JEE Main 2021 (27 August Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let us consider a curve, $y = f (x)$ passing through the point $(- 2, 2)$ and the slope of the tangent to the curve at any point $(x, f (x))$ is given by $f (x) + x f^{'} (x) = x^{2}$ . Then :

Q: How do I know when to use the Product Rule in reverse?

Look for the structure $f(x) + x f'(x)$. Whenever you see a function and its derivative multiplied by the independent variable, your brain should immediately scream 'Product Rule!' It is a classic pattern that examiners love to hide in plain sight.

Q: Do I really need the constant $C$?

Absolutely! Without $C$, you are describing an infinite family of curves. The point $(-2, 2)$ is your 'anchor'—it selects the one specific curve out of that infinite family that satisfies the problem. Forgetting $C$ is the most common way to lose marks in calculus problems.

Select Answer:

Visualized Solution

The Given Curve and Point

Let the curve be $y = f (x)$ .
It passes through the point $P (- 2, 2)$ .

The Differential Equation

The slope of the tangent at any point $(x, f (x))$ is $f^{'} (x)$ .
Given relation: $f (x) + x f^{'} (x) = x^{2}$ .

Recognizing the Pattern

Look closely at the left-hand side: $f (x) + x f^{'} (x)$ .
Recall the Product Rule of differentiation: $\frac{d}{d x} (u \cdot v) = u \frac{d v}{d x} + v \frac{d u}{d x}$ .

Applying the Product Rule

Let $u = x$ and $v = f (x)$ .
$\frac{d}{d x} (x \cdot f (x)) = x \cdot f^{'} (x) + f (x) \cdot (1)$ .
This perfectly matches our left-hand side!

Rewriting the Equation

Replace the LHS with the exact derivative.
$\frac{d}{d x} (x f (x)) = x^{2}$ .

Integrating Both Sides

To eliminate the derivative, we integrate both sides with respect to $x$ .
$\int \frac{d}{d x} (x f (x)) d x = \int x^{2} d x$ .

Performing the Integration

The integral of a derivative gives the original function: $x f (x)$ .
The integral of $x^{2}$ is $\frac{x ^{3}}{3}$ .
Don't forget the constant of integration, $C$ !
$x f (x) = \frac{x ^{3}}{3} + C$ .

Using the Initial Condition

We need to find the value of $C$ .
We know the curve passes through $P (- 2, 2)$ .
Substitute $x = - 2$ and $f (x) = 2$ into the equation.

Substituting the Values

$(- 2) (2) = \frac{( - 2 ) ^{3}}{3} + C$ .
$- 4 = \frac{- 8}{3} + C$ .

Calculating the Constant $C$

Isolate $C$ : $C = - 4 + \frac{8}{3}$ .
Find a common denominator: $C = \frac{- 12 + 8}{3}$ .
$C = - \frac{4}{3}$ .

Forming the Specific Equation

Substitute $C = - \frac{4}{3}$ back into the integrated equation.
$x f (x) = \frac{x ^{3}}{3} - \frac{4}{3}$ .

Simplifying the Equation

Notice the common denominator of $3$ on the right side.
Multiply the entire equation by $3$ to eliminate fractions.
$3 x f (x) = x^{3} - 4$ .

Final Rearrangement

Rearrange the terms to match the given options.
Move $3 x f (x)$ to the right side: $x^{3} - 3 x f (x) - 4 = 0$ .
This matches Option 3 perfectly!

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

We are tasked with solving the differential equation

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

for a curve

y = f (x)

that passes through the anchor point

(- 2, 2)

. This point serves as our boundary condition to determine the specific member of the family of curves.

The Master Equation

At first glance, the left-hand side of the equation

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

may appear complex. However, it is a direct application of the Product Rule of differentiation.

Recall that for two functions

u

and

v

, the derivative is given by:

\frac{d}{d x} (u \cdot v) = u \frac{d v}{d x} + v \frac{d u}{d x}

If we set

u = x

and

v = f (x)

, the derivative of

x \cdot f (x)

becomes:

\frac{d}{d x} (x \cdot f (x)) = x \cdot f^{'} (x) + f (x) \cdot 1

By recognizing this pattern, we can rewrite the differential equation in a much more manageable form:

\frac{d}{d x} (x f (x)) = x^{2}

Integration and the Constant

To isolate

f (x)

, we perform integration on both sides with respect to

x

\int \frac{d}{d x} (x f (x)) d x = \int x^{2} d x

The left side simplifies to

x f (x)

, while the right side yields the power rule result plus the constant of integration

C

x f (x) = \frac{x ^{3}}{3} + C

Determining the Constant

We now apply our anchor point

(- 2, 2)

by substituting

x = - 2

and

f (x) = 2

into the equation:

(- 2) (2) = \frac{( - 2 ) ^{3}}{3} + C

This simplifies to:

- 4 = - \frac{8}{3} + C

Solving for

C

C = - 4 + \frac{8}{3} = \frac{- 12 + 8}{3} = - \frac{4}{3}

Final Calculation

Substituting

C

back into our general solution, we obtain:

x f (x) = \frac{x ^{3}}{3} - \frac{4}{3}

To express this in a standard form, we multiply the entire equation by

3

3 x f (x) = x^{3} - 4

Rearranging the terms, we arrive at the final result: $x^{3} - 3 x f (x) - 4 = 0$

Common Traps & FAQs

Why can't I just separate the variables like

\frac{d y}{d x} = \frac{x ^{2} - f ( x )}{x}

You absolutely could! That would lead you to a linear differential equation of the form

\frac{d y}{d x} + \frac{1}{x} y = x

. While that is a valid path, recognizing the product rule is the 'JEE shortcut' that saves you precious minutes and reduces the chance of algebraic errors. In the exam, speed and elegance are your best friends.

How do I know when to use the Product Rule in reverse?

Look for the structure

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

. Whenever you see a function and its derivative multiplied by the independent variable, your brain should immediately scream 'Product Rule!' It is a classic pattern that examiners love to hide in plain sight.

Do I really need the constant

C

Absolutely! Without

C

, you are describing an infinite family of curves. The point

(- 2, 2)

is your 'anchor'—it selects the one specific curve out of that infinite family that satisfies the problem. Forgetting

C

is the most common way to lose marks in calculus problems.

Animated Solution for Mathematics - Differential Equations: Let us consider a curve, $y = f (x)$ passing through the point $(- 2, 2)$ and the slope of the tangent to the curve at any point $(x, f (x))$ is given by $f (x) + x f^{'} (x) = x^{2}$ . Then :

Select Answer:

Visualized Solution

The Given Curve and Point

The Differential Equation

Recognizing the Pattern

Applying the Product Rule

Rewriting the Equation

Integrating Both Sides

Performing the Integration

Using the Initial Condition

Substituting the Values

Calculating the Constant $C$

Forming the Specific Equation

Simplifying the Equation

Final Rearrangement

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

Integration and the Constant

Determining the Constant

Final Calculation

Similar Questions

If a curve passes through the point $(1, - 2)$ and has slope of the tangent at any point $(x, y)$ on it as $\frac{x ^{2} - 2 y}{x}$ , then the curve also passes through the point :

Let slope of the tangent line to a curve at any point P(x, y) be given by $\frac{x y ^{2} + y}{x}$ . If the curve intersects the line $x + 2 y = 4$ at $x = - 2$ , then the value of y, for which the point (3, y) lies on the curve, is :

Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $lim_{t \to x} \frac{t ^{2} y ( x ) - x ^{2} y ( t )}{x - t} = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to

Let $y^{'} (x) + y (x) g^{'} (x) = g (x) g^{'} (x)$ , $y (0) = 0$ , $x \in R$ , where $f^{'} (x)$ denotes $\frac{df ( x )}{d x}$ and $g (x)$ is a given non-constant differentiable function on $R$ with $g (0) = g (2) = 0$ . Then the value of $y (2)$ is

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If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is $\frac{x ^{2} - 4 x + y + 8}{x - 2}$ , then this curve also passes through the point :

A function $y = f (x)$ satisfies $f (x) sin 2 x + sin x - (1 + cos^{2} x) f^{'} (x) = 0$ with condition $f (0) = 0$ . Then $f (\frac{π}{2})$ is equal to

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

Let $y = y (x)$ be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x > 1$ , with $y (2) = - 2$ . Then $y (3)$ is equal to

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

Select Answer:

Visualized Solution

The Given Curve and Point

The Differential Equation

Recognizing the Pattern

Applying the Product Rule

Rewriting the Equation

Integrating Both Sides

Performing the Integration

Using the Initial Condition

Substituting the Values

Calculating the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Forming the Specific Equation

Simplifying the Equation

Final Rearrangement

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

Integration and the Constant

Determining the Constant

Final Calculation

Similar Questions

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Calculating the Constant $C$

If a curve passes through the point $(1, - 2)$ and has slope of the tangent at any point $(x, y)$ on it as $\frac{x ^{2} - 2 y}{x}$ , then the curve also passes through the point :

Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $lim_{t \to x} \frac{t ^{2} y ( x ) - x ^{2} y ( t )}{x - t} = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to

If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is $\frac{x ^{2} - 4 x + y + 8}{x - 2}$ , then this curve also passes through the point :

A function $y = f (x)$ satisfies $f (x) sin 2 x + sin x - (1 + cos^{2} x) f^{'} (x) = 0$ with condition $f (0) = 0$ . Then $f (\frac{π}{2})$ is equal to

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

Let $y = y (x)$ be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x > 1$ , with $y (2) = - 2$ . Then $y (3)$ is equal to

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is