Let - -a-b be the roots of the equation -If a x2-b x-c-0 - If lim x - m-a x2-b x-c-a x2-b x-c-1- then

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

Algebra of Limits

Let α, βa<b b...

Question

Let $α, β$ (a < b) be the roots of the equation .If $a x^{2} + b x + c = 0. I f {lim}_{x \to m} \frac{| a x^{2} + b x + c |}{a x^{2} + b x + c} = 1, t h e n$

, m <

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

a > 0, < m <

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

, m >

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

a < 0, m >

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

Open in App

Solution

The correct option is C
, m >

According to the given condition, we have

$| a m^{2} + b m + c | = a m^{2} + b m + c$

i.e. $a m^{2}$ + bm + c > 0

$\Rightarrow$ if a < 0, the m lies in $(α, β)$

and if a > 0, then m does not lies in $(α, β)$

Hence, option ( c) is correct, since

$\frac{| a |}{a} = 1 \Rightarrow a$ < 0

And in that case m does not lie in $(α, β)$

Suggest Corrections

Similar questions

Let $α$ and $β$ ( $α < β$ ) be roots of the quadratic equation $a x^{2} + b x + c = 0.$ If 0 < a < c and a + c = b, then which of the following statement(s) is/are correct?

Q. Let

cos (α + β) = \frac{4}{5}

and let

sin (α - β) = \frac{5}{13}

, where

0 \leq α, β \leq \frac{π}{4}

. Then

tan 2 α =

Q. Let

cos (α + β) = \frac{4}{5}

and let

sin (α - β) = \frac{5}{13}

where

0 \leq α, β \leq \frac{π}{4}

, then

tan 2 α =

Q. Let

α

and

β

be two numbers where

α < β

. The geometric mean of these numbers exceeds the smaller number

α

12

and the arithmetic mean of the same number is smaller by

24

than the larger number

β

, then the value of

| β - α |

is :

Q. Let

f (x) = \frac{x}{x - 1} and \frac{f (α)}{f (α + 1)} = f (α^{k}),

then k = ______________.

Join BYJU'S Learning Program

Let α,β (a < b) be the roots of the equation .If ax2+bx+c=0. If limx→m|ax2+bx+c|ax2+bx+c=1, then

The correct option is C , m > According to the given condition, we have |am2+bm+c|=am2+bm+c i.e. am2 + bm + c > 0 ⇒ if a < 0, the m lies in (α,β) and if a > 0, then m does not lies in (α,β) Hence, option ( c) is correct, since |a|a=1⇒a < 0 And in that case m does not lie in (α,β)

Let $α, β$ (a < b) be the roots of the equation .If $a x^{2} + b x + c = 0. I f {lim}_{x \to m} \frac{| a x^{2} + b x + c |}{a x^{2} + b x + c} = 1, t h e n$