is of the form Applying L'Ho ...

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lim x → 1 / α...

Question

$l i m x \to 1 / α \frac{1 - c o s (c x^{2} + b x + a)}{(1 - x α)^{2}}$ where $α$ is a root of $a x^{2} + b x + c = 0$ , is equal to

\frac{b^{2} - 4 a c}{2 α^{2}}

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\frac{b^{2} - 4 a c}{α^{2}}

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\frac{4 a c - b^{2}}{2 α^{2}}

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Solution

The correct option is C $\frac{4 a c - b^{2}}{2 α^{2}}$
$l i m x \to 1 / α \frac{1 - c o s (c x^{2} + b x + a)}{(1 - x α)^{2}}$
If $α$ is root of $a x^{2} + b x + c = 0$
$\frac{1}{α}$ is root of $a x^{2} + b x + α = 0$
This is of the form
Applying L'Hospital'r rule:
$l i m x \to 1 / α \frac{\frac{d}{d x} (1 - c o s (c x^{2} + b x + a))}{\frac{d}{d x} (1 - x α)^{2}} = l i m x \to 1 / α \frac{\frac{d}{d x} (1 - c o s (c x^{2} + b x + a))}{\frac{d}{d x} (1 - x α)^{2}} = l i m x \to 1 / α \frac{s i n (c x^{+} b x + a) (2 c x + b)}{2 (1 - x^{2}) (α)}$
Again apply LHR
$l i m x \to 1 / α \frac{c o s (c x^{2} + b x + a) (2 c x + b)^{2} + s i n (c x^{2} + b x + a) (2 x)}{- 2 α (- α)} = \frac{c o s (0) (2 c - \frac{1}{2} + b^{2}) + s i n (0) (2 c)}{2 α^{2}} = \frac{{(\frac{2 c}{2} + b)}^{2}}{2 α^{2}} = \frac{2 c + b α}{2 α^{4}}$

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limx→1/α1−cos(cx2+bx+a)(1−xα)2 where α is a root of ax2+bx+c=0, is equal to

$l i m x \to 1 / α \frac{1 - c o s (c x^{2} + b x + a)}{(1 - x α)^{2}}$ where $α$ is a root of $a x^{2} + b x + c = 0$ , is equal to