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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

જો \(f(x)\) એ અંતરાલ \((0,\infty )\) માં વિકલનીય વિધેય છે કે જેથી  \(f(1) = 1\) અને \(\mathop {\lim }\limits_{t \to x} \frac{{{t^2}f(x) - {x^2}f(t)}}{{t - x}} = 1,\) દરેક \(x > 0,\)  તો  \(f (\frac {3}{2})\) મેળવો.

  1. A \(\frac {23}{18}\)
  2. B \(\frac {13}{6}\)
  3. C \(\frac {25}{9}\)
  4. D \(\frac {31}{18}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac {31}{18}\)

Step-by-step Solution

Detailed explanation

\(\lim_{t \to x} \frac{t^2f(x) - x^2f(t)}{t - x} = \lim_{t \to x} \left[ f(x)\frac{t^2 - x^2}{t - x} - x^2\frac{f(t) - f(x)}{t - x} \right]\) \(= f(x)(2x) - x^2f'(x)\) \(2xf(x) - x^2f'(x) = 1\) \(x^2f'(x) - 2xf(x) = -1\) \(\frac{f'(x)}{x^2} - \frac{2f(x)}{x^3} = -\frac{1}{x^4}\)…
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