QJEEM16Maf820: JEE Main 2016 Mathematics - Class 12 JEEM16Maf820 Solution: If   f(x) is a differentiable function in the…

Question

Verified Solution • 2025–26 Session moderate

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 (3 / 2)

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}}

QJEEM16Maf820: JEE Main 2016 Mathematics - Class 12 JEEM16Maf820 Solution: If&nbsp;&nbsp; f(x) is a differentiable function in the…