QJEEM22Mac6a0: JEE Main 2022 Mathematics - Class 12 JEEM22Mac6a0 Solution: Let f be a differentiable function satisfying and . If passes…

$\because$ f(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {f\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda ,\,x > 0}

On differentiating both sides w.r.t., x, we get

$f'(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {{{{\lambda ^2}} \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda }$

$f'(x) = {1 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {\lambda \,.\,{{2\lambda } \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda }$

$\therefore$ $\sqrt 3 f'(x) = \left[ {{\lambda \over x}\,.\,f\left( {{{{\lambda ^2}x} \over 3}} \right)} \right]_0^{\sqrt 3 } - \int\limits_0^{\sqrt 3 } {{1 \over x}f\left( {{{{\lambda ^2}x} \over 3}} \right)dx}$

$\sqrt 3 x\,f'(x) = \sqrt 3 f(x) - {{\sqrt 3 } \over 2}f(x)$

$xf'(x) = {{f(x)} \over 2}$

On integrating we get : $\ln y = {1 \over 2}\ln x + \ln c$

$\because$ $f(1) = \sqrt 3$ then $c = \sqrt 3$

$\therefore$ ( $\alpha$ , 6) lies on

$\therefore$ $y = \sqrt {3x}$

$\therefore$ $6 = \sqrt {3\alpha } \Rightarrow \alpha = 12$ .