JEE MAIN - Mathematics (2022 - 26th July Morning Shift - No. 7)
If $$f(x) = \left\{ {\matrix{
{x + a} & , & {x \le 0} \cr
{|x - 4|} & , & {x > 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{x + 1} & , & {x < 0} \cr
{{{(x - 4)}^2} + b} & , & {x \ge 0} \cr
} } \right.$$ are continuous on R, then $$(gof)(2) + (fog)( - 2)$$ is equal to :
$$-$$10
10
8
$$-$$8
Explanation
$$f(x) = \left\{ {\matrix{ {x + a} & , & {x \le 0} \cr {|x - 4|} & , & {x > 0} \cr } } \right.$$ and $$g(x) = \left\{ {\matrix{ {x + 1} & , & {x < 0} \cr {{{(x - 4)}^2} + b} & , & {x \ge 0} \cr } } \right.$$
$$\because$$ $$f(x)$$ and $$g(x)$$ are continuous on R
$$\therefore$$ $$a = 4$$ and $$b = 1 - 16 = - 15$$
then $$(gof)(2) + (fog)( - 2)$$
$$ = g(2) + f( - 1)$$
$$ = - 11 + 3 = - 8$$
Comments (0)
