If - - are the distinct roots of x2 - bx - c - 0- then limx-e2x2-bx-c-1-2x2-bx-cx- 2 is equal to

Given: α, β are roots of x^2+bx+c =0 ∴ x^2+bx+c = (x α)(x β) ⋯ (i) nbsp; nbsp; Also β^2 + bβ + c = 0 ⋯ (ii ) nbsp; Now, L = lim_x→βe^2(x^2+bx+c) 1 2( ...

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Standard XII

Mathematics

Second Derivative Test for Local Maximum

If α, β are t...

Question

If $α, β$ 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

b^{2} - 4 c

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b^{2} + 4 c

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2 (b^{2} + 4 c)

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2 (b^{2} - 4 c)

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Solution

The correct option is D $2 (b^{2} - 4 c)$
Given: $α, β$ are roots of $x^{2} + b x + c = 0$
$∴ x^{2} + b x + c = (x - α) (x - β) \dots (i)$
Also $β^{2} + b β + c = 0 \dots (i i)$
Now, $L = lim x \to β \frac{e^{2 (x^{2} + b x + c)} - 1 - 2 (x^{2} + b x + c)}{{(x - β)}^{2} \times (x - α)^{2}} \times (x - α)^{2}$
$L = lim x \to β \frac{e^{2 (x^{2} + b x + c)} - 1 - 2 (x^{2} + b x + c)}{(x^{2} + b x + c)^{2}} \times lim x \to β (x - α)^{2}$ $(From (i))$
Let $x^{2} + b x + c = t$
Then, $x \to β \Rightarrow t \to 0$ $(From (i i))$
$∴ L = lim t \to 0 \frac{e^{2 t} - 1 - 2 t}{t^{2}} \times (β - α)^{2}$
Now, using expansion, we have:
$L = lim t \to 0 \frac{(1 + 2 t + \frac{(2 t)^{2}}{2!} + \frac{(2 t)^{3}}{3!} + \dots) - 1 - 2 t}{t^{2}} \times (α - β)^{2}$
$\Rightarrow L = 2 (α - β)^{2}$
$\Rightarrow L = 2 [(α + β)^{2} - 4 α β]$
$\Rightarrow L = 2 [(- b)^{2} - 4 c]$
$∴ L = 2 (b^{2} - 4 c)$

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Similar questions

Q. The value of

{lim}_{x \to β} \frac{1 - cos (a x^{2} + b x + c)}{(x^{2} - β^{2})}

where

α, β

are the distinct roots of

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

Q. If

α, β

are the roots of

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

then

{(α + β)}^{2} =

α, β

are the roots of

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

L_{1} = L t x \to α \frac{1 - c o s ({a x}^{2} + b x + c)}{{(x - α)}^{2}}

L_{2} = L t x \to β \frac{1 - c o s ({a x}^{2} + b x + c)}{{(x - β)}^{2}}

L_{3} = L t (x - α) (x - β) \to 0 \frac{1 - c o s ({a x}^{2} + b x + c)}{{(x - α)}^{2} {(x - β)}^{2}}

Q. Assertion :

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

is equal to

e^{a (α - β)}

Reason: If

α

β

are roots of

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

then

a x^{2} + b x + c = a (x - α) (x - β)

Q. Let

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

then

lim x \to a

\frac{1 - cos (a x^{2} + b x + c)}{(x - α)^{2}}

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If α,β are the distinct roots of x2+bx+c=0, then limx→βe2(x2+bx+c)−1−2(x2+bx+c)(x−β)2 is equal to

If $α, β$ 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