Let $f:\mathbb{R}-\{0\}\rightarrow \mathbb{R}$ be a function such that $f(x)-6f(\frac{1}{x})=\frac{35}{3x}-\frac{5}{2}.$ If the $\lim_{x\rightarrow0}(\frac{1}{\alpha x}+f(x))=\beta;$ $\alpha, \beta\in \mathbb{R},$ then $\alpha+2\beta$ is equal to Options: (A) 5 (B) 3 (C) 4 (D) 6

### Analyzing the Symmetry The given functional equation is: $$f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$$ To solve for $f(x)$, we utilize the substitution $x \to \frac{1}{x}$. This transformation yields a second equation: $$f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2}$$ ### Solving the System We now have a system of two linear equations. To isolate $f(x)$, we multiply the second equation by $6$: $$6f\left(\frac{1}{x}\right) - 36f(x) = 70x - 15$$ Adding this result to the original equation, the $f(\frac{1}{x})$ terms cancel out: $$-35f(x) = \frac{35}{3x} + 70x - \frac{35}{2}$$ Dividing the entire expression by $-35$, we obtain the explicit form of the function: $$f(x) = -\frac{1}{3x} - 2x + \frac{1}{2}$$ ### Navigating the Limit We are given the condition $\lim_{x\to 0} \left(\frac{1}{\alpha x} + f(x)\right) = \beta$. Substituting our derived $f(x)$ into this limit, we get: $$\lim_{x\to 0} \left(\frac{1}{\alpha x} - \frac{1}{3x} - 2x + \frac{1}{2}\right) = \beta$$ Grouping the terms involving $x$ in the denominator: $$\lim_{x\to 0} \left(\frac{3 - \alpha}{3\alpha x} - 2x + \frac{1}{2}\right) = \beta$$ For the limit to exist as a finite real number $\beta$, the term causing divergence must be eliminated. Thus, we set the numerator to zero: $$3 - \alpha = 0 \implies \alpha = 3$$ ### Final Calculation With $\alpha = 3$, the limit simplifies significantly: $$\beta = \lim_{x\to 0} \left(-2x + \frac{1}{2}\right) = \frac{1}{2}$$ We are tasked with finding the value of $\alpha + 2\beta$. Substituting our values: $$\alpha + 2\beta = 3 + 2\left(\frac{1}{2}\right) = 3 + 1 = 4$$ The final answer is **4**.

Why does the limit term $\frac{3 - \alpha}{3\alpha x}$ have to be zero? Can't it just be a very large number?

Remember the definition of a limit! The problem states that the limit equals $\beta$, where $\beta$ is a finite real number. If the term $\frac{3 - \alpha}{3\alpha x}$ does not vanish (i.e., if its numerator is non-zero), then as $x$ approaches $0$, the expression will blow up to $\pm\infty$. Since $\beta$ must be finite, the only way to prevent this divergence is to force the coefficient of the $\frac{1}{x}$ term to be exactly zero. It is a classic 'trap' in JEE problems—always check for conditions that ensure the limit remains finite!

JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R - {0} \to R$ be a function such that $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2} .$ If the $lim_{x \to 0} (\frac{1}{α x} + f (x)) = β;$ $α, β \in R,$ then $α + 2 β$ is equal to

Select Answer:

Visualized Solution

The Functional Equation

Given functional equation: $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2}$
Our goal is to isolate $f (x)$ by eliminating $f (\frac{1}{x})$ .
This is a classic functional equation problem where substitution is the key.

Substitution $x \to \frac{1}{x}$

Substitute $x$ with $\frac{1}{x}$ in the original equation.
This transformation swaps the roles of $f (x)$ and $f (\frac{1}{x})$ .
It also changes the terms on the right-hand side accordingly.

The Second Equation

Applying the substitution: $f (\frac{1}{x}) - 6 f (x) = \frac{35}{3 ( \frac{1}{x} )} - \frac{5}{2}$
Simplifying the right side gives: $f (\frac{1}{x}) - 6 f (x) = \frac{35 x}{3} - \frac{5}{2}$
Let's call this Equation 2.

Eliminating $f (\frac{1}{x})$

To eliminate $f (\frac{1}{x})$ , multiply Equation 2 by $6$ .
$6 \times [f (\frac{1}{x}) - 6 f (x)] = 6 \times [\frac{35 x}{3} - \frac{5}{2}]$
This yields: $6 f (\frac{1}{x}) - 36 f (x) = 70 x - 15$

Combining Equations

Add Equation 1 and Equation 3 together.
$(f (x) - 6 f (\frac{1}{x})) + (6 f (\frac{1}{x}) - 36 f (x)) = (\frac{35}{3 x} - \frac{5}{2}) + (70 x - 15)$
The $f (\frac{1}{x})$ terms cancel out perfectly.
Result: $- 35 f (x) = \frac{35}{3 x} + 70 x - \frac{35}{2}$

Solving for $f (x)$

Divide the entire equation by $- 35$ to isolate $f (x)$ .
$f (x) = \frac{\frac{35}{3 x}}{- 35} + \frac{70 x}{- 35} - \frac{\frac{35}{2}}{- 35}$
Simplifying gives: $f (x) = - \frac{1}{3 x} - 2 x + \frac{1}{2}$

The Limit Condition

We are given a limit condition: $lim_{x \to 0} (\frac{1}{α x} + f (x)) = β$
Here, $α$ and $β$ are real numbers ( $α, β \in R$ ).
We need to substitute our newly found $f (x)$ into this limit expression.

Substituting $f (x)$ into the Limit

Substitute $f (x) = - \frac{1}{3 x} - 2 x + \frac{1}{2}$ into the limit.
$lim_{x \to 0} (\frac{1}{α x} - \frac{1}{3 x} - 2 x + \frac{1}{2}) = β$
Group the terms with $\frac{1}{x}$ together:
$lim_{x \to 0} (\frac{3 - α}{3 α x} - 2 x + \frac{1}{2}) = β$

Condition for Existence

For the limit to evaluate to a finite real number $β$ , it must not diverge to $\pm \infty$ .
As $x \to 0$ , the term $\frac{3 - α}{3 α x}$ will tend to infinity unless its numerator is zero.
Therefore, we must enforce the condition: $3 - α = 0$ .

Finding $α$

Setting the numerator to zero: $3 - α = 0$
Solving this gives: $α = 3$
With $α = 3$ , the problematic term $\frac{3 - α}{3 α x}$ becomes $0$ , allowing the limit to exist.

Finding $β$

Now substitute $α = 3$ back into the limit expression.
The limit becomes: $β = lim_{x \to 0} (0 - 2 x + \frac{1}{2})$
Evaluate the limit by directly substituting $x = 0$ :
$β = - 2 (0) + \frac{1}{2} = \frac{1}{2}$

Final Calculation

We need to find the value of $α + 2 β$ .
Substitute the values we found: $α = 3$ and $β = \frac{1}{2}$ .
$α + 2 β = 3 + 2 (\frac{1}{2})$
$α + 2 β = 3 + 1 = 4$
The final answer is $4$ .

00:00 / 00:00

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

Analyzing the Symmetry

The given functional equation is:

f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2}

To solve for

f (x)

, we utilize the substitution

x \to \frac{1}{x}

. This transformation yields a second equation:

f (\frac{1}{x}) - 6 f (x) = \frac{35 x}{3} - \frac{5}{2}

Solving the System

We now have a system of two linear equations. To isolate

f (x)

, we multiply the second equation by

6

6 f (\frac{1}{x}) - 36 f (x) = 70 x - 15

Adding this result to the original equation, the

f (\frac{1}{x})

terms cancel out:

- 35 f (x) = \frac{35}{3 x} + 70 x - \frac{35}{2}

Dividing the entire expression by

- 35

, we obtain the explicit form of the function:

f (x) = - \frac{1}{3 x} - 2 x + \frac{1}{2}

Navigating the Limit

We are given the condition

lim_{x \to 0} (\frac{1}{α x} + f (x)) = β

. Substituting our derived

f (x)

into this limit, we get:

x \to 0 lim (\frac{1}{α x} - \frac{1}{3 x} - 2 x + \frac{1}{2}) = β

Grouping the terms involving

x

in the denominator:

x \to 0 lim (\frac{3 - α}{3 α x} - 2 x + \frac{1}{2}) = β

For the limit to exist as a finite real number

β

, the term causing divergence must be eliminated. Thus, we set the numerator to zero:

3 - α = 0 ⟹ α = 3

Final Calculation

With

α = 3

, the limit simplifies significantly:

β = x \to 0 lim (- 2 x + \frac{1}{2}) = \frac{1}{2}

We are tasked with finding the value of

α + 2 β

. Substituting our values:

α + 2 β = 3 + 2 (\frac{1}{2}) = 3 + 1 = 4

The final answer is 4.

Common Traps & FAQs

Wait, why do we substitute

x

with

\frac{1}{x}

? Is that always the trick?

That is a fantastic question! In functional equations involving terms like

f (x)

and

f (\frac{1}{x})

, this substitution is a standard, powerful technique. By replacing

x

with

\frac{1}{x}

, we transform the equation into a new one where the roles of

f (x)

and

f (\frac{1}{x})

are swapped. This creates a system of two linear equations with two unknowns, which we can easily solve using basic algebraic elimination. It is like having two different perspectives on the same function; combining them reveals the function's true identity.

Why does the limit term

\frac{3 - α}{3 α x}

have to be zero? Can't it just be a very large number?

Remember the definition of a limit! The problem states that the limit equals

β

, where

β

is a finite real number. If the term

\frac{3 - α}{3 α x}

does not vanish (i.e., if its numerator is non-zero), then as

x

approaches

0

, the expression will blow up to

\pm \infty

. Since

β

must be finite, the only way to prevent this divergence is to force the coefficient of the

\frac{1}{x}

term to be exactly zero. It is a classic 'trap' in JEE problems—always check for conditions that ensure the limit remains finite!

Can I use L'Hopital's rule for the limit part?

While L'Hopital's rule is a powerful tool, it is often overkill and can sometimes lead to messy derivatives in problems like this. Here, the algebraic approach of isolating the divergent term is much cleaner and less prone to calculation errors. By simplifying the expression first, we clearly see the condition required for the limit to exist, which is the core of the problem.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R - {0} \to R$ be a function such that $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2} .$ If the $lim_{x \to 0} (\frac{1}{α x} + f (x)) = β;$ $α, β \in R,$ then $α + 2 β$ is equal to

Select Answer:

Visualized Solution

The Functional Equation

Substitution $x \to \frac{1}{x}$

The Second Equation

Eliminating $f (\frac{1}{x})$

Combining Equations

Solving for $f (x)$

The Limit Condition

Substituting $f (x)$ into the Limit

Condition for Existence

Finding $α$

Finding $β$

Final Calculation

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

Analyzing the Symmetry

Solving the System

Navigating the Limit

Final Calculation

Similar Questions

Let $β = lim_{x \to 0} \frac{α x - ( e ^{3 x} - 1 )}{α x ( e ^{3 x} - 1 )}$ for some $α \in R$ . Then the value of $α + β$ is :

If $β = lim_{x \to 0} \frac{e ^{x^{3}} - ( 1 - x ^{3} ) ^{1/3} + ( ( 1 - x ^{2} ) ^{1/2} - 1 ) s i n x}{x s i n ^{2} x}$ , then the value of $6 β$ is ______.

For $α, β, γ, \in R$ , if $lim_{x \to 0} \frac{x ^{2} s i n α x + ( γ - 1 ) e ^{x^{2}}}{s i n 2 x - β x} = 3$ , then $β + γ - α$ is equal to:

If $lim_{x \to \infty} ((\frac{e}{1 - e}) (\frac{1}{e} - \frac{x}{1 + x}))^{x} = α$ , then the value of $\frac{l o g _{e} α}{1 + l o g _{e} α}$ equals

Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

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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

If $lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ , then $2 α - β$ is equal to :

For $x \in R, lim_{x \to \infty} (\frac{x - 3}{x + 2})^{x} =$

$lim_{x \to \infty} (\frac{x + 6}{x + 1})^{x + 4} = .........$

Select Answer:

Visualized Solution

The Functional Equation

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→x1​

The Second Equation

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x1​)

Combining Equations

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Limit Condition

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) into the Limit

Condition for Existence

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

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

Final Calculation

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

Analyzing the Symmetry

Solving the System

Navigating the Limit

Final Calculation

Similar Questions

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Substitution $x \to \frac{1}{x}$

Eliminating $f (\frac{1}{x})$

Solving for $f (x)$

Substituting $f (x)$ into the Limit

Finding $α$

Finding $β$

Let $β = lim_{x \to 0} \frac{α x - ( e ^{3 x} - 1 )}{α x ( e ^{3 x} - 1 )}$ for some $α \in R$ . Then the value of $α + β$ is :

If $β = lim_{x \to 0} \frac{e ^{x^{3}} - ( 1 - x ^{3} ) ^{1/3} + ( ( 1 - x ^{2} ) ^{1/2} - 1 ) s i n x}{x s i n ^{2} x}$ , then the value of $6 β$ is ______.

For $α, β, γ, \in R$ , if $lim_{x \to 0} \frac{x ^{2} s i n α x + ( γ - 1 ) e ^{x^{2}}}{s i n 2 x - β x} = 3$ , then $β + γ - α$ is equal to:

If $lim_{x \to \infty} ((\frac{e}{1 - e}) (\frac{1}{e} - \frac{x}{1 + x}))^{x} = α$ , then the value of $\frac{l o g _{e} α}{1 + l o g _{e} α}$ equals

Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

Let $f : [- \frac{π}{2}, \frac{π}{2}] \to R$ be a differentiable function such that $f (0) = \frac{1}{2}$ . If $lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) d t}{e ^{x^{2}} - 1} = α$ , then $8 α^{2}$ 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

If $lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ , then $2 α - β$ is equal to :

For $x \in R, lim_{x \to \infty} (\frac{x - 3}{x + 2})^{x} =$

$lim_{x \to \infty} (\frac{x + 6}{x + 1})^{x + 4} = .........$