Let \( f : \mathbb{R} - \{0\} \rightarrow \mathbb{R} \) be a function satisfying\[f\left( \frac{x}{y} \right) = \frac{f(x)}{f(y)}\]for all \( x, y \), \( f(y) \neq 0 \). If \( f'(1) = 2024 \),then
- \( x f'(x) - 2024 f(x) = 0 \)
- \( x f'(x) + 2024 f(x) = 0 \)
- \( x f'(x) + f(x) = 2024 \)
- \( x f'(x) - 2023 f(x) = 0 \)
The Correct Option is A
Approach Solution - 1
The functional equation given is:
\[ f\left( \frac{x}{y} \right) = \frac{f(x)}{f(y)}. \]
This suggests an exponential-like function. Assume \( f(x) = x^k \) for some constant \( k \).
Verify that \( f(x) = x^k \) satisfies the functional equation:
\[ f\left( \frac{x}{y} \right) = \left( \frac{x}{y} \right)^k = \frac{x^k}{y^k} = \frac{f(x)}{f(y)}. \]
So, \( f(x) = x^k \) is a valid solution.
Differentiating \( f(x) = x^k \):
\[ f'(x) = kx^{k-1}. \]
Given \( f'(1) = 2024 \):
\[ f'(1) = k = 2024. \]
Therefore, \( f(x) = x^{2024} \).
Check which option matches: Substitute \( f(x) = x^{2024} \) and \( f'(x) = 2024x^{2023} \) in each option.
Option (1):
\[ xf'(x) - 2024f(x) = x \times 2024x^{2023} - 2024x^{2024} = 0. \]
Therefore, the correct answer is:
\[ xf'(x) - 2024f(x) = 0. \]
Approach Solution -2
To solve this problem, we need to analyze the given functional equation and the information provided about the derivative of the function.
The function \( f : \mathbb{R} - \{0\} \rightarrow \mathbb{R} \) satisfies:
\(f\left( \frac{x}{y} \right) = \frac{f(x)}{f(y)}\) for all \( x, y \), with condition \( f(y) \neq 0 \).
This is a property of logarithmic functions or power functions. Let's explore the possibility of \( f(x) \) being a power function.
Assume \( f(x) = x^k \), where \( k \) is a constant. Check if the functional equation holds:
For \( f\left(\frac{x}{y}\right) = \left(\frac{x}{y}\right)^k \), and \( \frac{f(x)}{f(y)} = \frac{x^k}{y^k} = \left(\frac{x}{y}\right)^k \),
both expressions are equal, confirming that \( f(x) = x^k \) is indeed a valid assumption.
Next, we use the additional condition \( f'(1) = 2024 \) to find the exponent \( k \).
The derivative of \( f(x) = x^k \) is \( f'(x) = k x^{k-1} \).
Given \( f'(1) = 2024 \), substitute \( x = 1 \):
\(f'(1) = k \cdot 1^{k-1} = k = 2024\)
Hence, \( k = 2024 \), so \( f(x) = x^{2024} \).
Now, let's verify which differential equation matches \( f(x) = x^{2024} \).
Substitute \( f(x) = x^{2024} \) into the given options:
- \(x^{2024} \Rightarrow f'(x) = 2024 x^{2023}\)
Checking each option:
- \( x f'(x) - 2024 f(x) = x (2024 x^{2023}) - 2024 x^{2024} = 2024 x^{2024} - 2024 x^{2024} = 0 \). This satisfies the equation.
- \( x f'(x) + 2024 f(x) = x (2024 x^{2023}) + 2024 x^{2024} = 2024 x^{2024} + 2024 x^{2024} = 4048 x^{2024} \neq 0\). This is incorrect.
- \( x f'(x) + f(x) = 2024 = x (2024 x^{2023}) + x^{2024} = 2025 x^{2024} \neq 2024\). This is incorrect.
- \( x f'(x) - 2023 f(x) = x (2024 x^{2023}) - 2023 x^{2024} = 2024 x^{2024} - 2023 x^{2024} = x^{2024} \neq 0\). This is incorrect.
Thus, the correct option is:
\( x f'(x) - 2024 f(x) = 0 \).
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Differential equations Questions
Let $y=y(x)$ be the solution of the differential equation $\left(x^2-3 y^2\right) d x+3 x y d y=0, y(1)=1$.Then $6 y^2( e )$ is equal to
- If \((t+1)dx=(2x+(t+1)^3)dt\) and \(x(0)=2\), then \(x(1) \)is equal to:
- Let vertex \(A(2, 3, 1), B(3, 2, -1), C(-2, 1, 3)\). If \(AD\) is angle bisector of angle \(A\), then projection of \(\overrightarrow{AD}\) on \(\overrightarrow{AC}\) is equal to:
- Let \(y = y(x)\) be the solution of the differential equation \(\sec(x) \frac{dy}{dx} + [2(1 - x) \tan(x) + x(2 - x)] = 0\) such that \(y(0) = 2\). Then \(y(2)\) is equal to:
- Let \( y = y(x) \) be the solution of the differential equation \( (1 - x^2) \, dy = \left[ xy + \left( x^3 + 2 \right) \sqrt{3 \left( 1 - x^2 \right)} \right] dx \), \( -1 < x < 1, y(0) = 0 \). If \( y\left( \frac{1}{2} \right) = \frac{m}{n} \), \( m \) and \( n \) are coprime numbers, then \( m + n \) is equal to \(\_\_\_\_\_\_\_\_\_\).
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,