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If \(f(x) = \begin{cases} x + a,& x\le 0\\|x - 4|, &x >0 \end{cases}\) and\(g(x) = \begin{cases}x + 1, &x <0 \\(x - 4)^2 + b, & x \ge 0\end{cases}\) are continuous on R, then (gof) (2) + (fog) (–2) is equal to :(A) –10(B) 10(C) 8(D) –8 |
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Answer» Correct option is (D) –8 \(f(x) = \begin{cases} x + a,& x\le 0\\|x - 4|, &x >0 \end{cases}\) ; \(g(x) = \begin{cases}x + 1, &x <0 \\(x - 4)^2 + b, & x \ge 0\end{cases}\) For continuity a = 4 and b = –15 g(f(2)) + f(g(-2)) = g(2) + f(-1) = -8 |
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