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

### Analyzing the Setup To ensure the function $f(x)$ is continuous at $x=0$, the limit of the function as $x$ approaches $0$ must be equal to the defined value $f(0)$. Given the function: $$f(x) = \frac{1}{x} \ln\left(\frac{1+3x}{1-2x}\right) \text{ for } x \neq 0$$ And the condition $f(0) = k$, we must solve for $k$ using the following limit: $$k = \lim_{x \to 0} \frac{1}{x} \ln\left(\frac{1+3x}{1-2x}\right)$$ ### Breaking Down the Logarithm The expression inside the logarithm is a quotient. We apply the logarithmic property $\ln(A/B) = \ln A - \ln B$ to simplify the expression: $$\lim_{x \to 0} \frac{\ln(1+3x) - \ln(1-2x)}{x}$$ Next, we distribute the denominator $x$ to both terms to split the limit into two distinct parts: $$k = \lim_{x \to 0} \frac{\ln(1+3x)}{x} - \lim_{x \to 0} \frac{\ln(1-2x)}{x}$$ ### The Power of Standard Limits We utilize the fundamental standard limit identity: $$\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$$ For the first term, $\frac{\ln(1+3x)}{x}$, we multiply and divide by $3$ to match the argument of the logarithm: $$3 \cdot \lim_{x \to 0} \frac{\ln(1+3x)}{3x} = 3 \cdot 1 = 3$$ For the second term, $\frac{\ln(1-2x)}{x}$, we multiply and divide by $-2$ to match the argument: $$-2 \cdot \lim_{x \to 0} \frac{\ln(1-2x)}{-2x} = -2 \cdot 1 = -2$$ ### The Final Connection Now, we substitute these evaluated limits back into our master equation: $$k = 3 - (-2)$$ Simplifying the arithmetic, we find: $$k = 3 + 2 = 5$$ By setting **$k=5$**, we have successfully filled the hole in the function, ensuring it is continuous at $x=0$.

JEE Main 2020 (7 January Shift 1)

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, the $k$ is equal to ________.

Q: Why can't we just plug in $x=0$ directly?

If you try to plug in $x=0$ directly, you get $\frac{1}{0} \cdot \ln(1)$, which is undefined. This is an indeterminate form of the type $\infty \cdot 0$. We must use limits to find the value the function approaches as $x$ gets arbitrarily close to $0$.

Q: Do I really need to split the logarithm?

Splitting the logarithm using the property $\ln(A/B) = \ln A - \ln B$ is the most efficient way to transform the expression into standard limit forms. Without this, you would be stuck with a complex fraction that is difficult to evaluate.

Q: What happens to the negative sign in the second term?

Be very careful! The second term is $-\ln(1-2x)$. When you evaluate the limit, the coefficient is $-2$. Since there is a minus sign in the formula, you are subtracting a negative value, which becomes addition. This is a common trap, so always keep track of your signs.

Enter Numerical Value:

Visualized Solution

Understanding the Function $f (x)$

Given function: $f (x) = {\frac{1}{x} lo g_{e} (\frac{1 + 3 x}{1 - 2 x}) k x \neq = 0 x = 0$
Domain: $x \in (- \frac{1}{3}, \frac{1}{3})$
Objective: Find $k$ such that $f (x)$ is continuous at $x = 0$ .

The Condition for Continuity

For continuity at $x = 0$ :
$lim_{x \to 0} f (x) = f (0)$

Setting Up the Limit Equation

Substituting the given definitions:
$lim_{x \to 0} \frac{1}{x} lo g_{e} (\frac{1 + 3 x}{1 - 2 x}) = k$

Applying Logarithmic Properties

Using the quotient property of logarithms:
$lo g_{e} (\frac{A}{B}) = lo g_{e} A - lo g_{e} B$
The limit becomes:
$lim_{x \to 0} \frac{l o g _{e} ( 1 + 3 x ) - l o g _{e} ( 1 - 2 x )}{x} = k$

Splitting the Limit

Distributing the denominator $x$ :
$lim_{x \to 0} [\frac{l o g _{e} ( 1 + 3 x )}{x} - \frac{l o g _{e} ( 1 - 2 x )}{x}] = k$
Splitting into two separate limits:
$lim_{x \to 0} \frac{l o g _{e} ( 1 + 3 x )}{x} - lim_{x \to 0} \frac{l o g _{e} ( 1 - 2 x )}{x} = k$

The Standard Limit Formula

Recall the standard logarithmic limit:
$lim_{u \to 0} \frac{l o g _{e} ( 1 + u )}{u} = 1$
We need to manipulate our terms to match this exact form.

Adjusting the First Term

First term: $lim_{x \to 0} \frac{l o g _{e} ( 1 + 3 x )}{x}$
Multiply and divide by $3$ :
$lim_{x \to 0} (3 \cdot \frac{l o g _{e} ( 1 + 3 x )}{3 x})$
As $x \to 0$ , $3 x \to 0$ .
The limit evaluates to $3 \cdot 1 = 3$ .

Adjusting the Second Term

Second term: $lim_{x \to 0} \frac{l o g _{e} ( 1 - 2 x )}{x}$
Multiply and divide by $- 2$ :
$lim_{x \to 0} ((- 2) \cdot \frac{l o g _{e} ( 1 - 2 x )}{- 2 x})$
As $x \to 0$ , $- 2 x \to 0$ .
The limit evaluates to $(- 2) \cdot 1 = - 2$ .

Combining and Finding $k$

Substitute the evaluated limits back into our equation:
$k = (First Limit) - (Second Limit)$
$k = 3 - (- 2)$
$k = 3 + 2$
$k = 5$

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)

is continuous at

x = 0

, the limit of the function as

x

approaches

0

must be equal to the defined value

f (0)

Given the function:

f (x) = \frac{1}{x} ln (\frac{1 + 3 x}{1 - 2 x}) for x e q 0

And the condition

f (0) = k

, we must solve for

k

using the following limit:

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

Breaking Down the Logarithm

The expression inside the logarithm is a quotient. We apply the logarithmic property

ln (A / B) = ln A - ln B

to simplify the expression:

x \to 0 lim \frac{ln ( 1 + 3 x ) - ln ( 1 - 2 x )}{x}

Next, we distribute the denominator

x

to both terms to split the limit into two distinct parts:

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

The Power of Standard Limits

We utilize the fundamental standard limit identity:

u \to 0 lim \frac{ln ( 1 + u )}{u} = 1

For the first term,

\frac{l n ( 1 + 3 x )}{x}

, we multiply and divide by

3

to match the argument of the logarithm:

3 \cdot x \to 0 lim \frac{ln ( 1 + 3 x )}{3 x} = 3 \cdot 1 = 3

For the second term,

\frac{l n ( 1 - 2 x )}{x}

, we multiply and divide by

- 2

to match the argument:

- 2 \cdot x \to 0 lim \frac{ln ( 1 - 2 x )}{- 2 x} = - 2 \cdot 1 = - 2

The Final Connection

Now, we substitute these evaluated limits back into our master equation:

k = 3 - (- 2)

Simplifying the arithmetic, we find:

k = 3 + 2 = 5

By setting $k = 5$ , we have successfully filled the hole in the function, ensuring it is continuous at

x = 0

Common Traps & FAQs

Why can't we just plug in

x = 0

directly?

If you try to plug in

x = 0

directly, you get

\frac{1}{0} \cdot ln (1)

, which is undefined. This is an indeterminate form of the type

\infty \cdot 0

. We must use limits to find the value the function approaches as

x

gets arbitrarily close to

0

Do I really need to split the logarithm?

Splitting the logarithm using the property

ln (A / B) = ln A - ln B

is the most efficient way to transform the expression into standard limit forms. Without this, you would be stuck with a complex fraction that is difficult to evaluate.

What happens to the negative sign in the second term?

Be very careful! The second term is

- ln (1 - 2 x)

. When you evaluate the limit, the coefficient is

- 2

. Since there is a minus sign in the formula, you are subtracting a negative value, which becomes addition. This is a common trap, so always keep track of your signs.

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, the $k$ is equal to ________.

Enter Numerical Value:

Visualized Solution

Understanding the Function $f (x)$

The Condition for Continuity

Setting Up the Limit Equation

Applying Logarithmic Properties

Splitting the Limit

The Standard Limit Formula

Adjusting the First Term

Adjusting the Second Term

Combining and 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 Connection

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

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 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 :

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

The Condition for Continuity

Setting Up the Limit Equation

Applying Logarithmic Properties

Splitting the Limit

The Standard Limit Formula

Adjusting the First Term

Adjusting the Second Term

Combining and 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 Connection

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, 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(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 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; } 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 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; } 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 the Function $f (x)$

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

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 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 :

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