JEE Main 2024 (29 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $f (x) = lim_{r \to x} {\frac{2 r ^{2} [( f ( r ) ) ^{2} - f ( x ) f ( r )]}{r ^{2} - x ^{2}} - r^{3} e^{\frac{f ( r )}{r}}}$ be differentiable in $(- \infty, 0) \cup (0, \infty)$ and $f (1) = 1$ . Then the value of $e a$ , such that $f (a) = 0$ , is equal to ______.

Q: How did you know it was a homogeneous differential equation?

The presence of the term $e^{f(x)/x}$ (or $e^{y/x}$) is a strong signal. In differential equations, if you can express the equation in terms of the ratio $y/x$, it is almost certainly a homogeneous equation, which can be solved using the substitution $y = vx$.

Q: Why can we assume $x \neq 0$ when dividing by $x$?

The problem explicitly states that the function $f(x)$ is differentiable in $(-\infty, 0) \cup (0, \infty)$. This means the domain excludes $x = 0$, allowing us to safely divide by $x$ without worrying about division by zero.

Enter Numerical Value:

Visualized Solution

Squaring the Function

Square both sides to remove the radical:
$f^{2} (x) = lim_{r \to x} {\frac{2 r ^{2} [( f ( r ) ) ^{2} - f ( x ) f ( r )]}{r ^{2} - x ^{2}} - r^{3} e^{\frac{f ( r )}{r}}}$
Factor out $f (r)$ in the numerator of the first term:
$f^{2} (x) = lim_{r \to x} {\frac{2 r ^{2} f ( r ) [ f ( r ) - f ( x )]}{r ^{2} - x ^{2}} - r^{3} e^{\frac{f ( r )}{r}}}$

Evaluating the Limit

Focus on the first term inside the limit:
$lim_{r \to x} \frac{2 r ^{2} f ( r ) [ f ( r ) - f ( x )]}{( r - x ) ( r + x )}$
Recognize the derivative definition:
$lim_{r \to x} \frac{f ( r ) - f ( x )}{r - x} = f^{'} (x)$
Substitute $r = x$ in the remaining parts:
$\frac{2 x ^{2} f ( x )}{2 x} f^{'} (x) = x f (x) f^{'} (x)$

Forming the Differential Equation

Evaluate the second term of the limit:
$lim_{r \to x} r^{3} e^{\frac{f ( r )}{r}} = x^{3} e^{\frac{f ( x )}{x}}$
Combine to form the differential equation:
$f^{2} (x) = x f (x) f^{'} (x) - x^{3} e^{\frac{f ( x )}{x}}$
Let $y = f (x)$ , then:
$y^{2} = x y \frac{d y}{d x} - x^{3} e^{\frac{y}{x}}$

Rearranging the Equation

Divide the entire equation by $x^{2}$ :
$\frac{y ^{2}}{x ^{2}} = \frac{y}{x} \frac{d y}{d x} - x e^{\frac{y}{x}}$
Rewrite to highlight the homogeneous terms:
$(\frac{y}{x})^{2} = \frac{y}{x} \frac{d y}{d x} - x e^{\frac{y}{x}}$

Using Substitution $y = v x$

Apply the substitution for homogeneous equations:
Let $y = v x ⟹ \frac{y}{x} = v$
Differentiate with respect to $x$ :
$\frac{d y}{d x} = v + x \frac{d v}{d x}$

Simplifying the DE

Substitute $v$ and $\frac{d y}{d x}$ into the equation:
$v^{2} = v (v + x \frac{d v}{d x}) - x e^{v}$
Expand and simplify:
$v^{2} = v^{2} + v x \frac{d v}{d x} - x e^{v}$
$v x \frac{d v}{d x} = x e^{v}$
Divide by $x$ (assuming $x \neq = 0$ ):
$v \frac{d v}{d x} = e^{v}$

Separating Variables

Separate the variables $v$ and $x$ :
$v e^{- v} d v = d x$
Set up the integrals:
$\int v e^{- v} d v = \int d x$

Integration by Parts

Evaluate the left integral using integration by parts:
$\int v e^{- v} d v = - v e^{- v} - \int - e^{- v} d v$
$= - v e^{- v} - e^{- v} = - e^{- v} (v + 1)$
Equate to the right integral:
$- e^{- v} (v + 1) = x + C$

Applying Condition $f (1) = 1$

Use the initial condition $f (1) = 1$ :
At $x = 1, y = 1 ⟹ v = \frac{1}{1} = 1$
Substitute into the integrated equation:
$- e^{- 1} (1 + 1) = 1 + C$
Solve for $C$ :
$C = - 1 - \frac{2}{e}$

Applying Condition $f (a) = 0$

Use the condition $f (a) = 0$ to find $a$ :
At $x = a, y = 0 ⟹ v = \frac{0}{a} = 0$
Substitute $v = 0$ , $x = a$ , and $C$ into the equation:
$- e^{0} (0 + 1) = a - 1 - \frac{2}{e}$
$- 1 = a - 1 - \frac{2}{e}$
$a = \frac{2}{e}$

Final Answer

Calculate the final value for $e a$ :
$e a = e \cdot (\frac{2}{e}) = 2$
The final value is 2.

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The given equation is:

f (x) = r \to x lim {\frac{2 r ^{2} [( f ( r ) ) ^{2} - f ( x ) f ( r )]}{r ^{2} - x ^{2}} - r^{3} e^{\frac{f ( r )}{r}}}

To dismantle this structure, we square both sides to obtain:

f^{2} (x) = r \to x lim {\frac{2 r ^{2} f ( r ) [ f ( r ) - f ( x )]}{r ^{2} - x ^{2}} - r^{3} e^{\frac{f ( r )}{r}}}

The Derivative Revealed

Focusing on the first term inside the limit, we rewrite the denominator as

(r - x) (r + x)

r \to x lim \frac{2 r ^{2} f ( r ) [ f ( r ) - f ( x )]}{( r - x ) ( r + x )}

r \to x

, the expression

\frac{f ( r ) - f ( x )}{r - x}

converges to the derivative

f^{'} (x)

. Evaluating the remaining components at

r = x

yields:

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

Thus, the first term simplifies to

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

. The second term evaluates directly to

x^{3} e^{\frac{f ( x )}{x}}

, resulting in the differential equation:

f^{2} (x) = x f (x) f^{'} (x) - x^{3} e^{\frac{f ( x )}{x}}

The Homogeneous Strategy

Substituting

y = f (x)

and

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

, we have:

y^{2} = x y \frac{d y}{d x} - x^{3} e^{\frac{y}{x}}

Dividing by

x^{2}

reveals the homogeneous structure:

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

Applying the substitution

y = v x

, where

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

, the equation becomes:

v^{2} = v (v + x \frac{d v}{d x}) - x e^{v}

The

v^{2}

terms cancel, leaving

v x \frac{d v}{d x} = x e^{v}

. Dividing by

x

(assuming $x eq 0$), we obtain:

v \frac{d v}{d x} = e^{v}

The Final Integration

Separating the variables, we get

v e^{- v} d v = d x

. Integrating both sides:

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

Using integration by parts, the integral of

v e^{- v}

- e^{- v} (v + 1)

. Therefore:

- e^{- v} (v + 1) = x + C

Given

f (1) = 1

, we have

v = 1

x = 1

. Substituting these values:

- e^{- 1} (1 + 1) = 1 + C \Rightarrow C = - 1 - \frac{2}{e}

To find

a

such that

f (a) = 0

, we set

v = 0

x = a

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

- 1 = a - 1 - \frac{2}{e} \Rightarrow a = \frac{2}{e}

The final value of

e a

is: $e \cdot \frac{2}{e} = 2$

Common Traps & FAQs

Why do we square the entire equation at the beginning?

The equation starts with a square root covering the entire right-hand side. Squaring both sides is the most direct way to eliminate the radical, transforming the expression into a standard algebraic form that we can then manipulate using calculus.

How did you know it was a homogeneous differential equation?

The presence of the term

e^{f (x) / x}

(or

e^{y / x}

) is a strong signal. In differential equations, if you can express the equation in terms of the ratio

y / x

, it is almost certainly a homogeneous equation, which can be solved using the substitution

y = v x

Why can we assume

x \neq = 0

when dividing by

x

The problem explicitly states that the function

f (x)

is differentiable in

(- \infty, 0) \cup (0, \infty)

. This means the domain excludes

x = 0

, allowing us to safely divide by

x

without worrying about division by zero.

Enter Numerical Value:

Visualized Solution

Squaring the Function

Evaluating the Limit

Forming the Differential Equation

Rearranging the Equation

Using Substitution $y = v x$

Simplifying the DE

Separating Variables

Integration by Parts

Applying Condition $f (1) = 1$

Applying Condition $f (a) = 0$

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Derivative Revealed

The Homogeneous Strategy

The Final Integration

Similar Questions

Let a curve $y = f (x)$ pass through the point $(2, (lo g_{e} 2)^{2})$ and have slope $\frac{2 y}{x l o g _{e} x}$ for all positive real value of $x$ . Then the value of $f (e)$ is equal to ___

Let $f$ be a non-negative function in $[0, 1]$ and twice differentiable in $(0, 1)$ . If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ and $f (0) = 0$ , then $lim_{x \to 0} \frac{1}{x ^{2}} \int_{0}^{x} f (t) d t$ :

Let $f : [0, 2] \to R$ be a function which is continuous on $[0, 2]$ and is differentiable on $(0, 2)$ with $f (0) = 1$ . Let $F (x) = \int_{0}^{x^{2}} f (t) d t$ for $x \in [0, 2]$ . If $F^{'} (x) = f^{'} (x)$ for all $x \in (0, 2)$ , then $f (2)$ equals

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e$ . Then $lim_{x \to 0^{+}} f (x)$ is equal to

Let $f$ be a differentiable function defined on $[0, π /2]$ such that $f (x) > 0$ and $f (x) + \int_{0}^{x} f (t) 1 - (lo g_{e} f (t))^{2} d t = e, \forall x \in [0, π /2]$ . Then $(6 lo g_{e} f (π /6))^{2}$ is equal to ______.

Let $f : R \to R$ be a differentiable function with $f (0) = 0$ . If $y = f (x)$ satisfies the differential equation $\frac{d y}{d x} = (2 + 5 y) (y - 2)$ , then the value of $lim_{x \to \infty} f (x)$ is ________.

Let $f : R \to R$ be a thrice differentiable odd function satisfying $f^{'} (x) \geq 0, f^{'} (x) = f (x), f (0) = 0, f^{'} (0) = 3$ . Then $9 f (lo g_{e} 3)$ is equal to _______.

Let $f : R \to R$ and $g : R \to R$ be two non-constant differentiable functions. If $f^{'} (x) = (e^{f (x) - g (x)}) g^{'} (x)$ for all $x \in R$ , and $f (1) = g (2) = 1$ , then which of the following statement(s) is (are) TRUE ?

Let $f$ be a differentiable function satisfying $f (x) = \frac{2}{3} \int_{0}^{3} f (\frac{λ ^{2} x}{3}) d λ, x > 0$ and $f (1) = 3$ . If $y = f (x)$ passes through the point $(α, 6)$ , then $α$ is equal to \_\_\_\_.

Enter Numerical Value:

Visualized Solution

Squaring the Function

Evaluating the Limit

Forming the Differential Equation

Rearranging the Equation

Using Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=vx

Simplifying the DE

Separating Variables

Integration by Parts

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

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

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Derivative Revealed

The Homogeneous Strategy

The Final Integration

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=f(x) be the solution of the differential equation y(x+1)dx−x2dy=0,y(1)=e. Then limx→0+​f(x) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a differentiable function with f(0)=0. If y=f(x) satisfies the differential equation dxdy​=(2+5y)(y−2), then the value of limx→∞​f(x) is ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a thrice differentiable odd function satisfying f′(x)≥0,f′(x)=f(x),f(0)=0,f′(0)=3. Then 9f(loge​3) is equal to _______.

Using Substitution $y = v x$

Applying Condition $f (1) = 1$

Applying Condition $f (a) = 0$

Let a curve $y = f (x)$ pass through the point $(2, (lo g_{e} 2)^{2})$ and have slope $\frac{2 y}{x l o g _{e} x}$ for all positive real value of $x$ . Then the value of $f (e)$ is equal to ___

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e$ . Then $lim_{x \to 0^{+}} f (x)$ is equal to

Let $f : R \to R$ be a differentiable function with $f (0) = 0$ . If $y = f (x)$ satisfies the differential equation $\frac{d y}{d x} = (2 + 5 y) (y - 2)$ , then the value of $lim_{x \to \infty} f (x)$ is ________.

Let $f : R \to R$ be a thrice differentiable odd function satisfying $f^{'} (x) \geq 0, f^{'} (x) = f (x), f (0) = 0, f^{'} (0) = 3$ . Then $9 f (lo g_{e} 3)$ is equal to _______.