If the function $f$ defined on $(-\frac{1}{3}, \frac{1}{3})$ by $f(x) = \begin{cases} \frac{1}{x} \log_e \left( \frac{1+3x}{1-2x} \right), & \text{when } x \ne 0 \\ k, & \text{when } x = 0 \end{cases}$ is continuous, then $k$ is equal to . . . .

### Analyzing the Setup To ensure the function $f(x) = \frac{1}{x} \log_e \left( \frac{1+3x}{1-2x} \right)$ is continuous at $x=0$, we must satisfy the fundamental condition of continuity: $$\lim_{x \to 0} f(x) = f(0) = k$$ Our objective is to evaluate the limit of the function as $x$ approaches $0$ to determine the required value of $k$. ### Breaking Down the Logarithm The expression can be simplified using the logarithmic property $\log_e \left( \frac{a}{b} \right) = \log_e a - \log_e b$. Applying this to our function, the limit expression becomes: $$\lim_{x \to 0} \frac{\log_e(1+3x) - \log_e(1-2x)}{x}$$ We can now distribute the denominator to split this into two distinct, manageable limits: $$\lim_{x \to 0} \frac{\log_e(1+3x)}{x} - \lim_{x \to 0} \frac{\log_e(1-2x)}{x}$$ ### The Power of Standard Limits We utilize the standard calculus limit $\lim_{u \to 0} \frac{\log_e(1+u)}{u} = 1$. To apply this, we manipulate the terms to match the form of the standard limit. For the first term, we multiply and divide by $3$: $$3 \cdot \lim_{x \to 0} \frac{\log_e(1+3x)}{3x} = 3(1) = 3$$ For the second term, we adjust the denominator to match the argument $-2x$ by multiplying and dividing by $-2$: $$- \left( \lim_{x \to 0} \frac{\log_e(1-2x)}{-2x} \cdot (-2) \right) = -(-2)(1) = 2$$ ### The Final Victory By summing the results of these two limits, we find the value of the function at the point of discontinuity: $$3 + 2 = 5$$ Therefore, to make the function continuous at $x=0$, we must set **$k = 5$**.

JEE Main 2020 - 7 Jan (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f$ defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = {\frac{1}{x} lo g_{e} (\frac{1 + 3 x}{1 - 2 x}), k, when x \neq = 0 when x = 0$ is continuous, then $k$ is equal to . . . .

Q: Why can't I just substitute $x=0$ directly into the function?

Direct substitution gives $\frac{1}{0} \log_e(1)$, which is an indeterminate form of the type $\frac{0}{0}$. We need the limit to find the value that the function *should* have taken to be continuous.

Q: Why does the second term result in a positive value?

The second term is $-\lim_{x \to 0} \frac{\log_e(1-2x)}{x}$. Adjusting the denominator to match the argument $-2x$ gives $-(-2) \lim_{x \to 0} \frac{\log_e(1-2x)}{-2x} = -(-2)(1) = 2$. The negative sign from the formula and the negative sign from the adjustment cancel out to become positive.

Enter Numerical Value:

Visualized Solution

Understanding Continuity at $x = 0$

Function $f (x)$ is defined on $(- \frac{1}{3}, \frac{1}{3})$ .
For $f (x)$ to be continuous at $x = 0$ , there must be no breaks or holes in the graph.
Mathematically: $lim_{x \to 0} f (x) = f (0)$ .

The Continuity Equation

We are given $f (0) = k$ .
The limit as $x$ approaches $0$ must equal this value $k$ .
So, we need to evaluate: $lim_{x \to 0} \frac{1}{x} lo g_{e} (\frac{1 + 3 x}{1 - 2 x})$ .

Applying Logarithmic Properties

The expression contains a logarithm of a fraction.
Recall the quotient rule for logarithms: $lo g_{e} (\frac{a}{b}) = lo g_{e} a - lo g_{e} b$ .
Let's apply this to our numerator.

Expanding the Expression

Applying the property, the limit becomes:
$lim_{x \to 0} \frac{l o g _{e} ( 1 + 3 x ) - l o g _{e} ( 1 - 2 x )}{x}$
Now, we can split this single fraction into two separate terms.

Splitting into Two Limits

Separating the terms over the common denominator $x$ :
$lim_{x \to 0} (\frac{l o g _{e} ( 1 + 3 x )}{x} - \frac{l o g _{e} ( 1 - 2 x )}{x})$
We can evaluate the limit of each term individually.

Identifying the Standard Limit Tool

We need a standard limit formula to proceed.
Standard Limit: $lim_{u \to 0} \frac{l o g _{e} ( 1 + u )}{u} = 1$ .
Our goal is to manipulate each term to exactly match this structure.

Adjusting the First Term

First term: $lim_{x \to 0} \frac{l o g _{e} ( 1 + 3 x )}{x}$
The argument of the log is $1 + 3 x$ , so we need $3 x$ in the denominator.
Multiply and divide by $3$ : $3 \cdot lim_{x \to 0} \frac{l o g _{e} ( 1 + 3 x )}{3 x}$ .

Evaluating the First Term

As $x \to 0$ , $3 x \to 0$ as well.
The limit part becomes exactly $1$ .
So, the value of the first term is $3 \times 1 = 3$ .

Adjusting the Second Term

Second term: $lim_{x \to 0} \frac{l o g _{e} ( 1 - 2 x )}{x}$
The argument is $1 - 2 x$ , so we need $- 2 x$ in the denominator.
Multiply and divide by $- 2$ : $(- 2) \cdot lim_{x \to 0} \frac{l o g _{e} ( 1 - 2 x )}{- 2 x}$ .

Evaluating the Second Term

As $x \to 0$ , $- 2 x \to 0$ .
The limit part evaluates to $1$ .
So, the value is $- 2 \times 1 = - 2$ .
But remember, there is a minus sign between the two terms!

Calculating the Final Limit

Combining the results of both terms:
$lim_{x \to 0} f (x) = 3 - (- 2)$
$lim_{x \to 0} f (x) = 3 + 2 = 5$

Final Conclusion: Finding $k$

For continuity, the limit must equal the function's value at $x = 0$ .
$f (0) = lim_{x \to 0} f (x)$
Therefore, $k = 5$ .
The hole in the graph is perfectly filled!

00:00 / 00:00

The Sigma Insight: Continuity at a Point and in an Interval

Solution Diagram

Analyzing the Setup

To ensure the function

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

is continuous at

x = 0

, we must satisfy the fundamental condition of continuity:

x \to 0 lim f (x) = f (0) = k

Our objective is to evaluate the limit of the function as

x

approaches

0

to determine the required value of

k

Breaking Down the Logarithm

The expression can be simplified using the logarithmic property

lo g_{e} (\frac{a}{b}) = lo g_{e} a - lo g_{e} b

. Applying this to our function, the limit expression becomes:

x \to 0 lim \frac{lo g _{e} ( 1 + 3 x ) - lo g _{e} ( 1 - 2 x )}{x}

We can now distribute the denominator to split this into two distinct, manageable limits:

x \to 0 lim \frac{lo g _{e} ( 1 + 3 x )}{x} - x \to 0 lim \frac{lo g _{e} ( 1 - 2 x )}{x}

The Power of Standard Limits

We utilize the standard calculus limit

lim_{u \to 0} \frac{l o g _{e} ( 1 + u )}{u} = 1

. To apply this, we manipulate the terms to match the form of the standard limit.

For the first term, we multiply and divide by

3

3 \cdot x \to 0 lim \frac{lo g _{e} ( 1 + 3 x )}{3 x} = 3 (1) = 3

For the second term, we adjust the denominator to match the argument

- 2 x

by multiplying and dividing by

- 2

- (x \to 0 lim \frac{lo g _{e} ( 1 - 2 x )}{- 2 x} \cdot (- 2)) = - (- 2) (1) = 2

The Final Victory

By summing the results of these two limits, we find the value of the function at the point of discontinuity:

3 + 2 = 5

Therefore, to make the function continuous at

x = 0

, we must set $k = 5$ .

Common Traps & FAQs

Why can't I just substitute

x = 0

directly into the function?

Direct substitution gives

\frac{1}{0} lo g_{e} (1)

, which is an indeterminate form of the type

\frac{0}{0}

. We need the limit to find the value that the function *should* have taken to be continuous.

Why does the second term result in a positive value?

The second term is

- lim_{x \to 0} \frac{l o g _{e} ( 1 - 2 x )}{x}

. Adjusting the denominator to match the argument

- 2 x

gives

- (- 2) lim_{x \to 0} \frac{l o g _{e} ( 1 - 2 x )}{- 2 x} = - (- 2) (1) = 2

. The negative sign from the formula and the negative sign from the adjustment cancel out to become positive.

Can I use L'Hopital's Rule?

Yes, you can differentiate the numerator and denominator, but mastering standard limits is faster and more elegant for JEE. It builds a deeper intuition for how functions behave near zero.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f$ defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = {\frac{1}{x} lo g_{e} (\frac{1 + 3 x}{1 - 2 x}), k, when x \neq = 0 when x = 0$ is continuous, then $k$ is equal to . . . .

Enter Numerical Value:

Visualized Solution

Understanding Continuity at $x = 0$

The Continuity Equation

Applying Logarithmic Properties

Expanding the Expression

Splitting into Two Limits

Identifying the Standard Limit Tool

Adjusting the First Term

Evaluating the First Term

Adjusting the Second Term

Evaluating the Second Term

Calculating the Final Limit

Final Conclusion: Finding $k$

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

Breaking Down the Logarithm

The Power of Standard Limits

The Final Victory

Similar Questions

If the function $f$ defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = {(\frac{1}{x}) lo g_{e} (\frac{1 + 3 x}{1 - 2 x}) k when x \neq = 0 when x = 0$ is continuous, the $k$ is equal to ________.

If the function $f (x) = {\frac{l o g _{e} ( 1 - x + x ^{2} ) + l o g _{e} ( 1 + x + x ^{2} )}{s e c x - c o s x}, k, x \in (- \frac{π}{2}, \frac{π}{2}) - {0} x = 0$ is continuous at $x = 0$ , then $k$ is equal to :

If the function $f$ defined on $(π /6, π /3)$ by $f (x) = {\frac{2 c o s x - 1}{c o t x - 1}, k, x \neq = π /4 x = π /4$ is continuous, then $k$ is equal to

If the function $f (x) = ⎩ ⎨ ⎧ \frac{1}{x} lo g_{e} (\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}), k, \frac{c o s ^{2} x - s i n ^{2} x - 1}{x ^{2} + 1 - 1}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to :

Let the function $f (x) = {\frac{l o g _{e} ( 1 + 5 x ) - l o g _{e} ( 1 + α x )}{x}, 10, if x \neq = 0 if x = 0$ be continuous at $x = 0$ . The $α$ is equal to :

If the function $f (x) = ⎩ ⎨ ⎧ \frac{2}{x} {sin (k_{1} + 1) x + sin (k_{2} - 1) x}, 4, \frac{2}{x} lo g_{e} (\frac{2 + k _{1} x}{2 + k _{2} x}), x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then $k_{1}^{2} + k_{2}^{2}$ is equal to

Let $f (x) = {(x - 1)^{\frac{1}{2 - x}}, k, x > 1, x \neq = 2 x = 2$ . The value of k for which f is continuous at x = 2 is :-

Let $f (x) = {\frac{( x ^{3} + x ^{2} - 16 x + 20 )}{( x - 2 ) ^{2}} k if x \neq = 2 if x = 2$ . If $f (x)$ is continuous for all $x$ , then $k = .........$

If the function f defined as $f (x) = \frac{1}{x} - \frac{k - 1}{e ^{2 x} - 1}, x \neq = 0$ , is continuous at $x = 0$ , then the ordered pair $(k, f (0))$ is equal to :

If the function $f (x) = \frac{e ^{x} ( e ^{t a n x - x} - 1 ) + l o g _{e} ( s e c x + t a n x ) - x}{t a n x - x}$ is continuous at $x = 0$ , then the value of $f (0)$ is equal to

Enter Numerical Value:

Visualized Solution

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

The Continuity Equation

Applying Logarithmic Properties

Expanding the Expression

Splitting into Two Limits

Identifying the Standard Limit Tool

Adjusting the First Term

Evaluating the First Term

Adjusting the Second Term

Evaluating the Second Term

Calculating the Final Limit

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

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

Breaking Down the Logarithm

The Power of Standard Limits

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f defined on (−31​,31​) by f(x)={(x1​)loge​(1−2x1+3x​)k​when x=0when x=0​ is continuous, the k is equal to ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f(x)={secx−cosxloge​(1−x+x2)+loge​(1+x+x2)​,k,​x∈(−2π​,2π​)−{0}x=0​ is continuous at x=0, then k is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f defined on (π/6,π/3) by f(x)={cotx−12​cosx−1​,k,​x=π/4x=π/4​ is continuous, then k is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the function f(x)={xloge​(1+5x)−loge​(1+αx)​,10,​if x=0if x=0​ be continuous at x=0. The α is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f(x)=⎩⎨⎧​x2​{sin(k1​+1)x+sin(k2​−1)x},4,x2​loge​(2+k2​x2+k1​x​),​x<0x=0x>0​ is continuous at x=0, then k12​+k22​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={(x−1)2−x1​,k,​x>1,x=2x=2​. The value of k for which f is continuous at x = 2 is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={(x−2)2(x3+x2−16x+20)​k​if x=2if x=2​. If f(x) is continuous for all x, then k=.........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f defined as f(x)=x1​−e2x−1k−1​,x=0, is continuous at x=0, then the ordered pair (k,f(0)) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f(x)=tanx−xex(etanx−x−1)+loge​(secx+tanx)−x​ is continuous at x=0, then the value of f(0) is equal to

Understanding Continuity at $x = 0$

Final Conclusion: Finding $k$

If the function $f$ defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = {(\frac{1}{x}) lo g_{e} (\frac{1 + 3 x}{1 - 2 x}) k when x \neq = 0 when x = 0$ is continuous, the $k$ is equal to ________.

If the function $f (x) = {\frac{l o g _{e} ( 1 - x + x ^{2} ) + l o g _{e} ( 1 + x + x ^{2} )}{s e c x - c o s x}, k, x \in (- \frac{π}{2}, \frac{π}{2}) - {0} x = 0$ is continuous at $x = 0$ , then $k$ is equal to :

If the function $f$ defined on $(π /6, π /3)$ by $f (x) = {\frac{2 c o s x - 1}{c o t x - 1}, k, x \neq = π /4 x = π /4$ is continuous, then $k$ is equal to

Let the function $f (x) = {\frac{l o g _{e} ( 1 + 5 x ) - l o g _{e} ( 1 + α x )}{x}, 10, if x \neq = 0 if x = 0$ be continuous at $x = 0$ . The $α$ is equal to :

If the function $f (x) = ⎩ ⎨ ⎧ \frac{2}{x} {sin (k_{1} + 1) x + sin (k_{2} - 1) x}, 4, \frac{2}{x} lo g_{e} (\frac{2 + k _{1} x}{2 + k _{2} x}), x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then $k_{1}^{2} + k_{2}^{2}$ is equal to

Let $f (x) = {(x - 1)^{\frac{1}{2 - x}}, k, x > 1, x \neq = 2 x = 2$ . The value of k for which f is continuous at x = 2 is :-

Let $f (x) = {\frac{( x ^{3} + x ^{2} - 16 x + 20 )}{( x - 2 ) ^{2}} k if x \neq = 2 if x = 2$ . If $f (x)$ is continuous for all $x$ , then $k = .........$

If the function f defined as $f (x) = \frac{1}{x} - \frac{k - 1}{e ^{2 x} - 1}, x \neq = 0$ , is continuous at $x = 0$ , then the ordered pair $(k, f (0))$ is equal to :

If the function $f (x) = \frac{e ^{x} ( e ^{t a n x - x} - 1 ) + l o g _{e} ( s e c x + t a n x ) - x}{t a n x - x}$ is continuous at $x = 0$ , then the value of $f (0)$ is equal to