JEE Advanced 1987

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: The set of all $x$ for which $ln (1 + x) \leq x$ is equal to .........

Q: Is the derivative test enough to prove the inequality for all $x \ge 0$?

Absolutely. By showing that $f'(x) 0$, we prove that the function is strictly decreasing. Since we also know that $f(0) = 0$, any value of $x$ greater than zero must result in a function value less than $f(0)$. This confirms that $f(x) \le 0$ for all $x \ge 0$.

Visualized Solution

Visualizing the Functions

We want to find the set of all $x$ for which $ln (1 + x) \leq x$ .
Let's visualize this by plotting the curves $y = ln (1 + x)$ and $y = x$ .
The inequality holds where the logarithmic curve lies on or below the straight line.

Domain of the Logarithmic Function

For the term $ln (1 + x)$ to be defined, the argument must be strictly positive.
Therefore, we must have: $1 + x > 0 ⟹ x > - 1$ .
This establishes the domain of our function as $(- 1, \infty)$ .

Defining the Function $f (x)$

Let us define an auxiliary function: $f (x) = ln (1 + x) - x$
The domain of $f (x)$ is $x > - 1$ .
The inequality $ln (1 + x) \leq x$ is equivalent to finding where $f (x) \leq 0$ .

Differentiating $f (x)$

To find the intervals of increase or decrease, we differentiate $f (x)$ with respect to $x$ :
$f^{'} (x) = \frac{d}{d x} [ln (1 + x)] - \frac{d}{d x} [x]$
We will apply the chain rule to the logarithmic term.

Calculating $f^{'} (x)$

Using the standard derivative formulas:
$\frac{d}{d x} [ln (1 + x)] = \frac{1}{1 + x}$
$\frac{d}{d x} [x] = 1$
Thus, $f^{'} (x) = \frac{1}{1 + x} - 1$

Simplifying $f^{'} (x)$

Take the common denominator $(1 + x)$ :
$f^{'} (x) = \frac{1 - ( 1 + x )}{1 + x}$
Simplifying the numerator:
$f^{'} (x) = \frac{- x}{1 + x}$

Sign of $f^{'} (x)$ for $x > 0$

Let's analyze the sign of $f^{'} (x) = \frac{- x}{1 + x}$ for $x > 0$ :
Since $x > 0$ , the numerator $- x$ is negative.
The denominator $1 + x$ is positive.
Therefore, $f^{'} (x) < 0$ for all $x > 0$ .
This means $f (x)$ is strictly decreasing for $x > 0$ .

Evaluating $f (0)$

Let's find the value of the function at the boundary point $x = 0$ :
$f (0) = ln (1 + 0) - 0$
$f (0) = ln (1) - 0 = 0$
So, the curve passes through the origin $(0, 0)$ .

Final Interval for $ln (1 + x) \leq x$

Since $f (0) = 0$ and $f (x)$ is strictly decreasing for $x > 0$ :
$f (x) \leq f (0) ⟹ f (x) \leq 0$ for all $x \geq 0$ .
Substituting $f (x)$ back: $ln (1 + x) - x \leq 0$ .
Therefore, $ln (1 + x) \leq x$ for all $x \geq 0$ .

00:00 / 00:00

The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two paths. One is a straight, unwavering line,

y = x

, cutting through the origin at a perfect 45-degree angle.

The other is a gentle, curving path,

y = ln (1 + x)

, which starts near the vertical asymptote at

x = - 1

and climbs slowly, bending away from the straight line. Our mission is to find the set of all

x

where the curve

y = ln (1 + x)

stays below or touches the line

y = x

Phase 1

The Domain of Existence

Before we start our calculus, we must respect the boundaries. The function

ln (1 + x)

is only defined when its argument is strictly positive.

Thus, we require

1 + x > 0

, which means

x > - 1

. This is our playground; everything we do must happen to the right of the vertical line

x = - 1

Phase 2

The Auxiliary Function

To compare two functions, we use a powerful tool: the auxiliary function. Let us define:

f (x) = ln (1 + x) - x

Solving

ln (1 + x) \leq x

is identical to solving

f (x) \leq 0

. By shifting the problem to a single function, we can use the power of derivatives to see exactly where this function lives relative to the

x

-axis.

Phase 3

The Calculus of Change

Now, let us see how

f (x)

changes. We compute the derivative:

f^{'} (x) = \frac{d}{d x} [ln (1 + x)] - \frac{d}{d x} [x]

Applying the chain rule to the logarithmic term, we get:

f^{'} (x) = \frac{1}{1 + x} - 1

With a little algebraic finesse, we find a common denominator:

f^{'} (x) = \frac{1 - ( 1 + x )}{1 + x} = \frac{- x}{1 + x}

This is a beautiful, compact expression. It tells us everything we need to know about the slope of our function.

Phase 4

The Final Verdict

Let us look at the region

x > 0

. In this interval, the numerator

- x

is negative, and the denominator

1 + x

is positive.

A negative divided by a positive is always negative. Thus,

f^{'} (x) < 0

for all

x > 0

, meaning our function

f (x)

is strictly decreasing for all positive

x

We also know that at the origin:

f (0) = ln (1 + 0) - 0 = 0

Since the function starts at

0

x = 0

and then strictly decreases as

x

increases, it must be that

f (x) \leq 0

for all

x \geq 0

This confirms our inequality

ln (1 + x) \leq x

for all

x \geq 0

. The solution set is $x \in

Common Traps & FAQs

Why do we define

f (x) = ln (1 + x) - x

? Can't we just solve it directly?

Great question! Solving inequalities directly is often messy because you are comparing two different functions. By defining

f (x) = ln (1 + x) - x

, we turn an inequality problem into a function behavior problem. It allows us to use the power of derivatives to track the function's growth and find its maximum value, which is a much more robust approach.

Is the derivative test enough to prove the inequality for all

x \geq 0

Absolutely. By showing that

f^{'} (x) < 0

for all

x > 0

, we prove that the function is strictly decreasing. Since we also know that

f (0) = 0

, any value of

x

greater than zero must result in a function value less than

f (0)

. This confirms that

f (x) \leq 0

for all

x \geq 0

What if

x

is between

- 1

and

0

? Does the inequality still hold?

That is a brilliant observation! If you check the derivative

f^{'} (x) = \frac{- x}{1 + x}

for

x \in (- 1, 0)

, you will see that

f^{'} (x) > 0

, meaning the function is increasing. Since

f (0) = 0

, the function values for

x < 0

must be less than

0

. So, the inequality

ln (1 + x) \leq x

actually holds for all

x > - 1

. However, in many JEE contexts, we focus on the non-negative domain

x \geq 0

as the primary solution set.

Animated Solution for Mathematics - Differentiation: The set of all $x$ for which $ln (1 + x) \leq x$ is equal to .........

Visualized Solution

Visualizing the Functions

Domain of the Logarithmic Function

Defining the Function $f (x)$

Differentiating $f (x)$

Calculating $f^{'} (x)$

Simplifying $f^{'} (x)$

Sign of $f^{'} (x)$ for $x > 0$

Evaluating $f (0)$

Final Interval for $ln (1 + x) \leq x$

The Sigma Insight: Monotonicity

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The set of all x for which ln(1+x)≤x is equal to .........

Visualized Solution

Visualizing the Functions

Domain of the Logarithmic Function

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

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

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

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

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

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

Final Interval for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(1+x)≤x

The Sigma Insight: Monotonicity

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differentiation: The set of all $x$ for which $ln (1 + x) \leq x$ is equal to .........

Defining the Function $f (x)$

Differentiating $f (x)$

Calculating $f^{'} (x)$

Simplifying $f^{'} (x)$

Sign of $f^{'} (x)$ for $x > 0$

Evaluating $f (0)$

Final Interval for $ln (1 + x) \leq x$

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