JEE Advanced 2013

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f : [1/2, 1] \to R$ be a positive, non-constant and differentiable function such that $f^{'} (x) < 2 f (x)$ and $f (1/2) = 1$ . Then the value of $\int_{1/2}^{1} f (x) d x$ lies in the interval

Q: Why is $g(x)$ strictly decreasing? Does that mean $f(x)$ is also decreasing?

Not necessarily! $g(x) = e^{-2x}f(x)$ is decreasing, but $f(x) = g(x)e^{2x}$. Since $e^{2x}$ is increasing, $f(x)$ is the product of a decreasing function and an increasing function. We cannot immediately conclude the behavior of $f(x)$ just from $g(x)$, but we don't need to! We only need the bound provided by $g(x)$ to constrain the integral.

Q: Why can't we just use the mean value theorem or some other approximation?

You could try, but the inequality $f'(x) < 2f(x)$ is a global constraint that holds for all $x$ in the interval. The integrating factor method is the most robust way to convert this local derivative constraint into a global bound on the function's values. It is the 'JEE way' to ensure you capture the entire behavior of the function.

Q: Why is the interval $(0, \frac{e-1}{2})$ and not $[0, \frac{e-1}{2}]$?

The problem states that $f(x)$ is a non-constant function. If $f(x)$ were constant, the integral could potentially reach the boundary values. Because it is non-constant and strictly bounded by the exponential function derived from the inequality, the integral must be strictly less than the upper bound and strictly greater than the lower bound (since $f(x) > 0$).

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Look at the inequality

f^{'} (x) < 2 f (x)

. This is not just a random expression; it is a classic setup for a linear differential inequality. Think about how you would solve a linear differential equation of the form

\frac{d y}{d x} + P y = Q

. What tool do you use to force the left side into a product rule derivative?

Hint 2

Hint 2: The Arsenal. Once you multiply the inequality by the integrating factor

e^{- 2 x}

, you get

\frac{d}{d x} (e^{- 2 x} f (x)) < 0

. This tells you that the function

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

is strictly decreasing. How does this monotonicity help you bound

f (x)

on the interval

[1/2, 1]

Hint 3

Hint 3: The Execution. Since

g (x)

is decreasing, for any

x > 1/2

, we must have

g (x) < g (1/2)

. Substitute

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

and use the initial condition

f (1/2) = 1

to find an explicit upper bound for

f (x)

in terms of

x

. Then, simply integrate this bound from

1/2

1

Visualized Solution

Initial Conditions

Given function $f : [\frac{1}{2}, 1] \to R$ is positive, non-constant, and differentiable.
Initial condition: $f (\frac{1}{2}) = 1$ .

Differential Inequality

$f^{'} (x) < 2 f (x) ⟹ f^{'} (x) - 2 f (x) < 0$

Integrating Factor

This resembles $\frac{d y}{d x} + P y < Q$ .
Integrating Factor (I.F.) $= e^{\int - 2 d x} = e^{- 2 x}$ .

Exact Derivative Form

Multiply by $e^{- 2 x}$ : $e^{- 2 x} f^{'} (x) - 2 e^{- 2 x} f (x) < 0$ .
This is exactly $\frac{d}{d x} (e^{- 2 x} f (x)) < 0$ .

Decreasing Function

Let $g (x) = e^{- 2 x} f (x)$ .
Since $g^{'} (x) < 0$ , $g (x)$ is strictly decreasing on $[\frac{1}{2}, 1]$ .

Bounding $g (x)$

For $x > \frac{1}{2}$ , $g (x) < g (\frac{1}{2})$ .

Evaluating $g (\frac{1}{2})$

$g (\frac{1}{2}) = e^{- 2 (\frac{1}{2})} f (\frac{1}{2}) = e^{- 1} \cdot 1 = \frac{1}{e}$ .

Upper Bound of $f (x)$

$e^{- 2 x} f (x) < \frac{1}{e} ⟹ f (x) < e^{2 x - 1}$ .

Integral Bounds

Integrate from $\frac{1}{2}$ to $1$ : $\int_{1/2}^{1} f (x) d x < \int_{1/2}^{1} e^{2 x - 1} d x$ .
Also, $f (x) > 0 ⟹ \int_{1/2}^{1} f (x) d x > 0$ .

Calculating Maximum Area

$\int_{1/2}^{1} e^{2 x - 1} d x = [\frac{e ^{2 x - 1}}{2}]_{1/2}^{1}$ .

Final Interval

$\frac{e ^{2 (1) - 1}}{2} - \frac{e ^{2 (\frac{1}{2}) - 1}}{2} = \frac{e ^{1}}{2} - \frac{e ^{0}}{2} = \frac{e - 1}{2}$ .
Thus, $\int_{1/2}^{1} f (x) d x \in (0, \frac{e - 1}{2})$ .

00:00 / 00:00

The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Setup

We are given a function

f : [1/2, 1] \to R

that is positive, non-constant, and differentiable. We are provided with the initial condition

f (1/2) = 1

and the governing differential inequality

f^{'} (x) < 2 f (x)

This inequality acts as a constraint on the growth rate of the function. Our goal is to determine the bounds for the integral of

f (x)

over the interval

[1/2, 1]

The Art of the Integrating Factor

To solve the inequality

f^{'} (x) - 2 f (x) < 0

, we employ the method of the integrating factor. We define the integrating factor as:

I . F . = e^{\int - 2 d x} = e^{- 2 x}

Multiplying the entire inequality by

e^{- 2 x}

, we obtain:

e^{- 2 x} f^{'} (x) - 2 e^{- 2 x} f (x) < 0

This expression is the derivative of the product

e^{- 2 x} f (x)

. Thus, we can rewrite the inequality as:

\frac{d}{d x} (e^{- 2 x} f (x)) < 0

The Power of Monotonicity

Let

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

. Since the derivative of

g (x)

is strictly negative,

g (x)

is a strictly decreasing function on the interval

[1/2, 1]

For any

x \in [1/2, 1]

, it must hold that

g (x) < g (1/2)

. We calculate the value at the boundary:

g (1/2) = e^{- 2 (1/2)} f (1/2) = e^{- 1} \cdot 1 = \frac{1}{e}

Substituting back for

f (x)

, we find the upper bound:

e^{- 2 x} f (x) < \frac{1}{e} ⟹ f (x) < e^{2 x - 1}

The Final Integration

Given that

f (x) > 0

and

f (x) < e^{2 x - 1}

, the integral of

f (x)

over the interval

[1/2, 1]

is bounded by:

0 < \int_{1/2}^{1} f (x) d x < \int_{1/2}^{1} e^{2 x - 1} d x

We evaluate the upper bound integral as follows:

\int_{1/2}^{1} e^{2 x - 1} d x = [\frac{e ^{2 x - 1}}{2}]_{1/2}^{1}

\frac{e ^{2 (1) - 1}}{2} - \frac{e ^{2 (1/2) - 1}}{2} = \frac{e ^{1}}{2} - \frac{e ^{0}}{2} = \frac{e - 1}{2}

The integral is therefore trapped in the interval $(0, \frac{e - 1}{2})$ .

Common Traps & FAQs

Wait, why do we multiply by

e^{- 2 x}

? How did we know that was the integrating factor?

Great question! This is a standard technique for linear differential equations of the form

f^{'} (x) + P (x) f (x) = Q (x)

. By multiplying by the integrating factor

e^{\int P (x) d x}

, we force the left side to become the derivative of a product, specifically

\frac{d}{d x} (f (x) \cdot e^{\int P (x) d x})

. Here,

P (x) = - 2

, so the factor is

e^{\int - 2 d x} = e^{- 2 x}

. It is the 'magic key' that unlocks the derivative!

Why is

g (x)

strictly decreasing? Does that mean

f (x)

is also decreasing?

Not necessarily!

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

is decreasing, but

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

. Since

e^{2 x}

is increasing,

f (x)

is the product of a decreasing function and an increasing function. We cannot immediately conclude the behavior of

f (x)

just from

g (x)

, but we don't need to! We only need the bound provided by

g (x)

to constrain the integral.

Why can't we just use the mean value theorem or some other approximation?

You could try, but the inequality

f^{'} (x) < 2 f (x)

is a global constraint that holds for all

x

in the interval. The integrating factor method is the most robust way to convert this local derivative constraint into a global bound on the function's values. It is the 'JEE way' to ensure you capture the entire behavior of the function.

Why is the interval

(0, \frac{e - 1}{2})

and not

[0, \frac{e - 1}{2}]

The problem states that

f (x)

is a non-constant function. If

f (x)

were constant, the integral could potentially reach the boundary values. Because it is non-constant and strictly bounded by the exponential function derived from the inequality, the integral must be strictly less than the upper bound and strictly greater than the lower bound (since

f (x) > 0

Is there a shortcut? Can I just guess a function like

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

Testing a boundary function is a brilliant strategy for multiple-choice questions! If you assume

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

, you satisfy

f (1/2) = 1

and

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

. Since the problem says

f^{'} (x) < 2 f (x)

, the actual integral must be smaller than the integral of this boundary function. It is a perfect way to verify your result!

Animated Solution for Mathematics - Differentiation: Let $f : [1/2, 1] \to R$ be a positive, non-constant and differentiable function such that $f^{'} (x) < 2 f (x)$ and $f (1/2) = 1$ . Then the value of $\int_{1/2}^{1} f (x) d x$ lies in the interval

Select Answer:

Visualized Solution

Initial Conditions

Differential Inequality

Integrating Factor

Exact Derivative Form

Decreasing Function

Bounding $g (x)$

Evaluating $g (\frac{1}{2})$

Upper Bound of $f (x)$

Integral Bounds

Calculating Maximum Area

Final Interval

The Sigma Insight: Monotonicity

Analyzing the Setup

The Art of the Integrating Factor

The Power of Monotonicity

The Final Integration

Similar Questions

If $f : R \to R$ is a differentiable function such that $f^{'} (x) > 2 f (x)$ for all $x \in R$ , and $f (0) = 1$ , then

If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in

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

Let $g : (0, \infty) \to R$ be a differentiable function such that $\int (\frac{x ( c o s x - s i n x )}{e ^{x} + 1} + \frac{g ( x ) ( e ^{x} + 1 - x e ^{x} )}{( e ^{x} + 1 ) ^{2}}) d x = \frac{xg ( x )}{e ^{x} + 1} + c$ , for all $x > 0$ , where $c$ is an arbitrary constant. Then

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Let $f : (- 1, \infty) \to R$ be defined by $f (0) = 1$ and $f (x) = \frac{1}{x} lo g_{e} (1 + x)$ , $x \neq = 0$ . Then the function $f$ :

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Comprehension Passage

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 $2 f (x) + 1 = 2 x (1 + x)$

Which of the following is true?

Select Answer:

Visualized Solution

Initial Conditions

Differential Inequality

Integrating Factor

Exact Derivative Form

Decreasing Function

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

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(21​)

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

Integral Bounds

Calculating Maximum Area

Final Interval

The Sigma Insight: Monotonicity

Analyzing the Setup

The Art of the Integrating Factor

The Power of Monotonicity

The Final Integration

Similar Questions

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Comprehension Passage

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Upper Bound of $f (x)$

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If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in

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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

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 $2 f (x) + 1 = 2 x (1 + x)$