How do I know for sure that $1/3$ is less than $1/(2\sqrt{2})$ without a calculator?

This is a classic trap! Never guess. To compare two positive fractions $\frac{1}{A}$ and $\frac{1}{B}$, simply compare $A$ and $B$. Here, we compare $3$ and $2\sqrt{2}$. Square both sides: $3^2 = 9$ and $(2\sqrt{2})^2 = 8$. Since $9 > 8$, it follows that $3 > 2\sqrt{2}$. Therefore, the reciprocal $\frac{1}{3}$ must be less than $\frac{1}{2\sqrt{2}}$. Rigor beats intuition every time.

JEE Main 2010

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: 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 $x \in R$ .

Q: Wait, why can't we just differentiate $f(x)$ to find the maximum? Isn't that the standard way?

You absolutely can! Differentiating $f(x)$ and setting $f'(x) = 0$ will give you the same result. However, in JEE Advanced, time is your most precious currency. Using the AM-GM inequality allows you to find the minimum of the denominator in seconds without the tedious chain rule and quotient rule. Always look for the 'algebraic shortcut' before jumping into calculus.

Q: Why does Statement 2 explain Statement 1? I thought they were just two separate facts.

Think of Statement 2 as the 'boundary condition' and Statement 1 as the 'existence proof'. Statement 2 establishes the range of the function, specifically its maximum. Statement 1 asks if a specific value exists within that range. Because the function is continuous, the Intermediate Value Theorem (IVT) guarantees that if a value lies between the minimum and maximum, the function must hit it. Without the range provided by Statement 2, you wouldn't know for sure if $1/3$ is reachable. Thus, Statement 2 is the logical foundation for Statement 1.

Select Answer:

Visualized Solution

Analyzing the Function $f (x)$

Given function: $f (x) = \frac{1}{e ^{x} + 2 e ^{- x}}$
Observe that $e^{x} > 0$ and $2 e^{- x} > 0$ for all $x \in R$ .
Therefore, the denominator is always positive, and $f (x) > 0$ .

The AM-GM Inequality

To find the range of $f (x)$ , we must find the range of the denominator $D (x) = e^{x} + 2 e^{- x}$ .
We use the AM-GM Inequality: For any $a, b > 0$ , $\frac{a + b}{2} \geq ab$ .

Applying AM-GM to the Denominator

Let $a = e^{x}$ and $b = 2 e^{- x}$ .
Substitute into AM-GM: $\frac{e ^{x} + 2 e ^{- x}}{2} \geq e^{x} \cdot 2 e^{- x}$

Minimizing the Denominator

Simplify the product inside the square root: $e^{x} \cdot 2 e^{- x} = 2 e^{0} = 2$ .
The inequality becomes: $\frac{e ^{x} + 2 e ^{- x}}{2} \geq 2$
Multiplying by $2$ : $e^{x} + 2 e^{- x} \geq 22$

Maximizing the Function $f (x)$

Since the denominator $e^{x} + 2 e^{- x} \geq 22$ , taking the reciprocal reverses the inequality.
$\frac{1}{e ^{x} + 2 e ^{- x}} \leq \frac{1}{2 2}$
Thus, $f (x) \leq \frac{1}{2 2}$ for all $x \in R$ .

Verifying Statement-2

We established that $f (x) > 0$ and $f (x) \leq \frac{1}{2 2}$ .
Combining these: $0 < f (x) \leq \frac{1}{2 2}$ for all $x \in R$ .
Statement-2 is True.

Analyzing Statement-1

Statement-1 claims $f (c) = \frac{1}{3}$ for some $c \in R$ .
We need to check if $\frac{1}{3}$ lies within the range $(0, \frac{1}{2 2}]$ .
Let's compare the values $\frac{1}{3}$ and $\frac{1}{2 2}$ .

Comparing $\frac{1}{3}$ and $\frac{1}{2 2}$

To compare $\frac{1}{3}$ and $\frac{1}{2 2}$ , we can compare their denominators: $3$ and $22$ .
Square both numbers to remove the square root:
$3^{2} = 9$
$(22)^{2} = 4 \times 2 = 8$

Establishing the Inequality

Since $9 > 8$ , it follows that $3 > 22$ .
Taking reciprocals reverses the inequality: $\frac{1}{3} < \frac{1}{2 2}$ .
Thus, $\frac{1}{3}$ is strictly less than the maximum value of $f (x)$ .

The Intermediate Value Theorem (IVT)

$f (x)$ is a continuous function on $R$ .
The range of $f (x)$ is $(0, \frac{1}{2 2}]$ .
Since $\frac{1}{3} \in (0, \frac{1}{2 2}]$ , by the Intermediate Value Theorem, the function must take the value $\frac{1}{3}$ at some point $c$ .

Final Conclusion

Statement-1 is True.
Statement-2 is True.
Statement-2 provides the exact range that proves Statement-1 via IVT.
Therefore, Statement-2 is the correct explanation for Statement-1.

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at the function

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

. At first glance, it looks like a simple ratio, but it is actually a beautiful, symmetric bell-like curve.

The denominator is composed of two exponential terms:

e^{x}

, which grows rapidly, and

2 e^{- x}

, which decays rapidly. As

x

moves from negative infinity to positive infinity, the function rises from zero, hits a peak, and then descends back toward zero.

The Power of AM-GM

To find the maximum of

f (x)

, we must minimize the denominator

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

. We have two positive terms, and their product is a constant because

e^{x} \cdot 2 e^{- x} = 2 e^{0} = 2

This is the perfect setup for the Arithmetic Mean-Geometric Mean (AM-GM) inequality. We write:

\frac{e ^{x} + 2 e ^{- x}}{2} \geq e^{x} \cdot 2 e^{- x}

The right side simplifies beautifully to

2

. Multiplying by

2

, we get:

e^{x} + 2 e^{- x} \geq 22

This tells us that the denominator can never be smaller than

22

The Reciprocal Flip

Now, we take the reciprocal. Remember, when you take the reciprocal of an inequality, the sign flips!

\frac{1}{e ^{x} + 2 e ^{- x}} \leq \frac{1}{2 2}

This is the 'ceiling' of our function. We have just proven Statement 2: the function is always strictly positive and never exceeds

\frac{1}{2 2}

The IVT Bridge

Now, let us tackle Statement 1: Does

f (x) = 1/3

for some

c

? We need to know if

1/3

falls within our range

(0, \frac{1}{2 2}]

We compare the denominators:

3

and

22

. Squaring them gives

9

and

8

. Since

9 > 8

, we know

3 > 22

, which means:

\frac{1}{3} < \frac{1}{2 2}

Because our function is continuous—meaning it has no jumps or breaks—the Intermediate Value Theorem (IVT) tells us that if the function reaches a maximum of

\frac{1}{2 2}

and starts near zero, it must pass through every single value in between.

Since

1/3

is in that interval, the function must hit it. Statement 2 provided the range that made this conclusion possible, making it the correct explanation for Statement 1. You have successfully navigated the logic of the problem.

Common Traps & FAQs

Wait, why can't we just differentiate

f (x)

to find the maximum? Isn't that the standard way?

You absolutely can! Differentiating

f (x)

and setting

f^{'} (x) = 0

will give you the same result. However, in JEE Advanced, time is your most precious currency. Using the AM-GM inequality allows you to find the minimum of the denominator in seconds without the tedious chain rule and quotient rule. Always look for the 'algebraic shortcut' before jumping into calculus.

How do I know for sure that

1/3

is less than

1/ (22)

without a calculator?

This is a classic trap! Never guess. To compare two positive fractions

\frac{1}{A}

and

\frac{1}{B}

, simply compare

A

and

B

. Here, we compare

3

and

22

. Square both sides:

3^{2} = 9

and

(22)^{2} = 8

. Since

9 > 8

, it follows that

3 > 22

. Therefore, the reciprocal

\frac{1}{3}

must be less than

\frac{1}{2 2}

. Rigor beats intuition every time.

Why does Statement 2 explain Statement 1? I thought they were just two separate facts.

Think of Statement 2 as the 'boundary condition' and Statement 1 as the 'existence proof'. Statement 2 establishes the range of the function, specifically its maximum. Statement 1 asks if a specific value exists within that range. Because the function is continuous, the Intermediate Value Theorem (IVT) guarantees that if a value lies between the minimum and maximum, the function must hit it. Without the range provided by Statement 2, you wouldn't know for sure if

1/3

is reachable. Thus, Statement 2 is the logical foundation for Statement 1.

Animated Solution for Mathematics - Differentiation: 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 $x \in R$ .

Select Answer:

Visualized Solution

Analyzing the Function $f (x)$

The AM-GM Inequality

Applying AM-GM to the Denominator

Minimizing the Denominator

Maximizing the Function $f (x)$

Verifying Statement-2

Analyzing Statement-1

Comparing $\frac{1}{3}$ and $\frac{1}{2 2}$

Establishing the Inequality

The Intermediate Value Theorem (IVT)

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Power of AM-GM

The Reciprocal Flip

The IVT Bridge

Similar Questions

Comprehension Passage

Which of the following is true for $0 < x < 1$ ?

If the function $e^{- x} f (x)$ assumes its minimum in the interval $[0, 1]$ at $x = 1/4$ , which of the following is true?

Let $a, b \in R$ be such that the function $f$ given by $f (x) = ln ∣ x ∣ + b x^{2} + a x, x \neq = 0$ has extreme values at $x = - 1$ and $x = 2$ . \\ \textbf{Statement-1:} $f$ has local maximum at $x = - 1$ and at $x = 2$ . \\ \textbf{Statement-2:} $a = 1/2$ and $b = - 1/4$ .

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

Comprehension Passage

Which of the following is true?

Which of the following is true?

Let $g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t$ . Which of the following is true?

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

Let $f (x) = ⎩ ⎨ ⎧ e^{x}, 2 - e^{x - 1}, x - e, 0 \leq x \leq 1 1 < x \leq 2 2 < x \leq 3$ and $g (x) = \int_{0}^{x} f (t) d t, x \in [1, 3]$ then $g (x)$ has

Let $f (x) = 3^{(x^{2} - 2)^{3} + 4}, x \in R$ . Then which of the following statements are true ? $P : x = 0$ is a point of local minima of $f$ $Q : x = 2$ is a point of inflection of $f$ $R : f^{'}$ is increasing for $x > 2$

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

Let $f : R \to R$ be a function defined by $f (x) = (x - 3)^{n_{1}} (x - 5)^{n_{2}}, n_{1}, n_{2} \in N$ . Then, which of the following is NOT true?

Select Answer:

Visualized Solution

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

The AM-GM Inequality

Applying AM-GM to the Denominator

Minimizing the Denominator

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

Verifying Statement-2

Analyzing Statement-1

Comparing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 31​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 22​1​

Establishing the Inequality

The Intermediate Value Theorem (IVT)

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Power of AM-GM

The Reciprocal Flip

The IVT Bridge

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following is true for 0<x<1?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function e−xf(x) assumes its minimum in the interval [0,1] at x=1/4, which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫0x2​et​(t−2)(t−3)dt for all x∈(0,∞), then

Comprehension Passage

Which of the following is true?

Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=∫0ex​1+t2f′(t)​dt. Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=⎩⎨⎧​ex,2−ex−1,x−e,​0≤x≤11<x≤22<x≤3​ and g(x)=∫0x​f(t)dt,x∈[1,3] then g(x) has

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f(x)=(x1​)2x;x>0 attains the maximum value at x=e1​ then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function defined by f(x)=(x−3)n1​(x−5)n2​,n1​,n2​∈N. Then, which of the following is NOT true?

Analyzing the Function $f (x)$

Maximizing the Function $f (x)$

Comparing $\frac{1}{3}$ and $\frac{1}{2 2}$

Which of the following is true for $0 < x < 1$ ?

If the function $e^{- x} f (x)$ assumes its minimum in the interval $[0, 1]$ at $x = 1/4$ , which of the following is true?

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

Let $g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t$ . Which of the following is true?

Let $f (x) = ⎩ ⎨ ⎧ e^{x}, 2 - e^{x - 1}, x - e, 0 \leq x \leq 1 1 < x \leq 2 2 < x \leq 3$ and $g (x) = \int_{0}^{x} f (t) d t, x \in [1, 3]$ then $g (x)$ has

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

Let $f : R \to R$ be a function defined by $f (x) = (x - 3)^{n_{1}} (x - 5)^{n_{2}}, n_{1}, n_{2} \in N$ . Then, which of the following is NOT true?