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 :
Solution
<p>$$f(x) = \left\{ {\matrix{
{x + a} & , & {x \le 0} \cr
{|x - 4|} & , & {x > 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{x + 1} & , & {x < 0} \cr
{{{(x - 4)}^2} + b} & , & {x \ge 0} \cr
} } \right.$$</p>
<p>$\because$ $f(x)$ and $g(x)$ are continuous on R</p>
<p>$\therefore$ $a = 4$ and $b = 1 - 16 = - 15$</p>
<p>then $(gof)(2) + (fog)( - 2)$</p>
<p>$= g(2) + f( - 1)$</p>
<p>$= - 11 + 3 = - 8$</p>
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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