If $\alpha > \beta > 0$ are the roots of the equation $ax^2 + bx + 1 = 0$, and $\lim_{x \to \frac{1}{\alpha}} \left( \frac{1 - \cos(x^2 + bx + a)}{2(1 - \alpha x)^2} \right)^{\frac{1}{2}} = \frac{1}{k} \left( \frac{1}{\beta} - \frac{1}{\alpha} \right)$, then $k$ is equal to Options: (A) $2\beta$ (B) $\alpha$ (C) $2\alpha$ (D) $\beta$

### Analyzing the Setup Consider the quadratic equation $ax^2 + bx + 1 = 0$. When we swap the leading and constant coefficients to form $x^2 + bx + a = 0$, we invoke the property of **reciprocal roots**. If the original roots of $ax^2 + bx + 1 = 0$ are $\alpha$ and $\beta$, then the roots of the transformed equation $x^2 + bx + a = 0$ are $1/\alpha$ and $1/\beta$. This allows us to express the quadratic as a product of factors: $$x^2 + bx + a = (x - 1/\alpha)(x - 1/\beta)$$ ### The Trigonometric Bridge We are tasked with evaluating the following limit: $$\lim_{x \to 1/\alpha} \left( \frac{1 - \cos(x^2 + bx + a)}{2(1 - \alpha x)^2} \right)^{1/2}$$ Using the trigonometric identity $1 - \cos \theta = 2 \sin^2(\theta/2)$, we substitute $\theta = x^2 + bx + a$ into the numerator: $$1 - \cos(x^2 + bx + a) = 2 \sin^2 \left( \frac{x^2 + bx + a}{2} \right)$$ ### The Limit Dance As $x \to 1/\alpha$, the term $x^2 + bx + a \to 0$. Applying the small-angle approximation $\sin \theta \approx \theta$, the expression inside the limit simplifies significantly: $$\lim_{x \to 1/\alpha} \sqrt{ \frac{2 \sin^2 \left( \frac{x^2 + bx + a}{2} \right)}{2(1 - \alpha x)^2} } = \lim_{x \to 1/\alpha} \left| \frac{x^2 + bx + a}{2(1 - \alpha x)} \right|$$ Substituting the factorized form of the quadratic, we get: $$\lim_{x \to 1/\alpha} \left| \frac{(x - 1/\alpha)(x - 1/\beta)}{2\alpha(1/\alpha - x)} \right|$$ Since $(x - 1/\alpha) = -(1/\alpha - x)$, the terms cancel out, leaving: $$\left| \frac{-(x - 1/\beta)}{2\alpha} \right| = \left| \frac{1/\beta - 1/\alpha}{2\alpha} \right|$$ ### The Final Victory Given the condition $\alpha > \beta > 0$, it follows that $1/\beta > 1/\alpha$. Thus, the absolute value resolves to: $$\frac{1}{2\alpha} \left( \frac{1}{\beta} - \frac{1}{\alpha} \right)$$ Comparing this result to the form $\frac{1}{k} (1/\beta - 1/\alpha)$, we identify the constant as **$k = 2\alpha$**. This elegant symmetry confirms that recognizing the reciprocal nature of the roots is the key to unlocking the solution.

JEE Main 2023 (08 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $α > β > 0$ are the roots of the equation $a x^{2} + b x + 1 = 0$ , and $lim_{x \to \frac{1}{α}} (\frac{1 - c o s ( x ^{2} + b x + a )}{2 ( 1 - α x ) ^{2}})^{\frac{1}{2}} = \frac{1}{k} (\frac{1}{β} - \frac{1}{α})$ , then $k$ is equal to

Q: Why do we need absolute value signs when taking the square root?

This is a classic trap! The expression is $\sqrt{(f(x))^2}$, which is equal to $|f(x)|$, not just $f(x)$. Since we are taking a limit, the sign of the expression as $x$ approaches the value matters. Forgetting the absolute value can lead to a sign error in your final answer, which is a common way to lose marks.

Select Answer:

Visualized Solution

Analyze the Roots of the Given Equation

Given equation: $a x^{2} + b x + 1 = 0$ with roots $α$ and $β$ .
Condition: $α > β > 0$ .
We need to evaluate the limit as $x \to \frac{1}{α}$ for an expression involving $x^{2} + b x + a$ .

Reciprocal Roots Property

If roots of $a x^{2} + b x + 1 = 0$ are $α, β$ , then roots of $x^{2} + b x + a = 0$ are $\frac{1}{α}, \frac{1}{β}$ .
Proof: Replace $x$ with $\frac{1}{t}$ in $a x^{2} + b x + 1 = 0$ to get $a (\frac{1}{t})^{2} + b (\frac{1}{t}) + 1 = 0 ⟹ a + b t + t^{2} = 0$ .

Factorizing the Quadratic Expression

Let $f (x) = x^{2} + b x + a$ .
Using the roots $\frac{1}{α}$ and $\frac{1}{β}$ , we can factorize it as:
$f (x) = (x - \frac{1}{α}) (x - \frac{1}{β})$

Applying Trigonometric Identity

Use the identity: $1 - cos θ = 2 sin^{2} (\frac{θ}{2})$ .
Substitute $θ = x^{2} + b x + a$ :
$1 - cos (x^{2} + b x + a) = 2 sin^{2} (\frac{x ^{2} + b x + a}{2})$

Substituting into the Limit Expression

Substitute the identity into the limit $L$ :
$L = lim_{x \to \frac{1}{α}} (\frac{2 s i n ^{2} ( \frac{x ^{2} + b x + a}{2} )}{2 ( 1 - α x ) ^{2}})^{\frac{1}{2}}$
Cancel the constant $2$ .

Simplifying the Square Root

Simplify the square root: $y^{2} = ∣ y ∣$ .
$L = lim_{x \to \frac{1}{α}} \frac{∣ s i n ( \frac{x ^{2} + b x + a}{2} ) ∣}{∣1 - α x ∣}$

Applying the Standard Limit

As $x \to \frac{1}{α}$ , $(x^{2} + b x + a) \to 0$ .
Use the standard limit: $lim_{θ \to 0} \frac{s i n θ}{θ} = 1 ⟹ sin θ \approx θ$ .
$L = lim_{x \to \frac{1}{α}} \frac{∣ \frac{1}{2} ( x ^{2} + b x + a ) ∣}{∣1 - α x ∣}$

Substituting the Factors

Substitute $x^{2} + b x + a = (x - \frac{1}{α}) (x - \frac{1}{β})$ :
$L = lim_{x \to \frac{1}{α}} \frac{∣ \frac{1}{2} ( x - \frac{1}{α} ) ( x - \frac{1}{β} ) ∣}{∣ α ( \frac{1}{α} - x ) ∣}$

Canceling Common Terms

Cancel $∣ x - \frac{1}{α} ∣$ from numerator and denominator.
$L = \frac{∣ \frac{1}{α} - \frac{1}{β} ∣}{2 α}$

Evaluating the Absolute Value

Since $α > β > 0$ , then $\frac{1}{β} > \frac{1}{α}$ .
Thus, $∣ \frac{1}{α} - \frac{1}{β} ∣ = \frac{1}{β} - \frac{1}{α}$ .
$L = \frac{1}{2 α} (\frac{1}{β} - \frac{1}{α})$

Comparing to Find $k$

Given: $L = \frac{1}{k} (\frac{1}{β} - \frac{1}{α})$ .
Comparing with $L = \frac{1}{2 α} (\frac{1}{β} - \frac{1}{α})$ :
We get $k = 2 α$ .

Final Conclusion

Key Takeaway: For $a x^{2} + b x + 1 = 0$ , the expression $x^{2} + b x + a$ is related to the reciprocal roots.
Final Answer: $k = 2 α$ (Option 2).
Next Challenge: Try solving the same limit if the denominator was $(1 - β x)^{2}$ instead.

00:00 / 00:00

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Solution Diagram

Analyzing the Setup

Consider the quadratic equation

a x^{2} + b x + 1 = 0

. When we swap the leading and constant coefficients to form

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

, we invoke the property of reciprocal roots.

If the original roots of

a x^{2} + b x + 1 = 0

are

α

and

β

, then the roots of the transformed equation

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

are

1/ α

and

1/ β

. This allows us to express the quadratic as a product of factors:

x^{2} + b x + a = (x - 1/ α) (x - 1/ β)

The Trigonometric Bridge

We are tasked with evaluating the following limit:

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

Using the trigonometric identity

1 - cos θ = 2 sin^{2} (θ /2)

, we substitute

θ = x^{2} + b x + a

into the numerator:

1 - cos (x^{2} + b x + a) = 2 sin^{2} (\frac{x ^{2} + b x + a}{2})

The Limit Dance

x \to 1/ α

, the term

x^{2} + b x + a \to 0

. Applying the small-angle approximation

sin θ \approx θ

, the expression inside the limit simplifies significantly:

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

Substituting the factorized form of the quadratic, we get:

x \to 1/ α lim \frac{( x - 1/ α ) ( x - 1/ β )}{2 α ( 1/ α - x )}

Since

(x - 1/ α) = - (1/ α - x)

, the terms cancel out, leaving:

\frac{- ( x - 1/ β )}{2 α} = \frac{1/ β - 1/ α}{2 α}

The Final Victory

Given the condition

α > β > 0

, it follows that

1/ β > 1/ α

. Thus, the absolute value resolves to:

\frac{1}{2 α} (\frac{1}{β} - \frac{1}{α})

Comparing this result to the form

\frac{1}{k} (1/ β - 1/ α)

, we identify the constant as $k = 2 α$ . This elegant symmetry confirms that recognizing the reciprocal nature of the roots is the key to unlocking the solution.

Common Traps & FAQs

Wait, why do we swap the coefficients? Is there a trick?

It is not a trick, but a beautiful property of quadratic equations! If you replace

x

with

1/ t

a x^{2} + b x + 1 = 0

, you get

a (1/ t)^{2} + b (1/ t) + 1 = 0

. Multiplying by

t^{2}

gives

a + b t + t^{2} = 0

, which is

t^{2} + b t + a = 0

. This proves that the roots of the new equation are simply the reciprocals of the original roots. Always look for this symmetry in JEE problems!

Why do we need absolute value signs when taking the square root?

This is a classic trap! The expression is

(f (x))^{2}

, which is equal to

∣ f (x) ∣

, not just

f (x)

. Since we are taking a limit, the sign of the expression as

x

approaches the value matters. Forgetting the absolute value can lead to a sign error in your final answer, which is a common way to lose marks.

Can we use L'Hopital's rule instead of the standard limit?

Yes, you absolutely can! L'Hopital's rule is a powerful tool for

0/0

indeterminate forms. However, it can become algebraically messy with the chain rule and the square root. Using the standard limit

sin θ \approx θ

is often faster and more elegant, but if you are comfortable with differentiation, L'Hopital's is a perfectly valid path.

Select Answer:

Visualized Solution

Analyze the Roots of the Given Equation

Reciprocal Roots Property

Factorizing the Quadratic Expression

Applying Trigonometric Identity

Substituting into the Limit Expression

Simplifying the Square Root

Applying the Standard Limit

Substituting the Factors

Canceling Common Terms

Evaluating the Absolute Value

Comparing to Find $k$

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Trigonometric Bridge

The Limit Dance

The Final Victory

Similar Questions

Let $α$ and $β$ be the distinct roots of $a x^{2} + b x + c = 0$ , then $lim_{x \to α} \frac{1 - c o s ( a x ^{2} + b x + c )}{( x - α ) ^{2}}$ 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:

Let $α (a)$ and $β (a)$ be the roots of the equation $(3 1 + a - 1) x^{2} + (1 + a - 1) x + (6 1 + a - 1) = 0$ where $a > - 1$ . Then $lim_{a \to 0^{+}} α (a)$ and $lim_{a \to 0^{+}} β (a)$ are

Let $a > 0$ be a root of the equation $2 x^{2} + x - 2 = 0$ . If $lim_{x \to \frac{1}{a}} \frac{16 ( 1 - c o s ( 2 + x - 2 x ^{2} ))}{( 1 - a x ) ^{2}} = α + β 17$ , where $α, β \in Z$ , then $α + β$ is equal to _______

If $lim_{x \to 0} {(1 - cos \frac{x ^{2}}{2} - cos \frac{x ^{2}}{4} + cos \frac{x ^{2}}{2} cos \frac{x ^{2}}{4})} = 2^{k}$ , then the value of $k$ is

For $t > - 1$ , let $α_{t}$ and $β_{t}$ be the roots of the equation $((t + 2)^{\frac{1}{7}} - 1) x^{2} + ((t + 2)^{\frac{1}{6}} - 1) x + ((t + 2)^{\frac{1}{21}} - 1) = 0$ . If $lim_{t \to - 1^{+}} α_{t} = a$ and $lim_{t \to - 1^{+}} β_{t} = b$ , then $72 (a + b)^{2}$ is equal to ________.

If $a$ is the positive root of the equation, $p (x) = x^{2} - x - 2 = 0$ , then $lim_{x \to a^{+}} \frac{1 - c o s ( p ( x ))}{x + a - 4}$ is equal to :

If $α = lim_{x \to 0^{+}} (\frac{e ^{t a n x} - e ^{x}}{t a n x - x})$ and $β = lim_{x \to 0} (1 + sin x)^{\frac{1}{2} c o t x}$ are the roots of the quadratic equation $a x^{2} + b x - e = 0$ , then $12 lo g_{e} (a + b)$ is equal to__________

If $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$ , then $a^{2} + b^{2} + c^{2}$ is equal to :

Let $k \in R$ . If $lim_{x \to 0^{+}} (sin (sin k x) + cos x + x)^{2/ x} = e^{6}$ , then the value of $k$ is

Select Answer:

Visualized Solution

Analyze the Roots of the Given Equation

Reciprocal Roots Property

Factorizing the Quadratic Expression

Applying Trigonometric Identity

Substituting into the Limit Expression

Simplifying the Square Root

Applying the Standard Limit

Substituting the Factors

Canceling Common Terms

Evaluating the Absolute Value

Comparing to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Trigonometric Bridge

The Limit Dance

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α and β be the distinct roots of ax2+bx+c=0, then limx→α​(x−α)21−cos(ax2+bx+c)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α,β are the distinct roots of x2+bx+c=0, then limx→β​(x−β)2e2(x2+bx+c)−1−2(x2+bx+c)​ is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α(a) and β(a) be the roots of the equation (31+a​−1)x2+(1+a​−1)x+(61+a​−1)=0 where a>−1. Then lima→0+​α(a) and lima→0+​β(a) are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a>0 be a root of the equation 2x2+x−2=0. If limx→a1​​(1−ax)216(1−cos(2+x−2x2))​=α+β17​, where α,β∈Z, then α+β is equal to _______

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→0​{(1−cos2x2​−cos4x2​+cos2x2​cos4x2​)}=2k, then the value of k is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a is the positive root of the equation, p(x)=x2−x−2=0, then limx→a+​x+a−41−cos(p(x))​​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→0​xcosx−loge​(1+x)e(a−1)x+2cosbx+(c−2)e−x​=2, then a2+b2+c2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let k∈R. If limx→0+​(sin(sinkx)+cosx+x)2/x=e6, then the value of k is

Comparing to Find $k$

Let $α$ and $β$ be the distinct roots of $a x^{2} + b x + c = 0$ , then $lim_{x \to α} \frac{1 - c o s ( a x ^{2} + b x + c )}{( x - α ) ^{2}}$ 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:

Let $α (a)$ and $β (a)$ be the roots of the equation $(3 1 + a - 1) x^{2} + (1 + a - 1) x + (6 1 + a - 1) = 0$ where $a > - 1$ . Then $lim_{a \to 0^{+}} α (a)$ and $lim_{a \to 0^{+}} β (a)$ are

Let $a > 0$ be a root of the equation $2 x^{2} + x - 2 = 0$ . If $lim_{x \to \frac{1}{a}} \frac{16 ( 1 - c o s ( 2 + x - 2 x ^{2} ))}{( 1 - a x ) ^{2}} = α + β 17$ , where $α, β \in Z$ , then $α + β$ is equal to _______

If $lim_{x \to 0} {(1 - cos \frac{x ^{2}}{2} - cos \frac{x ^{2}}{4} + cos \frac{x ^{2}}{2} cos \frac{x ^{2}}{4})} = 2^{k}$ , then the value of $k$ is

If $a$ is the positive root of the equation, $p (x) = x^{2} - x - 2 = 0$ , then $lim_{x \to a^{+}} \frac{1 - c o s ( p ( x ))}{x + a - 4}$ is equal to :

If $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$ , then $a^{2} + b^{2} + c^{2}$ is equal to :

Let $k \in R$ . If $lim_{x \to 0^{+}} (sin (sin k x) + cos x + x)^{2/ x} = e^{6}$ , then the value of $k$ is