Question

If
$f(x) = \begin{cases} x + a, & x \leq 0 \\\ |x - 4|, & x > 0 \end{cases}$ and $g(x) = \begin{cases} x + 1, & x < 0 \\\ (x - 4)^2 + b, & x \geq 0 \end{cases}$
are continuous on R, then (gof) (2) + (fog) (–2) is equal to

• A. -10
• B. 10
• C. 8
• D. -8

Answer 1

Answer: (D)

-8

Answer 2

$f(x) = \begin{cases} x + a, & x \leq 0 \\\ |x - 4|, & x > 0 \end{cases}$ and$g(x) = \begin{cases} x + 1, & x < 0 \\\ (x - 4)^2 + b, & x \geq 0 \end{cases}$
∵ f(x) and g(x) are continuous on R
∴ a = 4 and b = 1 – 16 = –15
then (gof)(2) + (fog) (–2)
= g(2) + f(–1)
= –11 + 3 = – 8
So, the correct option is (D): -8