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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
lf  $f(x)$  is a differentiable function in the interval $(0,\infty )$  such that $f(1) = 1$ and $\mathop {\lim }\limits_{t \to x} \frac{{{t^2}f(x) - {x^2}f(t)}}{{t - x}} = 1,$ for each $x > 0,$  then $f (\frac {3}{2})$ is equal to
  • A
    $\frac {23}{18}$
  • B
    $\frac {13}{6}$
  • C
    $\frac {25}{9}$
  • ✓
    $\frac {31}{18}$
✓

Answer

Correct option: D.
$\frac {31}{18}$
d
Let $L = \mathop {\lim }\limits_{t \to x} \begin{array}{*{20}{c}}
{{t^2}f\left( x \right) - {x^2}f\left( t \right)}\\
{t - x}
\end{array} = 1$

Applying $L.H.$ rule

$L = \mathop {\lim }\limits_{t \to x} \begin{array}{*{20}{c}}
{2t\,f\left( x \right) - {x^2}f'\left( t \right)}\\
1
\end{array} = 1$

$2t\,f\left( x \right) - {x^2}f'\left( x \right) = 1$

solving above differential equation, we get

$f\left( x \right) = \frac{2}{3}{x^2} + \frac{1}{{3x}}$

Put $x = \frac{3}{2}$

$f\left( {\frac{3}{2}} \right) = \frac{2}{3} \times {\left( {\frac{3}{2}} \right)^2} + \frac{1}{3} \times \frac{2}{3}$

$ = \frac{3}{2} + \frac{2}{9} = \frac{{27 + 4}}{{18}} = \frac{{31}}{{18}}$

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