JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 7)
If $$\alpha$$, $$\beta$$ are the distinct roots of x2 + bx + c = 0, then
$$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$ is equal to :
$$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$ is equal to :
b2 + 4c
2(b2 + 4c)
2(b2 $$-$$ 4c)
b2 $$-$$ 4c
Explanation
$$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$
$$ = \mathop {\lim }\limits_{x \to \beta } {{1\left( {1 + {{2({x^2} + bx + c)} \over {1!}} + {{{2^2}{{({x^2} + bx + c)}^2}} \over {2!}} + ...} \right) - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$
$$ = \mathop {\lim }\limits_{x \to \beta } {{2{{({x^2} + bx + 1)}^2}} \over {{{(x - \beta )}^2}}}$$
$$ = \mathop {\lim }\limits_{x \to \beta } {{2{{(x - \alpha )}^2}{{(x - \beta )}^2}} \over {{{(x - \beta )}^2}}}$$
$$ = 2{(\beta - \alpha )^2} = 2({b^2} - 4c)$$
$$ = \mathop {\lim }\limits_{x \to \beta } {{1\left( {1 + {{2({x^2} + bx + c)} \over {1!}} + {{{2^2}{{({x^2} + bx + c)}^2}} \over {2!}} + ...} \right) - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$
$$ = \mathop {\lim }\limits_{x \to \beta } {{2{{({x^2} + bx + 1)}^2}} \over {{{(x - \beta )}^2}}}$$
$$ = \mathop {\lim }\limits_{x \to \beta } {{2{{(x - \alpha )}^2}{{(x - \beta )}^2}} \over {{{(x - \beta )}^2}}}$$
$$ = 2{(\beta - \alpha )^2} = 2({b^2} - 4c)$$
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