If f(x) is a differentiable function in the interval (0, [ ]) such that f (1) = 1 and [ _t…

Question

If    f(x) is a differentiable function in the interval (0, $$\infty $$) such that f (1) = 1 and

$$\mathop {\lim }\limits_{t \to x} $$   $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$ equal to :

Accepted Answer

$$\mathop {\lim }\limits_{t \to x} {{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1$$

It is in $${0 \over 0}$$ form

So, applying L' Hospital rule,

$$\mathop {\lim }\limits_{t \to x} {{2tf\left( x \right) - {x^2}f'\left( x \right)} \over 1} = 1$$

$$ \Rightarrow $$   2xf(x) - x2f '(x) = 1

$$ \Rightarrow $$   f '(x) - $${2 \over x}$$f(x) $=$ $${1 \over {{x^2}}}$$

$$ \therefore $$   I.F =  $${e^{\int {{{ - 2} \over x}dx} }} = {e^{ - 2\log x}} = {1 \over {{x^2}}}$$

$$ \therefore $$   Solution of equation,

f(x) ({1 \over {{x^2}}}\] = $$\int {{1 \over x}\left( { - {1 \over {{x^2}}}} \right)} \,dx$$

$$ \Rightarrow $$   $${{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + C$$

Given that,

f(1) = 1

$$ \therefore $$   $${1 \over 1}$$ = $${1 \over 3}$$ + C

$$ \Rightarrow $$   C $=$ $${2 \over 3}$$

$$ \therefore $$   f(x) $=$  $${2 \over 3}$$ x2 + $${1 \over {3x}}$$

$$ \therefore $$   f$$\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}$$

Solutions · 1