You visited us 2 times! Enjoying our articles? Unlock Full Access!

Standard Limits to Remove Indeterminate Form

If α and β ar...

Question

If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0,$ then $lim x \to α \frac{1 - cos (a x^{2} + b x + c)}{(x - α)^{2}}$ is :

\frac{(α - β)^{2}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 (α - β)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 a^{2} (α - β)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{a^{2} (α - β)^{2}}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{a^{2} (α - β)^{2}}{2}$
$a x^{2} + b x + c = a (x - α) (x - β)$

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

$= lim x \to α \frac{1 - cos (a (x - α) (x - β))}{(x - α)^{2}}$

$= lim x \to α \frac{2 {sin}^{2} (\frac{a (x - α) (x - β)}{2})}{(x - α)^{2}}$

$= lim x \to α \frac{2 {sin}^{2} (\frac{a (x - α) (x - β)}{2})}{\frac{a^{2} (x - α)^{2} (x - β)^{2}}{2^{2}}} \times \frac{a^{2} \times (x - β)^{2}}{4}$

$= 2 \times 1 \times \frac{a^{2} (α - β)^{2}}{4}$

$= \frac{a^{2} (α - β)^{2}}{2}$

Suggest Corrections

Similar questions

α

β

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

l i m x \to a \frac{1 - c o s (a x^{2} + b x + c)}{(x - α)^{2}}

α, β

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}}

α

β

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

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

α

β

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

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

α

β

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

lim x \to a

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

Join BYJU'S Learning Program