Let f(x) be a continuously differentiable function on the interval (0, ∞) such that f(1) = 2 and \(\lim\limits_{t→x}\frac{t^{10}f(x)-x^{10}f(t)}{t^9-x^9}=1\)
for each x > 0. Then, for all x > 0, f(x) is equal to
Let f(x) be a continuously differentiable function on the interval (0, ∞) such that f(1) = 2 and \(\lim\limits_{t→x}\frac{t^{10}f(x)-x^{10}f(t)}{t^9-x^9}=1\)
for each x > 0. Then, for all x > 0, f(x) is equal to
- \(\frac{31}{11x}-\frac{9}{11}x^{10}\)
- \(\frac{9}{11x}+\frac{13}{11}x^{10}\)
- \(\frac{-9}{11x}+\frac{31}{11}x^{10}\)
- \(\frac{13}{11x}+\frac{9}{11}x^{10}\)
The Correct Option is B
Approach Solution - 1
To solve the problem, we need to analyze the given limit condition and find the function \( f(x) \) satisfying it.
1. Understanding the limit expression:
Given:
\[ \lim_{t \to x} \frac{t^{10} f(x) - x^{10} f(t)}{t^9 - x^9} = 1 \] for each \( x > 0 \).
2. Rewrite the limit to use the derivative concept:
Notice the numerator and denominator both tend to zero as \( t \to x \). This suggests using the definition of derivative or applying algebraic manipulation.
3. Rewrite numerator and denominator:
\[ \frac{t^{10} f(x) - x^{10} f(t)}{t^9 - x^9} = \frac{t^{10} f(x) - x^{10} f(x) + x^{10} f(x) - x^{10} f(t)}{t^9 - x^9} = \frac{f(x)(t^{10} - x^{10}) - x^{10} (f(t) - f(x))}{t^9 - x^9} \]
4. Separate the fraction:
\[ = f(x) \frac{t^{10} - x^{10}}{t^9 - x^9} - x^{10} \frac{f(t) - f(x)}{t^9 - x^9} \]
5. Analyze the first fraction as \( t \to x \):
Since \( t \to x \), using derivative formulas:
\[ \lim_{t \to x} \frac{t^{10} - x^{10}}{t^9 - x^9} \] Both numerator and denominator tend to zero, apply L'Hôpital's Rule:
Derivative of numerator w.r.t. \( t \): \( 10 t^9 \)
Derivative of denominator w.r.t. \( t \): \( 9 t^8 \)
Thus, the limit is:
\[ \frac{10 x^9}{9 x^8} = \frac{10}{9} x \]
6. Analyze the second fraction as \( t \to x \):
\[ \lim_{t \to x} \frac{f(t) - f(x)}{t^9 - x^9} = \lim_{t \to x} \frac{f(t) - f(x)}{t - x} \cdot \frac{t - x}{t^9 - x^9} = f'(x) \cdot \lim_{t \to x} \frac{1}{9 t^8} = \frac{f'(x)}{9 x^8} \]
7. Substitute limits back:
\[ 1 = f(x) \times \frac{10}{9} x - x^{10} \times \frac{f'(x)}{9 x^8} = \frac{10}{9} x f(x) - \frac{x^2}{9} f'(x) \]
8. Multiply both sides by 9:
\[ 9 = 10 x f(x) - x^{2} f'(x) \]
9. Rearranged differential equation:
\[ x^{2} f'(x) + 9 = 10 x f(x) \] or \[ x^{2} f'(x) - 10 x f(x) = -9 \]
10. Divide both sides by \( x^2 \):
\[ f'(x) - \frac{10}{x} f(x) = -\frac{9}{x^2} \] This is a linear differential equation of the form:
\[ f'(x) + P(x) f(x) = Q(x) \] with \( P(x) = -\frac{10}{x} \) and \( Q(x) = -\frac{9}{x^2} \).
11. Find the integrating factor (IF):
\[ IF = e^{\int P(x) dx} = e^{\int -\frac{10}{x} dx} = e^{-10 \ln x} = x^{-10} \]
12. Multiply entire equation by integrating factor:
\[ x^{-10} f'(x) - \frac{10}{x} x^{-10} f(x) = -\frac{9}{x^2} x^{-10} \] which simplifies to \[ \frac{d}{dx} \left( x^{-10} f(x) \right) = -9 x^{-12} \]
13. Integrate both sides:
\[ \int \frac{d}{dx} \left( x^{-10} f(x) \right) dx = \int -9 x^{-12} dx \] \[ x^{-10} f(x) = -9 \int x^{-12} dx + C \] \[ = -9 \times \left( \frac{x^{-11}}{-11} \right) + C = \frac{9}{11} x^{-11} + C \]
14. Solve for \( f(x) \):
\[ f(x) = x^{10} \left( \frac{9}{11} x^{-11} + C \right) = \frac{9}{11} x^{-1} + C x^{10} \]
15. Use the initial condition \( f(1) = 2 \):
\[ f(1) = \frac{9}{11} (1)^{-1} + C (1)^{10} = \frac{9}{11} + C = 2 \] \[ C = 2 - \frac{9}{11} = \frac{22}{11} - \frac{9}{11} = \frac{13}{11} \]
Final form of \( f(x) \):
\[ \boxed{ f(x) = \frac{9}{11x} + \frac{13}{11} x^{10} } \]
Approach Solution -2
Top Questions on Differential equations
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
- JEE Advanced - 2025
- Mathematics
- Differential equations
- For all $ x>0 $, let $ y_1(x), y_2(x), y_3(x) $ be the functions satisfying $$ \frac{dy_1}{dx} - (\sin x)^2 y_1 = 0, \quad y_1(1) = 5, $$ $$ \frac{dy_2}{dx} - (\cos x)^2 y_2 = 0, \quad y_2(1) = \frac{1}{3}, $$ $$ \frac{dy_3}{dx} - \left(\frac{2 - x^3}{x^3}\right) y_3 = 0, \quad y_3(1) = \frac{3}{5e}, $$ respectively. Then $$ \lim_{x \to 0^+} \frac{y_1(x)y_2(x)y_3(x) + 2x}{e^{3x} \sin x} $$ is equal to __________.
- JEE Advanced - 2025
- Mathematics
- Differential equations
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f(x + y) = f(x) f(y) \) for all \( x, y \in \mathbb{R} \). If \( f'(0) = 4a \) and \( f \) satisfies \( f''(x) - 3a f'(x) - f(x) = 0 \), where \( a > 0 \), then the area of the region R = {(x, y) | 0 \(\leq\) y \(\leq\) f(ax), 0 \(\leq\) x \(\leq\) 2 is :
- JEE Main - 2025
- Mathematics
- Differential equations
- The value of \(\displaystyle \int \frac{1}{(x-5)^2}\,dx\) is:
- LPUNEST - 2025
- Mathematics
- Differential equations
- Let \( y = y(x) \) be the solution of the differential equation \( \frac{dy}{dx} + 3(\tan^2 x) y + 3y = \sec^2 x \), with \( y(0) = \frac{1}{3} + e^3 \). Then \( y\left(\frac{\pi}{4}\right) \) is equal to
- JEE Main - 2025
- Mathematics
- Differential equations
Questions Asked in JEE Advanced exam
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
- A linear octasaccharide (molar mass = 1024 g mol$^{-1}$) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26 % (w/w) of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is _____.
Use: Molar mass (in g mol$^{-1}$): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass (in amu): H = 1, O = 16- JEE Advanced - 2025
- Biomolecules
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
- JEE Advanced - 2025
- Conic sections
- Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g$^{-1}$ and 16 mg g$^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg g$^{-1}$ and 10 mg g$^{-1}$, respectively. At this temperature, the concentration (in mg g$^{-1}$) of adsorbed phenol from 20 mg g$^{-1}$ aqueous solution of phenol will be ____. Use: $\log_{10} 2 = 0.3$
- JEE Advanced - 2025
- Adsorption
- At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = 1.00 g cm$^{-3}$) where h is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 g dm$^{-3}$, the molar mass of the macromolecule is calculated to be $X \times 10^{4}$ g mol$^{-1}$. The value of $X$ is ____. Use: Universal gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$ and acceleration due to gravity (g) = 10 m s$^{-2}\}$
- JEE Advanced - 2025
- Colligative Properties