JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 20)
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 :
$$\mathop {\lim }\limits_{t \to x} $$ $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ 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 :
$${{13} \over 6}$$
$${{23} \over 18}$$
$${{25} \over 9}$$
$${{31} \over 18}$$
Explanation
$$\mathop {\lim }\limits_{t \to x} {{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1$$
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$$\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}$$
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$$\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}$$
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