Question:
Let f be a differentiable function such that \(f'(x) = 7 - \frac{3}{4}\frac{f(x)}{x}\), \((x \gt 0)\) and \(f(1) \ne 4\). Then \(\lim_{x \to 0^{+}} x\,f\left(\frac{1}{x}\right)\) :
Let f be a differentiable function such that \(f'(x) = 7 - \frac{3}{4}\frac{f(x)}{x}\), \((x \gt 0)\) and \(f(1) \ne 4\). Then \(\lim_{x \to 0^{+}} x\,f\left(\frac{1}{x}\right)\) :
Show Hint
Solve the linear ODE \(y' + \frac{3}{4x}y = 7\) with integrating factor \(x^{3/4}\) to get \(f(x) = 4x + Cx^{-3/4}\), then evaluate the limit.
Updated On: Nov 5, 2026
- does not exist
- exists and equals \(\frac{4}{7}\)
- exists and equals 4
- exists and equals 0
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The Correct Option is C
Solution and Explanation
\(f' + \frac{3}{4x}f = 7\). Integrating factor \(= e^{\int \frac{3}{4x}dx} = x^{3/4}\).
\(\frac{d}{dx}\left(x^{3/4}f\right) = 7x^{3/4} \Rightarrow x^{3/4}f = 4x^{7/4} + C\).
So \(f(x) = 4x + Cx^{-3/4}\) (the condition \(f(1)\ne 4\) means \(C \ne 0\)).
Then \(f\left(\frac1x\right) = \frac4x + Cx^{3/4}\) and \(x f\left(\frac1x\right) = 4 + Cx^{7/4}\).
As \(x \to 0^+\), \(Cx^{7/4} \to 0\), so the limit exists and equals 4.
Answer: (3) exists and equals 4
\(\frac{d}{dx}\left(x^{3/4}f\right) = 7x^{3/4} \Rightarrow x^{3/4}f = 4x^{7/4} + C\).
So \(f(x) = 4x + Cx^{-3/4}\) (the condition \(f(1)\ne 4\) means \(C \ne 0\)).
Then \(f\left(\frac1x\right) = \frac4x + Cx^{3/4}\) and \(x f\left(\frac1x\right) = 4 + Cx^{7/4}\).
As \(x \to 0^+\), \(Cx^{7/4} \to 0\), so the limit exists and equals 4.
Answer: (3) exists and equals 4
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