JEE Main 2021 (31 Aug Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $α = lim_{x \to π /4} \frac{t a n ^{3} x - t a n x}{c o s ( x + \frac{π}{4} )}$ and $β = lim_{x \to 0} (cos x)^{c o t x}$ are the roots of the equation, $a x^{2} + b x - 4 = 0$ , then the ordered pair $(a, b)$ is :

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Before diving into calculations, observe the indeterminate forms. For

α

, check the limit as

x \to π /4

. You will find a

0/0

form. For

β

, check the limit as

x \to 0

. You will encounter a

1^{\infty}

form. Identifying these forms is the first step to choosing your strategy.

Hint 2

Hint 2: The Arsenal. For

α

, L'Hopital's Rule is your primary weapon. Differentiate the numerator and denominator separately. For

β

, remember the standard limit identity:

lim_{x \to a} f (x)^{g (x)} = e^{l i m_{x \to a} g (x) (f (x) - 1)}

. This will transform the exponent into a product that is much easier to evaluate.

Hint 3

Hint 3: The Execution. Once you have calculated

α = - 4

and

β = 1

, you have the roots of the quadratic equation. Use the relationship between roots and coefficients:

x^{2} - (α + β) x + α β = 0

. Construct the equation and compare it with

a x^{2} + b x - 4 = 0

to find the values of

a

and

b

Visualized Solution

Problem Introduction

Given equation: $a x^{2} + b x - 4 = 0$
Roots: $α$ and $β$
$α = lim_{x \to \frac{π}{4}} \frac{t a n ^{3} x - t a n x}{c o s ( x + \frac{π}{4} )}$
$β = lim_{x \to 0} (cos x)^{c o t x}$

Analyzing $α$ Form

Check form for $α$ as $x \to \frac{π}{4}$ :
Numerator: $tan^{3} (\frac{π}{4}) - tan (\frac{π}{4}) = 1 - 1 = 0$
Denominator: $cos (\frac{π}{4} + \frac{π}{4}) = cos (\frac{π}{2}) = 0$
Form: $\frac{0}{0}$

Applying L'Hopital's Rule to $α$

Using L'Hopital's Rule:
$α = lim_{x \to \frac{π}{4}} \frac{\frac{d}{d x} ( t a n ^{3} x - t a n x )}{\frac{d}{d x} ( c o s ( x + \frac{π}{4} ))}$
$α = lim_{x \to \frac{π}{4}} \frac{3 t a n ^{2} x s e c ^{2} x - s e c ^{2} x}{- s i n ( x + \frac{π}{4} )}$

Evaluating $α$

Substitute $x = \frac{π}{4}$ :
$α = \frac{3 ( 1 ) ^{2} ( 2 ) ^{2} - ( 2 ) ^{2}}{- s i n ( \frac{π}{4} + \frac{π}{4} )}$
$α = \frac{3 ( 2 ) - 2}{- s i n ( \frac{π}{2} )} = \frac{4}{- 1} = - 4$

Analyzing $β$ Form

Check form for $β$ as $x \to 0$ :
Base: $cos (0) = 1$
Exponent: $cot (0) \to \infty$
Form: $1^{\infty}$

Transforming $β$

Using the formula $lim_{x \to a} f (x)^{g (x)} = e^{l i m_{x \to a} g (x) (f (x) - 1)}$ :
$β = e^{l i m_{x \to 0} c o t x (c o s x - 1)}$
$β = e^{l i m_{x \to 0} \frac{c o s x - 1}{t a n x}}$

Evaluating $β$ Exponent

Evaluate exponent limit using L'Hopital's Rule:
$lim_{x \to 0} \frac{- s i n x}{s e c ^{2} x} = \frac{- s i n 0}{s e c ^{2} 0} = \frac{0}{1} = 0$
$β = e^{0} = 1$

Forming the Quadratic Equation

Roots: $α = - 4$ , $β = 1$
Sum of roots: $α + β = - 4 + 1 = - 3$
Product of roots: $α β = (- 4) (1) = - 4$
Equation: $x^{2} - (α + β) x + α β = 0$
Equation: $x^{2} + 3 x - 4 = 0$

Comparing Coefficients

Compare $x^{2} + 3 x - 4 = 0$ with $a x^{2} + b x - 4 = 0$
Coefficient of $x^{2}$ : $a = 1$
Coefficient of $x$ : $b = 3$

Final Answer

Ordered pair $(a, b) = (1, 3)$
Correct Option: (1, 3)

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

The quadratic equation is given by

a x^{2} + b x - 4 = 0

, where the roots are

α

and

β

. Our objective is to determine the values of

α

and

β

using limit theory and subsequently identify the coefficients

a

and

b

Unmasking $α$

We begin by evaluating the limit for

α

α = x \to π /4 lim \frac{tan ^{3} x - tan x}{cos ( x + π /4 )}

Substituting

x = π /4

yields the indeterminate form

0/0

. Applying L'Hopital's Rule, we differentiate the numerator and denominator with respect to

x

Numerator derivative:

\frac{d}{d x} (tan^{3} x - tan x) = 3 tan^{2} x sec^{2} x - sec^{2} x

Denominator derivative:

\frac{d}{d x} (cos (x + π /4)) = - sin (x + π /4)

Evaluating at

x = π /4

, where

tan (π /4) = 1

and

sec^{2} (π /4) = 2

α = \frac{3 ( 1 ) ^{2} ( 2 ) - 2}{- sin ( π /2 )} = \frac{6 - 2}{- 1} = - 4

Thus, the first root is $α = - 4$ .

The $β$ Mystery

Next, we evaluate the limit for

β

β = x \to 0 lim (cos x)^{c o t x}

This is a

1^{\infty}

indeterminate form. We utilize the identity

lim_{x \to a} f (x)^{g (x)} = e^{l i m_{x \to a} g (x) (f (x) - 1)}

β = e^{l i m_{x \to 0} c o t x (c o s x - 1)}

Rewriting the exponent as

\frac{c o s x - 1}{t a n x}

and applying L'Hopital's Rule again:

x \to 0 lim \frac{- sin x}{sec ^{2} x} = \frac{0}{1} = 0

Therefore,

β = e^{0} = 1

. The second root is $β = 1$ .

The Quadratic Reunion

Given the roots

α = - 4

and

β = 1

, the sum and product of the roots are:

α + β = - 3

α β = - 4

The quadratic equation is expressed as

x^{2} - (α + β) x + α β = 0

. Substituting our values:

x^{2} - (- 3) x + (- 4) = 0 \Rightarrow x^{2} + 3 x - 4 = 0

Comparing this to the original equation

a x^{2} + b x - 4 = 0

, we identify

a = 1

and

b = 3

. The final result is the ordered pair $(a, b) = (1, 3)$ .

Common Traps & FAQs

Why can't I just use L'Hopital's rule directly on

(cos x)^{c o t x}

L'Hopital's rule is specifically designed for limits of the form

0/0

\infty/\infty

. Your expression is an exponential form

1^{\infty}

. You cannot differentiate the base and exponent directly. You must first use the logarithmic transformation or the standard identity

e^{l i m g (x) (f (x) - 1)}

to bring the exponent down into a multiplicative form.

I got

α = - 4

. Is it okay for a root to be negative?

Absolutely! Roots of a quadratic equation can be negative, positive, or even complex. Do not let the negative sign shake your confidence. If your calculus is correct, the negative sign is just a part of the answer. Trust your derivation.

How do I know when to stop differentiating in L'Hopital's rule?

You stop when the indeterminate form is resolved. After you differentiate once, plug the limit value back in. If you get a finite, defined number, you are done. If you still get

0/0

\infty/\infty

, you may need to differentiate again. In this problem, one application was sufficient.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $α = lim_{x \to π /4} \frac{t a n ^{3} x - t a n x}{c o s ( x + \frac{π}{4} )}$ and $β = lim_{x \to 0} (cos x)^{c o t x}$ are the roots of the equation, $a x^{2} + b x - 4 = 0$ , then the ordered pair $(a, b)$ is :

Select Answer:

Visualized Solution

Problem Introduction

Analyzing $α$ Form

Applying L'Hopital's Rule to $α$

Evaluating $α$

Analyzing $β$ Form

Transforming $β$

Evaluating $β$ Exponent

Forming the Quadratic Equation

Comparing Coefficients

Final Answer

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

Analyzing the Setup

Unmasking $α$

The $β$ Mystery

The Quadratic Reunion

Similar Questions

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Select Answer:

Visualized Solution

Problem Introduction

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α Form

Applying L'Hopital's Rule to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β Form

Transforming .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β Exponent

Forming the Quadratic Equation

Comparing Coefficients

Final Answer

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

Analyzing the Setup

Unmasking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β Mystery

The Quadratic Reunion

Similar Questions

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Analyzing $α$ Form

Applying L'Hopital's Rule to $α$

Evaluating $α$

Analyzing $β$ Form

Transforming $β$

Evaluating $β$ Exponent

Unmasking $α$

The $β$ Mystery

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