The minimum value of the twice differentiable function $f(x) = \int_{0}^{x} e^{x-t} f'(t) dt - (x^2 - x + 1) e^x, x \in \mathbb{R}$, is : Options: (A) $-\frac{2}{\sqrt{e}}$ (B) $-2\sqrt{e}$ (C) $-\sqrt{e}$ (D) $\frac{2}{\sqrt{e}}$

### Analyzing the Setup The given integral equation is: $$f(x) = \int_{0}^{x} e^{x-t} f'(t) dt - (x^2 - x + 1) e^x$$ The presence of $e^{x-t}$ inside the integral is the primary obstacle. Using the laws of exponents, we rewrite $e^{x-t}$ as $e^x \cdot e^{-t}$. Since the integration is with respect to $t$, $e^x$ acts as a constant and can be factored out. The equation becomes: $$f(x) = e^x \int_{0}^{x} e^{-t} f'(t) dt - (x^2 - x + 1) e^x$$ To simplify the expression, we divide the entire equation by $e^x$: $$f(x)e^{-x} = \int_{0}^{x} e^{-t} f'(t) dt - (x^2 - x + 1)$$ ### The Magic of Leibniz To eliminate the integral, we apply the **Newton-Leibniz Rule** by differentiating both sides with respect to $x$. On the left side, we apply the product rule to $f(x)e^{-x}$: $$\frac{d}{dx}[f(x)e^{-x}] = f'(x)e^{-x} - f(x)e^{-x}$$ On the right side, the derivative of the integral $\int_{0}^{x} e^{-t} f'(t) dt$ is simply $e^{-x} f'(x)$. The derivative of the polynomial $-(x^2 - x + 1)$ is $-(2x - 1)$. Combining these, we obtain: $$f'(x)e^{-x} - f(x)e^{-x} = e^{-x} f'(x) - (2x - 1)$$ ### The Elegant Cancellation Observe that the term $f'(x)e^{-x}$ appears on both sides of the equation. They cancel out perfectly, leaving us with: $$-f(x)e^{-x} = -(2x - 1)$$ Multiplying both sides by $-e^x$, we successfully isolate the function: $$f(x) = (2x - 1) e^x$$ ### Final Optimization With $f(x)$ determined, we find the critical points by calculating the derivative $f'(x)$: $$f'(x) = \frac{d}{dx}[(2x - 1)e^x] = 2e^x + (2x - 1)e^x = (2x + 1)e^x$$ Setting $f'(x) = 0$ and noting that $e^x \neq 0$ for all real $x$, we solve $2x + 1 = 0$ to find the critical point: $$x = -\frac{1}{2}$$ Finally, we substitute $x = -\frac{1}{2}$ into $f(x)$ to determine the minimum value: $$f\left(-\frac{1}{2}\right) = \left(2\left(-\frac{1}{2}\right) - 1\right) e^{-1/2} = (-1 - 1) e^{-1/2} = -2 e^{-1/2}$$ The minimum value of the function is: $$\mathbf{-\frac{2}{\sqrt{e}}}$$

JEE Main 2022 (28 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: The minimum value of the twice differentiable function $f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ , is :

Q: Why can't we just integrate the right side directly to solve for $f(x)$?

That is a great question! The problem is that $f'(t)$ is unknown. We are not integrating a known function; we are trying to find the function $f(x)$ itself. This is an integral equation, not a definite integral. By differentiating, we transform this integral equation into a differential equation, which is a standard and powerful technique to solve for the unknown function.

Select Answer:

Progressive Hints

Hint 1

The integral equation

f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}

contains a function

f (x)

trapped inside an integral. Notice that the term

e^{x - t}

can be factored as

e^{x} \cdot e^{- t}

. Since the integration is with respect to

t

e^{x}

acts as a constant and can be pulled out of the integral. This is your first step to liberation.

Hint 2

Once you have isolated the integral, you will have an expression involving an integral from

0

x

. The most powerful tool to handle such integral equations in JEE Advanced is the Newton-Leibniz Rule. Before applying it, divide the entire equation by

e^{x}

to simplify the structure, making the differentiation process much cleaner.

Hint 3

After dividing by

e^{x}

, you will have

f (x) e^{- x} = \int_{0}^{x} e^{- t} f^{'} (t) d t - (x^{2} - x + 1)

. Now, differentiate both sides with respect to

x

. Remember to use the product rule on the left side and the Leibniz rule on the right side. You will find that the integral term cancels out, leaving you with a simple algebraic expression for

f (x)

Visualized Solution

Analyze the Integral Equation

Given equation: $f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}$
Our goal is to isolate $f (x)$ by removing the integral sign.
Notice the term $e^{x - t}$ inside the integral.

Simplify the Expression

Rewrite the integral: $f (x) = e^{x} \int_{0}^{x} e^{- t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}$
Divide both sides by $e^{x}$ to isolate the integral term:
$f (x) e^{- x} = \int_{0}^{x} e^{- t} f^{'} (t) d t - (x^{2} - x + 1)$

Differentiate using Leibniz Rule

Differentiating both sides with respect to $x$ :
LHS (Product Rule): $\frac{d}{d x} [f (x) e^{- x}] = f^{'} (x) e^{- x} - f (x) e^{- x}$
RHS (Leibniz Rule): $\frac{d}{d x} [\int_{0}^{x} e^{- t} f^{'} (t) d t] = e^{- x} f^{'} (x)$
RHS (Polynomial): $\frac{d}{d x} [x^{2} - x + 1] = 2 x - 1$

Simplify and Solve for $f (x)$

Equating the derivatives: $f^{'} (x) e^{- x} - f (x) e^{- x} = e^{- x} f^{'} (x) - (2 x - 1)$
Cancel $f^{'} (x) e^{- x}$ from both sides:
$- f (x) e^{- x} = - (2 x - 1)$
Multiply by $- e^{x}$ : $f (x) = (2 x - 1) e^{x}$

Find the Critical Point

To find the minimum, find $f^{'} (x)$ :
$f^{'} (x) = \frac{d}{d x} [(2 x - 1) e^{x}]$
$f^{'} (x) = 2 e^{x} + (2 x - 1) e^{x}$
$f^{'} (x) = (2 x + 1) e^{x}$

Solve for $x$

Set $f^{'} (x) = 0$ :
$(2 x + 1) e^{x} = 0$
Since $e^{x} \neq = 0$ for all $x \in R$ :
$2 x + 1 = 0 ⟹ x = - \frac{1}{2}$

Calculate Minimum Value

Substitute $x = - \frac{1}{2}$ into $f (x) = (2 x - 1) e^{x}$ :
$f (- \frac{1}{2}) = (2 (- \frac{1}{2}) - 1) e^{- 1/2}$
$f (- \frac{1}{2}) = (- 1 - 1) e^{- 1/2}$
$f (- \frac{1}{2}) = - 2 e^{- 1/2} = - \frac{2}{e}$

Conclusion and Key Takeaways

Key Takeaway: Use Leibniz Rule to convert integral equations into differential equations.
Final Answer: The minimum value is $- \frac{2}{e}$ .
Next Challenge: Try solving if the integral was from $0$ to $x^{2}$ instead of $0$ to $x$ .

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

The given integral equation is:

f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}

The presence of

e^{x - t}

inside the integral is the primary obstacle. Using the laws of exponents, we rewrite

e^{x - t}

e^{x} \cdot e^{- t}

. Since the integration is with respect to

t

e^{x}

acts as a constant and can be factored out.

The equation becomes:

f (x) = e^{x} \int_{0}^{x} e^{- t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}

To simplify the expression, we divide the entire equation by

e^{x}

f (x) e^{- x} = \int_{0}^{x} e^{- t} f^{'} (t) d t - (x^{2} - x + 1)

The Magic of Leibniz

To eliminate the integral, we apply the Newton-Leibniz Rule by differentiating both sides with respect to

x

On the left side, we apply the product rule to

f (x) e^{- x}

\frac{d}{d x} [f (x) e^{- x}] = f^{'} (x) e^{- x} - f (x) e^{- x}

On the right side, the derivative of the integral

\int_{0}^{x} e^{- t} f^{'} (t) d t

is simply

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

. The derivative of the polynomial

- (x^{2} - x + 1)

- (2 x - 1)

Combining these, we obtain:

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

The Elegant Cancellation

Observe that the term

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

appears on both sides of the equation. They cancel out perfectly, leaving us with:

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

Multiplying both sides by

- e^{x}

, we successfully isolate the function:

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

Final Optimization

With

f (x)

determined, we find the critical points by calculating the derivative

f^{'} (x)

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

Setting

f^{'} (x) = 0

and noting that $e^x eq 0$ for all real

x

, we solve

2 x + 1 = 0

to find the critical point:

x = - \frac{1}{2}

Finally, we substitute

x = - \frac{1}{2}

into

f (x)

to determine the minimum value:

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

The minimum value of the function is:

- \frac{2}{e}

Common Traps & FAQs

Why can't we just integrate the right side directly to solve for

f (x)

That is a great question! The problem is that

f^{'} (t)

is unknown. We are not integrating a known function; we are trying to find the function

f (x)

itself. This is an integral equation, not a definite integral. By differentiating, we transform this integral equation into a differential equation, which is a standard and powerful technique to solve for the unknown function.

Why did we divide by

e^{x}

before differentiating? Can't we just differentiate as is?

You absolutely could differentiate as is, but it would be much messier. By dividing by

e^{x}

first, you simplify the expression

f (x) e^{- x}

. When you differentiate this using the product rule, you get terms that perfectly cancel out the result of the Leibniz rule on the right side. It is a strategic move to make the algebra elegant and error-free.

How do we know

x = - 1/2

is a minimum and not a maximum?

Excellent observation! We found the critical point by setting

f^{'} (x) = 0

. To confirm it is a minimum, you can look at the first derivative

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

. For

x < - 1/2

f^{'} (x)

is negative (the function is decreasing), and for

x > - 1/2

f^{'} (x)

is positive (the function is increasing). Since the function changes from decreasing to increasing at

x = - 1/2

, it must be a local minimum.

Animated Solution for Mathematics - Differentiation: The minimum value of the twice differentiable function $f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ , is :

Select Answer:

Visualized Solution

Analyze the Integral Equation

Simplify the Expression

Differentiate using Leibniz Rule

Simplify and Solve for $f (x)$

Find the Critical Point

Solve for $x$

Calculate Minimum Value

Conclusion and Key Takeaways

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Magic of Leibniz

The Elegant Cancellation

Final Optimization

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?

The sum of all the local minimum values of the twice differentiable function $f : R \to R$ defined by $f (x) = x^{3} - 3 x^{2} - \frac{3 f ^{''} ( 2 )}{2} x + f^{''} (1)$ is :

The function $f (x) = \int_{- 1}^{x} t (e^{t} - 1) (t - 1) (t - 2)^{3} (t - 3)^{5} d t$ has a local minimum at $x =$

Let $(2 α, α)$ be the largest interval in which the function $f (t) = \frac{∣ t + 1∣}{t ^{2}}, t < 0$ , is strictly decreasing. Then the local maximum value of the function $g (x) = 2 lo g_{e} (x - 2) + α x^{2} + 4 x - α, x > 2$ , is .........

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

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The local maximum value of the function $f (x) = (\frac{2}{x})^{x^{2}}, x > 0$ , is

Let $f (x) = x^{2} + \frac{1}{x ^{2}}$ and $g (x) = x - \frac{1}{x}, x \in R - {- 1, 0, 1}$ . If $h (x) = \frac{f ( x )}{g ( x )}$ , then the local minimum value of $h (x)$ is :

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?

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The minimum value of the twice differentiable function f(x)=∫0x​ex−tf′(t)dt−(x2−x+1)ex,x∈R, is :

Select Answer:

Visualized Solution

Analyze the Integral Equation

Simplify the Expression

Differentiate using Leibniz Rule

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

Find the Critical Point

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Calculate Minimum Value

Conclusion and Key Takeaways

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Magic of Leibniz

The Elegant Cancellation

Final Optimization

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?

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.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

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Comprehension Passage

Which of the following is true?

Which of the following is true?

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Animated Solution for Mathematics - Differentiation: The minimum value of the twice differentiable function $f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ , is :

Simplify and Solve for $f (x)$

Solve for $x$

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?

The sum of all the local minimum values of the twice differentiable function $f : R \to R$ defined by $f (x) = x^{3} - 3 x^{2} - \frac{3 f ^{''} ( 2 )}{2} x + f^{''} (1)$ is :

The function $f (x) = \int_{- 1}^{x} t (e^{t} - 1) (t - 1) (t - 2)^{3} (t - 3)^{5} d t$ has a local minimum at $x =$

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

If $f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}$ , for every real number $x$ , then the minimum value of $f$

The local maximum value of the function $f (x) = (\frac{2}{x})^{x^{2}}, x > 0$ , is

Let $f (x) = x^{2} + \frac{1}{x ^{2}}$ and $g (x) = x - \frac{1}{x}, x \in R - {- 1, 0, 1}$ . If $h (x) = \frac{f ( x )}{g ( x )}$ , then the local minimum value of $h (x)$ is :

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