JEE Mains · Maths · STD 12 - 1. relation and function
If \(f(x)=\left\{\begin{array}{ll}x+a, & x \leq 0 \\ |x-4|, & x>0\end{array}\right.\) and \(g(x)=\left\{\begin{array}{ll}x+1 & x<0 \\ (x-4)^{2}+b, & x \geq 0\end{array}\right.\) are continuous on \(R\), then \((gof) (2)+( fog) (-2)\) is equal to.
- A \(-10\)
- B \(10\)
- C \(8\)
- D \(-8\)
Answer & Solution
Correct Answer
(D) \(-8\)
Step-by-step Solution
Detailed explanation
\((x)=\left\{\begin{array}{l} x+a ; x \leq 0 \\ |x-4| ; x>0 \end{array} ; g(x)=\left\{\begin{array}{ll} x+1 & ; x<0 \\ (x-4)^{2}+b ; & x \geq 0 \end{array}\right.\right.\) For continuity \(a =4\) and \(b =-15\) \(g(f(2))+f(g(-2))\) \(=g(2)+f(-1)=-8\)
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