Let f : (–1, \infty ) \to R be defined by f(0) = 1 and f(x) = 1 \over x\log _e( 1 + x ) , x e 0. Then the function f :

decreases in (–1, 0) and increases in (0, \infty )

increases in (–1, 0) and decreases in (0, \infty )

Application of Derivatives PYQ — JEE Main 2020 · 2 Sep · Shift 2 · Q42 with Solution

View written solutionFree

Correct answer: A

Given function

We have

f(x)= \begin{cases} 1, & x=0,\\[4pt] \dfrac{\ln(1+x)}{x}, & x\neq 0, \end{cases}

with domain $(-1,\infty)$ .

We need to determine where $f$ is increasing/decreasing.

Differentiate for $x\neq 0$

For $x\neq 0$ ,

f(x)=\frac{\ln(1+x)}{x}.

Using the quotient rule,

f'(x)=\frac{x\cdot \frac{1}{1+x}-\ln(1+x)}{x^2} =\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}.

So the sign of $f'(x)$ depends on

\phi(x)=\frac{x}{1+x}-\ln(1+x).

Since $x^2>0$ for $x\neq 0$ , the sign of $f'(x)$ is the sign of $\phi(x)$ .

Study the sign of $\phi(x)$

Let

\phi(x)=\frac{x}{1+x}-\ln(1+x), \qquad x>-1.

Differentiate:

\phi'(x)=\frac{1}{(1+x)^2}-\frac{1}{1+x} =\frac{1-(1+x)}{(1+x)^2} =\frac{-x}{(1+x)^2}.

Now:

If $x\in(-1,0)$ , then $x<0$ , so $-x>0$ , hence $\phi'(x)>0.$ Thus $\phi$ is increasing on $(-1,0)$ .
If $x\in(0,\infty)$ , then $x>0$ , so $-x<0$ , hence $\phi'(x)<0.$ Thus $\phi$ is decreasing on $(0,\infty)$ .

Also,

\phi(0)=\frac{0}{1+0}-\ln(1)=0.

Since $\phi$ increases up to $0$ and decreases after $0$ , its maximum occurs at $x=0$ , with maximum value $0$ . Therefore,

\phi(x)\le 0 \quad \text{for all } x>-1,

and for $x\neq 0$ ,

\phi(x)<0.

Hence,

f'(x)=\frac{\phi(x)}{x^2}<0 \quad \text{for all } x\in(-1,0)\cup(0,\infty).

So $f$ is decreasing on both intervals $(-1,0)$ and $(0,\infty)$ .

Check behavior at $x=0$

We should verify that the function remains decreasing across $x=0$ .

Using the standard limit,

\lim_{x\to 0}\frac{\ln(1+x)}{x}=1=f(0).

So $f$ is continuous at $x=0$ .

Since it is decreasing on both sides of $0$ and continuous at $0$ , the function is decreasing on the entire domain $(-1,\infty)$ .

(Indeed, for $x<0$ , $f(x)>1$ , and for $x>0$ , $f(x)<1$ , consistent with overall decrease.)

Evaluate options

A: decreases in $(-1,\infty)$ ✅
B: decreases in $(-1,0)$ and increases in $(0,\infty)$ ❌
C: increases in $(-1,\infty)$ ❌
D: increases in $(-1,0)$ and decreases in $(0,\infty)$ ❌

Therefore, the correct option is

\boxed{\text{A}}

Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

More from Application of Derivatives

View written solutionFree

Correct answer: A

Given function

We have

f(x)= \begin{cases} 1, & x=0,\\[4pt] \dfrac{\ln(1+x)}{x}, & x\neq 0, \end{cases}

with domain $(-1,\infty)$ .

We need to determine where $f$ is increasing/decreasing.

Differentiate for $x\neq 0$

For $x\neq 0$ ,

f(x)=\frac{\ln(1+x)}{x}.

Using the quotient rule,

f'(x)=\frac{x\cdot \frac{1}{1+x}-\ln(1+x)}{x^2} =\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}.

So the sign of $f'(x)$ depends on

\phi(x)=\frac{x}{1+x}-\ln(1+x).

Since $x^2>0$ for $x\neq 0$ , the sign of $f'(x)$ is the sign of $\phi(x)$ .

Study the sign of $\phi(x)$

Let

\phi(x)=\frac{x}{1+x}-\ln(1+x), \qquad x>-1.

Differentiate:

\phi'(x)=\frac{1}{(1+x)^2}-\frac{1}{1+x} =\frac{1-(1+x)}{(1+x)^2} =\frac{-x}{(1+x)^2}.

Now:

If $x\in(-1,0)$ , then $x<0$ , so $-x>0$ , hence $\phi'(x)>0.$ Thus $\phi$ is increasing on $(-1,0)$ .
If $x\in(0,\infty)$ , then $x>0$ , so $-x<0$ , hence $\phi'(x)<0.$ Thus $\phi$ is decreasing on $(0,\infty)$ .

Also,

\phi(0)=\frac{0}{1+0}-\ln(1)=0.

Since $\phi$ increases up to $0$ and decreases after $0$ , its maximum occurs at $x=0$ , with maximum value $0$ . Therefore,

\phi(x)\le 0 \quad \text{for all } x>-1,

and for $x\neq 0$ ,

\phi(x)<0.

Hence,

f'(x)=\frac{\phi(x)}{x^2}<0 \quad \text{for all } x\in(-1,0)\cup(0,\infty).

So $f$ is decreasing on both intervals $(-1,0)$ and $(0,\infty)$ .

Check behavior at $x=0$

We should verify that the function remains decreasing across $x=0$ .

Using the standard limit,

\lim_{x\to 0}\frac{\ln(1+x)}{x}=1=f(0).

So $f$ is continuous at $x=0$ .

Since it is decreasing on both sides of $0$ and continuous at $0$ , the function is decreasing on the entire domain $(-1,\infty)$ .

(Indeed, for $x<0$ , $f(x)>1$ , and for $x>0$ , $f(x)<1$ , consistent with overall decrease.)

Evaluate options

A: decreases in $(-1,\infty)$ ✅
B: decreases in $(-1,0)$ and increases in $(0,\infty)$ ❌
C: increases in $(-1,\infty)$ ❌
D: increases in $(-1,0)$ and decreases in $(0,\infty)$ ❌

Therefore, the correct option is

\boxed{\text{A}}

Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

Application of Derivatives question

Application of Derivatives question

More from Application of Derivatives

Application of Derivatives question

Application of Derivatives question

More from Application of Derivatives