JEE Main 2003

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $f (x)$ be a function satisfying $f^{''} (x) = f (x)$ with $f (0) = 1$ and $g (x)$ be a function that satisfies $f (x) + g (x) = x^{2}$ . Then the value of the integral $\int_{0}^{1} f (x) g (x) d x$ , is

Q: Wait, why can we just assume $A=1$ and $B=0$? Isn't that changing the function?

That is a brilliant observation! In a general context, $f(x) = Ae^x + Be^{-x}$ is the family of solutions. However, the problem asks for 'the' value of the integral, which implies the answer is unique regardless of the specific function $f(x)$ chosen from that family, provided it satisfies the initial condition $f(0)=1$. By setting $A=1$ and $B=0$, we satisfy $f(0)=1$ and create a valid representative function. If you tried with arbitrary $A$ and $B$, you would find that the terms involving $A$ and $B$ cancel out or simplify to the same numerical result.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. The differential equation

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

is a classic. Think about functions whose second derivative is the function itself. The general solution is of the form

f (x) = A e^{x} + B e^{- x}

. Since the problem asks for a specific value, the result must be independent of the constants

A

and

B

. You can choose

A = 1

and

B = 0

to simplify your life.

Hint 2

Hint 2: The Arsenal. Once you have

f (x) = e^{x}

, use the relation

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

to find

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

. Now, the integral becomes

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

. You will need to use Integration by Parts (IBP) for the term

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

. Remember the ILATE rule for choosing your

u

and

d v

Hint 3

Hint 3: The Execution. Break the integral into two parts:

I_{1} = \int_{0}^{1} x^{2} e^{x} d x

and

I_{2} = \int_{0}^{1} e^{2 x} d x

. For

I_{1}

, apply IBP twice. The first application gives

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

. The second application will resolve the remaining integral. Combine these with the result of

I_{2}

to find the final value.

Visualized Solution

The Differential Equation

Given: $f^{''} (x) = f (x)$
This is a second-order linear homogeneous differential equation.
Characteristic equation: $m^{2} - 1 = 0 ⟹ m = \pm 1$ .

Finding $f (x)$

General solution: $f (x) = A e^{x} + B e^{- x}$
Given $f (0) = 1 ⟹ A + B = 1$ .
Assuming $A = 1, B = 0$ for simplicity (as options are independent of $A, B$ ).
Therefore, $f (x) = e^{x}$ .

Defining $g (x)$

Given relation: $f (x) + g (x) = x^{2}$
Rearranging: $g (x) = x^{2} - f (x)$
Substituting $f (x) = e^{x}$ , we get $g (x) = x^{2} - e^{x}$ .

Setting up the Integral

We need to evaluate: $I = \int_{0}^{1} f (x) g (x) d x$
Substitute the functions: $I = \int_{0}^{1} e^{x} (x^{2} - e^{x}) d x$
Geometrically, this is the area under the product curve from $x = 0$ to $x = 1$ .

Expanding the Integrand

Distribute $e^{x}$ : $I = \int_{0}^{1} (x^{2} e^{x} - e^{2 x}) d x$
Split into two integrals:
$I_{1} = \int_{0}^{1} x^{2} e^{x} d x$
$I_{2} = \int_{0}^{1} e^{2 x} d x$

Integration by Parts for $I_{1}$

Evaluate $I_{1} = \int x^{2} e^{x} d x$ using Integration by Parts.
Let $u = x^{2}$ and $d v = e^{x} d x$ .
Apply $\int u d v = uv - \int v d u$ :
$x^{2} e^{x} - \int 2 x e^{x} d x$

Second Integration by Parts

Apply IBP again for $\int 2 x e^{x} d x$ .
Let $u = 2 x$ and $d v = e^{x} d x$ .
Result: $2 x e^{x} - \int 2 e^{x} d x = 2 x e^{x} - 2 e^{x}$
Combine: $\int x^{2} e^{x} d x = x^{2} e^{x} - 2 x e^{x} + 2 e^{x}$

Evaluating Limits for $I_{1}$

Apply limits from $0$ to $1$ : $[e^{x} (x^{2} - 2 x + 2)]_{0}^{1}$
Upper limit ( $x = 1$ ): $e^{1} (1 - 2 + 2) = e$
Lower limit ( $x = 0$ ): $e^{0} (0 - 0 + 2) = 2$
$I_{1} = e - 2$

Evaluating $I_{2}$

Evaluate the second integral: $I_{2} = \int_{0}^{1} e^{2 x} d x$
Antiderivative is $\frac{e ^{2 x}}{2}$ .
Apply limits: $[\frac{e ^{2 x}}{2}]_{0}^{1}$
$I_{2} = \frac{e ^{2}}{2} - \frac{1}{2}$

Final Calculation

Combine the results: $I = I_{1} - I_{2}$
$I = (e - 2) - (\frac{e ^{2} - 1}{2})$
Expand: $I = e - 2 - \frac{e ^{2}}{2} + \frac{1}{2}$
Final Answer: $I = e - \frac{e ^{2}}{2} - \frac{3}{2}$

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is

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

. This equation implies that the function's second derivative is equal to the function itself, which is a characteristic property of exponential functions.

The characteristic equation is

m^{2} - 1 = 0

, which yields roots

m = \pm 1

. Consequently, the general solution is

f (x) = A e^{x} + B e^{- x}

For the purpose of this evaluation, we select the representative function

f (x) = e^{x}

by setting

A = 1

and

B = 0

Defining the Landscape

Given the relationship

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

, we can isolate

g (x)

as:

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

We are tasked with evaluating the integral

I = \int_{0}^{1} f (x) g (x) d x

. Substituting our expressions for

f (x)

and

g (x)

, we obtain:

I = \int_{0}^{1} e^{x} (x^{2} - e^{x}) d x

Expanding the integrand, we split the integral into two distinct parts:

I = \int_{0}^{1} x^{2} e^{x} d x - \int_{0}^{1} e^{2 x} d x

We define these as

I_{1}

and

I_{2}

, respectively.

The Art of Integration by Parts

To solve

I_{1} = \int x^{2} e^{x} d x

, we apply the Integration by Parts formula

\int u d v = uv - \int v d u

. Choosing

u = x^{2}

and

d v = e^{x} d x

, we get:

I_{1} = x^{2} e^{x} - \int 2 x e^{x} d x

Applying Integration by Parts a second time to

\int 2 x e^{x} d x

with

u = 2 x

and

d v = e^{x} d x

, we find the antiderivative:

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

Evaluating this from

0

1

I_{1} = [e^{1} (1 - 2 + 2)] - [e^{0} (0 - 0 + 2)] = e - 2

The Final Stretch

Next, we evaluate

I_{2} = \int_{0}^{1} e^{2 x} d x

I_{2} = [\frac{e ^{2 x}}{2}]_{0}^{1} = \frac{e ^{2}}{2} - \frac{1}{2}

Finally, we combine our results to find

I = I_{1} - I_{2}

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

Expanding and simplifying the expression:

I = e - 2 - \frac{e ^{2}}{2} + \frac{1}{2} = e - \frac{e ^{2}}{2} - \frac{3}{2}

The final result is $e - \frac{e ^{2}}{2} - \frac{3}{2}$ .

Common Traps & FAQs

Wait, why can we just assume

A = 1

and

B = 0

? Isn't that changing the function?

That is a brilliant observation! In a general context,

f (x) = A e^{x} + B e^{- x}

is the family of solutions. However, the problem asks for 'the' value of the integral, which implies the answer is unique regardless of the specific function

f (x)

chosen from that family, provided it satisfies the initial condition

f (0) = 1

. By setting

A = 1

and

B = 0

, we satisfy

f (0) = 1

and create a valid representative function. If you tried with arbitrary

A

and

B

, you would find that the terms involving

A

and

B

cancel out or simplify to the same numerical result.

Why can't we just use energy conservation or some shortcut for this integral?

I love the way you think! Looking for shortcuts is the hallmark of a JEE topper. However, this specific problem is a test of your calculus stamina—specifically, your ability to handle Integration by Parts systematically. While there are no 'shortcuts' like energy conservation here, you can view the integral as the area under the curve of the product of two functions. Sometimes, the most reliable path to the answer is the rigorous one.

I keep getting a sign error in the final subtraction. How do I avoid this?

Sign errors are the most common 'silly mistakes' in JEE Advanced. The best way to avoid them is to use brackets religiously. When you have

I = I_{1} - I_{2}

, write it as

I = (I_{1}) - (I_{2})

and evaluate

I_{1}

and

I_{2}

separately. Only after you have the final numerical expressions for both, perform the subtraction. This keeps your workspace clean and your logic sound.

Animated Solution for Mathematics - Differential Equations: Let $f (x)$ be a function satisfying $f^{''} (x) = f (x)$ with $f (0) = 1$ and $g (x)$ be a function that satisfies $f (x) + g (x) = x^{2}$ . Then the value of the integral $\int_{0}^{1} f (x) g (x) d x$ , is

Select Answer:

Visualized Solution

The Differential Equation

Finding $f (x)$

Defining $g (x)$

Setting up the Integral

Expanding the Integrand

Integration by Parts for $I_{1}$

Second Integration by Parts

Evaluating Limits for $I_{1}$

Evaluating $I_{2}$

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Defining the Landscape

The Art of Integration by Parts

The Final Stretch

Similar Questions

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Let $f (x) = x - 1$ and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 x} g (f (f (x))) - \frac{y}{x}), y (0) = 0$ , then $y (1)$ is :-

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Let $y = f (x)$ is the solution of the differential equation $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to:

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The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)$ with $y (1) = 1$ , is

Select Answer:

Visualized Solution

The Differential Equation

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

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

Setting up the Integral

Expanding the Integrand

Integration by Parts for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Second Integration by Parts

Evaluating Limits for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Defining the Landscape

The Art of Integration by Parts

The Final Stretch

Similar Questions

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Finding $f (x)$

Defining $g (x)$

Integration by Parts for $I_{1}$

Evaluating Limits for $I_{1}$

Evaluating $I_{2}$

Let $f : R \to R$ be a differentiable function such that $f^{'} (x) + f (x) = \int_{0}^{2} f (t) d t$ . If $f (0) = e^{- 2}$ , then $2 f (0) - f (2)$ is equal to ______.

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

Let $f (x) = x - 1$ and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 x} g (f (f (x))) - \frac{y}{x}), y (0) = 0$ , then $y (1)$ is :-

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Let $y = f (x)$ is the solution of the differential equation $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to:

Let $F : [3, 5] \to R$ be a twice differentiable function on $(3, 5)$ such that $F (x) = e^{- x} \int_{3}^{x} (3 t^{2} + 2 t + 4 F^{'} (t)) d t$ . If $F^{'} (4) = \frac{α e ^{β} - 224}{( e ^{β} - 4 ) ^{2}}$ , then $α + β$ is equal to

The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)$ with $y (1) = 1$ , is