MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relations between Roots and Coefficients : Higher Order Equations

α ,β are the ...

Question

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

A

L_{1} = L_{2}

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

B

L_{1} = L_{3} {(α - β)}^{2}

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

C

L_{2} = L_{3} {(α - β)}^{2}

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

D

L_{3} = \frac{a^{2}}{2}

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

Open in App

Solution

The correct options are
A $L_{1} = L_{2}$
B $L_{1} = L_{3} {(α - β)}^{2}$
C $L_{2} = L_{3} {(α - β)}^{2}$
D $L_{3} = \frac{a^{2}}{2}$

Suggest Corrections

thumbs-up

1

Similar questions

Q. The value of

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

α, β

are the distinct roots of

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

α

β

be the roots of

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

are the roots of the quadratic equation

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

be the roots of

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

be the distinct roots of

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

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Relation of Roots and Coefficients

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Relations between Roots and Coefficients : Higher Order Equations

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon