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STD 12 - 9. differential equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 9. differential equations4 Marks
MCQ
Let $f$ be a differential function such that $f'\left( x \right) = 7 - \frac{3}{4}\frac{{f\left( x \right)}}{x},\left( {x > 0} \right)$ and $f(1) \ne 4$. Then $\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {\frac{1}{x}} \right)$
  • A
    exists and equals $\frac{4}{7}$
  • ✓
    exists and equals $4$
  • C
    does not exist.
  • D
    exists and equals $0$
✓

Answer

Correct option: B.
exists and equals $4$
b
$f'\left( x \right) = 7 - \frac{3}{4}.\frac{{f\left( x \right)}}{x},x > 0$

$\therefore f'\left( x \right) + \frac{3}{{4x}}f\left( x \right) = 7$

$f\left( x \right).{e^{\int {\frac{3}{{4x}}dx} }} = \int {7.} {e^{\int {\frac{3}{{4x}}dx} }} + c$

$f\left( x \right).{x^{3/4}} = 7.{x^{3/4}} + c$

$ = 7\frac{{{x^{7/4}}}}{{\frac{7}{4}}} + c$

$\therefore f\left( x \right) = 4x + c{x^{ - 3/4}}$

$\therefore f\left( {\frac{1}{x}} \right) = \frac{4}{x} + c{x^{3/4}}$

$\therefore \mathop {Lt}\limits_{x \to {0^ + }} xf\left( {\frac{1}{x}} \right) = \mathop {Lt}\limits_{x \to {0^ + }} 4 + c{x^{7/4}} = 4$

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