JEE Advanced 2014

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{1/ x}^{x} e^{- (t + 1/ t)} \frac{d t}{t}$ . Then

Q: Why is $f(1/x) = -f(x)$? Can't we just substitute?

You absolutely can! When you substitute $1/x$ into the integral, the limits become $\int_{x}^{1/x} g(t) dt$. Recall the fundamental property of definite integrals: swapping the limits flips the sign. Thus, $\int_{x}^{1/x} = -\int_{1/x}^{x}$. It's not magic; it's the geometric symmetry of the area!

Q: How do I know if $f(2^x)$ is an odd function?

An odd function satisfies $h(-x) = -h(x)$. Let $h(x) = f(2^x)$. Then $h(-x) = f(2^{-x}) = f(1/2^x)$. Using our previous result that $f(1/y) = -f(y)$, we get $f(1/2^x) = -f(2^x) = -h(x)$. The logic holds perfectly!

Select Answer:

* Multiple Correct

Visualized Solution

Understanding the Function $f (x)$

Given function: $f (x) = \int_{1/ x}^{x} \frac{e ^{- (t + 1/ t)}}{t} d t$ for $x > 0$ .
The integrand is $g (t) = \frac{e ^{- (t + 1/ t)}}{t}$ .
The limits of integration are functions of $x$ : lower limit $\frac{1}{x}$ and upper limit $x$ .

Visualizing the Integral

The integral represents the area under $g (t)$ from $t = \frac{1}{x}$ to $t = x$ .
As $x$ varies, the boundaries of this area shift.

Applying Newton-Leibniz Rule

To find monotonicity, we need the derivative $f^{'} (x)$ .
Using Newton-Leibniz Rule: $\frac{d}{d x} \int_{u (x)}^{v (x)} g (t) d t = g (v (x)) \cdot v^{'} (x) - g (u (x)) \cdot u^{'} (x)$

Differentiating $f (x)$

Upper limit: $v (x) = x ⟹ v^{'} (x) = 1$
Lower limit: $u (x) = \frac{1}{x} ⟹ u^{'} (x) = - \frac{1}{x ^{2}}$
$f^{'} (x) = g (x) \cdot (1) - g (\frac{1}{x}) \cdot (- \frac{1}{x ^{2}})$

Substituting into $g (t)$

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

Simplifying $f^{'} (x)$

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

Analyzing Monotonicity

For $x \in (0, \infty)$ , $x > 0 ⟹ \frac{2}{x} > 0$ .
The exponential function $e^{anything} > 0$ always.
Therefore, $f^{'} (x) > 0$ for all $x > 0$ .
Conclusion: $f (x)$ is monotonically increasing on $(0, \infty)$ , and hence on $[1, \infty)$ .

Checking the Functional Equation

Option C asks about $f (x) + f (1/ x) = 0$ .
Let's evaluate $f (1/ x)$ by substituting $1/ x$ into the original integral definition.
$f (1/ x) = \int_{1/ (1/ x)}^{1/ x} \frac{e ^{- (t + 1/ t)}}{t} d t$

Evaluating $f (1/ x)$

$f (1/ x) = \int_{x}^{1/ x} \frac{e ^{- (t + 1/ t)}}{t} d t$
Using the definite integral property: $\int_{a}^{b} h (t) d t = - \int_{b}^{a} h (t) d t$
$f (1/ x) = - \int_{1/ x}^{x} \frac{e ^{- (t + 1/ t)}}{t} d t = - f (x)$

Concluding Option C

We found $f (1/ x) = - f (x)$ .
Rearranging this gives: $f (x) + f (1/ x) = 0$ .
This holds true for all $x \in (0, \infty)$ .
Conclusion: Option C is correct.

Testing Parity of $f (2^{x})$

Option D asks if $h (x) = f (2^{x})$ is an odd function.
A function $h (x)$ is odd if $h (- x) = - h (x)$ .
Let's evaluate $h (- x)$ .

Evaluating $h (- x)$

$h (- x) = f (2^{- x}) = f (\frac{1}{2 ^{x}})$
From our previous result, we know $f (1/ y) = - f (y)$ for any $y > 0$ .
Let $y = 2^{x}$ . Then $f (\frac{1}{2 ^{x}}) = - f (2^{x})$ .
Therefore, $h (- x) = - h (x)$ .

Final Answer

$h (- x) = - h (x) ⟹ f (2^{x})$ is an odd function. Option D is correct.
Option A is correct ( $f (x)$ is monotonically increasing).
Option C is correct ( $f (x) + f (1/ x) = 0$ ).
Final Correct Options: A, C, D

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The Sigma Insight: Newton-Leibniz & Reduction Formulas

Solution Diagram

Analyzing the Setup

The function is defined as:

f (x) = \int_{1/ x}^{x} e^{- (t + 1/ t)} \frac{d t}{t}

At first glance, the integrand

g (t) = \frac{e ^{- (t + 1/ t)}}{t}

appears complex, and the limits are functions of

x

. However, in JEE Advanced, complexity is often a mask for elegance. Let us strip away that mask.

The Newton-Leibniz Revelation

To understand the monotonicity of

f (x)

, we must determine its derivative

f^{'} (x)

. We invoke the Newton-Leibniz rule:

\frac{d}{d x} \int_{u (x)}^{v (x)} g (t) d t = g (v (x)) \cdot v^{'} (x) - g (u (x)) \cdot u^{'} (x)

Here, our upper limit is

v (x) = x

(so

v^{'} (x) = 1

) and our lower limit is

u (x) = 1/ x

(so

u^{'} (x) = - 1/ x^{2}

). Substituting these into the formula, we obtain:

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

Substituting the integrand

g (t) = \frac{e ^{- (t + 1/ t)}}{t}

, the first term becomes

\frac{e ^{- (x + 1/ x)}}{x}

. The second term becomes:

\frac{e ^{- (1/ x + x)}}{1/ x} \cdot (- \frac{1}{x ^{2}})

Notice the beauty here: the exponents

x + 1/ x

and

1/ x + x

are identical. When we simplify the second term, the

1/ x

in the denominator flips to the numerator, and the negative signs cancel out. We are left with:

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

Since

x > 0

, this derivative is always positive. Therefore, the function is strictly increasing.

The Symmetry of the Integral

Now, let us evaluate the property

f (x) + f (1/ x) = 0

. If we define

f (x) = \int_{1/ x}^{x} g (t) d t

, then

f (1/ x)

is the integral with the limits swapped:

f (1/ x) = \int_{x}^{1/ x} g (t) d t

Recall the fundamental property of definite integrals:

\int_{a}^{b} g (t) d t = - \int_{b}^{a} g (t) d t

. By swapping the limits, we introduce a negative sign:

f (1/ x) = - \int_{1/ x}^{x} g (t) d t = - f (x)

Rearranging this yields

f (x) + f (1/ x) = 0

. This is a perfect, symmetric cancellation.

The Parity of the Composition

Finally, let us tackle the parity of

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

. To check if a function is odd, we must verify if

h (- x) = - h (x)

We compute:

h (- x) = f (2^{- x}) = f (1/ 2^{x})

Using our result from the previous section, we know that

f (1/ y) = - f (y)

. Letting

y = 2^{x}

, we find:

f (1/ 2^{x}) = - f (2^{x}) = - h (x)

Thus, the function

h (x)

is odd. We have successfully navigated the derivative, the symmetry, and the parity of the function.

Common Traps & FAQs

Wait, why do we reject the negative root here?

Actually, in this problem, we don't have roots to reject! The confusion often stems from the derivative. We found

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

. Since

x > 0

, the term

\frac{2}{x}

is positive, and the exponential term

e^{anything}

is always positive. There are no negative values to worry about here—the function is strictly increasing!

Why is

f (1/ x) = - f (x)

? Can't we just substitute?

You absolutely can! When you substitute

1/ x

into the integral, the limits become

\int_{x}^{1/ x} g (t) d t

. Recall the fundamental property of definite integrals: swapping the limits flips the sign. Thus,

\int_{x}^{1/ x} = - \int_{1/ x}^{x}

. It's not magic; it's the geometric symmetry of the area!

How do I know if

f (2^{x})

is an odd function?

An odd function satisfies

h (- x) = - h (x)

. Let

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

. Then

h (- x) = f (2^{- x}) = f (1/ 2^{x})

. Using our previous result that

f (1/ y) = - f (y)

, we get

f (1/ 2^{x}) = - f (2^{x}) = - h (x)

. The logic holds perfectly!

Animated Solution for Mathematics - Definite Integration: Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{1/ x}^{x} e^{- (t + 1/ t)} \frac{d t}{t}$ . Then

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Visualizing the Integral

Applying Newton-Leibniz Rule

Differentiating $f (x)$

Substituting into $g (t)$

Simplifying $f^{'} (x)$

Analyzing Monotonicity

Checking the Functional Equation

Evaluating $f (1/ x)$

Concluding Option C

Testing Parity of $f (2^{x})$

Evaluating $h (- x)$

Final Answer

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Newton-Leibniz Revelation

The Symmetry of the Integral

The Parity of the Composition

Similar Questions

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(0,∞)→R be given by f(x)=∫1/xx​e−(t+1/t)tdt​. Then

Select Answer:

Visualized Solution

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

Visualizing the Integral

Applying Newton-Leibniz Rule

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

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

Analyzing Monotonicity

Checking the Functional Equation

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

Concluding Option C

Testing Parity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2x)

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

Final Answer

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Newton-Leibniz Revelation

The Symmetry of the Integral

The Parity of the Composition

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Animated Solution for Mathematics - Definite Integration: Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{1/ x}^{x} e^{- (t + 1/ t)} \frac{d t}{t}$ . Then

Understanding the Function $f (x)$

Differentiating $f (x)$

Substituting into $g (t)$

Simplifying $f^{'} (x)$

Evaluating $f (1/ x)$

Testing Parity of $f (2^{x})$

Evaluating $h (- x)$

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If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$ . Then $f (\frac{π ^{2}}{4})$ is equal to

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If $f (x) = \int_{0}^{x} t (sin x - sin t) d t$ then :