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
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