QJEEM22Mab1a7: JEE Main 2022 Mathematics - Class 12 JEEM22Mab1a7 Solution: Let f, g : R R be two real valued functions defined as f ( x )…

$\because$ gof is differentiable at x = 0

So R.H.D = L.H.D

${d \over {dx}}(4{e^x} + {k_2}) = {d \over {dx}}\left( {{{( - |x + 3|)}^2} - {k_1}|x + 3|} \right)$

$\Rightarrow 4 = 6 - {k_1} \Rightarrow {k_1} = 2$

Also $f(f({0^ + })) = g(f({0^ - }))$

$\Rightarrow 4 + {k_2} = 9 - 3{k_1} \Rightarrow {k_2} = - 1$

Now $g(f( - 4)) + g(f(4))$

$= g( - 1) + g({e^4}) = (1 - {k_1}) + (4{e^4} + {k_2})$

$= 4{e^4} - 2$

$= 2(2{e^4} - 1)$