Solveeit Logo

Question

Question: Let \(f\left( x \right)\) be a function whose domain is \(\left[ -5,7 \right]\). Let \(g\left( x \ri...

Let f(x)f\left( x \right) be a function whose domain is [5,7]\left[ -5,7 \right]. Let g(x)=2x+5g\left( x \right)=\left| 2x+5 \right|. The domain of f(g(x))f\left( g\left( x \right) \right) is
A. [4,1]\left[ -4,1 \right]
B. [5,1]\left[ -5,1 \right]
C. [6,1]\left[ -6,1 \right]
D. None of these

Explanation

Solution

We first find the relation between the range of g(x)g\left( x \right) and the domain of f(x)f\left( x \right) for the composite function f(g(x))f\left( g\left( x \right) \right). We use that to find the range of the function g(x)=2x+5g\left( x \right)=\left| 2x+5 \right|. We simplify the expression to find the required solution.

Complete answer:
f(g(x))f\left( g\left( x \right) \right) is a composite function for which the range of g(x)g\left( x \right) has to be the subset of the domain of f(x)f\left( x \right).
The domain of f(x)f\left( x \right) is [5,7]\left[ -5,7 \right]. Therefore, g(x)=2x+5g\left( x \right)=\left| 2x+5 \right| has to be in the range of [5,7]\left[ -5,7 \right].
This means 52x+57-5\le \left| 2x+5 \right|\le 7. We break it in two parts of 52x+5-5\le \left| 2x+5 \right| and 2x+57\left| 2x+5 \right|\le 7.
From 52x+5-5\le \left| 2x+5 \right|, we get xRx\in \mathbb{R} as the modulus value is always positive.
From 2x+57\left| 2x+5 \right|\le 7, we get
2x+57 72x+57 122x2 6x1 \begin{aligned} & \left| 2x+5 \right|\le 7 \\\ & \Rightarrow -7\le 2x+5\le 7 \\\ & \Rightarrow -12\le 2x\le 2 \\\ & \Rightarrow -6\le x\le 1 \\\ \end{aligned}
Therefore, x[6,1]x\in \left[ -6,1 \right].
The intersection of x[6,1]x\in \left[ -6,1 \right] and xRx\in \mathbb{R} gives x[6,1]x\in \left[ -6,1 \right]. Therefore, the correct option is C.

Note:
We need to remember that x[6,1]x\in \left[ -6,1 \right] is solely responsible for the expression to be defined. We also considered that complex values as the root are not allowed. In the first case the domain becomes the whole real line.