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
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
-10
10
8
-8
The Correct Option is D
Solution and Explanation
\(f(x) = \begin{cases} x + a, & x \leq 0 \\ |x - 4|, & x > 0 \end{cases}\) 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
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions