JEE Advanced 2014

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The largest value of non-negative integer $a$ for which $lim_{x \to 1} {\frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}}^{\frac{1 - x}{1 - x}} = \frac{1}{4}$ is .........

Q: Wait, why do we reject the negative root when solving the final quadratic?

We actually don't reject it! The equation $\left( \frac{1-a}{2} \right)^2 = \frac{1}{4}$ leads to $\frac{1-a}{2} = \pm \frac{1}{2}$. This gives two valid possibilities: $1-a = 1$ (so $a=0$) and $1-a = -1$ (so $a=2$). Both are non-negative integers. The question asks for the *largest* one, which is why we choose 2.

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: When you encounter a limit of the form

lim_{x \to c} [B (x)]^{E (x)}

, always evaluate the limit of the exponent

E (x)

first. If it converges to a finite value, you can simplify the problem by focusing entirely on the base

B (x)

and its limit.

Hint 2

The Arsenal: The expression contains trigonometric terms like

sin (x - 1)

. This is a massive hint to shift the origin. Use the substitution

t = x - 1

. As

x \to 1

t \to 0

. This will transform the base into a form where you can apply the standard limit

lim_{t \to 0} \frac{s i n t}{t} = 1

Hint 3

The Execution: After substituting

t = x - 1

, the base becomes

\frac{- a t + s i n t}{t + s i n t}

. Divide both the numerator and the denominator by

t

to reveal the standard limit. The resulting expression will be

\frac{- a + \frac{s i n t}{t}}{1 + \frac{s i n t}{t}}

. Evaluate this as

t \to 0

and equate it to the base of the final answer.

Visualized Solution

Analyze the Limit Structure

Given limit: $lim_{x \to 1} {\frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}}^{\frac{1 - x}{1 - x}} = \frac{1}{4}$
Identify the Base: $B (x) = \frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}$
Identify the Exponent: $E (x) = \frac{1 - x}{1 - x}$

Evaluate the Exponent Limit

Let's first evaluate the limit of the exponent as $x \to 1$ .
Exponent: $E (x) = \frac{1 - x}{1 - x}$
This is a $\frac{0}{0}$ form, so we can use factorization.

Factorize the Exponent

Factorize the numerator using $a^{2} - b^{2} = (a - b) (a + b)$ .
$1 - x = (1 - x) (1 + x)$
Substitute back: $E (x) = \frac{( 1 - x ) ( 1 + x )}{1 - x}$

Simplify and Evaluate $E (x)$

Cancel the common factor $(1 - x)$ .
$E (x) = 1 + x$
As $x \to 1$ , $E (x) \to 1 + 1 = 2$ .
The limit of the exponent is $2$ .

Focus on the Base $B (x)$

Now, let's look at the base: $B (x) = \frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}$
Group the algebraic terms together.
Numerator: $a (1 - x) + sin (x - 1)$
Denominator: $(x - 1) + sin (x - 1)$

Substitution for Simplification

To make the limit easier to evaluate, let's shift the origin.
Let $t = x - 1$ .
As $x \to 1$ , the new variable $t \to 0$ .

Rewrite Base in terms of $t$

Substitute $x - 1 = t$ and $1 - x = - t$ .
$B (t) = \frac{- a t + s i n t}{t + s i n t}$
We need to evaluate $lim_{t \to 0} B (t)$ .

Prepare for Standard Limit

We know the standard limit: $lim_{t \to 0} \frac{s i n t}{t} = 1$ .
To use this, divide the numerator and the denominator of $B (t)$ by $t$ .

Divide by $t$

Numerator divided by $t$ : $\frac{- a t}{t} + \frac{s i n t}{t} = - a + \frac{s i n t}{t}$
Denominator divided by $t$ : $\frac{t}{t} + \frac{s i n t}{t} = 1 + \frac{s i n t}{t}$
$B (t) = \frac{- a + \frac{s i n t}{t}}{1 + \frac{s i n t}{t}}$

Evaluate Limit of Base $L_{B}$

Apply $lim_{t \to 0} \frac{s i n t}{t} = 1$ .
Limit of numerator: $- a + 1$
Limit of denominator: $1 + 1 = 2$
Limit of Base $L_{B} = \frac{1 - a}{2}$

Combine Base and Exponent

We have the limit of the base $L_{B} = \frac{1 - a}{2}$ .
We have the limit of the exponent $L_{E} = 2$ .
The original limit is $(L_{B})^{L_{E}} = \frac{1}{4}$ .
So, $(\frac{1 - a}{2})^{2} = \frac{1}{4}$ .

Solve for $a$

Take the square root on both sides.
$\frac{1 - a}{2} = \pm \frac{1}{2}$
This gives two cases:
Case 1: $1 - a = 1 ⟹ a = 0$
Case 2: $1 - a = - 1 ⟹ a = 2$

Select the Final Answer

The possible values for $a$ are $0$ and $2$ .
The question asks for the largest non-negative integer $a$ .
Comparing $0$ and $2$ , the largest value is $2$ .
Final Answer: $2$ .

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

The given expression is a limit of the form

lim_{x \to 1} {B (x)}^{E (x)} = \frac{1}{4}

, where:

B (x) = \frac{- a x + sin ( x - 1 ) + a}{x + sin ( x - 1 ) - 1}

E (x) = \frac{1 - x}{1 - x}

The Exponent's Secret

Whenever you see a function raised to another function, your first instinct should be to separate them. Let us evaluate the exponent

E (x)

x \to 1

The expression

E (x) = \frac{1 - x}{1 - x}

yields the indeterminate form

\frac{0}{0}

x = 1

. We can simplify this by factoring the numerator:

E (x) = \frac{( 1 - x ) ( 1 + x )}{1 - x} = 1 + x

x \to 1

, the exponent simplifies to:

x \to 1 lim E (x) = 1 + 1 = 2

The Base Transformation

Now, we turn to the base

B (x)

. To simplify the expression, we perform a substitution. Let

t = x - 1

. As

x \to 1

t \to 0

Rewriting the base in terms of

t

B (t) = \frac{- a ( t + 1 ) + sin ( t ) + a}{( t + 1 ) + sin ( t ) - 1}

Simplifying the numerator and denominator:

B (t) = \frac{- a t - a + sin ( t ) + a}{t + sin ( t )} = \frac{- a t + sin ( t )}{t + sin ( t )}

The Standard Limit

We now evaluate the limit of the base as

t \to 0

. Dividing both the numerator and the denominator by

t

, we obtain:

t \to 0 lim \frac{- a + \frac{s i n ( t )}{t}}{1 + \frac{s i n ( t )}{t}}

Applying the standard limit

lim_{t \to 0} \frac{s i n ( t )}{t} = 1

, the base limit becomes:

\frac{- a + 1}{1 + 1} = \frac{1 - a}{2}

Final Calculation

We have the limit of the base as

\frac{1 - a}{2}

and the limit of the exponent as

2

. The original problem states the total limit is

\frac{1}{4}

, leading to the equation:

(\frac{1 - a}{2})^{2} = \frac{1}{4}

Taking the square root of both sides, we get:

\frac{1 - a}{2} = \pm \frac{1}{2}

This yields two possible paths:

1 - a = 1 \Rightarrow a = 0

1 - a = - 1 \Rightarrow a = 2

The question asks for the largest non-negative integer. Therefore, the final answer is 2.

Common Traps & FAQs

Wait, why do we reject the negative root when solving the final quadratic?

We actually don't reject it! The equation

(\frac{1 - a}{2})^{2} = \frac{1}{4}

leads to

\frac{1 - a}{2} = \pm \frac{1}{2}

. This gives two valid possibilities:

1 - a = 1

(so

a = 0

) and

1 - a = - 1

(so

a = 2

). Both are non-negative integers. The question asks for the *largest* one, which is why we choose 2.

Why can't we just use L'Hôpital's rule on the base immediately?

You absolutely could! L'Hôpital's rule is a powerful tool. However, using it on the base

\frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}

requires differentiating the numerator and denominator with respect to

x

. While it works, the substitution

t = x - 1

is often cleaner and less prone to algebraic errors, especially when dealing with trigonometric functions.

Is the exponent limit always independent of the base limit?

In this specific structure, yes. Because the limit of the exponent

E (x)

is a finite number (in this case, 2), we can use the property

lim [B (x)]^{E (x)} = [lim B (x)]^{l i m E (x)}

. This is a standard theorem in calculus, provided the base limit is positive, which it is here.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The largest value of non-negative integer $a$ for which $lim_{x \to 1} {\frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}}^{\frac{1 - x}{1 - x}} = \frac{1}{4}$ is .........

Enter Numerical Value:

Visualized Solution

Analyze the Limit Structure

Evaluate the Exponent Limit

Factorize the Exponent

Simplify and Evaluate $E (x)$

Focus on the Base $B (x)$

Substitution for Simplification

Rewrite Base in terms of $t$

Prepare for Standard Limit

Divide by $t$

Evaluate Limit of Base $L_{B}$

Combine Base and Exponent

Solve for $a$

Select the Final Answer

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

Analyzing the Setup

The Exponent's Secret

The Base Transformation

The Standard Limit

Final Calculation

Similar Questions

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :

The value of $lim_{x \to 0} ((sin x)^{1/ x} + (1/ x)^{s i n x})$ , where $x > 0$ is

The value of $lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} ( π x )}{x ^{4} - 2 x ^{3} + 2 x - 1}$ is equal to:

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If $f (x) = \frac{s i n [ x ]}{[ x ]}, [x] \neq = 0$ ; $f (x) = 0, [x] = 0$ . Where $[x]$ denotes the greatest integer less than or equal to $x$ . then $lim_{x \to 0} f (x)$ equals

Enter Numerical Value:

Visualized Solution

Analyze the Limit Structure

Evaluate the Exponent Limit

Factorize the Exponent

Simplify and Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E(x)

Focus on the Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B(x)

Substitution for Simplification

Rewrite Base in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Prepare for Standard Limit

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

Evaluate Limit of Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LB​

Combine Base and Exponent

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

Select the Final Answer

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

Analyzing the Setup

The Exponent's Secret

The Base Transformation

The Standard Limit

Final Calculation

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Simplify and Evaluate $E (x)$

Focus on the Base $B (x)$

Rewrite Base in terms of $t$

Divide by $t$

Evaluate Limit of Base $L_{B}$

Solve for $a$

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :

The value of $lim_{x \to 0} ((sin x)^{1/ x} + (1/ x)^{s i n x})$ , where $x > 0$ is

The value of $lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} ( π x )}{x ^{4} - 2 x ^{3} + 2 x - 1}$ is equal to:

For each $x \in R$ , let $[x]$ be the greatest integer less than or equal to $x$ . Then $lim_{x \to 0^{+}} \frac{x ([ x ] + ∣ x ∣ ) s i n [ x ]}{∣ x ∣}$ is equal to

For each $t \in R$ , let [t] be the greatest integer less than or equal to t. Then, $lim_{x \to 1^{+}} \frac{( 1 - ∣ x ∣ + s i n ∣1 - x ∣ ) s i n ( \frac{π}{2} [ 1 - x ])}{∣1 - x ∣ [ 1 - x ]}$

The integer $n$ for which $lim_{x \to 0} \frac{( c o s x - 1 ) ( c o s x - e ^{x} )}{x ^{n}}$ is a finite non-zero number is

Let $a$ be an integer such that $lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}$ exists, where $[t]$ is greatest integer $\leq t$ . Then $a$ is equal to :

If $f (x) = \frac{s i n [ x ]}{[ x ]}, [x] \neq = 0$ ; $f (x) = 0, [x] = 0$ . Where $[x]$ denotes the greatest integer less than or equal to $x$ . then $lim_{x \to 0} f (x)$ equals