JEE Main 2022 (28 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 1} \frac{s i n ( 3 x ^{2} - 4 x + 1 ) - x ^{2} + 1}{2 x ^{3} - 7 x ^{2} + a x + b} = - 2$ , then the value of $(a - b)$ is equal to

Q: I got confused with the chain rule in the derivative. How do I handle $\sin(3x^2 - 4x + 1)$?

The chain rule is your best friend! Remember the rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$. Here, $f(u) = \sin(u)$ and $g(x) = 3x^2 - 4x + 1$. So, the derivative is $\cos(3x^2 - 4x + 1)$ multiplied by the derivative of the inside, which is $(6x - 4)$. Always keep that inner derivative attached!

Enter Numerical Value:

Progressive Hints

Hint 1

Think like a detective. If you have a fraction

\frac{f ( x )}{g ( x )}

and the limit as

x \to 1

is a finite number like

- 2

, but the numerator

f (1)

is zero, what does that force the denominator

g (1)

to be? If

g (1)

were non-zero, the limit would be zero, not

- 2

Hint 2

You have two unknowns,

a

and

b

. This implies you need two independent equations. The first comes from the condition

g (1) = 0

. For the second, apply L'Hopital's Rule. Remember, the derivative of the numerator must also vanish at

x = 1

to maintain the

\frac{0}{0}

form required for a finite limit.

Hint 3

Start by setting the denominator to zero at

x = 1

2 (1)^{3} - 7 (1)^{2} + a (1) + b = 0

. Then, differentiate both numerator and denominator. Set the derivative of the numerator at

x = 1

to zero (or equate the limit of the derivatives to

- 2

) to find

a

Visualized Solution

Analyze the Limit at $x = 1$

Given limit: $lim_{x \to 1} \frac{s i n ( 3 x ^{2} - 4 x + 1 ) - x ^{2} + 1}{2 x ^{3} - 7 x ^{2} + a x + b} = - 2$
Let's check the value of the numerator as $x$ approaches $1$ .

Evaluate the Numerator

Substitute $x = 1$ into the numerator:
$sin (3 (1)^{2} - 4 (1) + 1) - (1)^{2} + 1$
$= sin (3 - 4 + 1) - 1 + 1$
$= sin (0) = 0$

Condition for a Finite Limit

The numerator approaches $0$ .
The overall limit is finite ( $- 2$ ).
Therefore, the limit must be of the $\frac{0}{0}$ indeterminate form.
The denominator must also approach $0$ as $x \to 1$ .

Establish Equation for $a$ and $b$

Set the denominator to $0$ at $x = 1$ :
$2 (1)^{3} - 7 (1)^{2} + a (1) + b = 0$
$2 - 7 + a + b = 0$
$a + b = 5$

Apply L'Hopital's Rule: Numerator

Since it's a $\frac{0}{0}$ form, apply L'Hopital's Rule.
Differentiate the numerator with respect to $x$ :
$\frac{d}{d x} [sin (3 x^{2} - 4 x + 1) - x^{2} + 1]$
$= cos (3 x^{2} - 4 x + 1) \cdot (6 x - 4) - 2 x$

Apply L'Hopital's Rule: Denominator

Differentiate the denominator with respect to $x$ :
$\frac{d}{d x} [2 x^{3} - 7 x^{2} + a x + b]$
$= 6 x^{2} - 14 x + a$
The new limit expression is:
$lim_{x \to 1} \frac{c o s ( 3 x ^{2} - 4 x + 1 ) \cdot ( 6 x - 4 ) - 2 x}{6 x ^{2} - 14 x + a} = - 2$

Evaluate the New Numerator

Substitute $x = 1$ into the new numerator:
$cos (3 (1)^{2} - 4 (1) + 1) \cdot (6 (1) - 4) - 2 (1)$
$= cos (0) \cdot (2) - 2$

Simplify the New Numerator

$cos (0) = 1$
The expression becomes: $1 \cdot 2 - 2$
$= 2 - 2 = 0$
The new numerator is again $0$ .

Second Application of the $\frac{0}{0}$ Condition

The new numerator is $0$ at $x = 1$ .
The limit is still finite ( $- 2$ ).
Therefore, the new denominator must also be $0$ at $x = 1$ .
$6 (1)^{2} - 14 (1) + a = 0$

Solve for $a$

Set the new denominator to $0$ :
$6 - 14 + a = 0$
$- 8 + a = 0$
$a = 8$

Solve for $b$

Recall Equation 1: $a + b = 5$
Substitute $a = 8$ :
$8 + b = 5$
$b = 5 - 8 = - 3$

Calculate Final Value of $(a - b)$

We have $a = 8$ and $b = - 3$ .
We need to find the value of $(a - b)$ .
$a - b = 8 - (- 3)$
$a - b = 8 + 3 = 11$
Final Answer: 11

00:00 / 00:00

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

Analyzing the Setup

We are given the limit:

x \to 1 lim \frac{sin ( 3 x ^{2} - 4 x + 1 ) - x ^{2} + 1}{2 x ^{3} - 7 x ^{2} + a x + b} = - 2

When we substitute

x = 1

into the numerator, we obtain:

sin (3 (1)^{2} - 4 (1) + 1) - (1)^{2} + 1 = sin (0) - 1 + 1 = 0

Since the limit is a non-zero value (

- 2

), the denominator must also evaluate to

0

x = 1

to satisfy the

\frac{0}{0}

indeterminate form.

Establishing the First Constraint

Setting the denominator to zero at

x = 1

2 (1)^{3} - 7 (1)^{2} + a (1) + b = 0

This simplifies to:

2 - 7 + a + b = 0

a + b = 5 \dots (Equation 1)

Applying L'Hopital's Rule

To resolve the indeterminate form, we differentiate the numerator and the denominator with respect to

x

Numerator derivative:

\frac{d}{d x} [sin (3 x^{2} - 4 x + 1) - x^{2} + 1] = cos (3 x^{2} - 4 x + 1) \cdot (6 x - 4) - 2 x

Denominator derivative:

\frac{d}{d x} [2 x^{3} - 7 x^{2} + a x + b] = 6 x^{2} - 14 x + a

Determining the Constants

Evaluating the derivative of the numerator at

x = 1

cos (0) \cdot (6 - 4) - 2 (1) = 1 \cdot 2 - 2 = 0

Since the derivative of the numerator is

0

x = 1

, the derivative of the denominator must also be

0

x = 1

for the limit to exist as a finite value:

6 (1)^{2} - 14 (1) + a = 0

6 - 14 + a = 0 \Rightarrow a = 8

Now, substitute

a = 8

into Equation 1:

8 + b = 5 \Rightarrow b = - 3

Final Calculation

The problem asks for the value of

(a - b)

a - b = 8 - (- 3) = 8 + 3 = 11

The final answer is 11.

Common Traps & FAQs

Wait, why do we assume the denominator is zero? Can't the limit be

- 2

if the denominator is something else?

That is a fantastic question! Think about the definition of a limit. If we have a fraction

\frac{N}{D}

and

N

approaches

0

, then if

D

approaches any non-zero constant

C

, the limit would be

\frac{0}{C} = 0

. But our limit is

- 2

. The only way to get a non-zero finite result from a numerator that is zero is if the denominator is also zero, creating an indeterminate form

\frac{0}{0}

that allows for cancellation or further simplification.

Do I really need to use L'Hopital's Rule twice? Is there a faster way?

L'Hopital's Rule is the most reliable path here, but you could technically use Taylor Series expansions for

sin (u)

around

u = 0

. However, for JEE purposes, L'Hopital's Rule is the standard, robust tool. Using it twice is not a sign of a long method; it is a sign of a problem that requires two layers of 'unmasking' to reveal the hidden constants

a

and

b

I got confused with the chain rule in the derivative. How do I handle

sin (3 x^{2} - 4 x + 1)

The chain rule is your best friend! Remember the rule:

\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)

. Here,

f (u) = sin (u)

and

g (x) = 3 x^{2} - 4 x + 1

. So, the derivative is

cos (3 x^{2} - 4 x + 1)

multiplied by the derivative of the inside, which is

(6 x - 4)

. Always keep that inner derivative attached!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 1} \frac{s i n ( 3 x ^{2} - 4 x + 1 ) - x ^{2} + 1}{2 x ^{3} - 7 x ^{2} + a x + b} = - 2$ , then the value of $(a - b)$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Limit at $x = 1$

Evaluate the Numerator

Condition for a Finite Limit

Establish Equation for $a$ and $b$

Apply L'Hopital's Rule: Numerator

Apply L'Hopital's Rule: Denominator

Evaluate the New Numerator

Simplify the New Numerator

Second Application of the $\frac{0}{0}$ Condition

Solve for $a$

Solve for $b$

Calculate Final Value of $(a - b)$

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

Analyzing the Setup

Establishing the First Constraint

Applying L'Hopital's Rule

Determining the Constants

Final Calculation

Similar Questions

If $lim_{x \to 1} \frac{x ^{2} - a x + b}{x - 1} = 5$ , then $a + b$ is equal to :-

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Enter Numerical Value:

Visualized Solution

Analyze the Limit at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Evaluate the Numerator

Condition for a Finite Limit

Establish Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Apply L'Hopital's Rule: Numerator

Apply L'Hopital's Rule: Denominator

Evaluate the New Numerator

Simplify the New Numerator

Second Application of the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 00​ Condition

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

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

Calculate Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a−b)

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

Analyzing the Setup

Establishing the First Constraint

Applying L'Hopital's Rule

Determining the Constants

Final Calculation

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