JEE Advanced 2010

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f, g$ and $h$ be real-valued functions defined on the interval $[0, 1]$ by $f (x) = e^{x^{2}} + e^{- x^{2}}, g (x) = x e^{x^{2}} + e^{- x^{2}}$ and $h (x) = x^{2} e^{x^{2}} + e^{- x^{2}}$ . If $a, b$ and $c$ denote, respectively, the absolute maximum of $f, g$ and $h$ on $[0, 1]$ , then

Q: Why do we always check the endpoints? Can't the maximum be in the middle?

Great question! The maximum of a continuous function on a closed interval $[0, 1]$ can occur at critical points (where the derivative is zero) or at the endpoints. If the derivative is strictly positive, the function is 'climbing' the entire time, so the peak must be at the very end!

Q: Why can't I just test $x=0.5$ and see which one is bigger?

Testing values is a great way to get a 'feel' for the problem, but in JEE Advanced, we need rigorous proof. Testing a single point doesn't tell us about the behavior across the whole interval. We need to prove the trend for all $x$ in $[0, 1]$.

Q: Wait, if $x^2 < x < 1$, shouldn't $h(x)$ be smaller than $g(x)$? How can they have the same maximum?

You are absolutely right that $h(x) < g(x) < f(x)$ for any $x$ strictly between $0$ and $1$. But remember, the maximum is a specific point. As $x$ approaches $1$, the difference between $x^2$, $x$, and $1$ vanishes. They all converge to the same destination!

Select Answer:

Visualized Solution

Introduction to the Functions

We are given three real-valued functions defined on the closed interval $[0, 1]$ :
$f (x) = e^{x^{2}} + e^{- x^{2}}$
$g (x) = x e^{x^{2}} + e^{- x^{2}}$
$h (x) = x^{2} e^{x^{2}} + e^{- x^{2}}$
Our goal is to find and compare their absolute maximum values $a, b,$ and $c$ on $[0, 1]$ .

Derivative of $f (x)$

Function: $f (x) = e^{x^{2}} + e^{- x^{2}}$
Differentiating with respect to $x$ using the chain rule:
$f^{'} (x) = \frac{d}{d x} (e^{x^{2}}) + \frac{d}{d x} (e^{- x^{2}})$
$f^{'} (x) = 2 x e^{x^{2}} - 2 x e^{- x^{2}}$
Factoring out $2 x$ gives: $f^{'} (x) = 2 x (e^{x^{2}} - e^{- x^{2}})$

Monotonicity and Maximum of $f (x)$

For $x \in [0, 1]$ , we have $x \geq 0$ .
Since $x^{2} \geq - x^{2}$ , we have $e^{x^{2}} \geq e^{- x^{2}} ⟹ e^{x^{2}} - e^{- x^{2}} \geq 0$ .
Therefore, $f^{'} (x) = 2 x (e^{x^{2}} - e^{- x^{2}}) \geq 0$ for all $x \in [0, 1]$ .
This means $f (x)$ is monotonically increasing on $[0, 1]$ .
The absolute maximum $a$ occurs at the right endpoint, $x = 1$ :
$a = f (1) = e^{1^{2}} + e^{- 1^{2}} = e + \frac{1}{e}$

Derivative of $g (x)$

Function: $g (x) = x e^{x^{2}} + e^{- x^{2}}$
Using the product rule on the first term:
$g^{'} (x) = \frac{d}{d x} (x) \cdot e^{x^{2}} + x \cdot \frac{d}{d x} (e^{x^{2}}) + \frac{d}{d x} (e^{- x^{2}})$
$g^{'} (x) = 1 \cdot e^{x^{2}} + x \cdot (2 x e^{x^{2}}) - 2 x e^{- x^{2}}$
Simplifying: $g^{'} (x) = (1 + 2 x^{2}) e^{x^{2}} - 2 x e^{- x^{2}}$

Monotonicity and Maximum of $g (x)$

We check the sign of $g^{'} (x) = (1 + 2 x^{2}) e^{x^{2}} - 2 x e^{- x^{2}}$ on $[0, 1]$ .
For $x \in (0, 1]$ , we have $e^{x^{2}} > e^{- x^{2}} > 0$ .
Also, $(1 + 2 x^{2}) > 2 x$ because $2 x^{2} - 2 x + 1 = 2 (x - \frac{1}{2})^{2} + \frac{1}{2} > 0$ .
Therefore, $(1 + 2 x^{2}) e^{x^{2}} > 2 x e^{- x^{2}} ⟹ g^{'} (x) > 0$ for $x \in (0, 1]$ .
At $x = 0$ , $g^{'} (0) = 1 > 0$ . Thus, $g (x)$ is strictly increasing on $[0, 1]$ .
The absolute maximum $b$ occurs at $x = 1$ :
$b = g (1) = 1 \cdot e^{1^{2}} + e^{- 1^{2}} = e + \frac{1}{e}$

Derivative of $h (x)$

Function: $h (x) = x^{2} e^{x^{2}} + e^{- x^{2}}$
Using the product rule on the first term:
$h^{'} (x) = \frac{d}{d x} (x^{2}) \cdot e^{x^{2}} + x^{2} \cdot \frac{d}{d x} (e^{x^{2}}) + \frac{d}{d x} (e^{- x^{2}})$
$h^{'} (x) = 2 x e^{x^{2}} + x^{2} (2 x e^{x^{2}}) - 2 x e^{- x^{2}}$
Factoring out $2 x$ : $h^{'} (x) = 2 x [(1 + x^{2}) e^{x^{2}} - e^{- x^{2}}]$

Monotonicity and Maximum of $h (x)$

We analyze the sign of $h^{'} (x) = 2 x [(1 + x^{2}) e^{x^{2}} - e^{- x^{2}}]$ on $[0, 1]$ .
For $x \in (0, 1]$ , we have $1 + x^{2} > 1$ and $e^{x^{2}} \geq 1 ⟹ (1 + x^{2}) e^{x^{2}} > 1$ .
Also, $e^{- x^{2}} \leq 1$ for all $x \in [0, 1]$ .
Therefore, $(1 + x^{2}) e^{x^{2}} - e^{- x^{2}} > 1 - 1 = 0$ .
Since $2 x > 0$ for $x \in (0, 1]$ , we have $h^{'} (x) > 0$ on $(0, 1]$ .
At $x = 0$ , $h^{'} (0) = 0$ . Thus, $h (x)$ is strictly increasing on $[0, 1]$ .
The absolute maximum $c$ occurs at $x = 1$ :
$c = h (1) = 1^{2} \cdot e^{1^{2}} + e^{- 1^{2}} = e + \frac{1}{e}$

Visualizing the Convergence

For $x \in (0, 1)$ , we have $x^{2} < x < 1$ .
Multiplying by $e^{x^{2}} > 0$ , we get: $x^{2} e^{x^{2}} < x e^{x^{2}} < e^{x^{2}}$ .
Adding $e^{- x^{2}}$ to all terms:
$h (x) < g (x) < f (x)$ for all $x \in (0, 1)$ .
At $x = 1$ , all three functions converge to the exact same value:
$f (1) = g (1) = h (1) = e + \frac{1}{e}$

Final Comparison of Maximums

We found the absolute maximums on $[0, 1]$ :
$a = e + \frac{1}{e}$
$b = e + \frac{1}{e}$
$c = e + \frac{1}{e}$
Therefore, $a = b = c$ .
This matches Option (D).

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, looking out across the interval

[0, 1]

. You have three paths before you:

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

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

h (x) = x^{2} e^{x^{2}} + e^{- x^{2}}

At first glance, they look like a tangled mess of exponentials. But as a JEE aspirant, you know that chaos is just order waiting to be discovered. Let us embark on this journey to find their absolute maximums.

The Slope of the Journey

First, we must understand the 'slope' of our journey. We calculate the derivative of

f (x)

. Using the chain rule, we find:

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

Because

x

is positive and

e^{x^{2}}

dominates

e^{- x^{2}}

on this interval,

f^{'} (x)

is always non-negative. This means

f (x)

is a climber; it is monotonically increasing. It starts at

f (0) = 2

and marches steadily upward until it reaches its peak at

x = 1

Analyzing the Complex Paths

Now, consider

g (x)

and

h (x)

. They look more complex because of the

x

and

x^{2}

coefficients. But do not fear the product rule!

When we differentiate

g (x)

, we get:

g^{'} (x) = (1 + 2 x^{2}) e^{x^{2}} - 2 x e^{- x^{2}}

By analyzing the signs, we find that this, too, is strictly positive. The same logic applies to

h (x)

. We find:

h^{'} (x) = 2 x [(1 + x^{2}) e^{x^{2}} - e^{- x^{2}}]

This derivative is also positive for

x > 0

. What does this tell us? It tells us that all three functions are strictly increasing on the interval

[0, 1]

The Convergence at the Summit

They are all climbing toward the same destination! Even though

h (x)

stays below

g (x)

, and

g (x)

stays below

f (x)

for all

x \in (0, 1)

, they are all racing toward the same finish line.

When we evaluate them at

x = 1

, we get:

f (1) = e + \frac{1}{e}

g (1) = e + \frac{1}{e}

h (1) = e + \frac{1}{e}

The beauty of this problem lies in the convergence. Despite their different paths, they meet at the exact same summit. Thus, $a = b = c$ . You have conquered the functions!

Common Traps & FAQs

Why do we always check the endpoints? Can't the maximum be in the middle?

Great question! The maximum of a continuous function on a closed interval

[0, 1]

can occur at critical points (where the derivative is zero) or at the endpoints. If the derivative is strictly positive, the function is 'climbing' the entire time, so the peak must be at the very end!

Why can't I just test

x = 0.5

and see which one is bigger?

Testing values is a great way to get a 'feel' for the problem, but in JEE Advanced, we need rigorous proof. Testing a single point doesn't tell us about the behavior across the whole interval. We need to prove the trend for all

x

[0, 1]

Wait, if

x^{2} < x < 1

, shouldn't

h (x)

be smaller than

g (x)

? How can they have the same maximum?

You are absolutely right that

h (x) < g (x) < f (x)

for any

x

strictly between

0

and

1

. But remember, the maximum is a specific point. As

x

approaches

1

, the difference between

x^{2}

x

, and

1

vanishes. They all converge to the same destination!

Select Answer:

Visualized Solution

Introduction to the Functions

Derivative of $f (x)$

Monotonicity and Maximum of $f (x)$

Derivative of $g (x)$

Monotonicity and Maximum of $g (x)$

Derivative of $h (x)$

Monotonicity and Maximum of $h (x)$

Visualizing the Convergence

Final Comparison of Maximums

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Slope of the Journey

Analyzing the Complex Paths

The Convergence at the Summit

Similar Questions

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Let $f : R \to R$ be a continuous function defined by $f (x) = \frac{1}{e ^{x} + 2 e ^{- x}}$ . \\ \textbf{Statement-1:} $f (c) = 1/3$ for some $c \in R$ . \\ \textbf{Statement-2:} $0 < f (x) \leq 1/ (22)$ for all $x \in R$ .

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

Select Answer:

Visualized Solution

Introduction to the Functions

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

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

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

Monotonicity and Maximum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

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

Monotonicity and Maximum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

Visualizing the Convergence

Final Comparison of Maximums

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Slope of the Journey

Analyzing the Complex Paths

The Convergence at the Summit

Similar Questions

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Derivative of $f (x)$

Monotonicity and Maximum of $f (x)$

Derivative of $g (x)$

Monotonicity and Maximum of $g (x)$

Derivative of $h (x)$

Monotonicity and Maximum of $h (x)$

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If the function $f (x) = (\frac{1}{x})^{2 x}; x > 0$ attains the maximum value at $x = \frac{1}{e}$ then :

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