Can we just plug in values for $\alpha$ to check the options?

While testing values is a great way to verify your answer in an exam, it is dangerous to rely on it as your primary method. The question asks for the range of $\alpha$, which implies a continuous interval. By solving the inequality algebraically, we ensure we capture the entire domain of validity, not just a few lucky points. Always trust the calculus!

JEE Main 2022 (25 June Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ and $g : R \to R$ be two functions defined by $f (x) = lo g_{e} (x^{2} + 1) - e^{- x} + 1$ and $g (x) = \frac{1 - 2 e ^{2 x}}{e ^{x}}$ . Then, for which of the following range of $α$ , the inequality $f (g (\frac{( α - 1 ) ^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

Q: Wait, for $f'(x) = \frac{2x}{x^2 + 1} + e^{-x}$, how can we be sure it's positive when $x$ is negative? Doesn't the first term become negative?

That is a brilliant observation! When $x$ is negative, the first term is indeed negative. However, consider the magnitude. The term $\frac{2x}{x^2 + 1}$ is bounded between $-1$ and $1$. Specifically, for $x < 0$, it is greater than $-1$. Meanwhile, $e^{-x}$ for $x < 0$ is $e^{|x|}$, which is always greater than $1$. Since the positive exponential term grows faster than the negative rational term, the sum remains strictly positive. It is a beautiful balance of functions!

Q: Why do we flip the inequality sign when removing $g$ but not when removing $f$?

Think of a function as a machine. An increasing function preserves the order of inputs—if you put in a larger number, you get a larger output. A decreasing function reverses the order—if you put in a larger number, you get a smaller output. Since $f$ is increasing, it respects the inequality. Since $g$ is decreasing, it flips the inequality. It is not just a rule; it is the geometric nature of the function!

Select Answer:

Progressive Hints

Hint 1

Look at the outermost function

f (x)

first. Before diving into the algebra, ask yourself: is this function always increasing or always decreasing? Calculate the derivative

f^{'} (x)

and check its sign for all real

x

. Don't be intimidated by the exponential terms; they are your friends.

Hint 2

Now, turn your attention to the inner function

g (x)

. Simplify it by writing it as

e^{- x} - 2 e^{x}

. Find its derivative

g^{'} (x)

. Notice that both terms in the derivative are negative. This tells you something crucial about the direction of the inequality when you 'remove'

g

from both sides.

Hint 3

Remember the golden rule of inequalities: if

f

is strictly increasing,

f (A) > f (B)

implies

A > B

. However, if

g

is strictly decreasing,

g (A) > g (B)

implies

A < B

. Apply this to your expression:

\frac{( α - 1 ) ^{2}}{3} < α - \frac{5}{3}

. Now, solve this quadratic inequality for

α

Visualized Solution

Understand the Functions

Given functions:
$f (x) = lo g_{e} (x^{2} + 1) - e^{- x} + 1$
$g (x) = \frac{1 - 2 e ^{2 x}}{e ^{x}}$
We need to solve: $f (g (\frac{( α - 1 ) ^{2}}{3})) > f (g (α - \frac{5}{3}))$

Differentiate $f (x)$

$f^{'} (x) = \frac{d}{d x} (lo g_{e} (x^{2} + 1)) - \frac{d}{d x} (e^{- x})$
$f^{'} (x) = \frac{2 x}{x ^{2} + 1} + e^{- x}$

Analyze $f^{'} (x)$ for $x \geq 0$

For $x \geq 0$ :
$\frac{2 x}{x ^{2} + 1} \geq 0$
$e^{- x} > 0$
$⟹ f^{'} (x) > 0$

Analyze $f^{'} (x)$ for $x < 0$

For $x < 0$ , let $x = - t$ ( $t > 0$ ):
$f^{'} (- t) = \frac{- 2 t}{t ^{2} + 1} + e^{t}$
Since $t^{2} + 1 \geq 2 t ⟹ \frac{2 t}{t ^{2} + 1} \leq 1$
Since $t > 0 ⟹ e^{t} > 1$
$⟹ f^{'} (x) > 0$

Conclusion for $f (x)$

$f^{'} (x) > 0\forall x \in R$
$⟹ f (x)$ is strictly increasing.
Property: $f (A) > f (B) ⟹ A > B$

Simplify and Differentiate $g (x)$

$g (x) = \frac{1}{e ^{x}} - \frac{2 e ^{2 x}}{e ^{x}} = e^{- x} - 2 e^{x}$
$g^{'} (x) = - e^{- x} - 2 e^{x}$

Conclusion for $g (x)$

$g^{'} (x) = - (e^{- x} + 2 e^{x})$
Since $e^{- x} > 0$ and $2 e^{x} > 0 ⟹ g^{'} (x) < 0\forall x \in R$
$⟹ g (x)$ is strictly decreasing.
Property: $g (A) > g (B) ⟹ A < B$

Apply Monotonicity of $f$

$f (g (\frac{( α - 1 ) ^{2}}{3})) > f (g (α - \frac{5}{3}))$
Removing $f$ (increasing, sign remains):
$g (\frac{( α - 1 ) ^{2}}{3}) > g (α - \frac{5}{3})$

Apply Monotonicity of $g$

Removing $g$ (decreasing, sign flips):
$\frac{( α - 1 ) ^{2}}{3} < α - \frac{5}{3}$

Simplify the Inequality

Multiply by $3$ :
$(α - 1)^{2} < 3 (α - \frac{5}{3})$
$α^{2} - 2 α + 1 < 3 α - 5$

Form the Quadratic Inequality

$α^{2} - 5 α + 6 < 0$
$(α - 2) (α - 3) < 0$

Solve for $α$

Roots are $α = 2$ and $α = 3$
$α \in (2, 3)$

00:00 / 00:00

The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Setup

To solve the inequality

f (g (\frac{( α - 1 ) ^{2}}{3})) > f (g (α - \frac{5}{3}))

, we must first understand the behavior of the constituent functions. We are given:

f (x) = lo g_{e} (x^{2} + 1) - e^{- x} + 1

g (x) = \frac{1 - 2 e ^{2 x}}{e ^{x}}

The key to unlocking this problem lies in the concept of monotonicity. If we determine whether these functions are strictly increasing or strictly decreasing, the inequality will simplify significantly.

Analyzing the Outer Function $f (x)$

To understand the motion of

f (x)

, we calculate its derivative:

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

For

x \geq 0

, both terms are clearly positive. For

x < 0

, let

x = - t

where

t > 0

. The derivative becomes:

f^{'} (- t) = e^{t} - \frac{2 t}{t ^{2} + 1}

Since

t^{2} + 1 \geq 2 t

(by AM-GM inequality), we know that

\frac{2 t}{t ^{2} + 1} \leq 1

. Because

e^{t} > 1

for all

t > 0

, the derivative

f^{'} (x)

is always positive. Thus, $f (x)$ is strictly increasing for all real numbers.

Simplifying the Inequality

Because

f

is strictly increasing, we can remove it from both sides of the inequality without changing the direction of the inequality sign:

g (\frac{( α - 1 ) ^{2}}{3}) > g (α - \frac{5}{3})

Now, we analyze

g (x)

. Simplifying the expression, we get:

g (x) = e^{- x} - 2 e^{x}

The derivative is:

g^{'} (x) = - e^{- x} - 2 e^{x}

Since both terms are negative,

g^{'} (x) < 0

for all

x

. Therefore, $g (x)$ is strictly decreasing.

The Final Calculation

When we remove a strictly decreasing function from an inequality, the inequality sign must flip. Our inequality transforms as follows:

\frac{( α - 1 ) ^{2}}{3} < α - \frac{5}{3}

Multiplying by

3

yields:

(α - 1)^{2} < 3 α - 5

Expanding the left side and rearranging the terms:

α^{2} - 2 α + 1 < 3 α - 5

α^{2} - 5 α + 6 < 0

Factoring the quadratic expression:

(α - 2) (α - 3) < 0

Since the quadratic is less than zero,

α

must lie between the roots

2

and

3

. Thus, the final solution is:

α \in (2, 3)

Common Traps & FAQs

Wait, for

f^{'} (x) = \frac{2 x}{x ^{2} + 1} + e^{- x}

, how can we be sure it's positive when

x

is negative? Doesn't the first term become negative?

That is a brilliant observation! When

x

is negative, the first term is indeed negative. However, consider the magnitude. The term

\frac{2 x}{x ^{2} + 1}

is bounded between

- 1

and

1

. Specifically, for

x < 0

, it is greater than

- 1

. Meanwhile,

e^{- x}

for

x < 0

e^{∣ x ∣}

, which is always greater than

1

. Since the positive exponential term grows faster than the negative rational term, the sum remains strictly positive. It is a beautiful balance of functions!

Why do we flip the inequality sign when removing

g

but not when removing

f

Think of a function as a machine. An increasing function preserves the order of inputs—if you put in a larger number, you get a larger output. A decreasing function reverses the order—if you put in a larger number, you get a smaller output. Since

f

is increasing, it respects the inequality. Since

g

is decreasing, it flips the inequality. It is not just a rule; it is the geometric nature of the function!

Can we just plug in values for

α

to check the options?

While testing values is a great way to verify your answer in an exam, it is dangerous to rely on it as your primary method. The question asks for the range of

α

, which implies a continuous interval. By solving the inequality algebraically, we ensure we capture the entire domain of validity, not just a few lucky points. Always trust the calculus!

Select Answer:

Visualized Solution

Understand the Functions

Differentiate $f (x)$

Analyze $f^{'} (x)$ for $x \geq 0$

Analyze $f^{'} (x)$ for $x < 0$

Conclusion for $f (x)$

Simplify and Differentiate $g (x)$

Conclusion for $g (x)$

Apply Monotonicity of $f$

Apply Monotonicity of $g$

Simplify the Inequality

Form the Quadratic Inequality

Solve for $α$

The Sigma Insight: Monotonicity

Analyzing the Setup

Analyzing the Outer Function $f (x)$

Simplifying the Inequality

The Final Calculation

Similar Questions

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Comprehension Passage

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Select Answer:

Visualized Solution

Understand the Functions

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Simplify the Inequality

Form the Quadratic Inequality

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The Sigma Insight: Monotonicity

Analyzing the Setup

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Simplifying the Inequality

The Final Calculation

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Analyze $f^{'} (x)$ for $x \geq 0$

Analyze $f^{'} (x)$ for $x < 0$

Conclusion for $f (x)$

Simplify and Differentiate $g (x)$

Conclusion for $g (x)$

Apply Monotonicity of $f$

Apply Monotonicity of $g$

Solve for $α$

Analyzing the Outer Function $f (x)$

Let $f (x) = 2 x + tan^{- 1} x$ and $g (x) = lo g_{e} (1 + x^{2} + x), x \in [0, 3]$ . Then

Let $f (x) = e^{x} - x$ and $g (x) = x^{2} - x, x \in R$ . Then the set of all $x \in R$ , where the function $h (x) = (f o g) (x)$ is increasing, is :

Consider the statements: $P$ : There exists some $x \in R$ such that $f (x) + 2 x = 2 (1 + x^{2})$ , $Q$ : There exists some $x \in R$ such that $2 f (x) + 1 = 2 x (1 + x)$

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval