CAT Exam 2021 Slot 3 Solved Paper

Now we have :
$f(g(x)) - 3\x$
So we get $f ( x + 3 ) - 3x $
$= (x + 3)^2 - 7(x + 3) - 3x $
$x^2 - 4x -12 $
Now minimum value of expression = $ \D/{-4a} × {4ac - b^2}/ {4a} $
We get - $(16 + 48)/4 =-16 $