If α, β are the distinct roots of x ^ 2 + bx + c = 0, then _ x → β e ^ 2( x ^ 2 + bx + c) - 1 - 2( x ^ 2 + bx + c) ( x - β ) ^ 2 is equal to

Correct answer: C. 2( b ^ 2 - 4c) _ x → β e ^ 2( x ^ 2 + bx + c) - 1 - 2( x ^ 2 + bx + c) (x - β) ^ 2 _ x → β 1(1 + 2( x ^ 2 + bx + c) 1 ! + 2 ^ 2 ( x ^ 2 + bx + c) ^ 2 2 ! + . . .) - 1 - 2( x ^ 2 + bx + c) (x - β) ^ 2 _ x → β 2 ( x ^ 2 + bx + c) ^ 2 (x - β) ^ 2 _ x → β 2 (x - α) ^ 2 (x - β) ^ 2 (x - β) ^ 2 2 (β - α) ^ 2 = 2( b ^ 2 - 4c)

JEE Main 202127 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2021 · 27 August · Morning Shift · Mathematics · Limits

If α, β are the distinct roots of x ^ 2 + bx + c = 0, then _ x → β...

\alpha, \beta

are the distinct roots of

{x}^{2} + bx + c = 0,

then

\lim_{x \to \beta}\dfrac{{e}^{2\left({x}^{2} + bx + c\right)} - 1 - 2\left({x}^{2} + bx + c\right)}{( x - \beta{)}^{2}}

is equal to

Options

A $2\left({b}^{2} + 4c\right)$
B ${b}^{2} - 4c$
C $2\left({b}^{2} - 4c\right)$
D ${b}^{2} + 4c$

Correct answer

C. $2\left({b}^{2} - 4c\right)$

Step-by-step solution

\lim_{x \to \beta}\dfrac{{e}^{2\left({x}^{2} + bx + c\right)} - 1 - 2\left({x}^{2} + bx + c\right)}{{\left(x - \beta\right)}^{2}}\lim_{x \to \beta}\dfrac{1\left(1 + \dfrac{2\left({x}^{2} + bx + c\right)}{1 !} + \dfrac{{2}^{2}{\left({x}^{2} + bx + c\right)}^{2}}{2 !} + . . .\right) - 1 - 2\left({x}^{2} + bx + c\right)}{{\left(x - \beta\right)}^{2}}\lim_{x \to \beta}\dfrac{2{\left({x}^{2} + bx + c\right)}^{2}}{{\left(x - \beta\right)}^{2}}\lim_{x \to \beta}\dfrac{2{\left(x - \alpha\right)}^{2}{\left(x - \beta\right)}^{2}}{{\left(x - \beta\right)}^{2}}2{\left(\beta - \alpha\right)}^{2} = 2\left({b}^{2} - 4c\right)

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