If $\alpha, \beta$ are the distinct roots of $x^{2}+bx+c=0$, then $\lim_{x \rightarrow \beta} \frac{e^{2(x^{2}+bx+c)}-1-2(x^{2}+bx+c)}{(x-\beta)^{2}}$ is equal to

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Limits

α, β

are the distinct roots of

x^{2} + b x + c = 0

, then

lim_{x \to β} \frac{e ^{2 (x^{2} + b x + c)} - 1 - 2 ( x ^{2} + b x + c )}{( x - β ) ^{2}}

is equal to

Answer: (D)

Since

α, β

are roots of

x^{2} + b x + c = 0

, we have:

x^{2} + b x + c = (x - α) (x - β)

Expanding

e^{2 (x^{2} + b x + c)}

using Taylor series:

e^{2 (x^{2} + b x + c)} = 1 + 2 (x^{2} + b x + c) + \frac{4 ( x ^{2} + b x + c ) ^{2}}{2 !} + \frac{8 ( x ^{2} + b x + c ) ^{3}}{3 !} + \dots

Substituting into the limit:

L = x \to β lim \frac{2 ( x ^{2} + b x + c ) + 2 ( x ^{2} + b x + c ) ^{2} + \frac{4 ( x ^{2} + b x + c ) ^{3}}{3} + \dots}{( x - β ) ^{2}}

= x \to β lim \frac{2 ( x - α ) ^{2} ( x - β ) ^{2} + \frac{8 ( x - α ) ^{3} ( x - β ) ^{3}}{6} + \dots}{( x - β ) ^{2}}

= 2 (β - α)^{2} = 2 [(α + β)^{2} - 4 α β]

Using Vieta's formulas:

α + β = - b

and

α β = c

L = 2 [b^{2} - 4 c] = 2 (b^{2} - 4 c)

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option