The value of limx-1-cos ax2-bx-c-x- 2 where - - - are the distinct roots of ax2-bx-c-0 is

The correct option is C (α β )^2/2 lim_x→β1 cos (ax^2+bx+c)(x β )^2=lim_x→ #160;β2sin ^2ax^2+bx+c2(x β )^2 =lim_x→βsin ^2(x α )(x β )2 ( #1 ...

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

Cos(A+B)Cos(A-B)

The value of ...

Question

The value of $lim x \to β \frac{1 - cos (a x^{2} + b x + c)}{(x - β)^{2}}$ where $α, β$ are the distinct roots of $a x^{2} + b x + c = 0$ is

(α - β)^{2}

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

\frac{(α - β)^{2}}{2}

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

(α \frac{(α - β)^{2}}{2})

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

none of these

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

Open in App

Solution

Suggest Corrections

Similar questions

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

(\frac{α}{β} - \frac{α}{β})^{2}

α, β

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

(α - β)^{2}, (α + β)^{2}

α, β

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

\frac{α^{2} + β^{2}}{α^{- 2} + β^{- 2}}

α

β

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

a, b

c

\frac{a α^{2} + b α + 6 c}{a β^{2} + b β^{2} + 9 c} + \frac{a β^{2} + b β^{2} + 19 c}{a α^{2} + b α + 13 c}

Join BYJU'S Learning Program